CMPhysBench: A Benchmark for Evaluating Large Language Models in Condensed Matter Physics
Paper
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2508.18124
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Published
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49
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int64 | question
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1
|
For an oscillator with charge $q$, its energy operator without an external field is
\begin{equation*}
H_{0}=\frac{p^{2}}{2 m}+\frac{1}{2} m \omega^{2} x^{2}
\end{equation*}
If a uniform electric field $\mathscr{E}$ is applied, causing an additional force on the oscillator $f= q \mathscr{E}$, the total energy operator becomes
\begin{equation*}
H=\frac{p^{2}}{2 m}+\frac{1}{2} m \omega^{2} x^{2}-q \mathscr{E} x
\end{equation*}
Find the expression for the new energy levels $E_{n}$.
|
In $H_{0}$ and $H$, $p$ is the momentum operator,
$p=-\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} x}$
The potential energy term in equation (2) can be rewritten as
$\frac{1}{2} m \omega^{2} x^{2}-q \mathscr{E} x=\frac{1}{2} m \omega^{2}[(x-x_{0})^{2}-x_{0}^{2}]$
where
\begin{equation*}
x_{0}=\frac{q \mathscr{E}}{m \omega^{2}} \tag{3}
\end{equation*}
By performing a coordinate shift, let
\begin{equation*}
x^{\prime}=x-x_{0} \tag{4}
\end{equation*}
Because
\begin{equation*}
p=-\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} x}=-\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} x}=p^{\prime} \tag{5}
\end{equation*}
$H$ can be expressed as
\begin{equation*}
H=\frac{p^{\prime 2}}{2 m}+\frac{1}{2} m \omega^{2} x^{\prime 2}-\frac{1}{2} m \omega^{2} x_{0}^{2} \tag{6}
\end{equation*}
Comparing equations (1) and (6), it is evident that the difference between $H$ and $H_{0}$ is that the variable changes from $x$ to $x^{\prime}$, with an added constant term $(-\frac{1}{2} m \omega^{2} x_{0}^{2})$. Hence, we get
\begin{gather*}
E_{n}=E_{n}^{(0)}-\frac{1}{2} m \omega^{2} x_{0}^{2} \tag{7}\\
\varphi_{n}(x)=\psi_{n}(x^{\prime})=\psi_{n}(x-x_{0}) \tag{8}
\end{gather*}
It is well-known that the energy levels of the free oscillator are
\begin{equation*}
E_{n}^{(0)}=(n+\frac{1}{2}) \hbar \omega, \quad n=0,1,2, \cdots \tag{9}
\end{equation*}
Thus,
\begin{align*}
E_{n} & =(n+\frac{1}{2}) \hbar \omega-\frac{1}{2} m \omega^{2} x_{0}^{2} \\
& =(n+\frac{1}{2}) \hbar \omega-\frac{q^{2} \mathscr{E}^{2}}{2 m \omega^{2}} \tag{10}
\end{align*}
Introducing the coordinate shift operator
\begin{equation*}
D_{x}(x_{0})=\mathrm{e}^{-\mathrm{i} x_{0} p / \hbar}=\mathrm{e}^{-x_{0} \frac{d}{d x}} \tag{11}
\end{equation*}
Its effect on the wave function is
\begin{equation*}
D_{x}(x_{0}) \psi(x)=\psi(x-x_{0}) \tag{11'}
\end{equation*}
Then the eigenfunctions of $H$ and $H_{0}$ can be related through the shift operator:
\begin{equation*}
\varphi_{n}(x)=\psi_{n}(x-x_{0})=D_{x}(x_{0}) \psi_{n}(x) \tag{12}
\end{equation*}
Conversely,
\begin{equation*}
\psi_{n}(x)=\varphi_{n}(x+x_{0})=D_{x}(-x_{0}) \varphi_{n}(x) \tag{$\prime$}
\end{equation*}
Solution two, using the raising and lowering operators of the oscillator
\begin{equation*}
a=\sqrt{\frac{m \omega}{2 \hbar}}(x+\frac{\mathrm{i}}{m \omega} p), \quad a^{+}=\sqrt{\frac{m \omega}{2 \hbar}}(x-\frac{\mathrm{i}}{m \omega} p) \tag{11}
\end{equation*}
Expressing $H_{0}$ and $H$ as
\begin{gather*}
H_{0}=(a^{+} a+\frac{1}{2}) \hbar \omega \tag{14}\\
H=(a^{+} a+\frac{1}{2}) \hbar \omega-q \mathscr{E} \sqrt{\frac{\hbar}{2 m \omega}}(a+a^{+}) \tag{15}
\end{gather*}
Introducing
\begin{equation*}
x_{0}=q \varepsilon / m \omega^{2}
\end{equation*}
Then
\begin{align*}
H & =\hbar \omega[a^{+} a+\frac{1}{2}-x_{0} \sqrt{\frac{m \omega}{2 \hbar}}(a+a^{+})] \\
& =\hbar \omega[(a^{+}-x_{0} \sqrt{\frac{m \omega}{2 \hbar}})(a-x_{0} \sqrt{\frac{m \omega}{2 \hbar}})+\frac{1}{2}-\frac{m \omega x_{0}^{2}}{2 \hbar}] \\
& =\hbar \omega[(a^{+}-\alpha_{0})(a-\alpha_{0})+\frac{1}{2}]-\frac{1}{2} m \omega^{2} x_{0}^{2} \tag{16}
\end{align*}
where
\begin{equation*}
\alpha_{0}=x_{0} \sqrt{m \omega / 2 \hbar} \tag{17}
\end{equation*}
Comparing equations (14), (16), the difference between $H$ and $H_{0}$ is $a \rightarrow a-\alpha_{0}, a^{+} \rightarrow a^{+}-\alpha_{0}$, and an added constant term $(-\frac{1}{2} m \omega^{2} x_{0}^{2})$.
Starting from the fundamental commutation relation
\begin{equation*}
[a, a^{+}]=1 \tag{18}
\end{equation*}
It was proved that the energy level formula (9) and the recursion relations between eigenstates
\begin{equation*}
a \psi_{n}=\sqrt{n} \psi_{n-1}, \quad a^{-} \psi_{n}=\sqrt{n+1} \psi_{n+1} \tag{19}
\end{equation*}
And the ground state wave function satisfies
\begin{equation*}
a \psi_{0}=0 \tag{20}
\end{equation*}
As
$[a-\alpha_{0}, a^{+}-\alpha_{0}]=[a, a^{+}]=1$
So the same reasoning logically leads to similar conclusions for the eigenvalues and eigenfunctions of $H$, simply substituting $(a-\alpha_{0})$ for $a$ in the entire derivation, $(a^{+}-\alpha_{0})$ for $a^{+}$. The eigenvalue of $H$ is clearly equation (10). The recursion relations and ground state equation for the eigenfunctions are
\begin{gather*}
(a-\alpha_{0}) \varphi_{n}(x)=\sqrt{n} \varphi_{n-1}(x) \tag{21}\\
(a^{+}-\alpha_{0}) \varphi_{n}(x)=\sqrt{n+1} \varphi_{n+1}(x) \\
(a-\alpha_{0}) \varphi_{0}(x)=0 \tag{22}
\end{gather*}
$a \rightarrow(a-\alpha_{0})$ (and $a^{+} \rightarrow a^{+}-\alpha_{0})$ are equivalent to $x \rightarrow(x-x_{0})$, thus replacing $x$ with $(x-x_{0})$ in $\psi_{n}(x)$ gives
\begin{equation*}
\varphi_{n}(x)=\psi_{n}(x-x_{0})=D_{x}(x_{0}) \psi_{n}(x) \tag{12}
\end{equation*}
$\varphi_{n}$ and $\varphi_{0}$ can be related through the operator $(a^{+}-\alpha_{0})$:
\begin{equation*}
\varphi_{n}(x)=\frac{1}{\sqrt{n!}}(a^{+}-\alpha_{0})^{n} \varphi_{0}(x) \tag{23}
\end{equation*}
|
[
"E_{n} =(n+\\frac{1}{2}) \\hbar \\omega-\\frac{q^{2} \\mathscr{E}^{2}}{2 m \\omega^{2}}"
] |
Expression
|
Theoretical Foundations
|
$E_n$: New energy levels of the oscillator in the electric field
$n$: Quantum number, $n=0,1,2, \cdots$
$\hbar$: Reduced Planck's constant
$\omega$: Angular frequency of the oscillator
$q$: Charge of the oscillator
$\mathscr{E}$: Uniform electric field
$m$: Mass of the oscillator
|
2
|
A particle of mass $m$ is in the ground state of a one-dimensional harmonic oscillator potential
\begin{equation*}
V_{1}(x)=\frac{1}{2} k x^{2}, \quad k>0
\end{equation*}
When the spring constant $k$ suddenly changes to $2k$, the potential then becomes
\begin{equation*}
V_{2}(x)=k x^{2}
\end{equation*}
Immediately measure the energy of the particle, and find the expression for the probability of the particle being in the ground state of the new potential $V_{2}$.
|
(a) The wave function of the particle $\psi(x, t)$ should satisfy the time-dependent Schrödinger equation
\begin{equation*}
\mathrm{i} \hbar \frac{\partial}{\partial t} \psi=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} \psi+V \psi \tag{3}
\end{equation*}
When $V$ undergoes a sudden change (from $V_{1} \rightarrow V_{2}$) but with a finite change quantity, $\psi$ remains a continuous function of $t$, implying that $\psi$ does not change when $V$ changes abruptly.
Denote $\psi_{0}(x)$ and $\phi_{0}(x)$ as the ground state wave functions of the potential $V_{1}$ and $V_{2}$, respectively. After the potential suddenly changes from $V_{1}$ to $V_{2}$, the wave function of the particle remains $\psi_{0}$. The probability of measuring the particle in the state $\phi_{0}$ is $|\langle\psi_{0} \mid \phi_{0}\rangle|^{2}$.
Rewrite $V_{1}$ and $V_{2}$ in standard form:
\begin{gather*}
V_{1}(x)=\frac{1}{2} k x^{2}=\frac{1}{2} m \omega_{1}^{2} x^{2} \tag{$\prime$}\\
V_{2}(x)=k x^{2}=\frac{1}{2} m \omega_{2}^{2} x^{2} \tag{$\prime$}
\end{gather*}
It is clear that
\begin{equation*}
\omega_{2}=\sqrt{2} \omega_{1} \tag{4}
\end{equation*}
$\psi_{0}$ and $\phi_{0}$ can be expressed as in formulas (3) and (5) from problem 3.2, namely
\begin{align}
\psi_{0}(x) &= \left( \frac{\alpha}{\sqrt{\pi}} \right)^{\frac{1}{2}} \mathrm{e}^{-\alpha^2 x^2 / 2}, &
\alpha^{2} &= m \omega_{1} / \hbar \notag \\
\phi_{0}(x) &= \left( \frac{\beta}{\sqrt{\pi}} \right)^{\frac{1}{2}} \mathrm{e}^{-\beta^2 x^2 / 2}, &
\beta^{2} &= m \omega_{2} / \hbar \tag{6}
\end{align}
where
\begin{equation*}
\beta^{2} / \alpha^{2}=\omega_{2} / \omega_{1}=\sqrt{2} \tag{7}
\end{equation*}
Thus
\begin{align*}
\langle\psi_{0} \mid \phi_{0}\rangle & =\sqrt{\frac{\alpha \beta}{\pi}} \int_{-\infty}^{+\infty} \mathrm{e}^{-\frac{1}{2}(\alpha^{2}+\beta^{2}) x^{2}} \mathrm{~d} x=(\frac{2 \alpha \beta}{\alpha^{2}+\beta^{2}})^{\frac{1}{2}} \\
|\langle\psi_{0} \mid \phi_{0}\rangle|^{2} & =\frac{2 \alpha \beta}{\alpha^{2}+\beta^{2}}=\frac{2 \beta / \alpha}{1+\beta^{2} / \alpha^{2}}=\frac{2^{5 / 4}}{1+\sqrt{2}}=0.9852 \tag{8}
\end{align*}
This is the required probability.
(b) Consider the time when the potential changes for the first time $(V_{1} \rightarrow V_{2})$ as $t=0$, then the wave function is
\begin{equation*}
\psi(x, 0)=\psi_{0}(x) \tag{9}
\end{equation*}
Let $\phi_{n}(x)$ denote the energy eigenstates of the potential $V_{2}$, corresponding to energy levels
$E_{n}=(n+\frac{1}{2}) \hbar \omega_{2}$
Expand $\psi_{0}$ as a linear combination of $\phi_{n}$,
\begin{equation*}
\psi_{0}(x)=\sum_{n} C_{n} \phi_{n}(x), \quad(n \text { can only take even values }) \tag{10}
\end{equation*}
For $0<t<\tau$, in Schrödinger equation (3) $V=V_{2}(x)$, its solution is
\begin{align*}
\psi(x, t) & =\sum_{n} C_{n} \phi_{n}(x) \mathrm{e}^{-\mathrm{i} E_{n} / / \hbar} \\
& =\mathrm{e}^{-i \omega_{2} t \cdot 2} \sum_{n} C_{n} \phi_{n}(x) \mathrm{e}^{-i \omega_{\omega_{2}} t} \tag{11}
\end{align*}
To have $\psi(x, \tau)=A \psi_{0}(x)$, it must hold that
\begin{equation*}
\mathrm{e}^{-\mathrm{in} \omega_{2} \tau}=1, \quad n=0,2,4, \cdots \tag{12}
\end{equation*}
That is
\begin{equation*}
\mathrm{e}^{\mathrm{i} \omega_{2} \tau}= \pm 1 \tag{12'}
\end{equation*}
The $\tau$ that satisfies this condition is
\begin{equation*}
\tau=l \pi / \omega_{2}=l \pi \sqrt{\frac{m}{2 k}}, \quad l=1,2,3, \cdots \tag{13}
\end{equation*}
When $t=\tau$, after the potential changes from $V_{2}$ back to $V_{1}$, the particle remains in the state $\psi_{0}$, with energy $E=\hbar \omega_{1} / 2$.
|
[
"\\frac{2^{5 / 4}}{1+\\sqrt{2}}"
] |
Expression
|
Theoretical Foundations
| |
3
|
A harmonic oscillator with charge $q$ is in a free vibration state at $t<0$ and $t>\tau$, with the total energy operator given by
\begin{equation*}
H_{0}=\frac{1}{2 m} p^{2}+\frac{1}{2} m \omega^{2} x^{2}
\end{equation*}
The energy eigenstates are denoted by $\psi_{n}$, and the energy levels $E_{n}^{(0)}=(n+\frac{1}{2}) \hbar \omega$ . When $0 \leqslant t \leqslant \tau$, a uniform electric field $\mathscr{E}$ is applied, and the total energy operator becomes
\begin{equation*}
H^{\prime}=\frac{1}{2 m} p^{2}+\frac{1}{2} m \omega^{2} x^{2}-q \mathscr{\varepsilon} x
\end{equation*}
Assuming the oscillator is in the ground state $\psi_{0}$ at $t \leqslant 0$, find the expression for the probability $P_n$ that the system is in the energy eigenstate $\psi_{n}$ at $t>\tau$.
|
At $t>\tau$, the external electric field has vanished, and the wavefunction satisfies the Schrödinger equation
\begin{equation*}
\mathrm{i} \hbar \frac{\partial}{\partial t} \psi(x, t)=H_{0} \psi(x, t) \tag{3}
\end{equation*}
The general solution is
\begin{equation*}
\psi(x, t)=\sum_{n} f_{n} \psi_{n}(x) \mathrm{e}^{-\mathrm{i} E_{n}^{(0)}(t-\tau) / \hbar} \tag{4}
\end{equation*}
where the components of the $\psi_{n}$ states are
\begin{equation*}
|\langle\psi_{n} \mid \psi\rangle|^{2}=|f_{n}|^{2} \tag{5}
\end{equation*}
Each coefficient $f_{n}$ depends on the wavefunction at $t=\tau$
\begin{equation*}
\psi(x, \tau)=\sum_{n} f_{n} \psi_{n}(x) \tag{6}
\end{equation*}
So the key is to find $\psi(x, \tau)$.
In the interval $0 \leqslant t \leqslant \tau$, the Schrödinger equation is
\begin{equation*}
\mathrm{i} \hbar \frac{\partial}{\partial t} \psi(x, t)=H \psi(x, t) \tag{7}
\end{equation*}
The solution is
\begin{equation*}
\psi(x, t)=\sum_{n} C_{n} \varphi_{n}(x) \mathrm{e}^{-\mathrm{i} \mathrm{E}_{n} t / \hbar} \tag{8}
\end{equation*}
where $C_{n}$ depends on the initial wavefunction
\begin{equation*}
\psi(x, 0)=\psi_{0}(x)=\sum_{n} C_{n} \varphi_{n}(x) \tag{9}
\end{equation*}
It has been proven that
\begin{equation*}
\psi_{0}(x)=\varphi_{0}(x+x_{0})=D_{x}(-x_{0}) \varphi_{0}(x) \tag{10}
\end{equation*}
where $x_{0}=q \mathscr{E} / m \omega^{2}$. If we express the displacement operator $D_{x}(-x_{0})$ in terms of the ladder operators,
\begin{equation*}
D_{x}(-x_{0})=\mathrm{e}^{\mathrm{i} x_{0} p / \hbar}=\mathrm{e}^{-a_{0}(a^{+}-a)} \tag{11}
\end{equation*}
Utilizing Glauber's formula
$\mathrm{e}^{A+B}=\mathrm{e}^{A} \mathrm{e}^{B} \mathrm{e}^{-\frac{1}{2}[A, B]}$
We obtain
\begin{align*}
D_{x}(-x_{0}) & =\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \mathrm{e}^{-\alpha_{0} a^{+}} \mathrm{e}^{\alpha_{0} a} \\
& =\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \mathrm{e}^{-\alpha_{0}(a^{+}-\alpha_{0})} \mathrm{e}^{\alpha_{0}(a-\alpha_{0})} \tag{$\prime$}
\end{align*}
where $\alpha_{0}=x_{0} \sqrt{m \omega / 2 \hbar}$. Substituting equation ( $11^{\prime}$ ) into equation (10), we obtain
\begin{equation*}
\psi_{0}(x)=\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \mathrm{e}^{-\alpha_{0}(a^{+}-\alpha_{0})} \mathrm{e}^{\alpha_{0}(a-\alpha_{0})} \varphi_{0}(x) \tag{12}
\end{equation*}
It has been shown in item 3.4
\begin{gather*}
(a-\alpha_{0}) \varphi_{0}=0 \tag{13}\\
(a^{+}-\alpha_{0})^{n} \varphi_{0}=\sqrt{n!} \varphi_{n} \tag{14}
\end{gather*}
Thus,
\begin{align*}
& \mathrm{e}^{\alpha_{0}(a-\alpha_{0})} \varphi_{0}=\sum_{n=0}^{\infty} \frac{\alpha_{0}^{n}}{n!}(a-\alpha_{0})^{n} \varphi_{0}=\varphi_{0} \tag{$\prime$}\\
& \begin{aligned}
\psi_{0}(x) & =\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \mathrm{e}^{-\alpha_{0}(a^{+}-\alpha_{0})} \varphi_{0} \\
& =\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \sum_{n} \frac{(-\alpha_{0})^{n}}{n!}(a^{+}-\alpha_{0})^{n} \varphi_{0} \\
& =\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \sum_{n} \frac{(-\alpha_{0})^{n}}{\sqrt{n!}} \varphi_{n}(x)
\end{aligned}
\end{align*}
namely, $\psi_{0}(x)$ is a coherent state wavefunction composed of the basis vectors $\varphi_{n}$. Comparing equations (9) and (15), we obtain
\begin{equation*}
C_{n}=\frac{(-\alpha_{0})^{n}}{\sqrt{n!}} \mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \tag{16}
\end{equation*}
Substituting equation (16) into equation (8), and using the energy formula derived in item 3.4
$E_{n}=(n+\frac{1}{2}) \hbar \omega-\frac{1}{2} m \omega^{2} x_{0}^{2}$
we obtain
\begin{equation*}
\psi(x, \tau)=\mathrm{e}^{\mathrm{i} \delta} \mathrm{e}^{-\frac{1}{2} a_{0}^{2}} \sum_{n} \frac{[-\alpha(\tau)]^{n}}{\sqrt{n!}} \varphi_{n}(x) \tag{17}
\end{equation*}
where
\begin{gather*}
\alpha(\tau)=\alpha_{0} \mathrm{e}^{-\mathrm{i} \omega \tau} \tag{18}\\
\delta=\frac{m \omega^{2} x_{0}^{2} \tau}{2 \hbar}-\frac{\omega \tau}{2} \tag{19}
\end{gather*}
$\psi(x, \tau)$ is also a coherent state wavefunction composed of the $\varphi_{n}$ states.
Using equation (14), equation (17) can be expressed as
\begin{equation*}
\psi(x, \tau)=\mathrm{e}^{\mathrm{i} \delta} \mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \mathrm{e}^{-\alpha(\tau)(a^{+}-\alpha_{0})} \varphi_{0}(x) \tag{20}
\end{equation*}
From equation (12 ${ }^{\prime}$ ), it follows that
$\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \varphi_{0}=\mathrm{e}^{\alpha_{0}(a^{+}-\alpha_{0})} \psi_{0}$
Substituting into equation (20), we obtain
\begin{equation*}
\psi(x, \tau)=\mathrm{e}^{\mathrm{i} \delta} \mathrm{e}^{[\alpha_{0}-\alpha(\tau)](a^{+}-a_{0})} \psi_{0}=\mathrm{e}^{\wp \delta} \mathrm{e}^{-\alpha_{0}^{2}+\alpha_{0} \alpha(\tau)} \mathrm{e}^{[\alpha_{0}-\alpha(\tau)] a^{+}} \psi_{0} \tag{21}
\end{equation*}
To express $\psi(x, \tau)$ as a linear superposition of the $\psi_{n}$, we use the formula
\begin{equation*}
(a^{+})^{n} \psi_{0}=\sqrt{n!} \psi_{n} \tag{22}
\end{equation*}
Thus, we have
\begin{equation*}
\psi(x, \tau)=\mathrm{e}^{\mathrm{i} \delta} \mathrm{e}^{\alpha_{0} \alpha(\tau)-\alpha_{0}^{2}} \sum_{n} \frac{[\alpha_{0}-\alpha(\tau)]^{n}}{\sqrt{n!}} \psi_{n}(x) \tag{23}
\end{equation*}
where the component fraction of the $\psi_{n}$ state is
\begin{align*}
|\langle\psi_{n} \mid \psi(x, \tau)\rangle|^{2} & =|\mathrm{e}^{\alpha_{0} \alpha(\tau)-\alpha_{0}^{2}}|^{2} \frac{|\alpha_{0}-\alpha(\tau)|^{2 n}}{n!} \\
& =\frac{1}{n!}(2 \alpha_{0} \sin \frac{\omega \tau}{2})^{2 n} \mathrm{e}^{-(2 \alpha_{0} \sin \frac{\omega \tau}{2})^{2}} \tag{24}
\end{align*}
It can be easily verified that
\begin{equation*}
\sum_{n}|\langle\psi_{n} \mid \psi(x, \tau)\rangle|^{2}=1 \tag{25}
\end{equation*}
This is a specific manifestation of the conservation of total probability.
By changing the external electric field duration $\tau$, each time
$\tau=k 2 \pi / \omega, \quad k=1,2,3, \cdots$
In $\psi(x, \tau)$, the components of each excited state $(n \geqslant 1)$ become 0, and the ground state $(\psi_{0})$ component is 1. Each time
$\tau=(2 k+1) \pi / \omega, \quad k=0,1,2, \cdots$
the ground state component in $\psi(x, \tau)$ reaches a minimum value of $\mathrm{e}^{-4 \alpha_{0}^{2}}, \psi_{n}$ state components are
$$|f_{n}|^{2}=\frac{(4 \alpha_{0}^{2})^{n}}{n!} \mathrm{e}^{-4 \alpha_{0}^{2}}$$
|
[
"P_n = \\frac{1}{n!}(2 \\alpha_{0} \\sin \\frac{\\omega \\tau}{2})^{2 n} \\mathrm{e}^{-(2 \\alpha_{0} \\sin \\frac{\\omega \\tau}{2})^{2}}"
] |
Expression
|
Theoretical Foundations
|
$P_n$: Probability that the system is in the energy eigenstate $\psi_n$ at $t>\tau$.
$n$: Quantum number, representing the energy level.
$\alpha_0$: Dimensionless parameter related to the displacement, $\alpha_{0}=x_{0} \sqrt{m \omega / 2 \hbar}$.
$\omega$: Angular frequency of the harmonic oscillator.
$\tau$: Duration for which the uniform electric field is applied.
$\mathrm{e}$: Base of the natural logarithm.
|
4
|
Calculate the result of the commutator $[\boldsymbol{p}, \frac{1}{r}]$.
|
Using the commutator
\begin{equation*}
[\boldsymbol{p}, F(\boldsymbol{r})]=-\mathrm{i} \hbar \nabla F \tag{1}
\end{equation*}
We obtain
\begin{equation*}
[p, \frac{1}{r}]=-\mathrm{i} \hbar(\nabla \frac{1}{r})=\mathrm{i} \hbar \frac{r}{r^{3}} \tag{2}
\end{equation*}
Using the formula (see question 4.2)
$[\boldsymbol{A} \cdot \boldsymbol{B}, F]=[\boldsymbol{A}, F] \cdot \boldsymbol{B}+\boldsymbol{A} \cdot[\boldsymbol{B}, F]$
We obtain
\[
\begin{aligned}
{[\boldsymbol{p}^{2}, \frac{1}{r}] } & =[\boldsymbol{p}, \frac{1}{r}] \cdot \boldsymbol{p}+\boldsymbol{p} \cdot[\boldsymbol{p}, \frac{1}{r}]=\mathrm{i} \hbar(\frac{1}{r^{3}} \boldsymbol{r} \cdot \boldsymbol{p}+\boldsymbol{p} \cdot \frac{\boldsymbol{r}}{r^{3}}) \\
& =2 \mathrm{i} \hbar \frac{1}{r^{3}} \boldsymbol{r} \cdot \boldsymbol{p}+\hbar^{2}(\nabla \cdot \frac{\boldsymbol{r}}{r^{3}})
\end{aligned}
\]
However, since
$\nabla \cdot \frac{\boldsymbol{r}}{r^{3}}=\frac{1}{r^{3}} \nabla \cdot \boldsymbol{r}+\boldsymbol{r} \cdot(\nabla \frac{1}{r^{3}})=\frac{3}{r^{3}}-\boldsymbol{r} \cdot \frac{3 \boldsymbol{r}}{r^{5}}=0$
So
\begin{equation*}
[\boldsymbol{p}^{2}, \frac{1}{r}]=2 \mathrm{i} \hbar \frac{1}{r^{3}} \boldsymbol{r} \cdot \boldsymbol{p}=2 \hbar^{2} \frac{1}{r^{2}} \frac{\partial}{\partial r} \tag{3}
\end{equation*}
Subsequently, using equation (1), we get
\begin{align*}
{[\boldsymbol{p}, r^{2}] } & =-i \hbar \nabla(r^{2})=-2 i \hbar \boldsymbol{r} \tag{4}\\
{[\boldsymbol{p}^{2}, r^{2}] } & =[\boldsymbol{p}, r^{2}] \cdot \boldsymbol{p}+\boldsymbol{p} \cdot[\boldsymbol{p}, r^{2}]=-2 i \hbar(\boldsymbol{r} \cdot \boldsymbol{p}+\boldsymbol{p} \cdot \boldsymbol{r}) \\
& =-4 i \hbar \boldsymbol{r} \cdot \boldsymbol{p}-2 \hbar^{2} \nabla \cdot \boldsymbol{r}=-4 i \hbar \boldsymbol{r} \cdot \boldsymbol{p}-6 \hbar^{2} \\
& =-4 \hbar^{2} r \frac{\partial}{\partial r}-6 \hbar^{2} \tag{5}
\end{align*}
Finally, using the commutator
\begin{equation*}
[\boldsymbol{p}^{2}, \boldsymbol{r}]=-\mathrm{i} \hbar \frac{\partial p^{2}}{\partial \boldsymbol{p}}=-2 \mathrm{i} \hbar \boldsymbol{p} \tag{6}
\end{equation*}
And equation (3), we get
\begin{align*}
{[\boldsymbol{p}^{2}, \frac{\boldsymbol{r}}{r}] } & =[\boldsymbol{p}^{2}, \boldsymbol{r}] \frac{1}{r}+\boldsymbol{r}[\boldsymbol{p}^{2}, \frac{1}{r}] \\
& =-2 \mathrm{i} \hbar \boldsymbol{p} \frac{1}{r}+2 \mathrm{i} \hbar \boldsymbol{r} \frac{1}{r^{3}}(\boldsymbol{r} \cdot \boldsymbol{p}) \\
& =-2 \mathrm{i} \hbar[\frac{1}{r} \boldsymbol{p}-\mathrm{i} \hbar(\nabla \frac{1}{r})]+2 \mathrm{i} \hbar \frac{\boldsymbol{r}}{r^{3}}(\boldsymbol{r} \cdot \boldsymbol{p}) \\
& =2 \hbar^{2} \frac{\boldsymbol{r}}{r^{3}}+2 \mathrm{i} \hbar[\frac{\boldsymbol{r}}{r^{3}}(\boldsymbol{r} \cdot \boldsymbol{p})-\frac{1}{r} \boldsymbol{p}] \\
& =2 \hbar^{2}(\frac{\boldsymbol{r}}{r^{3}}+\boldsymbol{r} \frac{1}{r^{2}} \frac{\partial}{\partial r}-\frac{1}{r} \nabla) \tag{7}
\end{align*}
|
[
"\\mathrm{i} \\hbar \\frac{\\boldsymbol{r}}{r^{3}}"
] |
Expression
|
Theoretical Foundations
|
$\mathrm{i}$: Imaginary unit.
$\hbar$: Reduced Planck's constant.
$\boldsymbol{r}$: Position vector.
$r$: Magnitude of the position vector $\boldsymbol{r}$.
|
5
|
For the hydrogen-like ion (nuclear charge $Z e$) with the $(H, l^{2}, l_{z})$ common eigenstate $\psi_{n l m}$, it is known that the various $\langle r^{\lambda}\rangle$ satisfy the following recursion relation (Kramers' formula):
\begin{equation*}
\frac{\lambda+1}{n^{2}}\langle r^{\lambda}\rangle-(2 \lambda+1) \frac{a_{0}}{Z}\langle r^{\lambda-1}\rangle+\frac{\lambda}{4}[(2 l+1)^{2}-\lambda^{2}] \frac{a_{0}^{2}}{Z^{2}}\langle r^{\lambda-2}\rangle=0
\end{equation*}
It is known that $\langle r^{0}\rangle=1$. Use this formula to calculate the expression for $\langle r\rangle_{n l m}$.
|
The spherical coordinate expression of $\psi_{n l m}$ is
\begin{equation*}
\psi_{n l m}=R_{n l}(r) \mathrm{Y}_{l m}(\theta, \varphi)=\frac{1}{r} u_{n l}(r) \mathrm{Y}_{l m}(\theta, \varphi) \tag{2}
\end{equation*}
The expectation value of $r^{\lambda}$ is
\begin{equation*}
\langle r^{\lambda}\rangle_{n l m}=\int r^{\lambda}|\psi_{n l m}|^{2} \mathrm{~d}^{3} x=\int_{0}^{\infty} r^{\lambda}(u_{n l})^{2} \mathrm{~d} r \tag{3}
\end{equation*}
$u_{n l}$ satisfies the radial equation
\begin{equation*}
-\frac{\hbar^{2}}{2 \mu} u^{\prime \prime}+[l(l+1) \frac{\hbar^{2}}{2 \mu r^{2}}-\frac{Z e^{2}}{r}] u=E_{n} u \tag{4}
\end{equation*}
Because
\begin{equation*}
E_{n}=-\frac{Z^{2} e^{2}}{2 n^{2} a_{0}}, \quad a_{0}=\frac{\hbar^{2}}{\mu e^{2}} \tag{5}
\end{equation*}
Equation (4) can be rewritten as
\begin{equation*}
u^{\prime \prime}+[\frac{2 Z}{a_{0} r}-\frac{l(l+1)}{r^{2}}-(\frac{Z}{n a_{0}})^{2}] u=0 \tag{$\prime$}
\end{equation*}
Multiply each term of equation ($4^{\prime}$) by $r^{\lambda} u$ and integrate $\int_{0}^{\infty} \cdots \mathrm{d} r$, the last three terms clearly yield the expectation values of $r^{\lambda-1}, ~ r^{\lambda-2}$ and $r^{\lambda}$, while
The first term gives
\begin{align*}
\int_{0}^{\infty} r^{\lambda} u u^{\prime \prime} \mathrm{d} r & =r^{\lambda} u u^{\prime}|_{0} ^{\infty}-\int_{0}^{\infty}(r^{\lambda} u^{\prime}+\lambda r^{\lambda-1} u) u^{\prime} \mathrm{d} r \\
& =(r^{\lambda} u u^{\prime}-\frac{\lambda}{2} r^{\lambda-1} u^{2})|_{0} ^{\infty}+\frac{\lambda(\lambda-1)}{2}\langle r^{\lambda-2}\rangle-\int_{0}^{\infty} r^{\lambda}(u^{\prime})^{2} \mathrm{~d} r \tag{6}
\end{align*}
If the value of $\lambda$ ensures that
\begin{equation*}
r^{\lambda} u u^{\prime}|_{0} ^{\infty}=0,\quad r^{\lambda-1} u^{2}|_{0} ^{\infty}=0 \tag{7}
\end{equation*}
We obtain the following preliminary result:
\begin{equation*}
[\frac{\lambda(\lambda-1)}{2}-l(l+1)]\langle r^{\lambda-2}\rangle+\frac{2 Z}{a_{0}}\langle r^{\lambda-1}\rangle-(\frac{Z}{n a_{0}})^{2}\langle r^{\lambda}\rangle=\int_{0}^{\infty} r^{\lambda}(u^{\prime})^{2} \mathrm{~d} r \tag{8}
\end{equation*}
Moreover, multiply each term of equation ($4^{\prime}$) by $2 r^{\lambda+1} u^{\prime}$ and integrate, successively obtaining
\[
\begin{aligned}
& \int_{0}^{\infty} 2 r^{\lambda+1} u^{\prime} u^{\prime \prime} \mathrm{d} r=r^{\lambda+1}(u^{\prime})^{2}|_{0} ^{\infty}-\int_{0}^{\infty}(\lambda+1) r^{\lambda}(u^{\prime})^{2} \mathrm{~d} r \\
& \int_{0}^{\infty} 2 r^{\lambda+1} u^{\prime} u \mathrm{~d} r=r^{\lambda+1} u^{2}|_{0} ^{\infty}-(\lambda+1)\langle r^{\lambda}\rangle \\
& \int_{0}^{\infty} 2 r^{\lambda+1} u^{\prime} \frac{u}{r} \mathrm{~d} r=r^{\lambda} u^{2}|_{0} ^{\infty}-\lambda\langle r^{\lambda-1}\rangle \\
& \int_{0}^{\infty} 2 r^{\lambda+1} u^{\prime} \frac{u}{r^{2}} \mathrm{~d} r=r^{\lambda-1} u^{2}|_{0} ^{\infty}-(\lambda-1)\langle r^{\lambda-2}\rangle
\end{aligned}
\]
Under the conditions guaranteed by equation (7), all first terms in the above equations are zero, thus
\begin{equation*}
(\lambda-1) l(l+1)\langle r^{\lambda-2}\rangle-2 \lambda \frac{Z}{a_{0}}\langle r^{\lambda-1}\rangle+(\lambda+1)(\frac{Z}{n a_{0}})^{2}\langle r^{\lambda}\rangle=(\lambda+1) \int_{0}^{\infty} r^{\lambda}(u^{\prime})^{2} \mathrm{~d} r \tag{9}
\end{equation*}
Combine equations (8) and (9), eliminating the integrals on the right to obtain equation (1). The condition for the validity of equation (1) is equation (7). Given ${ }^{(1)}$
\begin{align*}
& r \rightarrow 0, \quad u \sim r^{l+1} \\
& r \rightarrow \infty, \quad u \sim r^{n} \mathrm{e}^{-Z r / n a_{0}} \tag{10}
\end{align*}
It is evident that the necessary and sufficient condition for equation (7) to hold is
\begin{equation*}
\lambda>-(2 l+1) \tag{11}
\end{equation*}
In equation (1), taking $\lambda=0$ and noting $\langle r^{0}\rangle=1$ immediately yields
\begin{equation*}
\langle\frac{1}{r}\rangle_{n l m}=\frac{Z}{n^{2} a_{0}} \tag{12}
\end{equation*}
\footnotetext{
(1) Refer to Zeng Jin-Yan. Quantum Mechanics Volume I. Beijing: Science Press, 1997. §6.3.
}
This result has been obtained using the virial theorem.
Sequentially taking $\lambda=1, 2$, leads to
\begin{gather*}
\langle r\rangle_{n l m}=\frac{1}{2}[3 n^{2}-l(l+1)] \frac{a_{0}}{Z} \tag{13}\\
\langle r^{2}\rangle_{n l m}=\frac{n^{2}}{2}[1+5 n^{2}-3 l(l+1)](\frac{a_{0}}{Z})^{2} \tag{14}
\end{gather*}
For example
\[
\begin{array}{rc}
1 \mathrm{~s} \text { state (ground state), } & \langle r\rangle_{100}=\frac{3}{2},\langle r^{2}\rangle_{100}=3 \\
2 \mathrm{~s} \text { state, } & \langle r\rangle_{200}=6,\langle r^{2}\rangle_{200}=42 \tag{15}\\
2 \mathrm{p} \text { state, } & \langle r\rangle_{21 m}=5,\langle r^{2}\rangle_{21 m}=30
\end{array}
\]
In equation (15), $\langle r\rangle$ is measured in units of $a_{0} / Z$ and $\langle r^{2}\rangle$ is measured in units of $a_{0}^{2} / Z^{2}$.
Note that in the context of this problem, equation (1) cannot be used to calculate $\langle r^{-2}\rangle$, but if the result from the previous problem on $\langle r^{-2}\rangle$ is used, then by substituting $\lambda=-1$ into equation (1), $\langle r^{-3}\rangle$ can be calculated, with results consistent with the previous problem. Furthermore, by taking $\lambda=-2(l \geqslant 1)$, it is possible to calculate
\begin{equation*}
\langle r^{-4}\rangle=(\frac{Z}{a_{0}})^{4} \frac{3 n^{2}-l(l+1)}{2 n^{5}(l-\frac{1}{2}) l(l+\frac{1}{2})(l+1)(l+\frac{3}{2})} \tag{16}
\end{equation*}
Calculating other expectations $\langle r^{\lambda}\rangle$ can follow this analogy ${ }^{(1)}$.
|
[
"\\langle r\\rangle_{n l m}=\\frac{1}{2}[3 n^{2}-l(l+1)] \\frac{a_{0}}{Z}"
] |
Expression
|
Theoretical Foundations
|
$\langle r\rangle_{nlm}$: Expectation value of the radial coordinate $r$ for the state $\psi_{nlm}$.
$n$: Principal quantum number.
$l$: Azimuthal (orbital) quantum number.
$a_0$: Bohr radius, defined as $a_0 = \frac{\hbar^2}{\mu e^2}$.
$Z$: Nuclear charge number of the hydrogen-like ion.
|
6
|
Three-dimensional isotropic harmonic oscillator, the total energy operator is
\begin{equation*}
H=\frac{\boldsymbol{p}^{2}}{2 \mu}+\frac{1}{2} \mu \omega^{2} r^{2}=-\frac{\hbar^{2}}{2 \mu} \nabla^{2}+\frac{1}{2} \mu \omega^{2} r^{2}
\end{equation*}
For the common eigenstates of $(H, l^{2}, l_{z})$
\begin{equation*}
\psi_{n_{r} l m}=R_{n_{r} l}(r) \mathrm{Y}_{l m}(\theta, \varphi)=\frac{1}{r} u_{n_{r} l}(r) \mathrm{Y}_{l m}(\theta, \varphi)
\end{equation*}
Calculate the expression for $\langle r^{-2}\rangle_{n_{r} l m}$.
|
The energy levels of the three-dimensional isotropic harmonic oscillator are
\begin{equation*}
E_{n, l m}=E_{N}=\left(N+\frac{3}{2}\right) \hbar \omega, \quad N=l+2 n_{r} . \tag{3}
\end{equation*}
For $\psi_{n_{r}, l m}, H$ is equivalent to
\footnotetext{
(1) Refer to H. A. Kramers. Quantum Mechanics. Amsterdam: North-Holland, 1958. § 59.
}
\begin{equation*}
H \rightarrow-\frac{\hbar^{2}}{2 \mu} \frac{1}{r} \frac{\partial^{2}}{\partial r^{2}} r+l(l+1) \frac{\hbar^{2}}{2 \mu r^{2}}+\frac{1}{2} \mu \omega^{2} r^{2} \tag{4}
\end{equation*}
According to the Hellmann theorem, there should be
\begin{equation*}
\frac{\partial E_{n_{l} l m}}{\partial l}=\langle\frac{\partial H}{\partial l}\rangle_{n_{r} l m}=(l+\frac{1}{2}) \frac{\hbar^{2}}{\mu}\langle\frac{1}{r^{2}}\rangle_{n_{r}, l m} \tag{5}
\end{equation*}
From equation (3) it is evident
\begin{equation*}
\frac{\partial E_{n, l m}}{\partial l}=\frac{\partial E_{N}}{\partial N}=\hbar \omega \tag{6}
\end{equation*}
Substituting into equation (5), we get
\begin{equation*}
\langle\frac{1}{r^{2}}\rangle_{n, l m}=\frac{1}{l+\frac{1}{2}} \frac{\mu \omega}{\hbar}=\frac{\alpha^{2}}{l+\frac{1}{2}}, \quad \alpha=\sqrt{\frac{\mu \omega}{\hbar}} \tag{7}
\end{equation*}
The average value of the centrifugal potential energy is
\begin{equation*}
\langle\frac{l^{2}}{2 \mu r^{2}}\rangle_{n_{r}, m}=\frac{l(l+1)}{2 l+1} \hbar \omega \tag{8}
\end{equation*}
Note that the average value of the centrifugal potential energy is directly determined by the angular quantum number $l$ and is independent of the principal quantum number $N$. Among the states with the same energy level $E_{N}$, the centrifugal potential energy is highest in the state $l=N$.
\begin{equation*}
\langle\frac{l^{2}}{2 \mu r^{2}}\rangle_{l=N}=\frac{N(N+1)}{2 N+1} \hbar \omega \tag{9}
\end{equation*}
The corresponding radial kinetic energy is only
\begin{align*}
\langle\frac{p_{r}^{2}}{2 \mu}\rangle_{l=N} & =\langle\frac{\boldsymbol{p}^{2}}{2 \mu}-\frac{l^{2}}{2 \mu r^{2}}\rangle_{l=N}=\frac{E_{N}}{2}-\frac{N(N+1)}{2 N+1} \hbar \omega \\
& =(1+\frac{1}{4 N+2}) \frac{\hbar \omega}{2} \tag{10}
\end{align*}
When $N \gg 1$, it follows
\begin{equation*}
\langle\frac{p_{r}^{2}}{2 \mu}\rangle_{l=N} \approx \frac{1}{2} \hbar \omega \tag{$\prime$}
\end{equation*}
This situation corresponds to the circular orbit in Bohr's quantum theory. The general formula for the average value of radial kinetic energy is
\begin{equation*}
\langle\frac{p_{r}^{2}}{2 \mu}\rangle_{n_{r} l m}=\frac{E_{N}}{2}-\langle\frac{l^{2}}{2 \mu r^{2}}\rangle_{n_{r} l m}=[N+\frac{3}{2}-\frac{l(l+1)}{l+1 / 2}] \frac{\hbar \omega}{2} \tag{11}
\end{equation*}
If $n_{r} \ggg 1$ (quasi-classical case), the approximate treatment can be made as follows:
$$\frac{l(l+1)}{l+\frac{1}{2}} \approx l+\frac{1}{2}$$
Equations (8) and (11) become
\begin{align*}
& \langle\frac{l^{2}}{2 \mu r^{2}}\rangle_{n_{r} l m} \approx(l+\frac{1}{2}) \frac{\hbar \omega}{2}=(\frac{l}{2}+\frac{1}{4}) \hbar \omega \tag{$\prime$}\\
& \langle\frac{p_{r}^{2}}{2 \mu}\rangle_{n_{r} l m} \approx(N+1-l) \frac{\hbar \omega}{2}=(n_{r}+\frac{1}{2}) \hbar \omega \tag{11'}
\end{align*}
|
[
"\\langle\\frac{1}{r^{2}}\\rangle_{n, l m}=\\frac{1}{l+\\frac{1}{2}} \\frac{\\mu \\omega}{\\hbar}"
] |
Expression
|
Theoretical Foundations
|
$\langle\frac{1}{r^{2}}\rangle_{n, l m}$: Expectation value of $1/r^2$ for the state $\psi_{n_r l m}$ (where `n` is the radial quantum number)
$l$: Orbital angular momentum quantum number
$\mu$: Reduced mass
$\omega$: Angular frequency of the harmonic oscillator
$\hbar$: Reduced Planck's constant
|
7
|
A particle with mass $\mu$ moves in a "spherical square well" potential, $$V(r)= \begin{cases}0, & r<a \\ V_{0}>0, & r \geqslant a.\end{cases}$$ Consider only the bound state $(0<E<V_{0})$. As $V_{0}$ gradually increases from small to large, find the value of $V_{0} a^{2}$ when the first bound state (with angular quantum number $l=0$ and energy $E \approx V_0$) appears.
|
As a central force problem, the bound state wave function can be taken as the common eigenfunction of $(H, l^{2}, l_{z})$, written as
\begin{equation*} \psi=R(r) \mathrm{Y}_{l m}(\theta, \varphi) \tag{2} \end{equation*}
The radial equation is
\begin{equation*} R^{\prime \prime}+\frac{2}{r} R^{\prime}+[\frac{2 \mu}{\hbar^{2}}(E-V)-\frac{l(l+1)}{r^{2}}] R=0 \tag{3} \end{equation*}
When $V_{0}$ gradually increases, and a new bound state $E \approx V_{0}$ appears, the above equation becomes
\begin{align*} & R^{\prime \prime}+\frac{2}{r} R^{\prime}+[\frac{2 \mu V_{0}}{\hbar^{2}}-\frac{l(l+1)}{r^{2}}] R=0, \quad r<a(\text { inside the well }) \tag{4a}\\ & R^{\prime \prime}+\frac{2}{r} R^{\prime}-\frac{l(l+1)}{r^{2}} R=0, \quad r>a(\text { outside the well }) \tag{4b} \end{align*}
The equation inside the well (4a) is precisely the spherical Bessel equation, and the physically allowed solution is the spherical Bessel function
\begin{equation*} R(r)=j_{l}(k_{0} r), \quad k_{0}=\sqrt{2 \mu V_{0}} / \hbar \tag{5a} \end{equation*}
The solution outside the well (4b) is
\begin{equation*} R(r)=C / r^{l+1} \tag{5b} \end{equation*}
(The other solution $r^{l}$ does not satisfy the bound state boundary condition $R \rightarrow 0$ as $r \rightarrow \infty$, so it is discarded.) For the wave function outside the well (5b), it is evident that
\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} r}[r^{l+1} R(r)]=0, \quad r \geqslant a \tag{6} \end{equation*}
At $r=a$, $R$ and $R^{\prime}$ should both be continuous. Therefore, for the wave function inside the well (5a), as $r \rightarrow a$, it should also satisfy condition (6), that is,
\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} r}[r^{l+1} j_{l}(k_{0} r)]_{r=a}=0 \tag{$\prime$} \end{equation*}
Using the formula
$\frac{\mathrm{d}}{\mathrm{~d} x}[x^{l+1} j_{l}(x)]=x^{l+1} j_{l-1}(x)$ Equation $(6^{\prime})$
can be transformed into
\begin{equation*} j_{l-1}(k_{0} a)=0 \tag{7} \end{equation*}
This is the condition for a new bound state (angular quantum number $l$, energy level $E_{n l} \approx V_{0}$) to appear. For the first bound state $l=0$, considering that $j_{-1}(k_{0} a)=\frac{\cos k_{0} a}{k_{0} a}$ The condition for the appearance of a new s-state $(l=0)$ energy level $(E \approx V_{0})$ is \begin{equation*} \cos k_{0} a=0, \quad k_{0} a=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \cdots \tag{8} \end{equation*} When the first bound state appears, $k_{0} a=\pi / 2$, which means \begin{equation*} V_{0} a^{2}=\frac{\pi^{2} \hbar^{2}}{8 \mu} \tag{9} \end{equation*} As $V_{0}$ gradually increases, whenever condition (7) is satisfied, a new energy level $E_{n l} \approx V_{0}$ appears. The order of appearance of each energy level can be determined based on the zeros of the spherical Bessel function $j_{l}(x)$:
\[
\begin{aligned}
&1~\mathrm{s},\quad 1~\mathrm{p},\quad 1~\mathrm{d},\quad 2~\mathrm{s},\quad 1~\mathrm{f},\quad 2~\mathrm{p},\quad 1~\mathrm{g},\quad 2~\mathrm{d},\quad 3~\mathrm{s},\\
&1~\mathrm{h},\quad 2~\mathrm{f},\quad 1~\mathrm{i},\quad 3~\mathrm{p},\quad 2~\mathrm{g},\quad 1~\mathrm{k},\quad 3~\mathrm{d},\quad 4~\mathrm{s},\ \cdots
\end{aligned}
\]
The $l$ values corresponding to each spectral notation are:
\[
\begin{array}{ccccccccc}
l & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\text{Letter} & \mathrm{s} & \mathrm{p} & \mathrm{d} & \mathrm{f} & \mathrm{g} & \mathrm{h} & \mathrm{i} & \mathrm{j}
\end{array}
\]
When $V_0$ is large, the total number of bound states can be approximated by “one bound state per $h^3$ volume in phase space.” The maximum momentum for a particle in a bound state inside the well is:
\begin{equation*} p_{0}=\hbar k_{0}=\sqrt{2 \mu V_{0}} \tag{10} \end{equation*}
Therefore, the total phase space volume occupied by the bound states is $\frac{4 \pi}{3} a^{3} \cdot \frac{4 \pi}{3} p_{0}^{3}=\frac{16}{9} \pi^{2}(a \hbar k_{0})^{3}=\Omega$ The total number of bound states is \begin{align*} N & \approx \Omega /(2 \pi \hbar)^{3}=\frac{2 \pi^{2}}{9}(\frac{a k_{0}}{\pi})^{3} \\ & =\frac{2 \pi^{2}}{9}(\frac{a}{\pi \hbar})^{3}(2 \mu V_{0})^{3 / 2} \tag{11} \end{align*} For example, when $a k_{0}=7 \pi / 2$, the highest energy level is 4 s, and counting all the states from 1 s to 4 s (consider the degeneracy of energy levels $E_{n l}$ as $2 l+1$) gives 99, while equation (11) yields $N \approx 94$, which is indeed very close ${ }^{(1)}$.
|
[
"V_{0} a^{2}=\\frac{\\pi^{2} \\hbar^{2}}{8 \\mu}"
] |
Expression
|
Theoretical Foundations
|
$V_0$: Height of the potential barrier
$a$: Radius of the spherical square well
$\hbar$: Reduced Planck's constant
$\mu$: Mass of the particle
|
8
|
For the common eigenstate $|l m\rangle$ of $l^{2}$ and $l_{z}$, calculate the expectation value $\overline{l_{n}}$ of $l_{n}=\boldsymbol{n} \cdot \boldsymbol{l}$. Here, $\boldsymbol{n}$ is a unit vector in an arbitrary direction, and its angle with the $z$-axis is $\gamma$.
|
Using the basic commutation relation $\boldsymbol{l} \times \boldsymbol{l}=\mathrm{i} \hbar \boldsymbol{l}$, we have
\[
\begin{aligned}
& \mathrm{i} \hbar l_{x} l_{y}=(l_{y} l_{z}-l_{z} l_{y}) l_{y}=l_{y} l_{z} l_{y}-l_{z} l_{y}^{2} y \\
& \mathrm{i} \hbar l_{y} l_{x}=l_{y}(l_{y} l_{z}-l_{z} l_{y})=l_{y}^{2} l_{z}-l_{y} l_{z} l_{y}
\end{aligned}
\]
Calculating the expectation value in the state $|l m\rangle$, since
$\overline{l_{z} l_{y}^{2}}=\overline{l_{y}^{2} l_{z}}=m \hbar \overline{l_{y}^{2}}$
Therefore
$\overline{l_{y} l_{x}}=-\overline{l_{x} l_{y}}$
Thus
\begin{align*}
& \overline{l_{x} l_{y}}-\overline{l_{y} l_{x}}=2 \overline{l_{x} l_{y}}=\mathrm{i} \hbar \overline{l_{z}}=\mathrm{i} \hbar^{2} m \\
& \overline{l_{x} l_{y}}=\mathrm{i}^{2} m / 2, \quad \overline{l_{y} l_{x}}=-\mathrm{i}^{2} m / 2 \tag{1}
\end{align*}
The projection operator of $\boldsymbol{l}$ in the direction of $\boldsymbol{n}$ is
$l_{n}=\boldsymbol{n} \cdot \boldsymbol{l}=l_{x} \cos \alpha+l_{y} \cos \beta+l_{z} \cos \gamma$
For the state $|l m\rangle$, since $\overline{l_{x}}=0, \overline{l_{y}}=0$, we have
\begin{gather*}
\overline{l_{n}}=m \hbar \cos \gamma \tag{2}\\
l_{n}^{2}=l_{x}^{2} \cos ^{2} \alpha+l_{y}^{2} \cos ^{2} \beta+l_{z}^{2} \cos ^{2} \gamma \\
\\
+(l_{x} l_{y}+l_{y} l_{x}) \cos \alpha \cos \beta+\cdots \text { (similar terms with cyclic permutations) }
\end{gather*}
Calculating the expectation value in the state $|l m\rangle$, since
\begin{gather*}
\overline{l_{x} l_{y}}+\overline{l_{y} l_{x}}=0, \quad \overline{l_{y} l_{z}}+\overline{l_{z} l_{y}}=2 m \hbar \overline{l_{y}}=0, \cdots \\
\overline{l_{x}^{2}}=\overline{l_{y}^{2}}=\frac{1}{2}(\overline{l^{2}}-\overline{l_{z}^{2}})=\frac{\hbar^{2}}{2}{l(l+1)-m^{2}} \tag{3}
\end{gather*}
Thus
\begin{align*}
\overline{l_{n}^{2}} & =m^{2} \hbar^{2} \cos ^{2} \gamma+\frac{\hbar^{2}}{2}{l(l+1)-m^{2}}(\cos ^{2} \alpha+\cos ^{2} \beta) \\
& =m^{2} \hbar^{2} \cos ^{2} \gamma+\frac{\hbar^{2}}{2}{l(l+1)-m^{2}}(1-\cos ^{2} \gamma) \\
& =\frac{\hbar^{2}}{2}{l(l+1)(1-\cos ^{2} \gamma)+m^{2}(3 \cos ^{2} \gamma-1)} \tag{4}
\end{align*}
If $\boldsymbol{n}$ is orthogonal to the $z$-axis, that is, $\cos \gamma=0$, then $\overline{l_{n}^{2}}=\overline{l_{x}^{2}}=\overline{l_{y}^{2}}, \overline{l_{n}}=0$.
|
[
"m \\hbar \\cos \\gamma"
] |
Expression
|
Theoretical Foundations
|
$m$: Magnetic quantum number, associated with the z-component of angular momentum.
$\hbar$: Reduced Planck's constant.
$\gamma$: Angle of the unit vector $\boldsymbol{n}$ with the $z$-axis.
|
9
|
For an electron's spin state $\chi_{\frac{1}{2}}(\sigma_{z}=1)$ (i.e., the state where the Pauli matrix $\sigma_z$ has an eigenvalue of $+1$), if we measure its spin projection in an arbitrary direction $\boldsymbol{n}$, $\sigma_n = \boldsymbol{\sigma} \cdot \boldsymbol{n}$, where $\boldsymbol{n}$ is a unit vector and $n_z$ is its component along the $z$-axis. Find the expression for the probability of measuring $\sigma_n = +1$ (expressed in terms of $n_z$).
|
One Using the eigenfunctions of $\sigma_{n}$ obtained from the previous problem, it is easy to find
(a) In the spin state $\chi_{\frac{1}{2}}=[\begin{array}{l}1 \\ 0\end{array}]$,
the probability of $\sigma_{n}=1$ is
\begin{equation*}
|\langle\phi_{1} \lvert\, \chi_{\frac{1}{2}}\rangle|^{2}=\cos ^{2} \frac{\theta}{2}=\frac{1}{2}(1+n_{z}) \tag{1}
\end{equation*}
the probability of $\sigma_{n}=-1$ is
\begin{equation*}
|\langle\phi_{-1} \lvert\, \chi_{\frac{1}{2}}\rangle|^{2}=\sin ^{2} \frac{\theta}{2}=\frac{1}{2}(1-n_{z}) \tag{2}
\end{equation*}
(b) In the spin state $\phi_{1}(\sigma_{n}=1)$,
the probability of $\sigma_{z}=1$ is
\begin{equation*}
|\langle\chi_{\frac{1}{2}} \rvert\, \phi_{1}\rangle|^{2}=\frac{1}{2}(1+n_{z}) \tag{3}
\end{equation*}
the probability of $\sigma_{z}=-1$ is
\begin{equation*}
1-\frac{1}{2}(1+n_{z})=\frac{1}{2}(1-n_{z}) \tag{4}
\end{equation*}
\begin{equation*}
\langle\sigma_{z}\rangle=\frac{1}{2}(1+n_{z})-\frac{1}{2}(1-n_{z})=n_{z} \tag{5}
\end{equation*}
Considering
$\sigma_{n}=\sigma_{x} n_{x}+\sigma_{y} n_{y}+\sigma_{z} n_{z}$
The components of $\boldsymbol{\sigma}$ and $\boldsymbol{n}$ have symmetric roles in the construction of $\sigma_{n}$, so using formulas (3), (4), (5), and cyclically permuting $x, ~ y$, $z$, the following can be deduced:
\begin{gather*}
\sigma_{x}= \pm 1 \text { with probability } \frac{1}{2}(1 \pm n_{x}) \tag{6}\\
\langle\sigma_{x}\rangle=n_{x} \tag{7}\\
\sigma_{y}= \pm 1 \text { with probability } \frac{1}{2}(1 \pm n_{y}) \tag{8}\\
\langle\sigma_{y}\rangle=n_{y} \tag{9}
\end{gather*}
By combining equations (5), (7), (9) in vector form as follows:
In the spin state $\phi_{1}(\sigma_{n}=1)$,
\begin{equation*}
\langle\boldsymbol{\sigma}\rangle=\boldsymbol{n} \tag{10}
\end{equation*}
Similarly, it is easy to calculate
In the spin state $\phi_{-1}(\sigma_{n}=-1)$,
\begin{equation*}
\langle\boldsymbol{\sigma}\rangle=-\boldsymbol{n} \tag{11}
\end{equation*}
Solution two (a) In the spin state $\chi_{\frac{1}{2}}$ where $\sigma_{z}=1$, the possible measured values of $\sigma_{n}$ are the eigenvalues $\pm 1$; let the corresponding probabilities be $w_{+}$ and $w_{-}$, then
\begin{equation*}
\langle\sigma_{n}\rangle=w_{+} \times 1+w_{-} \times(-1)=w_{+}-w_{-} \tag{12}
\end{equation*}
Since
\begin{equation*}
\sigma_{n}=\sigma_{x} n_{x}+\sigma_{y} n_{y}+\sigma_{z} n_{z} \tag{13}
\end{equation*}
Considering that in the eigenstate of $\sigma_{z}$ the average value of $\sigma_{x}$ and $\sigma_{y}$ is zero, and the average value of $\sigma_{z}$ is the eigenvalue, hence in the state $\chi_{\frac{1}{2}}$,
\begin{equation*}
\langle\sigma_{n}\rangle=\langle\sigma_{z}\rangle n_{z}=n_{z}=\cos \theta \tag{14}
\end{equation*}
From equations (12), (14), and using $w_{+}+w_{-}=1$, it can be found that
\begin{equation*}
w_{+}=\frac{1}{2}(1+n_{z}), \quad w_{-}=\frac{1}{2}(1-n_{z}) \tag{15}
\end{equation*}
These are the equations (1), (2) in solution one.
(b) In equation (14), $\theta$ is the angle parameter in the $z$-axis and $\boldsymbol{n}$. The choices of the $z$-axis and $\boldsymbol{n}$ are arbitrary, and the original $z$-axis can be taken as the new $\boldsymbol{n}$, while the original $\boldsymbol{n}$ is taken as the new $z$-axis. Thus, it can be known that in the spin state where $\sigma_{n}=1$
The average value of $\sigma_{z}$ remains $\cos \theta$, which is $n_{z}$. By letting $x, ~ y, ~ z$ permute, we obtain
\begin{equation*}
\text { In the spin state } \phi_{1}(\sigma_{n}=1) \text {, }\langle\boldsymbol{\sigma}\rangle=\boldsymbol{n} \tag{10}
\end{equation*}
In the state $\phi_{1}$, the values of each component of $\boldsymbol{\sigma}$ are of course all $\pm 1$, and their probabilities can be written similarly to those in (a), thus
\begin{align*}
\sigma_{x} & = \pm 1 \text { with probability } \frac{1}{2}(1 \pm n_{x}) \tag{6}\\
\sigma_{y} & = \pm 1 \text { with probability } \frac{1}{2}(1 \pm n_{y}) \tag{8}\\
\sigma_{z} & = \pm 1 \text { with probability } \frac{1}{2}(1 \pm n_{z}) \tag{3,4}
\end{align*}
|
[
"\\frac{1}{2}(1+n_z)"
] |
Expression
|
Theoretical Foundations
|
$n_z$: Component of the unit vector $\boldsymbol{n}$ along the $z$-axis.
|
10
|
Express the operator $(I+\sigma_{x})^{1 / 2}$ (where $\sigma_x$ is the Pauli matrix) as a linear combination of the $2 \times 2$ identity matrix (denoted as $1$ in the expression) and $\sigma_x$. The operation takes the principal square root.
|
(a) $(I+\sigma_{x})^{I / 2}$. The eigenvalues of $\sigma_{x}$ are $\pm 1$, and for each eigenvalue, $(I+\sigma_{x})^{1 / 2}$ gives a clear value (principal root is taken), so it can be concluded that $(I+\sigma_{x})^{1 / 2}$ exists, and it is a function of $\sigma_{x}$. According to the argument in problem 6.14, we can set
\begin{equation*}
(I+\sigma_{x})^{1 / 2}=C_{0}+C_{1} \sigma_{x} \tag{1}
\end{equation*}
Squaring this equation, we get
$I+\sigma_{x}=(C_{0}+C_{1} \sigma_{x})^{2}=C_{0}^{2}+C_{1}^{2}+2 C_{0} C_{1} \sigma_{x}$
Therefore
\[
\begin{gathered}
C_{0}^{2}+C_{1}^{2}=1 \\
2 C_{0} C_{1}=1
\end{gathered}
\]
Adding and subtracting the two equations, we get
\[
\begin{aligned}
& (C_{0}+C_{1})^{2}=2 \\
& (C_{0}-C_{1})^{2}=0
\end{aligned}
\]
If we set $(C_{0}+C_{1})$ to take positive value, we can solve
$C_{0}=C_{1}=1 / \sqrt{2}$
Substituting into equation (1), we obtain
\begin{equation*}
(1+\sigma_{z})^{1 / 2}=\frac{1}{\sqrt{2}}(I+\sigma_{x}) \tag{2}
\end{equation*}
It is easy to verify that for any eigenvalue $( \pm 1)$ of $\sigma_{x}$, this equation holds true.
|
[
"\\frac{1}{\\sqrt{2}}(1+\\sigma_x)"
] |
Expression
|
Theoretical Foundations
|
$1$: The $2 \times 2$ identity matrix
$\sigma_x$: Pauli matrix
|
11
|
For a spin $1/2$ particle, $\langle\boldsymbol{\sigma}\rangle$ is often called the polarization vector, denoted as $\boldsymbol{P}$, which is the spatial orientation of the spin angular momentum. Given the initial spin wave function at $t=0$ (in the $\sigma_{z}$ representation) as:
\chi(0)=[\begin{array}{c}
\cos \delta \mathrm{e}^{-\mathrm{i} \alpha} \\
\sin \delta \mathrm{e}^{\mathrm{i} \alpha}
\end{array}]
where $\delta, ~ \alpha$ are positive real numbers (or 0), $\delta \leqslant \pi / 2, \alpha \leqslant \pi$. Determine the azimuthal angle $\theta_{0}$ (polar angle) of the initial polarization vector $\boldsymbol{P}(t=0)$.
|
Any definite spin state is an eigenstate (with eigenvalue 1) of the projection $\sigma_{n}$ of $\boldsymbol{\sigma}$ in some direction $(\theta, \varphi)$, and
\begin{equation*}
\boldsymbol{P}=\langle\boldsymbol{\sigma}\rangle=\boldsymbol{n} \tag{3}
\end{equation*}
where $\boldsymbol{n}$ is the unit vector in the direction $(\theta, \varphi)$, the eigenfunction of $\sigma_{n}=1$ is
\begin{equation}
\phi_{1}(\theta, \varphi) =[\begin{array}{l}
\cos \frac{\theta}{2} \mathrm{e}^{-\mathrm{i} \varphi / 2} \tag{4}\\
\sin \frac{\theta}{2} \mathrm{e}^{\mathrm{i} \varphi / 2}
\end{array}]
\end{equation}
By comparing Equation (1) and (4), we find the azimuthal angle of the initial polarization vector $\boldsymbol{P}(t=0)$
\begin{equation*}
\theta_{0}=2 \delta, \quad \varphi_{0}=2 \alpha \tag{5}
\end{equation*}
|
[
"2 \\delta"
] |
Expression
|
Theoretical Foundations
|
$\delta$: Angle parameter in the initial spin wave function $\chi(0)$, a positive real number or 0.
|
12
|
For a system composed of two spin $1 / 2$ particles, let $s_{1}$ and $s_{2}$ denote their spin angular momentum operators. Calculate and simplify the triple product $s_{1} \cdot (s_{1} \times s_{2})$ (take $\hbar=1$ ).
|
The basic relationships are as follows (the single particle formula is only written for particle 1)
(a) $s_{1}^{2}=\frac{3}{4}, \boldsymbol{\sigma}_{1}^{2}=3 ;(s_{1 x})^{2}=\frac{1}{4},(\sigma_{1 x})^{2}=1$, and so on.
(b) $s_{1} \times s_{1}=\mathrm{i} s_{1}, \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{1}=2 \mathrm{i} \boldsymbol{\sigma}_{1}$
(c) $\sigma_{1 x} \sigma_{1 y}=-\sigma_{1}, \sigma_{1 x}=\mathrm{i} \sigma_{1 z}$, and so on.
(d) $s_{1}$ commutes with $s_{2}$, $s_{1} \cdot s_{2}=s_{2} \cdot s_{1}, s_{1} \times s_{2}=-s_{2} \times s_{1}$
For the triple product, there are the following types:
\begin{equation*}
1^{\circ} \tag{5}
\end{equation*}
\begin{align*}
s_{1} \cdot(s_{1} \times s_{2}) & =(s_{1} \times s_{1}) \cdot s_{2}=\mathrm{i} s_{1} \cdot s_{2} \\
s_{2} \cdot(s_{1} \times s_{2}) & =-s_{2} \cdot(s_{2} \times s_{1}) \\
& =-(s_{2} \times s_{2}) \cdot s_{1}=-\mathrm{i} s_{1} \cdot s_{2} \tag{6}\\
\boldsymbol{\sigma}_{1} \cdot(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) & =2 \mathrm{i} \boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2} \tag{7}\\
\boldsymbol{\sigma}_{2} \cdot(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) & =-2 \mathrm{i} \boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2} \tag{8}
\end{align*}
$\mathbf{2}^{\circ} \boldsymbol{\sigma}_{1}(\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2})$ type. Since $\boldsymbol{\sigma}_{1}$ and $\boldsymbol{\sigma}_{2}$ commute, using the formula proven in question 6.21
\begin{aligned}
\boldsymbol{\sigma}(\boldsymbol{\sigma} \cdot \boldsymbol{A})-\boldsymbol{A} & =\boldsymbol{A}-(\boldsymbol{\sigma} \cdot \boldsymbol{A}) \boldsymbol{\sigma} \\
& =\mathrm{i} \boldsymbol{A} \times \boldsymbol{\sigma}, \quad(\boldsymbol{\sigma}, \boldsymbol{A} \text { commute })
\end{aligned}
Let $\boldsymbol{\sigma}$ and $\boldsymbol{A}$ be equal to $\boldsymbol{\sigma}_{1}$ and $\boldsymbol{\sigma}_{2}$ respectively, and obtain
\begin{gather*}
\boldsymbol{\sigma}_{1}(\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2})=\boldsymbol{\sigma}_{2}-\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2} \tag{9}\\
\boldsymbol{\sigma}_{2}(\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2})=\boldsymbol{\sigma}_{1}+\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2} \tag{10}\\
(\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2}) \boldsymbol{\sigma}_{1}=\boldsymbol{\sigma}_{2}+\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2} \tag{11}\\
(\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2}) \boldsymbol{\sigma}_{2}=\boldsymbol{\sigma}_{1}-\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2} \tag{12}\\
s_{1}(s_{1} \cdot s_{2})=\frac{1}{4} s_{2}-\frac{\mathrm{i}}{2} s_{1} \times s_{2}, \text { etc. } \tag{13}\\
{[s_{1} \cdot s_{2}, s_{1}]=\mathrm{i} s_{1} \times s_{2}} \tag{14}\\
{[s_{1} \cdot s_{2}, s_{2}]=-\mathrm{i} s_{1} \times s_{2}} \tag{15}
\end{gather*}
$3^{\circ} s_{1} \times(s_{1} \times s_{2})$ type. Using the vector operator formula (see question 4.1)
\boldsymbol{A} \times(\boldsymbol{B} \times \boldsymbol{C})=\overparen{\boldsymbol{A} \cdot(\boldsymbol{B C})}-(\boldsymbol{A} \cdot \boldsymbol{B}) \boldsymbol{C}
(\boldsymbol{A} \times \boldsymbol{B}) \times \boldsymbol{C}=\boldsymbol{A \cdot ( B C )}-\boldsymbol{A}(\boldsymbol{B} \cdot \boldsymbol{C})
Hence, we obtain
\begin{align*}
s_{1} \times(s_{1} \times s_{2}) & =(s_{1} \cdot s_{2}) s_{1}-\frac{3}{4} s_{2}=\frac{\mathrm{i}}{2} s_{1} \times s_{2}-\frac{1}{2} s_{2} \tag{16}\\
(s_{1} \times s_{2}) \times s_{1} & =\frac{3}{4} s_{2}-s_{1}(s_{1} \cdot s_{2})=\frac{1}{2} s_{2}+\frac{\mathrm{i}}{2} s_{1} \times s_{2} \tag{17}\\
s_{2} \times(s_{1} \times s_{2}) & =\frac{1}{2} s_{1}+\frac{\mathrm{i}}{2} s_{1} \times s_{2} \tag{18}\\
(s_{1} \times s_{2}) \times s_{2} & =\frac{\mathrm{i}}{2} s_{1} \times s_{2}-\frac{1}{2} s_{1} \tag{19}\\
\boldsymbol{\sigma}_{1} \times(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) & =\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}-2 \boldsymbol{\sigma}_{2} \tag{20}\\
(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) \times \boldsymbol{\sigma}_{1} & =\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}+2 \boldsymbol{\sigma}_{2} \tag{21}\\
\boldsymbol{\sigma}_{2} \times(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) & =\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}+2 \boldsymbol{\sigma}_{1} \tag{22}\\
(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) \times \boldsymbol{\sigma}_{2} & =\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}-2 \boldsymbol{\sigma}_{1} \tag{23}
\end{align*}
$4^{\circ}$ Total spin $\boldsymbol{S}=s_{1}+s_{2}$,
\begin{align*}
& \boldsymbol{S}^{2}=2 s_{1} \cdot s_{2}+\frac{3}{2} \tag{24}\\
& {[\boldsymbol{S}^{2}, s_{1}]=2[s_{1} \cdot s_{2}, s_{1}]=2 \mathrm{i} s_{1} \times s_{2}} \tag{25}\\
& {[\boldsymbol{S}^{2}, s_{2}]=-2 \mathrm{i} s_{1} \times s_{2}} \tag{26}\\
& \boldsymbol{S} \cdot(s_{1} \times s_{2})=(s_{1} \times s_{2}) \cdot \boldsymbol{S}=0 \tag{27}\\
& \boldsymbol{S}(s_{1} \cdot s_{2})=(s_{1} \cdot s_{2}) \boldsymbol{S}=\frac{1}{4} \boldsymbol{S} \tag{28}\\
& \boldsymbol{S S}^{2}=\boldsymbol{S}^{2} \boldsymbol{S}=2 \boldsymbol{S} \tag{29}
\end{align*}
|
[
"\\mathrm{i} s_{1} \\cdot s_{2}"
] |
Expression
|
Theoretical Foundations
|
$\mathrm{i}$: Imaginary unit
$s_1$: Spin angular momentum operator for particle 1
$s_2$: Spin angular momentum operator for particle 2
|
13
|
Two localized non-identical particles with spin $1/2$ (ignoring orbital motion) have an interaction energy given by (setting $\hbar=1)$
\begin{equation*}
H=A \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}
\end{equation*}
At $t=0$, particle 1 has spin 'up' $(s_{1 z}=1 / 2)$, and particle 2 has spin 'down' $(s_{2 z}=-\frac{1}{2})$. Find the probability that particle 1 has spin 'up' $(s_{1 z}=1 / 2)$ at any time $t>0$.
|
Start by finding the spin wave function of the system. Since
\begin{equation*}
H=A s_{1} \cdot s_{2}=\frac{A}{2}(\boldsymbol{S}^{2}-\frac{3}{2}) \tag{$\prime$}
\end{equation*}
It is evident that the total spin $\boldsymbol{S}$ is a conserved quantity, so the stationary wave function can be chosen as a common eigenfunction of $\boldsymbol{S}^{2}, ~ S_{z}$. According to different values of the total spin quantum number $S$, the eigenfunctions and energy levels are
\begin{array}{ll}
S=1, & \chi_{1 M_{s}}, \quad E_{1}=A / 4 \tag{2}\\
S=0, & \chi_{00}, \quad E_{0}=-3 A / 4
\end{array}}
At $t=0$, the spin state of the system is
\begin{equation*}
\chi(0)=\alpha(1) \beta(2)=\frac{1}{\sqrt{2}}(\chi_{10}+\chi_{00}) \tag{3}
\end{equation*}
Therefore, the wave function at $t>0$ is
\begin{equation*}
\chi(t)=\frac{1}{\sqrt{2}} \chi_{10} \mathrm{e}^{-\mathrm{i} E_{1} t}+\frac{1}{\sqrt{2}} \chi_{00} \mathrm{e}^{-\mathrm{i} E_{0} t} \tag{4}
\end{equation*}
That is
\begin{align*}
\chi(t) & =\frac{1}{2}[\alpha(1) \beta(2)+\beta(1) \alpha(2)] \mathrm{e}^{-\mathrm{i} A / 4}+\frac{1}{2}[\alpha(1) \beta(2)-\beta(1) \alpha(2)] \mathrm{e}^{3 \mathrm{i} t / 4} \\
& =[\alpha(1) \beta(2) \cos \frac{A t}{2}-\mathrm{i} \beta(1) \alpha(2) \sin \frac{A t}{2}] \mathrm{e}^{\mathrm{i} A / 4} \tag{$4^\prime$}
\end{align*}
From formula ($4^{\prime}$), it can be seen that at time $t$, the probability of particle 1 having spin 'up' [while particle 2 has spin 'down', corresponding to the $\alpha(1) \beta(2)$ term] is $\cos ^{2}(\frac{A t}{2})$.
|
[
"\\cos^{2}(\\frac{A t}{2})"
] |
Expression
|
Theoretical Foundations
|
$A$: Constant in the interaction energy.
$t$: Time.
|
14
|
Consider a system consisting of three distinguishable particles each with spin $1/2$, with the Hamiltonian given by
\begin{equation*}
H=A(s_{1} \cdot s_{2}+s_{2} \cdot s_{3}+s_{3} \cdot s_{1}) \quad \text { ( } A \text { is real) }
\end{equation*}
Let $S$ denote the total spin quantum number of the system. Determine the expression for the energy level $E_{3/2}$ when $S=3/2$ (taking $\hbar=1$).
|
The Hamiltonian $H$ can be written as
\begin{equation*}
H=\frac{A}{2}(\boldsymbol{S}_{123}^{2}-3 \times \frac{3}{4}) \tag{$\prime$}
\end{equation*}
Therefore, the energy levels (taking $\hbar=1$) are
\begin{equation*}
E_{S}=\frac{A}{2}[S(S+1)-\frac{9}{4}] \tag{2}
\end{equation*}
A complete set of conserved quantities can be taken as $[S_{123}^{2}, ~(S_{123})_{z}, S_{12}^{2}]$, with eigenvalues
\begin{array}{l}
\boldsymbol{S}_{12}^{2}=S^{\prime}(S^{\prime}+1), \quad S^{\prime}=1,0 \tag{3}\\
\boldsymbol{S}_{123}^{2}=S(S+1), \quad S=\frac{3}{2}, \frac{1}{2}, \frac{1}{2} \\
(S_{123})_{z}=M=S, S-1, \cdots,(-S)
\end{array}}
The possible combinations for the quantum numbers $S^{\prime}, ~ S$ are $S=S^{\prime} \pm \frac{1}{2}>0$, i.e.,
\begin{array}{l}
S=3 / 2, \quad S^{\prime}=1 \tag{4}\\
S=1 / 2, \quad S^{\prime}=1,0
\end{array}}
For each pair $(S, S^{\prime})$, the degeneracy of the energy level is $(2 S+1)$, so
\begin{align*}
& E_{3 / 2}=\frac{3}{4} A, \quad S=\frac{3}{2}, \quad S^{\prime}=1, \quad \text { Degeneracy }=4 \\
& E_{1 / 2}=-\frac{3}{4} A, \quad S=\frac{1}{2}, \quad S^{\prime}=1,0, \quad \text { Degeneracy }=4 \tag{5}
\end{align*}
We now aim to determine the common eigenstates of $[\boldsymbol{S}_{123}^{2},(S_{123})_{z}, \boldsymbol{S}_{12}^{2}]$. The spin "up" and "down" states of the $k$-th particle $(k=1,2,3)$ are denoted as $\alpha(k), ~ \beta(k)$, corresponding to $s_{k z}=\frac{1}{2}, ~-\frac{1}{2}$. As is known, the common eigenstates of $S_{12}^{2}$ and $(S_{12})_{z}$ are given by:
\begin{aligned}
& \chi_{11}(1,2)=\alpha(1)_{\alpha}(2), \quad S^{\prime}=1, \quad(S_{12})_{z}=1 \\
& \chi_{10}(1,2)=\frac{1}{\sqrt{2}}[\alpha(1) \beta(2)+\beta(1)_{\alpha}(2)], \quad S^{\prime}=1, \quad(S_{12})_{z}=0 \\
& \chi_{1,-1}(1,2)=\beta(1) \beta(2), \quad S^{\prime}=1, \quad(S_{12})_{z}=-1 \\
& \chi_{00}(1,2)=\frac{1}{\sqrt{2}}[\alpha(1) \beta(2)-\beta(1) \alpha(2)], \quad S^{\prime}=0, \quad(S_{12})_{z}=0
\end{aligned}
The common eigenstates of $\boldsymbol{S}_{12}^{2}, ~ \boldsymbol{S}_{123}^{2}, ~(S_{123})_{z}$ are denoted as $\chi_{S_{S M}}(1,2,3)$. When all three quantum numbers take their maximum values, the eigenstate is clearly
\begin{equation*}
\chi_{1 \frac{3}{2} \frac{3}{2}}=\chi_{11}(1,2)_{\alpha}(3)=\alpha(1)_{\alpha}(2) \alpha(3) \tag{6a}
\end{equation*}
When $S=3 / 2$, $M$ has 4 possible values; corresponding eigenstates for $M=\frac{1}{2},-\frac{1}{2},-\frac{3}{2}$ can be obtained by repeatedly applying the ladder operator $[(S_{123})_{x}-\mathrm{i}(S_{123})_{y}]$ to $\chi_{1 \frac{3}{2}} \frac{3}{2}$. Since the ladder operator and $\chi_{1 \frac{3}{2} \frac{3}{2}}$ are symmetric under permutation of the particles, each $\chi_{1 \frac{3}{2} M}$ obtained from this is a symmetric function. Based on this observation and considering the values of $M$, the only possible structure of these functions can be immediately written:
\begin{align*}
\chi_{1 \frac{3}{2} \frac{1}{2}} & =\frac{1}{\sqrt{3}}[\alpha(1) \alpha(2) \beta(3)+\beta(1) \alpha(2) \alpha(3)+\alpha(1) \beta(2) \alpha(3)] \tag{6b}\\
\chi_{1 \frac{3}{2},-\frac{1}{2}} & =\frac{1}{\sqrt{3}}[\alpha(1) \beta(2) \beta(3)+\beta(1) \alpha(2) \beta(3)+\beta(1) \beta(2) \alpha(3)] \tag{6c}\\
\chi_{1 \frac{3}{2},-\frac{3}{2}} & =\beta(1) \beta(2) \beta(3) \tag{6d}
\end{align*}
When $S^{\prime}=0$, the part of the system wave function concerning particles $1, ~ 2$ can only be $\chi_{00}$, and considering the values of $M$, it can be immediately concluded that
\begin{align*}
\chi_{0 \frac{1}{2} \frac{1}{2}} & =\chi_{00}(1,2) \alpha(3) \\
& =\frac{1}{\sqrt{2}}[\alpha(1) \beta(2) \alpha(3)-\beta(1) \alpha(2) \alpha(3)] \tag{7a}\\
\chi_{0 \frac{1}{2},-\frac{1}{2}} & =\chi_{00}(1,2) \beta(3) \\
& =\frac{1}{\sqrt{2}}[\alpha(1) \beta(2) \beta(3)-\beta(1) \alpha(2) \beta(3)] \tag{7b}
\end{align*}
Now only $\chi_{1 \frac{1}{2} \frac{1}{2}}$ and $\chi_{1 \frac{1}{2},-\frac{1}{2}}$ remain to be determined. In the construction of $\chi_{1 \frac{1}{2} \frac{1}{2}}$, each term should include two $\alpha$ states and one $\beta$ state. It must be a linear combination of $\chi_{11}(1,2) \beta(3)$ and $\chi_{10}(1,2) \alpha(3)$, and should be orthogonal to $\chi_{1 \frac{3}{2} \frac{1}{2}}$ since they have different $S$ values. Expression (6b) for $\chi_{1 \frac{3}{2} \frac{1}{2}}$ can be written as
\begin{equation*}
\chi_{1 \frac{3}{2} \frac{1}{2}}=\frac{1}{\sqrt{3}}[\chi_{11}(1,2) \beta(3)+\sqrt{2} \chi_{10}(1,2)_{\alpha}(3)] \tag{$\prime$}
\end{equation*}
Thus, the construction of $\chi_{1 \frac{1}{2} \frac{1}{2}}$ can only be
\begin{align*}
\chi_{1 \frac{1}{2} \frac{1}{2}} & =\frac{1}{\sqrt{3}}[\sqrt{2} \chi_{11}(1,2) \beta(3)-\chi_{10}(1,2) \alpha(3)] \\
& =\frac{1}{\sqrt{6}}[2 \alpha(1) \alpha(2) \beta(3)-\alpha(1) \beta(2) \alpha(3)-\beta(1) \alpha(2) \beta(3)] \tag{8a}
\end{align*}
Similarly, expression (6c) can be written as
\begin{equation*}
\chi_{1 \frac{3}{2},-\frac{1}{2}}=\frac{1}{\sqrt{3}}[\sqrt{2} \chi_{10}(1,2) \beta(3)+\chi_{1-1}(1,2)_{\alpha}(3)] \tag{$\prime$}
\end{equation*}
$\chi_{1 \frac{1}{2},-\frac{1}{2}}$ should be formed with the same terms but orthogonal to the above expression, so
$$\chi_{1 \frac{1}{2},-\frac{1}{2}}=\frac{1}{\sqrt{3}}[\chi_{\mathrm{i} 0}(1,2) \beta(3)-\sqrt{2} \chi_{1,-1}(1,2) \alpha(3)]$$
\begin{equation*}
=\frac{1}{\sqrt{6}}[\alpha(1) \beta(2) \beta(3)+\beta(1) \alpha(2) \beta(3)-2 \beta(1) \beta(2) \alpha(3)] \tag{8b}
\end{equation*}
Expressions (6), (7), and (8) above are the complete set of common eigenfunctions for $\boldsymbol{S}_{12}^{2}, ~ \boldsymbol{S}_{123}^{2}, ~(S_{123})_{z}$. It is easy to verify that they are indeed orthogonal to each other. Since each particle has two spin states $\alpha$ and $\beta$, the system of three particles has a total of 8 independent states, thus expressions (6) to (8) form the sought orthogonal and complete set of energy eigenstates.
The reader can easily verify that according to the theory of angular momentum coupling, similar results would be obtained using the C.G. coefficient table.
|
[
"E_{3/2} = \\frac{3}{4}A"
] |
Expression
|
Theoretical Foundations
|
$E_{3/2}$: Energy level of the system when the total spin quantum number $S=3/2$.
$A$: Real constant in the Hamiltonian.
|
15
|
An electron moves freely in a one-dimensional region $-L/2 \leqslant x \leqslant L/2$, with the wave function satisfying periodic boundary conditions $\psi(x)=\psi(x+L)$. Its unperturbed energy eigenvalue is $E_n^{(0)}$. A perturbation $H^{\prime}=\varepsilon \cos q x$ is applied to this system where $L q=4 \pi N$ ($N$ is a large positive integer). Consider the degenerate state where the electron's momentum is $|p|=q \hbar / 2$ (corresponding to energy $E_N^{(0)}$) without perturbation. Find the expression for the second-order energy correction $E_N^{(2)}$ of these degenerate states caused by the perturbation $H^{\prime}$.
|
(a) For a free particle, the common eigenfunction of the energy $(H_{0})$ and momentum $(p)$ satisfying the periodic condition is
\begin{equation*}
\psi_{n}^{(0)}=\frac{1}{\sqrt{L}} \mathrm{e}^{\mathrm{i} 2 \pi x / L}, \quad n=0, \pm 1, \pm 2, \cdots \tag{1}
\end{equation*}
The eigenvalue is
\begin{align*}
p_{n} & =\hbar k_{n}=2 \pi n \hbar / L \\
E_{n}^{(0)} & =\frac{p_{n}^{2}}{2 m}=\frac{(2 \pi n \hbar)^{2}}{2 m L^{2}} \tag{2}
\end{align*}
(b) The perturbation operator is
\begin{equation*}
H^{\prime}=\varepsilon \cos q x=\frac{\varepsilon}{2}(\mathrm{e}^{\mathrm{i} \pi \lambda N_{x} / L}+\mathrm{e}^{-\mathrm{i} 4 \pi N_{x} / L}) \tag{3}
\end{equation*}
For $|p|=q \hbar / 2=2 \pi N \hbar / L$ , the corresponding zeroth-order wave function is
\begin{equation*}
\psi^{(0)}=C_{N} \psi_{N}^{(0)}+C_{-, ~} \psi_{-, ~}^{(0)} \tag{4}
\end{equation*}
It can be easily calculated that
\begin{equation*}
H_{\mathrm{vv}}^{\prime}=H_{-\mathrm{v},-\mathrm{N}}^{\prime}=0, \quad H_{v,-\mathrm{v}}^{\prime}=H_{-\mathrm{N}, \mathrm{v}}^{\prime}=\frac{\varepsilon}{2} \tag{5}
\end{equation*}
Therefore, in the subspace ${\psi_{N}^{(0)}, \psi_{-N}^{(0)}}$, the matrix representation of $H^{\prime}$ is
H^{\prime}=\frac{\varepsilon}{2}[\begin{array}{ll}
0 & 1 \tag{$\prime$}\\
1 & 0
\end{array}]
In this subspace, $\psi^{(0)}$ should satisfy the eigenvalue equation
$H^{\prime} \psi^{(0)}=E_{N}^{(1)} \psi^{(0)}
That is
\begin{equation}
\frac{\varepsilon}{2}\left[\begin{array}{ll}
0 & 1 \tag{6}\\
1 & 0
\end{array}\right]\left[\begin{array}{l}
C_{\mathrm{N}} \\
C_{-\mathrm{N}}
\end{array}\right]=E_{N}^{(1)}\left[\begin{array}{l}
C_{\mathrm{N}} \\
C_{-N}
\end{array}\right]
\end{equation}
It is easy to solve
\begin{align*}
& E_{\mathrm{N}+}^{(1)}=\varepsilon / 2 \tag{7}\\
& \psi^{(0)}=\frac{1}{\sqrt{2}}[\psi_{N}^{(0)}+\psi_{-\mathrm{N}}^{(0)}]=\psi_{N+}^{(0)} \\
& E_{\mathrm{N}-}^{(1)}=-\varepsilon / 2 \\
& \psi^{(0)}=\frac{1}{\sqrt{2}}[\psi_{\mathrm{N}}^{(0)}-\psi_{-\mathrm{N}}^{(0)}]=\psi_{N-}^{(0)}
\end{align*}
The first-order correction to the wave function is
\begin{equation*}
\psi^{(1)}=\sum_{n}^{\prime} \frac{1}{E_{N}^{(0)}-E_{n}^{(0)}}\langle\psi_{n}^{(0)}| H^{\prime}|\psi^{(0)}\rangle \psi_{n}^{(0)} \quad(n \neq \pm N) \tag{8}
\end{equation*}
The matrix elements contributing to $\psi^{(1)}$ are (corresponding to $n= \pm 3 N$)
$$ H_{3, \mathrm{~V} . \mathrm{N}}^{\prime}=H_{-3, \mathrm{~V},-\mathrm{N}}^{\prime}=\varepsilon / 2$$
The corresponding energy difference is
$$ E_{N}^{(0)}-E_{3 N}^{(0)}=-8 E_{N}^{(0)}=-\frac{8(2 \pi N \hbar)^{2}}{2 m L^{2}}$$
Thus, the first-order correction to the wave function is
\begin{align}
E^{(1)} & =E_{N+}^{(1)}=\frac{\varepsilon}{2} \\
\psi^{(1)} & =-\frac{\varepsilon}{2} \frac{2 m L^{2}}{8(2 \pi N \hbar)^{2}} \frac{1}{\sqrt{2}}[\psi_{3 N}^{(0)}+\psi_{-3 N}^{(0)}]
\end{align}
\begin{align*}
E^{(1)} & =E_{N-}^{(1)}=-\frac{\varepsilon}{2} \\
\psi^{(1)} & =-\frac{\varepsilon}{2} \frac{2 m L^{2}}{8(2 \pi N \hbar)^{2}} \frac{1}{\sqrt{2}}[\psi_{3 N}^{(0)}-\psi_{-3 N}^{(0)}] \tag{9}
\end{align*}
(c) The second-order correction to the energy is
\begin{equation*}
E_{N}^{(2)}=\langle\psi^{(0)}| H^{\prime}|\psi^{(1)}\rangle \tag{10}
\end{equation*}
Substituting equations (7) and (9) into the above formula, we obtain
\begin{equation*}
E_{N}^{(2)}=-\frac{\varepsilon^{2}}{32} \frac{2 m L^{2}}{(2 \pi N \hbar)^{2}}=-\frac{\varepsilon^{2}}{32 E_{N}^{(0)}} \tag{11}
\end{equation*}
Combining equations (7) and (11), the conclusion for the energy level up to $\varepsilon^{2}$ order is:
\begin{equation*}
E_{N}=E_{N}^{(0)}+E_{N}^{(1)}+E_{N}^{(2)}=E_{\mathrm{N}}^{(0)} \pm \frac{\varepsilon}{2}-\frac{\varepsilon^{2}}{32 E_{N}^{(0)}} \tag{12}
\end{equation*}
Corresponding to
$$\psi^{(0)}=\frac{1}{\sqrt{2}}[\psi_{N}^{(0)} \pm \psi_{-, .}^{(0)}] $$
|
[
"-\\frac{\\varepsilon^{2}}{32 E_{N}^{(0)}}"
] |
Expression
|
Theoretical Foundations
|
$\varepsilon$: Strength of the perturbation.
$E_N^{(0)}$: Unperturbed energy eigenvalue for the $n$-th state, $E_{n}^{(0)} = \frac{(2 \pi n \hbar)^{2}}{2 m L^{2}}$.
$m$: Mass of the electron.
$\hbar$: Reduced Planck's constant.
$q$: Wave number parameter of the perturbation, defined by $L q=4 \pi N$.
|
16
|
For the $n$-th bound state $\psi_{n}, ~ E_{n}$ of a square well (depth $V_{0}$, width $a$), under the condition $V_{0} \gg E_{n}$, calculate the probability of the particle appearing outside the well.
|
Take the even parity state as an example. The energy eigenvalue equation can be written as
\begin{equation}
\begin{array}{lll}
\psi^{\prime \prime}+k^{2} \psi=0, & |x| \leqslant a / 2 & \text { (inside the well) } \tag{1}\\
\psi^{\prime \prime}-\beta^{2} \psi=0, & |x| \geqslant a / 2 & \text { (outside the well) }
\end{array}
\end{equation}
where
\begin{equation*}
k=\sqrt{2 m E} / \hbar, \quad \beta=\sqrt{2 m(V_{0}-E)} / \hbar \tag{2}
\end{equation*}
Note that under the condition $V_{0} \gg E$, $\beta \gg k$.
The even parity solution of equation (1) is
\begin{array}{ll}
\psi=\cos k x, & |x| \leqslant a / 2 \\
\psi=C \mathrm{e}^{-\beta|x|}, & |x| \geqslant a / 2 \tag{3}
\end{array}
At $x=a / 2$, $\psi$ should be continuous, thus yielding
\begin{equation*}
C=\mathrm{e}^{\beta a / 2} \cos \frac{k a}{2} \tag{4}
\end{equation*}
At $x=a / 2$, $\psi^{\prime}$ should also be continuous, thus yielding
$$ C=\frac{k}{\beta} \mathrm{e}^{\beta a / 2} \sin \frac{k a}{2} $$
Dividing by equation (4), we obtain the energy level formula
\begin{equation*}
\tan \frac{k a}{2}=\frac{\beta}{k} \tag{5}
\end{equation*}
Under the condition $\beta \gg k$, the solution of equation (5) is
\begin{equation*}
k a \approx n \pi, \quad n=1,3,5, \cdots \tag{6}
\end{equation*}
Substituting into equation (2), the energy levels are
\begin{equation*}
E_{n}=\frac{\hbar^{2} k^{2}}{2 m} \approx \frac{1}{2 m}(\frac{n \pi \hbar}{a})^{2} \tag{7}
\end{equation*}
This is precisely the energy level formula for an infinitely deep potential well.
Now calculate the probability of the particle appearing inside and outside the well. From equations (3) and (4), it is easy to find
\begin{align*}
& \int_{\text {outside }}|\psi|^{2} \mathrm{~d} x=2 C^{2} \int_{a / 2}^{\infty} \mathrm{e}^{-2 \beta x} \mathrm{~d} x=\frac{C^{2}}{\beta} \mathrm{e}^{-\beta a}=\frac{1}{\beta} \cos ^{2} \frac{k a}{2} \tag{8}\\
& \int_{\text {inside }}|\psi|^{2} \mathrm{~d} x=2 \int_{0}^{a / 2} \cos ^{2} k x \mathrm{~d} x=\frac{a}{2}(1+\frac{\sin k a}{k a}) \tag{9}
\end{align*}
Considering $k a \approx n \pi(n=1,3,5, \cdots), \sin k a$ and $\cos (k a / 2)$ are both close to zero, it is understood that the probability of the particle appearing outside the well is much smaller than that inside the well. Additionally,
\begin{gather*}
\int_{-\infty}^{+\infty}|\psi|^{2} \mathrm{~d} x \approx \int_{\text {inside }}|\psi|^{2} \mathrm{~d} x \approx \frac{a}{2} \\
\text { Outside probability }=\frac{\int_{\text {outside }}|\psi|^{2} \mathrm{~d} x}{\int_{-\infty}^{+\infty}|\psi|^{2} \mathrm{~d} x} \approx \frac{2}{\beta a} \cos ^{2} \cdot \frac{k a}{2} \tag{10}
\end{gather*}
Using the energy level formula (5), it is easy to obtain
\begin{gather*}
1+\tan ^{2} \frac{k a}{2}=\frac{k^{2}+\beta^{2}}{k^{2}}=\frac{V_{0}}{E} \\
\cos ^{2} \frac{k a}{2}=\frac{E}{V_{0}} \tag{11}
\end{gather*}
Substituting into equation (10), we get
\begin{equation*}
\text { Outside probability }=\frac{2 E}{a \beta V_{0}} \approx \frac{2 \hbar}{a \sqrt{2 m V_{0}}} \frac{E}{V_{0}} \tag{12}
\end{equation*}
Considering $V_{0} \gg E$ and the energy level formula (7), it is easily seen
\begin{equation*}
\sqrt{2 m V_{0}} \gg n \pi \hbar / a \tag{13}
\end{equation*}
Thus,
\begin{equation*}
\text { Outside probability } \ll \frac{2 E}{n \pi V_{0}} \tag{14}
\end{equation*}
|
[
"\\frac{2 \\hbar E_n}{a V_0 \\sqrt{2 m V_0}}"
] |
Expression
|
Theoretical Foundations
|
$\hbar$: Reduced Planck's constant.
$E_n$: Energy of the $n$-th bound state.
$a$: Width of the square well.
$V_0$: Depth of the square well potential.
$m$: Mass of the particle.
|
17
|
A particle moves freely, and the initial wave function at $t=0$ is given as
$$ \psi(x, 0)=(2 \pi a^{2})^{-1 / 4} \exp [\mathrm{i} k_{0}(x-x_{0})-(\frac{x-x_{0}}{2 a})^{2}], \quad a>0 $$
Find the wave function $\varphi(p)$ in the $p$ representation at $t=0$;
|
First, consider the shape of the wave packet at $t=0$
\begin{equation*}
|\psi(x, 0)|^{2}=(2 \pi a^{2})^{-1 / 2} \mathrm{e}^{-(x-x_{0})^{2} / 2 a^{2}} \tag{1}
\end{equation*}
which is a Gaussian distribution. According to the mean value formula
\begin{equation}
\overline{f(x)}=\int_{-\infty}^{+\infty}|\psi|^{2} f(x) \mathrm{d} x
\end{equation}
it is easy to calculate that at $t=0$
\begin{align*}
& \bar{x}=x_{0}, \quad \bar{x}^{2}=a^{2}+x_{0}^{2} \\
& \Delta x=(\bar{x}^{2}-\bar{x}^{2})^{1 / 2}=a \tag{2}
\end{align*}
which means at $t=0$, the center of the wave packet is at $x_{0}$, and the width of the wave packet is $a$. We first calculate the wave function in terms of the wave number $k$. Let
\begin{equation*}
\psi(x, 0)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \varphi(k) \mathrm{e}^{\mathrm{i} k x} \mathrm{~d} k \tag{3}
\end{equation*}
where $\varphi(k)$ is the initial wave function in the $k$ representation. According to the Fourier transform formula
\begin{align*}
\varphi(k)&=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \psi(x, 0) \mathrm{e}^{-\mathrm{i} k x} \mathrm{~d} x\\
& =(\frac{1}{2 \pi})^{3 / 4} \frac{1}{\sqrt{a}} \mathrm{e}^{-\mathrm{i} k x_{0}} \int_{-\infty}^{+\infty} \mathrm{d} x \exp [-(\frac{x-x_{0}}{2 a})^{2}-\mathrm{i}(k-k_{0})(x-x_{0})] \\
& =(\frac{2 a^{2}}{\pi})^{\frac{1}{4}} \exp [-\mathrm{i} k x_{0}-a^{2}(k-k_{0})^{2}] \tag{4}
\end{align*}
The integration formula is used in the calculation (refer to the note at the end of the question). From equation (4)
\begin{equation*}
|\varphi(k)|^{2}=\sqrt{\frac{2}{\pi}} a \mathrm{e}^{-2 a^{2}(k-k_{0})^{2}} \tag{5}
\end{equation*}
which indicates that the probability distribution of $k$ is also a Gaussian distribution. According to the mean value formula
\begin{equation}
\overline{f(k)}=\int_{-\infty}^{+\infty}|\varphi(k)|^{2} f(k) \mathrm{d} k
\end{equation}
it is readily calculated that
\begin{equation*}
\bar{k}=k_{0}, \quad \overline{k^{2}}=k_{0}^{2}+\frac{1}{4 a^{2}} \tag{6}
\end{equation*}
Since $p=\hbar k$, thus
\begin{align*}
& \bar{p}=\hbar k_{0}, \quad \bar{p}^{2}=\hbar^{2} k_{0}^{2}+\frac{\hbar^{2}}{4 a^{2}} \tag{7}\\
& \Delta p=(\overline{p^{2}}-\bar{p}^{2})^{1 / 2}=\hbar / 2 a
\end{align*}
These are the characteristics of the momentum distribution at $t=0$. Since it is a free particle, momentum conservation implies that the probability distribution of momentum is also conserved. Therefore, equation (7) applies to any time. From equations (2) and (7), it follows that
\begin{equation*}
\Delta x \cdot \Delta p=\hbar / 2 \quad(t=0 \text{ instant}) \tag{8}
\end{equation*}
Finally, the wave function in the momentum representation $\varphi(p)$ is related to $\varphi(k)$ by the normalization condition $\int |\varphi(p)|^2 dp = \int |\varphi(k)|^2 dk = 1$. Since $dp = \hbar dk$, we have $\varphi(p) = \frac{1}{\sqrt{\hbar}}\varphi(k)|_{k=p/\hbar}$.
|
[
"\\varphi(p) = (\\frac{2 a^{2}}{\\pi \\hbar^2})^{\\frac{1}{4}} \\exp [-\\mathrm{i} \\frac{p x_{0}}{\\hbar}-a^{2}(\\frac{p}{\\hbar}-k_{0})^{2}]"
] |
Expression
|
Theoretical Foundations
|
$\varphi(p)$: Wave function in the momentum representation at $t=0$.
$a$: Parameter defining the width of the initial wave packet, $a>0$. It also represents the uncertainty in position at $t=0$, $\Delta x = a$.
$\mathrm{i}$: Imaginary unit.
$p$: Momentum.
$x_0$: Initial position of the center of the wave packet. It also represents the mean position at $t=0$, $\bar{x} = x_0$.
$\hbar$: Reduced Planck's constant.
$k_0$: Initial wave number, representing the center of the wave number distribution.
|
18
|
The particle moves freely, with the initial wave function at $t=0$ given as
$$ \psi(x, 0)=(2 \pi a^{2})^{-1 / 4} \exp [\mathrm{i} k_{0}(x-x_{0})-(\frac{x-x_{0}}{2 a})^{2}], \quad a>0 $$
Find the wave function $\psi(x, t)$ for $t>0$
|
In equation \begin{equation*}
\psi(x, 0)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \varphi(k) \mathrm{e}^{\mathrm{i} k x} \mathrm{~d} k, \tag{3}
\end{equation*} let
$$ \mathrm{e}^{\mathrm{i} k x} \rightarrow \mathrm{e}^{\mathrm{i}(k x-\omega t)}, \quad \omega=\frac{\hbar k^{2}}{2 m}=\frac{k^{2}}{2 m} \quad(\hbar=1) $$
thus obtaining the wave function $\psi(x, t)$ for $t>0$, i.e.,
\begin{equation*}
\psi(x, t)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \varphi(k) \exp (\mathrm{i} k x-\frac{\mathrm{i} k^{2} t}{2 m}) \mathrm{d} k \tag{9}
\end{equation*}
Substitute equation \begin{align*}
\varphi(k)&= (\frac{2 a^{2}}{\pi})^{\frac{1}{4}} \exp [-\mathrm{i} k x_{0}-a^{2}(k-k_{0})^{2}]
\end{align*} into the above expression and we obtain
\begin{align*}
& \psi(x, t) \\
= & \frac{\exp [\mathrm{i} k_{0}(x-x_{0})-\mathrm{i} t k_{0}^{2} / 2 m]}{(2 \pi)^{1 / 4}(a+\mathrm{i} t / 2 m a)^{1 / 2}} \exp [-\frac{1}{4}(x-x_{0}-\frac{k_{0} t}{m})^{2} \frac{1-\mathrm{i} t / 2 m a^{2}}{a^{2}+(t / 2 m a)^{2}}] \tag{10}\\
\end{align*}
|
[
"\\psi(x, t) = \\frac{\\exp [\\mathrm{i} k_{0}(x-x_{0})-\\mathrm{i} t k_{0}^{2} / 2 m]}{(2 \\pi)^{1 / 4}(a+\\mathrm{i} t / 2 m a)^{1 / 2}} \\exp [-\\frac{1}{4}(x-x_{0}-\\frac{k_{0} t}{m})^{2} \\frac{1-\\mathrm{i} t / 2 m a^{2}}{a^{2}+(t / 2 m a)^{2}}]"
] |
Expression
|
Theoretical Foundations
|
$\psi(x, t)$: Wave function of the particle at time $t$
$\mathrm{i}$: Imaginary unit
$k_0$: Initial wave number
$x$: Position coordinate
$x_0$: Initial position offset
$t$: Time
$m$: Mass of the particle
$\pi$: Mathematical constant pi
$a$: A positive constant related to the width of the initial wave packet
|
19
|
Using the raising and lowering operators $a^{+}, ~ a$, find the energy eigenfunctions of the harmonic oscillator (in the $x$ representation), and briefly discuss their mathematical properties.
|
Start with the ground state wave function $\psi_{0}(x)$. We have,
\begin{equation*}
a|0\rangle=0 \tag{1}
\end{equation*}
In the $x$ representation this reads as
\begin{equation*}
(\mathrm{i} \hat{p}+m \omega x) \psi_{0}(x)=(\hbar \frac{\mathrm{d}}{\mathrm{~d} x}+m \omega x) \psi_{0}(x)=0 \tag{2}
\end{equation*}
Let $\alpha=\sqrt{m \omega / \hbar}$, the above equation can be written as
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} x} \psi_{0}+\alpha^{2} x \psi_{0}=0 \tag{2'}
\end{equation*}
It is easy to solve
\begin{equation*}
\psi_{0}(x)=N_{0} \mathrm{e}^{-\alpha^{2} x^{2} / 2} \tag{3}
\end{equation*}
where $N_{0}$ is the normalization constant. From the normalization condition
\begin{equation*}
\int_{-\infty}^{+\infty}|\psi_{n}(x)|^{2} \mathrm{~d} x=1 \tag{4}
\end{equation*}
Find (taking as real)
\begin{equation*}
N_{0}=\sqrt{\alpha} / \pi^{1 / 4} \tag{5}
\end{equation*}
Secondly, we have
$$ a^{+}|0\rangle=|1\rangle $$
That is
\begin{equation*}
\psi_{1}(x)=a^{+} \psi_{0}(x)=\frac{1}{\sqrt{2}}(\alpha x-\frac{1}{\alpha} \frac{\mathrm{~d}}{\mathrm{~d} x}) \psi_{0}(x) \tag{6}
\end{equation*}
Substituting equation (3) and (5) into equation (6), it's easy to find
\begin{equation*}
\psi_{1}=\sqrt{2} \alpha x \psi_{0}=\frac{\sqrt{2 \alpha}}{\pi^{1 / 4}} \alpha x \mathrm{e}^{-\alpha^{2} x^{2} / 2} \tag{7}
\end{equation*}
Generally, we have
\begin{equation*}
\psi_{n}=\frac{1}{\sqrt{n}} a^{+} \psi_{n-1}=\frac{1}{\sqrt{2 n}}(\alpha x-\frac{1}{\alpha} \frac{\mathrm{~d}}{\mathrm{~d} x}) \psi_{n-1} \tag{8}
\end{equation*}
Introducing the dimensionless variable
\begin{equation*}
\xi=\alpha x, \tag{9}
\end{equation*}
we thus obtain
\begin{equation*}
\psi_{n}=\frac{1}{\sqrt{2 n}}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi}) \psi_{n-1} \tag{10}.
\end{equation*}
Recursively, we get
\begin{equation*}
\psi_{n}=(\frac{1}{2^{n} n!})^{\frac{1}{2}}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{n} \psi_{0}=(\frac{\alpha}{\sqrt{\pi} 2^{n} n!})^{\frac{1}{2}}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{n} \mathrm{e}^{-\xi^{2} / 2}=N_{n} H_{n}(\xi) \mathrm{e}^{-\xi^{2} / 2} \tag{11}
\end{equation*}
Where
\begin{equation*}
N_{n}=(\frac{\alpha}{\sqrt{\pi} 2^{n} n!})^{\frac{1}{2}} \tag{12}
\end{equation*}
is the normalization constant. Notice
\begin{equation*}
N_{n}=N_{n-1} / \sqrt{2 n} \tag{13}
\end{equation*}
In equation (11),
\begin{equation*}
H_{n}(\xi)=\mathrm{e}^{\xi^{2} / 2}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{n} \mathrm{e}^{-\xi^{2} / 2} \tag{1}
\end{equation*}
It is obvious that $H_{n}(\xi)$ is an $n$-th degree polynomial in $\xi$, called the Hermite polynomial.
Below is a brief discussion of the mathematical properties of $H_{n}(\xi)$.
{Parity}
It is obvious from equation (14)
\begin{equation*}
H_{n}(-\xi)=(-1)^{n} H_{n}(\xi) \tag{15}
\end{equation*}
When $n$ is even, $H_{n}(\xi)$ has even parity; when $n$ is odd, $H_{n}(\xi)$ has odd parity.
{Recursion relations}
We also have,
$$ a \psi_{n}=\frac{1}{\sqrt{2}}(\xi+\frac{\mathrm{d}}{\mathrm{~d} \xi}) \psi_{n}=\sqrt{n} \psi_{n-1}$$
Substituting equation (11) into this equation, and using equation (13) to eliminate the normalization constant, we get
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} \xi} H_{n}(\xi)=2 n H_{n-1}(\xi) \tag{16},
\end{equation*}
We also have:
$$ \xi \psi_{n}=\sqrt{\frac{n+1}{2}} \psi_{n+1}+\sqrt{\frac{n}{2}} \psi_{n-1} $$
Substituting equations (11), (13) into this equation, we get
\begin{equation*}
2 \xi H_{n}(\xi)=H_{n+1}(\xi)+2 n H_{n-1}(\xi) \tag{17}
\end{equation*}
{Differential equation}
Combining equations (16) and (17), eliminate $H_{n-1}$, then differentiate once more, and using equation (16) to eliminate $\mathrm{d} H_{n+1} / \mathrm{d} \xi$, we find that $H_{n}$ satisfies the following differential equation:
\begin{equation*}
\frac{\mathrm{d}^{2}}{\mathrm{~d} \xi^{2}} H_{n}(\xi)-2 \xi \frac{\mathrm{~d}}{\mathrm{~d} \xi} H_{n}(\xi)+2 n H_{n}(\xi)=0 \tag{18}
\end{equation*}
This equation is known as the Hermite equation. Equation (14) is the unique polynomial solution to this equation.
$4{ }^{\circ}$ Differential expression
Equation (14) is equivalent to the following:
\begin{equation*}
H_{n}(\xi)=(-1)^{n} e^{\xi^{2}}(\frac{d}{d \xi})^{n} e^{-\xi^{2}} \tag{$\prime$}
\end{equation*}
This can be proved by mathematical induction. First, for $n=0,1$, both equations (14) and ($14'$) give
$$H_{0}(\xi)=1, \quad H_{1}(\xi)=2 \xi $$
Assume the proposition holds for $n=k$, i.e., assume
\mathrm{e}^{\xi^{2}}(-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k} \mathrm{e}^{-\xi^{2}}=\mathrm{e}^{\xi^{2} / 2}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k} \mathrm{e}^{-\xi^{2} / 2}=H_{k}(\xi)
Then
\begin{gathered}
(-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k+1} \mathrm{e}^{-\xi^{2}}=-\frac{\mathrm{d}}{\mathrm{~d} \xi}(\mathrm{e}^{-\xi^{2}} H_{k})=(2 \xi H_{k}-H_{k}^{\prime}) \mathrm{e}^{-\xi^{2}} \\
(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k+1} \mathrm{e}^{-\xi^{2} / 2}=(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})(\mathrm{e}^{-\xi^{2} / 2} H_{k})=(2 \xi H_{k}-H_{k}^{\prime}) \mathrm{e}^{-\xi^{2} / 2}
\end{gathered}
Thus
\mathrm{e}^{\xi^{2}}(-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k+1} \mathrm{e}^{-\xi^{2}}=2 \xi H_{k}-H_{k}^{\prime}=\mathrm{e}^{\xi^{2} / 2}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k+1} \mathrm{e}^{-\xi^{2} / 2}
The proposition holds for $n=k+1$. Proof complete.
$5^{\circ} H_{n}(0)$ and $\psi_{n}(0)$
When $n=2 k+1$ (odd), from equation (15), we have
H_{2 k+1}(-\xi)=-H_{2 k+1}(\xi)
Let $\xi \rightarrow 0$, we get
\begin{equation*}
H_{2 k+1}(0)=0, \quad k=1,2,3, \cdots \tag{19}
\end{equation*}
When $n=2 k$ (even), in equation (17) let $\xi \rightarrow 0$, and change $n$ to ( $n-1$ ), we get
H_{2 k}(0)=-2(2 k-1) H_{2 k-2}(0)
Using this repeatedly, and noting that $H_{0}=1$, we get
\begin{equation*}
H_{2 k}(0)=(-1)^{k} 2^{k}(2 k-1)!!=(-1)^{k}(2 k)!/ k! \tag{20}
\end{equation*}
Using equations (11), (12) again, we get
\begin{gather*}
\psi_{2 k+1}(0)=0 \tag{21}\\
\psi_{2 k}(0)=(-1)^{k} \frac{\sqrt{\alpha}}{\pi^{1 / 4}} \frac{\sqrt{(2 k)!}}{(2 k)!!} \tag{22}
\end{gather*}
\begin{equation*}
[\psi_{2 k}(0)]^{2}=\frac{\alpha}{\sqrt{\pi}} \frac{(2 k)!}{[(2 k)!!]^{2}}=\frac{\alpha}{\sqrt{\pi}} \frac{(2 k-1)!!}{(2 k)!!} \tag{23}
\end{equation*}
|
[
"\\psi_n(x) = N_n H_n(\\alpha x) e^{-\\frac{1}{2}(\\alpha x)^2}"
] |
Expression
|
Theoretical Foundations
|
$\psi_n(x)$: Energy eigenfunction for the $n$-th state in the x representation.
$N_n$: Normalization constant for the $n$-th energy eigenfunction.
$H_n(\alpha x)$: Hermite polynomial of degree $n$ evaluated at $\alpha x$.
$\alpha$: Dimensionless constant defined as $\sqrt{m \omega / \hbar}$.
$x$: Position variable in the x representation.
$\pi$: Pi, a mathematical constant.
$n$: Quantum number representing the energy level or state index.
|
20
|
The emission of recoil-free $\gamma$ radiation by nuclei bound in a lattice is a necessary condition for the Mössbauer effect. The potential acting on the nuclei in the lattice can be approximated as a harmonic oscillator potential
$$ V(x)=\frac{1}{2} M \omega^{2} x^{2} $$
where $M$ is the mass of the nucleus, $x$ is the coordinate of the nucleus's center of mass, and $\omega$ is the vibration frequency. Assume that initially, the nucleus's center of mass motion (harmonic vibration) is in its ground state, and at $t=0$, due to a transition of energy levels within the nucleus, a photon is emitted along the $x$-axis with energy $E_{\gamma}$, and momentum $E_{\gamma} / c$. Since the $\gamma$ radiation occurs suddenly, the only effect on the nucleus's center of mass motion is that its momentum eigenvalue changes from $p$ to $(p-E_{\gamma} / c)$. Determine the expression for the probability that the nucleus's center of mass motion remains in the ground state after the photon is emitted.
|
Due to the harmonic oscillator potential, the center of mass motion of the nucleus is harmonic, initially (for $t<0$) in the ground state $\psi_{0}(x)$. Expanding $\psi_{0}(x)$ using momentum eigenfunctions gives:
\begin{equation*}
\psi_{0}(x)=(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \varphi(p) \mathrm{e}^{\mathrm{i} p x / \hbar} \mathrm{d} p \tag{1}
\end{equation*}
Due to the emission of $\gamma$ photon, momentum changes (total momentum is conserved!) from $p$ to $p-p_{0},(p_{0}=E_{\gamma} / c)$, which means the wave function $\varphi(p)$ transforms as follows:
\begin{equation*}
\varphi(p) \rightarrow \varphi(p+p_{0}) \tag{2}
\end{equation*}
Therefore, after $\gamma$ emission, the wave function of the nucleus's center of mass motion becomes
\begin{align*}
\psi(x) & =(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \varphi(p+p_{0}) \mathrm{e}^{\mathrm{i} p x / \hbar} \mathrm{d} p \\
& =(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \varphi(p^{\prime}) \mathrm{e}^{\mathrm{i}(p^{\prime}-p_{0}) x / \hbar} \mathrm{d} p^{\prime} \\
& =\psi_{0}(x) \mathrm{e}^{-\mathrm{i} p_{0} x / \hbar} \tag{3}
\end{align*}
Here, the exponential factor indicates that the nucleus's center of mass momentum is reduced by $p_{0}$. The probability that the nucleus's center of mass vibration remains in the ground state after photon emission is
\begin{equation*}
P=|\langle\psi_{0} \mid \psi\rangle|^{2}=|\int_{-\infty}^{+\infty} \psi_{0}^{2}(x) \mathrm{e}^{-i p_{0} x / \hbar} \mathrm{d} x|^{2} \tag{4}
\end{equation*}
Using the explicit form of the ground state wave function $\psi_{0}$
\begin{equation*}
\psi_{0}(x)=\frac{\sqrt{\alpha}}{\pi^{1 / 4}} \mathrm{e}^{-\alpha^{2} x^{2} / 2}, \quad \alpha=\sqrt{\frac{M_{\omega}}{\hbar}} \tag{5}
\end{equation*}
It is straightforward to calculate
\begin{align*}
\langle\psi_{0} \mid \psi\rangle & =\frac{\alpha}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \mathrm{e}^{-\alpha^{2} x^{2}-i p_{0} x / \hbar} \mathrm{d} x \\
& =\frac{\alpha}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \cos (\frac{p_{0} x}{\hbar}) \mathrm{e}^{-\alpha^{2} x^{2}} \mathrm{~d} x \\
& =\exp (-\frac{p_{0}^{2}}{4 \hbar^{2} \alpha^{2}})=\exp (-\frac{E_{\gamma}^{2}}{4 \hbar \omega M c^{2}}) \tag{6}
\end{align*}
Substituting into equation (4), we obtain the probability, which is
\begin{equation*}
P=|\langle\psi_{0} \mid \psi\rangle|^{2}=\exp (-\frac{E_{\gamma}^{2}}{2 \hbar \omega M c^{2}}) \tag{7}
\end{equation*}
Calculated numerical values are as follows:
\begin{gathered}
{ }^{57} \mathrm{Fe} \text { nucleus, } M c^{2} \approx 57 m_{\mathrm{p}} c^{2} \approx 57 \times 938 \mathrm{MeV}=5.35 \times 10^{10} \mathrm{eV} \\
E_{\gamma}=18 \mathrm{keV}=1.8 \times 10^{4} \mathrm{eV} \\
\hbar \omega=\hbar c \cdot \frac{\omega}{c} \approx 1.97 \times 10^{-5} \mathrm{eV} \cdot \mathrm{~cm} \times \frac{2 \pi \times 10^{12}}{3 \times 10^{10}} \mathrm{~cm}^{-1} \\
=4.13 \times 10^{-3} \mathrm{eV} \\
P=\exp [-\frac{(1.8 \times 10^{4})^{2}}{2 \times 4.13 \times 10^{-3} \times 5.35 \times 10^{10}}] \\
=\mathrm{e}^{-0.733}=0.48
\end{gathered}
|
[
"P=\\exp (-\\frac{E_{\\gamma}^{2}}{2 \\hbar \\omega M c^{2}})"
] |
Expression
|
Theoretical Foundations
|
$P$: Probability that the nucleus's center of mass motion remains in the ground state after photon emission.
$\mathrm{e}$: Base of the natural logarithm.
$E_{\gamma}$: Energy of the emitted photon.
$\hbar$: Reduced Planck's constant.
$\omega$: Vibration frequency of the nucleus in the lattice.
$M$: Mass of the nucleus.
$c$: Speed of light.
|
21
|
Calculate the probability that the nucleus is in various energy eigenstates after $\gamma$ radiation, as in the previous problem.
|
From the previous problem, the center-of-mass wave function of the nucleus after $\gamma$ radiation is
\begin{equation*}
\psi(x)=\mathrm{e}^{-\mathrm{i} p_{0} x / \hbar} \psi_{0}(x), \quad p_{0}=E_{\gamma} / c \tag{1}
\end{equation*}
The probability of being in the ${ }^{\prime} n$-th vibrational excited state $\psi_{n}(x)$ is
\begin{equation*}
P_{n}=|\langle\psi_{n} \mid \psi\rangle|^{2}=|\langle\psi_{n} \mid \mathrm{e}^{-\mathrm{i} p_{0} x / \hbar} \psi_{0}\rangle|^{2} \tag{2}
\end{equation*}
Let $\lambda=-p_{0} / \hbar$, calculate $\langle\psi_{n} \mid \mathrm{e}^{\mathrm{i} \lambda x} \psi_{0}\rangle$ where $\mathrm{e}^{\mathrm{i} \lambda x}$ can be viewed as an operator, expressed in terms of creation and annihilation operators, with the following relations:
\begin{align*}
& \mathrm{i} \lambda x=\alpha a^{+}-\alpha^{*} a, \quad \alpha=\mathrm{i} \lambda \sqrt{\frac{\hbar}{2 M \omega}} \tag{3}\\
& \mathrm{e}^{\mathrm{i} \lambda x}=\mathrm{e}^{\alpha a^{+}-\alpha^{*} a}=\mathrm{e}^{\alpha a^{+}} \mathrm{e}^{-\alpha^{*} a} \mathrm{e}^{-\frac{1}{2}|\alpha|^{2}} \tag{4}\\
& \mathrm{e}^{\mathrm{i} \lambda x}|\psi_{0}\rangle=\mathrm{e}^{-\frac{1}{2}|\alpha|^{2}} \mathrm{e}^{\alpha a^{+}} \mathrm{e}^{-\alpha^{*} a}|\psi_{0}\rangle=|\psi_{\alpha}\rangle \\
&= \mathrm{e}^{-\frac{1}{2}|\alpha|^{2}} \sum_{n} \frac{\alpha^{n}}{\sqrt{n!}}|\psi_{n}\rangle \tag{5}
\end{align*}
where $\psi_{\alpha}$ is the coherent state of the harmonic oscillator. Utilizing equation (5), we get
\begin{gather*}
\langle\psi_{n} \mid \mathrm{e}^{\mathrm{i} \lambda x} \psi_{0}\rangle=\frac{\alpha^{n}}{\sqrt{n!}} \mathrm{e}^{-\frac{1}{2}|\alpha|^{2}} \tag{6}\\
P_{n}=\frac{|\alpha|^{2 n}}{n!} \mathrm{e}^{-|\alpha|^{2}}=\frac{1}{n!}(\frac{\lambda^{2} \hbar}{2 M \omega})^{n} \mathrm{e}^{-\frac{\lambda^{2} \hbar}{2 M \omega}} \\
= \frac{1}{n!}(\frac{E_{\gamma}^{2}}{2 \hbar \omega M c^{2}})^{n} \exp (-\frac{E_{\gamma}^{2}}{2 \hbar \omega M c^{2}}) \tag{7}
\end{gather*}
It is easy to verify $\sum_{n} P_{n}=1$. For the example of the ${ }^{57} \mathrm{Fe}$ nucleus, the probabilities for the first few states are
$$ P_{0}=0.48, \quad P_{1}=0.35, \quad P_{2}=0.13 $$
|
[
"P_n = \\frac{1}{n!}(\\frac{E_{\\gamma}^{2}}{2 \\hbar \\omega M c^{2}})^{n} \\exp (-\\frac{E_{\\gamma}^{2}}{2 \\hbar \\omega M c^{2}})"
] |
Expression
|
Theoretical Foundations
|
$P_n$: Probability of the nucleus being in the $n$-th vibrational excited state.
$n$: Quantum number for the $n$-th vibrational excited state.
$E_{\gamma}$: Energy of the $\gamma$ radiation (gamma ray photon).
$\hbar$: Reduced Planck's constant.
$\omega$: Angular frequency of the vibrational excited states (harmonic oscillator).
$M$: Mass of the nucleus.
$c$: Speed of light in vacuum.
$\mathrm{e}$: Base of the natural logarithm.
|
22
|
Suppose the operator $\hat{H}$ has continuous eigenvalues $\omega$, and its eigenfunctions $u_{\omega}(\boldsymbol{x})$ form an orthonormal complete system, i.e.
\begin{gather*}
\hat{H} u_{\omega}(\boldsymbol{x})=\omega u_{\omega}(\boldsymbol{x}) \tag{1}\\
\int u_{\omega^{*}}^{*}(\boldsymbol{x}) u_{\omega}(\boldsymbol{x}) \mathrm{d}^{3} x=\delta(\omega-\omega^{\prime}) \tag{2}\\
\int u_{\omega}^{*}(\boldsymbol{x}^{\prime}) u_{\omega}(\boldsymbol{x}) \mathrm{d} \omega=\delta(\boldsymbol{x}-\boldsymbol{x}^{\prime}) \tag{3}
\end{gather*}
Solve the equation
\begin{equation*}
(\hat{H}-\omega_{0}) u(\boldsymbol{x})=F(\boldsymbol{x}) \tag{4}
\end{equation*}
where $F(\boldsymbol{x})$ is a known function, and $\omega_{0}$ is a specific eigenvalue of $H$.
If the answer exists in an integral, then find the integrand
|
Since $u_{\omega}(\boldsymbol{x})$ is a complete system, the solution of equation (4) can always be expressed as
\begin{equation*}\nu(x)=\int C_{\omega} u_{\omega}(x) \mathrm{d} \omega \tag{5}
\end{equation*}
Substitute into equation (4), we obtain
\begin{equation*}
\int(\omega-\omega_{0}) C_{\omega} u_{\omega}(\boldsymbol{x}) \mathrm{d} \omega=F(\boldsymbol{x}) \tag{6}
\end{equation*}
Expand $F(\boldsymbol{x})$ as
\begin{align*}
F(\boldsymbol{x}) & =\int F(\boldsymbol{x}^{\prime}) \delta(\boldsymbol{x}-\boldsymbol{x}^{\prime}) \mathrm{d}^{3} x^{\prime} \\
& =\iint F(\boldsymbol{x}^{\prime}) u_{\omega}^{*}(\boldsymbol{x}^{\prime}) u_{\omega}(\boldsymbol{x}) \mathrm{d}^{3} x^{\prime} \mathrm{d} \omega \\
& =\int F_{\omega} u_{\omega}(\boldsymbol{x}) \mathrm{d} \omega \tag{7}
\end{align*}
where
\begin{equation*}
F_{\omega}=\int F(\boldsymbol{x}^{\prime}) u_{\omega}^{*}(\boldsymbol{x}^{\prime}) \mathrm{d}^{3} x^{\prime}=\int F(\boldsymbol{x}) u_{\omega}^{*}(\boldsymbol{x}) \mathrm{d}^{3} x \tag{8}
\end{equation*}
Substitute equation (7) into equation (6), we obtain
\begin{equation*}
C_{\omega}=F_{\omega} /(\omega-\omega_{0}) \tag{9}
\end{equation*}
Substitute back into equation (5), the solution to equation (4) is
\begin{equation*}\nu(\boldsymbol{x})=\int \frac{F_{\omega}}{\omega-\omega_{0}} u_{\omega}(\boldsymbol{x}) \mathrm{d} \omega \tag{10}
\end{equation*}
|
[
"\\frac{F_{\\omega}}{\\omega-\\omega_{0}} u_{\\omega}(\\boldsymbol{x})"
] |
Expression
|
Theoretical Foundations
|
$F_{\omega}$: Expansion coefficient for the known function $F(\boldsymbol{x})$ in the basis of eigenfunctions $u_{\omega}(\boldsymbol{x})$, defined as $F_{\omega}=\int F(\boldsymbol{x}^{\prime}) u_{\omega}^{*}(\boldsymbol{x}^{\prime}) \mathrm{d}^{3} x^{\prime}$.
$\omega$: Continuous eigenvalue of the operator $\hat{H}$.
$\omega_{0}$: A specific, constant eigenvalue of the operator $\hat{H}$.
$u_{\omega}(\boldsymbol{x})$: Eigenfunction of the operator $\hat{H}$ corresponding to eigenvalue $\omega$ at position $\boldsymbol{x}$.
|
23
|
For a hydrogen-like ion (nuclear charge $Z e$ ), an electron is in the bound state $\psi_{n l m}$, calculate $\langle r^{\lambda}\rangle, \lambda=-3．
|
Known energy levels of a hydrogen-like ion are
\begin{equation*}
E_{n l m}=E_{n}=-\frac{Z^{2} e^{2}}{2 n^{2} a_{0}}, \quad n=n_{r}+l+1 \tag{1}
\end{equation*}
where $a_{0}=\hbar^{2} / \mu e^{2}$ is the Bohr radius. According to the virial theorem,
\begin{equation*}
\langle\frac{p^{2}}{2 \mu}\rangle_{n l m}=\langle\frac{r}{2} \frac{\mathrm{~d} V}{\mathrm{~d} r}\rangle_{n l m}=\frac{Z e^{2}}{2}\langle\frac{1}{r}\rangle_{n l m}=-\frac{1}{2}\langle V\rangle_{n l m} \tag{2}
\end{equation*}
so
E_{n}=\frac{1}{2}\langle V\rangle_{n l m}=-\frac{Z e^{2}}{2}\langle\frac{1}{r}\rangle_{n l m}
\begin{equation*}
\langle\frac{1}{r}\rangle_{n l m}=-\frac{2 E_{n}}{Z e^{2}}=\frac{Z}{n^{2} a_{0}} . \quad n=1,2,3, \cdots \tag{3}
\end{equation*}
Further, the spherical coordinate representation of $\psi_{n l m}$ is
\begin{equation*}
\psi_{n l m}(r, \theta, \varphi)=R_{n l}(r) Y_{l m}(\theta, \varphi) \tag{4}
\end{equation*}
It is a common eigenfunction of $(H, l^{2}, l_{z})$, satisfying the energy eigenvalue equation
\begin{equation*}
-\frac{\hbar^{2}}{2 \mu} \frac{1}{r} \frac{\partial^{2}}{\partial r^{2}} r \psi_{n l m}+[\frac{l(l+1) \hbar^{2}}{2 \mu r^{2}}-\frac{Z e^{2}}{r}] \psi_{n l m}=E_{n} \psi_{n l m} \tag{5}
\end{equation*}
The total energy operator is equivalent to
\begin{equation*}
H \rightarrow-\frac{\hbar^{2}}{2 \mu} \frac{1}{r} \frac{\partial^{2}}{\partial r^{2}} r+l(l+1) \frac{\hbar^{2}}{2 \mu r^{2}}-\frac{Z e^{2}}{r} \tag{6}
\end{equation*}
Considering $l$ as a parametric variable, differentiate equation (6) with respect to $l$, using the Hellmann theorem (refer to Chapter 8), we get
\begin{equation*}
\frac{\partial E_{n}}{\partial l}=\langle\frac{\partial H}{\partial l}\rangle_{n l m}=(l+\frac{1}{2}) \frac{\hbar^{2}}{\mu}\langle\frac{1}{r^{2}}\rangle_{n l m} \tag{7}
\end{equation*}
Since $n=n_{r}+l+1$, we find
\begin{equation*}
\frac{\partial E_{n}}{\partial l}=\frac{\partial E_{n}}{\partial n}=\frac{Z^{2} e^{2}}{n^{3} a_{0}} \tag{8}
\end{equation*}
Substituting into equation (7) and using $a_{0}=\hbar^{2} / \mu e^{2}$, we obtain
\begin{equation*}
\langle\frac{1}{r^{2}}\rangle_{n l m}=\frac{1}{(l+\frac{1}{2}) n^{3}} \frac{Z^{2}}{a_{0}^{2}} \tag{9}
\end{equation*}
Finally, compute $(r^{-3})$．
For the s state $(l=0), r \rightarrow 0$, $\psi \rightarrow C$ (constant), therefore
\begin{equation*}
\langle r^{-3}\rangle_{n 00} \rightarrow \infty \tag{10}
\end{equation*}
When $l \neq 0$, using formula (7b) in problem (5.7), we get
\begin{equation*}
\langle\frac{1}{r^{3}}\rangle_{n l m}=\frac{Z}{l(l+1) a_{0}}\langle\frac{1}{r^{2}}\rangle_{n l m} \tag{11}
\end{equation*}
Thus
\begin{equation*}
\langle\frac{1}{r^{3}}\rangle_{n l m}=\frac{1}{n^{3} l(l+\frac{1}{2})(l+1)}(\frac{Z}{a_{0}})^{3} \tag{12}
\end{equation*}
As $l \rightarrow 0$, the right side of the equation $\rightarrow \infty$, so this formula is actually applicable for all $l$ values.
Discussion: Since both the total energy operator and radial equation are independent of the magnetic quantum number $m$, $\langle r^{\lambda}\rangle$ is independent of $m$. However, $\langle r^{-1}\rangle$ is also independent of the angular quantum number $l$, depending only on the principal quantum number $n.$ $\langle r^{-2}\rangle$ and $\langle r^{-3}\rangle$ depend on both $n$ and $l$, meaning for states with the same energy level but different "orbital shapes" (different $l$), $\langle r^{-2}\rangle$ or $\langle r^{-3}\rangle$ have different values.
Using formula (9), it can be easily obtained that the average value of the centrifugal potential energy in the state $\psi_{n l m}$ is
\begin{equation*}
\langle\frac{l^{2}}{2 \mu r^{2}}\rangle_{n l m}=\frac{l(l+1) Z^{2} e^{2}}{(2 l+1) n^{3} a_{0}}=-\frac{l(l+1)}{(l+\frac{1}{2}) n} E_{n} \tag{13}
\end{equation*}
Since $(-E_{n})$ is the average kinetic energy, the proportion of centrifugal potential energy within kinetic energy is $l(l+1) /(l+\frac{1}{2}) n$. When $n$ is fixed, as $l$ increases, this proportion grows. When $l$ takes the maximum value $(l=n-1)$, this proportion is $(n-1) /$ $(n-\frac{1}{2})$, and the radial kinetic energy occupies only $1 /(2 n-1)$ of the kinetic energy in this case. Therefore, if $n \gg 1,(n, n-1, m)$ implies small radial kinetic energy, corresponding to circular orbits in Bohr's quantum theory.
|
[
"\\langle\\frac{1}{r^{3}}\\rangle_{n l m}=\\frac{1}{n^{3} l(l+\\frac{1}{2})(l+1)}(\\frac{Z}{a_{0}})^{3}"
] |
Expression
|
Theoretical Foundations
|
$\langle\frac{1}{r^{3}}\rangle_{n l m}$: Expectation value of $1/r^3$ for the state $\psi_{n l m}$.
$n$: Principal quantum number, $n=n_{r}+l+1$.
$l$: Azimuthal (orbital angular momentum) quantum number.
$Z$: Nuclear charge number (atomic number) of the hydrogen-like ion.
$a_{0}$: Bohr radius, $a_{0}=\hbar^{2} / \mu e^{2}$.
|
24
|
The potential acting on the valence electron (outermost electron) of a monovalent atom by the atomic nucleus (atomic nucleus and inner electrons) can be approximately expressed as
\begin{equation*}
V(r)=-\frac{e^{2}}{r}-\lambda \frac{e^{2} a_{0}}{r^{2}}, \quad 0<\lambda \ll 1 \tag{1}
\end{equation*}
where $a_{0}$ is the Bohr radius. Find the energy level of the valence electron and compare it with the energy level of the hydrogen atom.
|
Take the complete set of conserved quantities as $(H, l^{2}, l_{z})$, whose common eigenfunctions are
\begin{equation*}
\psi(r, \theta, \varphi)=R(r) \mathrm{Y}_{l m}(\theta, \varphi)=\frac{u(r)}{r} \mathrm{Y}_{l m}(\theta, \varphi) \tag{2}
\end{equation*}
$u(r)$ satisfies the radial equation
\begin{equation*}
-\frac{\hbar^{2}}{2 \mu} u^{\prime \prime}+[l(l+1) \frac{\hbar^{2}}{2 \mu r^{2}}-\frac{e^{2}}{r}-\lambda \frac{e^{2} a_{0}}{r^{2}}] u=E u \tag{3}
\end{equation*}
Let
\begin{equation*}
l(l+1)-2 \lambda=l^{\prime}(l^{\prime}+1) \tag{4}
\end{equation*}
Equation (3) can then be transformed into
\begin{equation*}
-\frac{\hbar^{2}}{2 \mu} u^{\prime \prime}+[l^{\prime}(l^{\prime}+1) \frac{\hbar^{2}}{2 \mu r^{2}}-\frac{e^{2}}{r}] u=E u \tag{3'}
\end{equation*}
This is equivalent to the radial equation of the hydrogen atom with $l$ replaced by $l^{\prime}$. Therefore, the solution process for equation ($3^{\prime}$) is entirely analogous to the hydrogen atom problem. The latter's energy levels are
\begin{equation*}
E_{n}=-\frac{e^{2}}{2 n^{2} a_{0}}, \quad n=n_{r}+l+1, \quad n_{r}=0,1,2, \cdots \tag{5}
\end{equation*}
Replacing $l$ with $l^{\prime}$ gives the energy levels of the valence electron:
\begin{equation*}
E_{n l}=-\frac{e^{2}}{2(n^{\prime})^{2} a_{0}}, \quad n^{\prime}=n_{r}+l^{\prime}+1 \tag{6}
\end{equation*}
It is generally assumed that
\begin{equation*}
l^{\prime}=l+\Delta_{l} \tag{7}
\end{equation*}
\begin{equation*}
n^{\prime}=n_{r}+l+\Delta_{l}+1=n+\Delta_{l} \tag{8}
\end{equation*}
$\Delta_{l}$ is referred to as the 'correction number' of the quantum numbers $l$ and $n$. Since $\lambda \ll 1$, equation (4) can be approximated as follows:
\begin{aligned}
l(l+1)-2 \lambda & =l^{\prime}(l^{\prime}+1)=(l+\Delta_{l})(l+\Delta_{l}+1) \\
& =l(l+1)+(2 l+1) \Delta_{l}+(\Delta_{l})^{2}
\end{aligned}
Neglecting $(\Delta_{l})^{2}$, we get
\begin{equation*}
\Delta_{l} \approx-\lambda /(l+\frac{1}{2}) \tag{9}
\end{equation*}
Since $\lambda \ll 1$, $|\Delta_{l}| \ll 1$. Thus, the energy level $E_{n l}$ obtained in this problem has only a small difference from the hydrogen atomic energy level, but the 'degeneracy in $l$' of the energy levels has been removed. Equation (6) is broadly consistent with experimental data on alkali metal spectra; in particular, the correction number $|\Delta_{l}|$ decreases as $l$ increases, which agrees well with experiments.
The exact solution for equation (4) is
\begin{equation*}
l^{\prime}=-\frac{1}{2}+(l+\frac{1}{2})[1-\frac{8 \lambda}{(2 l+1)^{2}}]^{\frac{1}{2}} \tag{10}
\end{equation*}
Expanding the above equation as a binomial series and retaining the $\lambda$ term while neglecting terms of order $\lambda^{2}$ and higher, yields equation (9).
|
[
"E_{n l}=-\\frac{e^{2}}{2(n^{\\prime})^{2} a_{0}}"
] |
Expression
|
Theoretical Foundations
|
$E_{nl}$: Energy levels of the valence electron.
$e$: Elementary charge.
$n'$: Modified principal quantum number for the valence electron.
$a_0$: Bohr radius.
|
25
|
For a particle of mass $\mu$ moving in a spherical shell $\delta$ potential well
\begin{equation*}
V(r)=-V_{0} \delta(r-a), \quad V_{0}>0, a>0 \tag{1}
\end{equation*}
find the minimum value of $V_{0}$ required for bound states to exist.
|
The ground state is an s-state $(l=0)$, and the wave function can be expressed as
\begin{equation*}
\psi(r)=u(r) / r \tag{2}
\end{equation*}
$u(r)$ satisfies the radial equation
\begin{equation*}\nu^{\prime \prime}+\frac{2 \mu E}{\hbar^{2}} u+\frac{2 \mu V_{0}}{\hbar^{2}} \delta(r-a) u=0 \tag{3}
\end{equation*}
As $r \rightarrow \infty$, $V(r) \rightarrow 0$, so for a bound state $E<0$. Define
\begin{equation*}
\beta=\sqrt{-2 \mu E} / \hbar \tag{4}
\end{equation*}
Equation (3) can be rewritten as
\begin{equation*}
\nu^{\prime \prime} - \beta^{2} u + \frac{2 \mu V_{0}}{\hbar^{2}} \delta(r-a) u=0 \tag{3^{\prime}}
\end{equation*}
The boundary conditions are
$$r \rightarrow 0, \infty \text {, } u \rightarrow 0$$
Integrating equation ($3^{\prime}$) around $r \sim a$, we obtain the jump condition for $u^{\prime}$ (see Problem 2.1)
\begin{equation*}
\nu^{\prime}(a+0)-u^{\prime}(a-0)=-\frac{2 \mu V_{0}}{\hbar^{2}} u(a) \tag{5}
\end{equation*}
This means
\begin{equation*}
\frac{u^{\prime}}{u}|_{r=a-0} ^{r=a+0}=-\frac{2 \mu V_{0}}{\hbar^{2}} \tag{$\prime$}
\end{equation*}
For $r \neq a$, equation ($3^{\prime}$) becomes
\begin{equation*}\nu^{\prime \prime}-\beta^{2} u=0 \tag{\prime\prime}
\end{equation*}
In the region $r>a$, the solution satisfying the boundary condition at infinity is
\begin{equation*}\nu=C \mathrm{e}^{-\beta r}, \quad r>a \tag{6}
\end{equation*}
Thus
\begin{equation*}
(\frac{u^{\prime}}{u})_{r=a+0}=-\beta \tag{7}
\end{equation*}
If the value of $V_{0}$ is just sufficient to form the first bound state, the energy level must be $E=0^{-}$, at which point $\beta=0$, and equation ($3^{\prime\prime}$) becomes
\begin{equation*}\nu^{\prime \prime}=0 \quad(E \rightarrow 0^{-}) \tag{8}
\end{equation*}
Equation (7) becomes
\begin{equation*}
(\frac{u^{\prime}}{u})_{r=a+0}=0 \quad(E \rightarrow 0^{-}) \tag{7'}
\end{equation*}
When $E \rightarrow 0^{-}$, the solution of equation (8) in the region $r<a$, which satisfies the boundary condition $u(0)=0$, is
\begin{equation*}\nu=A r, \quad r<a \quad(E \rightarrow 0^{-}) \tag{9}
\end{equation*}
Thus
\begin{equation*}
(\frac{u^{\prime}}{u})_{r=a-0}=\frac{1}{a} \quad(E \rightarrow 0^{-}) \tag{10}
\end{equation*}
Substituting equations ($7^{\prime}$) and (10) into equation ($5^{\prime}$) gives
\begin{equation*}
V_{0}=\hbar^{2} / 2 \mu a \tag{11}
\end{equation*}
This is the minimum value of $V_{0}$ required for the existence of bound states, with the corresponding energy level being $E=0^{-}$.
|
[
"V_{0}=\\frac{\\hbar^{2}}{2 \\mu a}"
] |
Expression
|
Theoretical Foundations
|
$V_0$: Strength of the spherical shell potential well
$\hbar$: Reduced Planck's constant
$\mu$: Mass of the particle
$a$: Radius of the spherical shell
|
26
|
Simplify $\mathrm{e}^{\mathrm{i} \lambda_{\sigma_{2}}} \sigma_{\alpha} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}, \alpha=x, ~ y, \lambda$ are constants.
|
Solution 1: Utilize the formula:
\begin{equation*}
\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}}=\cos \lambda+\mathrm{i} \sigma_{z} \sin \lambda \tag{1}
\end{equation*}
We get
$$\mathrm{e}^{\mathrm{i} \lambda \sigma_{z} \sigma_{x} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=(\cos \lambda+\mathrm{i} \sigma_{z} \sin \lambda) \sigma_{x}(\cos \lambda-\mathrm{i} \sigma_{z} \sin \lambda) .}$$
Further using
\begin{equation*}
\sigma_{z}^{2}=1, \quad \sigma_{z} \sigma_{x}=-\sigma_{x} \sigma_{z}=\mathrm{i} \sigma_{y}, \tag{2}
\end{equation*}
we obtain
\begin{equation*}
\mathrm{e}^{\mathrm{i} \lambda_{\sigma_{z}} \sigma_{y} \mathrm{e}^{-\mathrm{i} \alpha_{z}}}=\sigma_{x} \sin 2 \lambda+\sigma_{y} \cos 2 \lambda \tag{4}
\end{equation*}
Solution 2: let
\begin{equation*}
\mathrm{e}^{\mathrm{i} \lambda \sigma_{z} \sigma_{x} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=C_{0} I+C_{1} \sigma_{x}+C_{2} \sigma_{y}+C_{3} \sigma_{z} .} \tag{5}
\end{equation*}
Applying the above expression to the eigenstate $\chi_{\frac{1}{2}}$ of $\sigma_{z}$, using
\begin{equation*}
\sigma_{z} \chi_{\frac{1}{2}}=\chi_{\frac{1}{2}}, \quad \sigma_{x} \chi_{\frac{1}{2}}=\chi_{-\frac{1}{2}}, \quad \sigma_{y} \chi_{\frac{1}{2}}=\mathrm{i} \chi_{-\frac{1}{2}} \tag{6}
\end{equation*}
we get
$$ \mathrm{e}^{-2 \mathrm{a}} \chi_{-\frac{1}{2}}=(C_{0}+C_{3}) \chi_{\frac{1}{2}}+(C_{1}+\mathrm{i} C_{2}) \chi_{-\frac{1}{2}}, $$
Since $\chi_{\frac{1}{2}}$ and $\chi_{-\frac{1}{2}}$ are linearly independent, comparing the coefficients on both sides, we have
\begin{align*}
& C_{0}+C_{3}=0 \tag{7}\\
& C_{1}+\mathrm{i}_{2}=\mathrm{e}^{-2 \mathrm{u}}
\end{align*}
Now apply expression (5) to the eigenstate $\chi_{-\frac{1}{2}}$ of $\sigma_{z}$, using
\begin{equation*}
\sigma_{z} \chi_{-\frac{1}{2}}=-\chi_{-\frac{1}{2}}, \quad \sigma_{x} \chi_{-\frac{1}{2}}=\chi_{\frac{1}{2}}, \quad \sigma_{y} \chi_{-\frac{1}{2}}=-\mathrm{i} \chi_{\frac{1}{2}} \tag{8}
\end{equation*}
which gives
$$ \mathrm{e}^{2 i \mathrm{i}} \chi_{\frac{1}{2}}=(C_{1}-\mathrm{i} C_{2}) \chi_{\frac{1}{2}}+(C_{0}-C_{3}) \chi_{-\frac{1}{2}}$$
Comparing the coefficients, we have
\begin{gather*}
C_{0}-C_{3}=0 \\
C_{1}-\mathrm{i} C_{2}=\mathrm{e}^{2 \mathrm{i}} \tag{9}
\end{gather*}
From equations (7) and (9), it is easy to solve for
\begin{equation*}
C_{0}=C_{3}=0, \quad C_{1}=\cos 2 \lambda, \quad C_{2}=-\sin 2 \lambda \tag{10}
\end{equation*}
Substitute back into expression (5), thus we have
\begin{equation*}
\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \sigma_{x} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=\sigma_{x} \cos 2 \lambda-\sigma_{y} \sin 2 \lambda \tag{3}
\end{equation*}
Similarly, it can be proven that
\begin{equation*}
\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \sigma_{y} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=\sigma_{x} \sin 2 \lambda+\sigma_{y} \cos 2 \lambda \tag{4}
\end{equation*}
Solution 3: Treat $\lambda$ as a parameter, let
\begin{align*}
& f(\lambda)=\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \sigma_{x} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}} \tag{11}\\
& g(\lambda)=\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \sigma_{y} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}. \tag{12}
\end{align*}
Note that
\begin{equation*}
f(0)=\sigma_{x}, \quad g(0)=\sigma_{y} \tag{13}
\end{equation*}
Differentiate equations (11) and (12) with respect to $\lambda$, we have
\begin{align*}
& f^{\prime}(\lambda)=\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \mathrm{i}(\sigma_{z} \sigma_{x}-\sigma_{x} \sigma_{z}) \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=-2 g(\lambda) \tag{14}\\
& g^{\prime}(\lambda)=\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \mathrm{i}(\sigma_{z} \sigma_{y}-\sigma_{y} \sigma_{z}) \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=2 f(\lambda)
\end{align*}
In equation (14), multiply the second equation by i, then add it to the first, yielding
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} \lambda}[f(\lambda)+\mathrm{i} g(\lambda)]=2 \mathrm{i}[f(\lambda)+\mathrm{i} g(\lambda)] \tag{$\prime$}
\end{equation*}
As a first-order differential equation, the solution is evidently
\begin{equation*}
f(\lambda)+\mathrm{i} g(\lambda)=[f(0)+\mathrm{i} g(0)] \mathrm{e}^{2 i \lambda}=(\sigma_{x}+\mathrm{i} \sigma_{y}) \mathrm{e}^{2 i \lambda} \tag{15}
\end{equation*}
In equation (14), multiply the second equation by (-i), then add it to the first, yielding
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} \lambda}[f(\lambda)-\mathrm{i} g(\lambda)]=-2 \mathrm{i}[f(\lambda)-\mathrm{i} g(\lambda)] \tag{$\prime\prime$}
\end{equation*}
The solution is
\begin{equation*}
f(\lambda)-\mathrm{i} g(\lambda)=[f(0)-\mathrm{i} g(0)] \mathrm{e}^{-2 \mathrm{i}}=(\sigma_{x}-\mathrm{i} \sigma_{y}) \mathrm{e}^{-2 \mathrm{i} \lambda} \tag{16}
\end{equation*}
Adding and subtracting equations (15) and (16), we obtain
\begin{align*}
& f(\lambda)=\sigma_{x} \cos 2 \lambda-\sigma_{y} \sin 2 \lambda \tag{17}\\
& g(\lambda)=\sigma_{x} \sin 2 \lambda+\sigma_{y} \cos 2 \lambda
\end{align*}
This is exactly the equations (3) and (4).
(Note) If $\lambda$ is a real number, then both $f(\lambda)$ and $g(\lambda)$ are Hermitian operators. At this time, equation (16) is the conjugate equation of equation (15), and equation (17) can be derived directly from equation (15).
|
[
"\\mathrm{e}^{\\mathrm{i} \\lambda \\sigma_{z}} \\sigma_{x} \\mathrm{e}^{-\\mathrm{i} \\lambda \\sigma_{z}} = \\sigma_{x} \\cos 2 \\lambda-\\sigma_{y} \\sin 2 \\lambda"
] |
Expression
|
Theoretical Foundations
|
$\mathrm{e}$: Euler's number, the base of the natural logarithm.
$\mathrm{i}$: Imaginary unit, defined as $\sqrt{-1}$.
$\lambda$: A constant parameter.
$\sigma_{z}$: Pauli matrix in the z-direction.
$\sigma_{x}$: Pauli matrix in the x-direction.
$\sigma_{y}$: Pauli matrix in the y-direction.
|
27
|
A particle with spin $\hbar / 2$ and magnetic moment $\boldsymbol{\mu}=\mu_{0} \boldsymbol{\sigma}$ is placed in a uniform magnetic field $\boldsymbol{B}$ that points in an arbitrary direction $(\theta, \varphi)$ (where $n_3 = \cos \theta$ is the projection of the direction vector of the magnetic field on the $z$ axis). Neglecting orbital motion, assume at $t=0$ the particle is polarized along the positive $z$ direction, namely $\sigma_{z}=1,\langle\boldsymbol{\sigma}\rangle_{t=0}=$ $\boldsymbol{e}_{3}$. Find the expression for the $z$ component of the polarization direction of the particle for $t>0$, i.e., $\langle\sigma_{z}\rangle_{t}$.
|
First, determine the spin wave function, then compute $\langle\boldsymbol{\sigma}\rangle$. Using $\boldsymbol{n}$ to denote the unit vector in the $(\theta, \varphi)$ direction, $\boldsymbol{e}_{1}, ~ \boldsymbol{e}_{2}, ~ \boldsymbol{e}_{3}$ denote the unit vectors in the $x, ~ y, ~ z$ directions, respectively:
\begin{align*}
\boldsymbol{n} & =n_{1} \boldsymbol{e}_{1}+n_{2} \boldsymbol{e}_{2}+n_{3} \boldsymbol{e}_{3} \\
& =\sin \theta \cos \varphi \boldsymbol{e}_{1}+\sin \theta \sin \varphi \boldsymbol{e}_{2}+\cos \theta \boldsymbol{e}_{3} \tag{1}
\end{align*}
The Hamiltonian associated with the spin motion is
\begin{equation*}
H=-\boldsymbol{\mu} \cdot \boldsymbol{B}=-\mu_{0} B \boldsymbol{\sigma} \cdot \boldsymbol{n}=-\mu_{0} B_{\sigma_{n}} \tag{2}
\end{equation*}
In the $\sigma_{z}$ representation, the matrix representation of $\sigma_{n}$ is
\sigma_{n}=\sigma_{x} n_{1}+\sigma_{y} n_{2}+\sigma_{z} n_{3}=[\begin{array}{cc}
n_{3} & n_{1}-\mathrm{i} n_{2} \tag{3}\\
n_{1}+\mathrm{i} n_{2} & -n_{3}
\end{array}]
Assume the spin wave function of the particle is
\chi(t)=[\begin{array}{l}
a(t) \tag{4}\\
b(t)
\end{array}]
$\chi(t)$ satisfies the Schrödinger equation
\begin{equation*}
\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} t} \chi(t)=H \chi(t)=-\mu_{0} B \sigma_{n} \chi(t) \tag{5}
\end{equation*}
The eigenvalues of $\sigma_{n}$ are $\pm 1$, the eigenvalues of $H$ (i.e., stationary energy levels) are
\begin{gather*}
E=\mp \mu_{0} B=\mp \hbar \omega \\
\omega=\mu_{0} B / \hbar \tag{6}
\end{gather*}
Since $\omega$ and $\mu_{0}$ have the same sign, if $\mu_{0}<0$ then $\omega<0$. The eigenfunctions of $\sigma_{n}$ and $H$ are
\begin{gather*}
\phi_{1}=\frac{1}{\sqrt{2(1+n_{3})}}[\begin{array}{c}
1+n_{3} \\
n_{1}+\mathrm{i} n_{2}
\end{array}] \quad(\sigma_{n}=1, E=-\hbar \omega) \\
\phi_{-1}=\frac{1}{\sqrt{2(1+n_{3})}}[\begin{array}{c}
n_{1}-\mathrm{i} n_{2} \\
-1-n_{3}
\end{array}] \quad(\sigma_{n}=-1, E=\hbar \omega) \tag{7}
\end{gather*}
The general solution of equation (5) is
\begin{equation*}
\chi(t)=C_{1} \phi_{1} \mathrm{e}^{\mathrm{i} \omega t}+C_{-1} \phi_{-1} \mathrm{e}^{-\mathrm{i} \omega t} \tag{8}
\end{equation*}
$C_{1}, ~ C_{-1}$ are determined by initial conditions:
\begin{equation*}
C_{1}=\phi_{1}^{+} \chi(0), \quad C_{-1}=\phi_{-1}^{+} \chi(0) \tag{9}
\end{equation*}
The initial wave function for this problem is the eigenfunction of $\sigma_{z}=1$, that is
\chi(0)=\chi_{\frac{1}{2}}=[\begin{array}{l}
1 \tag{10}\\
0
\end{array}]
Thus
\begin{equation*}
C_{1}=\sqrt{\frac{1+n_{3}}{2}}, \quad C_{-1}=\frac{n_{1}+\mathrm{i} n_{2}}{\sqrt{2(1+n_{3})}} \tag{11}
\end{equation*}
Substituting into equation (8), we get
\begin{align*}
\chi(t) & =\frac{1}{\sqrt{2(1+n_{3})}}[\begin{array}{l}
(1+n_{3}) C_{1} \mathrm{e}^{\mathrm{i} \omega t}+(n_{1}-\mathrm{i} n_{2}) C_{-1} \mathrm{e}^{-\mathrm{i} \omega t} \\
(n_{1}+\mathrm{i} n_{2}) C_{1} \mathrm{e}^{\mathrm{i} \omega t}-(1+n_{3}) C_{-1} \mathrm{e}^{-\mathrm{i} \omega t}
\end{array}] \\
& =[\begin{array}{l}
\cos \omega t+\mathrm{i} n_{3} \sin \omega t \\
(\mathrm{i} n_{1}-n_{2}) \sin \omega t
\end{array}]=[\begin{array}{l}
a(t) \\
b(t)
\end{array}] \tag{12}
\end{align*}
The expectation values of $\boldsymbol{\sigma}$ in the $\chi(t)$ state are
\begin{align*}
\langle\sigma_{x}\rangle & =\chi^{+} \sigma_{x} \chi=[a^{*} b^{*}][\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}][\begin{array}{l}
a \\
b
\end{array}]=a^{*} b+b^{*} a \\
& =n_{1} n_{3}(1-\cos 2 \omega t)-n_{2} \sin 2 \omega t \tag{13a}\\
\langle\sigma_{y}\rangle & =\chi^{+} \sigma_{y} \chi=\mathrm{i}(b^{*} a-a^{*} b) \\
& =n_{2} n_{3}(1-\cos 2 \omega t)+n_{1} \sin 2 \omega t \tag{1;b}\\
\langle\sigma_{z}\rangle & =\chi^{-} \sigma_{*} \chi=a^{*} a-b^{*} b \\
& =n_{3}^{2}+(1-n_{3}^{2}) \cos 2 \omega t \tag{13c}
\end{align*}
If the magnetic field $\boldsymbol{B}$ points in the positive $x$ direction, then
n_{1}=1, \quad n_{2}=n_{3}=0
Then equation (13) becomes
\langle\sigma_{x}\rangle=0,\langle\sigma_{y}\rangle=\sin 2 \omega t, \quad\langle\sigma_{z}\rangle=\cos 2 \omega t
This is precisely the result obtained in the previous problem (note that $\omega$ in the previous problem corresponds to $-\omega$ in this problem).
|
[
"n_{3}^{2}+(1-n_{3}^{2}) \\cos 2 \\omega t"
] |
Expression
|
Theoretical Foundations
|
$n_3$: z-component of the unit vector in the magnetic field direction, $n_3 = \cos \theta$.
$\omega$: Larmor frequency, $\omega=\mu_{0} B / \hbar$.
$t$: Time.
$\theta$: Polar angle of the magnetic field direction.
|
28
|
In a system with a spin of $\hbar / 2$, the magnetic moment $\boldsymbol{\mu}=\mu_{0} \boldsymbol{\sigma}$ is placed in a uniform magnetic field $\boldsymbol{B}_{0}$ directed along the positive $z$ direction for $t<0$. At $t \geqslant 0$, an additional rotating magnetic field $\boldsymbol{B}_{1}(t)$, perpendicular to the $z$ axis, is applied:
$$ \boldsymbol{B}_{1}(t)=B_{1} \cos 2 \omega_{0} t e_{1}-B_{1} \sin 2 \omega_{0} t e_{2},$$
where $\omega_{0}=\mu_{0} B_{0} / \hbar$. It is known that for $t \leqslant 0$, the system is in the eigenstate $\chi_{\frac{1}{2}}$ with $s_{z}=\hbar / 2$. Find the expression for the time $\Delta t$ it takes for the system's spin to first reverse from $s_z = \hbar/2$ (along the positive $z$ direction) to $s_z = -\hbar/2$ (along the negative $z$ direction) starting from $t=0$. Express this in terms of $\omega_1 = \mu_0 B_1 / \hbar$ and relevant constants.
|
The Hamiltonian related to the spin motion of the system is
\begin{equation*}
H=-\mu \cdot[\boldsymbol{B}_{0}+\boldsymbol{B}_{1}(t)], \quad t \geqslant 0 \tag{1}
\end{equation*}
In the $s_{z}$ representation, the matrix form of $H$ is
\begin{align}
H & =-\mu_{0} B_{1}(\sigma_{x} \cos 2 \omega_{0} t-\sigma_{y} \sin 2 \omega_{0} t)-\mu_{0} B_{0} \sigma_{z} \\
& =-\mu_{0} \begin{pmatrix}
B_{0} & B_{1} \mathrm{e}^{2 i \omega_{0} t} \\
B_{1} \mathrm{e}^{-2 i \omega_{0} t} & -B_{0}
\end{pmatrix}
\tag{\prime}
\end{align}
Assuming the wave function for $t \geqslant 0$ is
$$\chi(t)=\begin{pmatrix}
a(t) \tag{2}\\
b(t)
\end{pmatrix}.$$
Substituting into the Schrödinger equation
\begin{equation*}
\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} t} \chi(t)=H \chi(t) \tag{3},
\end{equation*}
we have
\begin{gather*}
\frac{\mathrm{d}}{\mathrm{~d} t} a(t)=\mathrm{i} \omega_{0} a(t)+\mathrm{i} \omega_{1} b(t) \mathrm{e}^{2 \mathrm{i} \omega_{0} t} \\
\frac{\mathrm{~d}}{\mathrm{~d} t} b(t)=-\mathrm{i} \omega_{0} b(t)+\mathrm{i} \omega_{1} a(t) \mathrm{e}^{-2 \mathrm{i} \omega_{0} t}. \tag{4}
\end{gather*}
where
\begin{equation*}
\omega_{0}=\frac{\mu_{0} B_{0}}{\hbar}, \quad \omega_{1}=\frac{\mu_{0} B_{1}}{\hbar} \tag{5}
\end{equation*}
Let
\begin{equation*}
a(t)=c_{1}(t) \mathrm{e}^{\mathrm{i} \omega_{0} t}, \quad b(t)=c_{2}(t) \mathrm{e}^{-\mathrm{i} \omega_{0} t} \tag{6}
\end{equation*}
Substituting into equation (4), yields equations for $c_{1}$ and $c_{2}$
\begin{align*}
& \frac{\mathrm{d}}{\mathrm{~d} t} c_{1}(t)=\mathrm{i} \omega_{1} c_{2}(t) \\
& \frac{\mathrm{d}}{\mathrm{~d} t} c_{2}(t)=\mathrm{i} \omega_{1} c_{1}(t) \tag{7}
\end{align*}
The initial conditions are
\begin{equation*}
\chi(0)=\begin{pmatrix}{l}
a(0) \\
b(0)
\end{pmatrix}=\begin{pmatrix}
1 \\
0
\end{pmatrix}=\chi_{\frac{1}{2}}, \quad \text { i.e., } \quad \left\{\begin{array}{l}
c_{1}(0)=1 \\
c_{2}(0)=0
\end{array}\right.
\tag{8}
\end{equation*}
By adding and subtracting equations in (7), we obtain
\begin{gather*}
\frac{\mathrm{d}}{\mathrm{~d} t}(c_{1}+c_{2})=\mathrm{i} \omega_{1}(c_{1}+c_{2}) \\
\frac{\mathrm{d}}{\mathrm{~d} t}(c_{1}-c_{2})=-\mathrm{i} \omega_{1}(c_{1}-c_{2}) \tag{9}
\end{gather*}
The solution is
\begin{align*}
c_{1}(t)+c_{2}(t) & =[c_{1}(0)+c_{2}(0)] \mathrm{e}^{\mathrm{i} \omega_{1} t} = \mathrm{e}^{\mathrm{i} \omega_{1} t}\\
c_{1}(t)-c_{2}(t) &= (c_{1}(0)-c_{2}(0)) \mathrm{e}^{-\mathrm{i} \omega_{1} t}=\mathrm{e}^{-\mathrm{i} \omega_{1} t} .
\end{align*}
Adding and subtracting these, we find
\begin{equation*}
c_{1}(t)=\cos \omega_{1} t, \quad c_{2}(t)=\mathrm{i} \sin \omega_{1} t \tag{11}
\end{equation*}
Substituting into equation (6), we obtain
\begin{equation*}
a(t)=\cos \omega_{1} t \mathrm{e}^{\mathrm{i} \omega_{0} t}, \quad b(t)=\mathrm{i} \sin \omega_{1} t \mathrm{e}^{-\mathrm{i} \omega_{0} t} \tag{12}
\end{equation*}
Substituting into equation (2), we find
\begin{equation}
\begin{split}
\chi(t)&=\begin{pmatrix}
\cos \omega_{1} t \mathrm{e}^{\mathrm{i} \omega_{0} t} \\
\mathrm{i} \sin \omega_{1} t \mathrm{e}^{-\mathrm{i} \omega_{0} t}
\end{pmatrix} \\
&=\cos \omega_{1} t \mathrm{e}^{\mathrm{i} \omega_{0} t} \chi_{\frac{1}{2}}+\mathrm{i} \sin \omega_{1} t \mathrm{e}^{-\mathrm{i} \omega_{0} t} \chi_{-\frac{1}{2}} \tag{13}
\end{split}
\end{equation}
Clearly
\begin{gathered}
t=0, \quad \chi=\chi_{\frac{1}{2}}=[\begin{array}{l}
1 \\
0
\end{array}] \\
s_{z}=\frac{\hbar}{2}, \quad\langle s\rangle=\frac{\hbar}{2} e_{3} \\
t=\frac{\pi}{2 \omega_{1}}, \quad \chi=\mathrm{ie}^{-\mathrm{i} \omega_{0} t} \chi_{-\frac{1}{2}}=\mathrm{ie}^{-\mathrm{i} \omega_{0} t}[\begin{array}{l}
0 \\
1
\end{array}] \\
s_{\tilde{z}}=-\frac{\hbar}{2}, \quad\langle\boldsymbol{s}\rangle=-\frac{\hbar}{2} e_{3} \\
t=\frac{\pi}{\omega_{1}}, \quad \chi=-\mathrm{e}^{\mathrm{i} \omega_{0} t} \chi_{\frac{1}{2}}=-\mathrm{e}^{\mathrm{i} \omega_{0} t}[\begin{array}{l}
1 \\
0
\end{array}] \\
s_{z}=\frac{\hbar}{2}, \quad\langle\boldsymbol{s}\rangle=\frac{\hbar}{2} e_{3}
\end{gathered}
In other words, the system's spin direction changes once every $\Delta t=\pi / 2 \omega_{1}=\pi \hbar / 2 \mu_{0} B_{1}$ . The spin state of the system undergoes periodic oscillation between $\chi_{\frac{1}{2}}$ and $\chi_{-\frac{1}{2}}$, with a period $T=2 \Delta t=\pi \hbar / \mu_{0} B_{1}$.
This problem illustrates the basic principle of magnetic resonance.
|
[
"\\pi \\hbar / 2 \\mu_{0} B_{1}"
] |
Expression
|
Theoretical Foundations
|
$\pi$: Mathematical constant, approximately 3.14159.
$\hbar$: Reduced Planck's constant.
$\mu_{0}$: Constant relating magnetic moment to spin.
$B_{1}$: Amplitude of the rotating magnetic field.
|
29
|
The magnetic moment (operator) of an electron is
\begin{equation*}
\boldsymbol{\mu}=\boldsymbol{\mu}_{l}+\boldsymbol{\mu}_{s}=-\frac{e}{2 m_{\mathrm{e}} c}(\boldsymbol{l}+2 \boldsymbol{s}) \tag{1}
\end{equation*}
Try to calculate the expectation value of $\mu_{z}$ for the $|l j m_{j}\rangle$ state.
|
If we use the Bohr magneton $\mu_{\mathrm{B}}=e \hbar / 2 m_{\mathrm{e}} c$ as the unit of magnetic moment, then the magnetic moment operator of an electron can be written as (here $\hbar=1$ is taken)
\begin{equation*}
\boldsymbol{\mu}=-(\boldsymbol{l}+2 \boldsymbol{s})=-(\boldsymbol{j}+\boldsymbol{s})=-(\boldsymbol{j}+\frac{1}{2} \boldsymbol{\sigma}) \tag{2}
\end{equation*}
Thus, we have
\begin{equation*}
\langle l j m_{j}| \mu_{z}|l j m_{j}\rangle=-g m_{j} \tag{3}
\end{equation*}
where
\begin{equation*}
g=1+\frac{j(j+1)-l(l+1)+3 / 4}{2 j(j+1)} \quad \text { (Landè } g \text { factor) } \tag{4}
\end{equation*}
The average value of $\mu_{z}$ for $m_{j}=j$ (its maximum value) is usually taken as the definition of the magnetic moment observable, denoted as $\mu$. For an electron,
\begin{equation*}
\mu=\langle l j j| \mu_{z}|l j j\rangle=-g j \tag{5}
\end{equation*}
that is
\mu= \begin{cases}-(j+\frac{1}{2}), & j=l+\frac{1}{2} \tag{$\prime$}\\ -j(2 j+1) /(2 j+2), & j=l-\frac{1}{2}\end{cases}
|
[
"\\langle l j m_{j}| \\mu_{z}|l j m_{j}\\rangle=-(1+\\frac{j(j+1)-l(l+1)+3 / 4}{2 j(j+1)}) m_{j}"
] |
Expression
|
Theoretical Foundations
|
$l$: Orbital angular momentum quantum number
$j$: Total angular momentum quantum number
$m_{j}$: Z-component of total angular momentum quantum number
$\mu_{z}$: Z-component of the magnetic moment operator
|
30
|
A system composed of two spin-$1 / 2$ particles is placed in a uniform magnetic field, with the magnetic field direction as the $z$-axis. The Hamiltonian of the system related to spin is given by
\begin{equation*}
H=a \sigma_{1 z}+b \sigma_{2 z}+c_{0} \boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2} \tag{1}
\end{equation*}
where $a, ~ b$ terms arise from the interaction between the magnetic field and the particles' intrinsic magnetic moments, and the $c_{0}$ term arises from the interaction between the two particles. $a, ~ b, ~ c_{0}$ are real constants. If the system is in the common eigenstate $\chi_{11}=\alpha(1) \alpha(2)$ of the total spin operators $(\boldsymbol{S}^{2}, S_z)$, find the energy level of the system in this case.
|
We will solve using matrix methods in spin state vector space. The basis vectors can be chosen as the common eigenstates of $(\sigma_{1 z}, \sigma_{2 z})$
$$\alpha(1) \alpha(2), \quad \alpha(1) \beta(2), \quad \beta(1) \alpha(2), \quad \beta(1) \beta(2) $$
or as the common eigenstates $\chi_{S M_{s}}$ of the total spin operators $(\boldsymbol{S}^{2}, S_{z})$. For this problem, considering the diagonalization of $\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2}$, it is more convenient to use $\chi_{S M_{S}}$ as the basis vectors. For convenience, the basis vector order is as follows:
\begin{array}{l}
\chi_{1}=\chi_{11}=\alpha(1) \alpha(2) \tag{2}\\
\chi_{2}=\chi_{1-1}=\beta(1) \beta(2) \\
\chi_{3}=\chi_{10}=\frac{1}{\sqrt{2}}[\alpha(1) \beta(2)+\beta(1) \alpha(2)] \\
\chi_{4}=\chi_{00}=\frac{1}{\sqrt{2}}[\alpha(1) \beta(2)-\beta(1) \alpha(2)]
\end{array}}
The Hamiltonian operator can be rewritten as
\begin{gather*}
H=c_{0} \boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2}+c_{1}(\sigma_{1 z}+\sigma_{2 z})+c_{2}(\sigma_{1 z}-\sigma_{2 z}) \tag{1'}\\
c_{1}=\frac{1}{2}(a+b), \quad c_{2}=\frac{1}{2}(a-b) \tag{3}
\end{gather*}
All four basis vectors are common eigenstates of $(\boldsymbol{S}^{2}, S_{z})$, and also common eigenstates of $\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2}$ and $(\sigma_{1 z}+\sigma_{2 z})$. It is easy to see that $\chi_{1}$ and $\chi_{2}$ are also eigenstates of $(\sigma_{1 z}-\sigma_{2 z})$, thus $\chi_{1}$ and $\chi_{2}$ are already eigenstates of $H$.
\begin{align*}
& H \chi_{1}=(c_{0}+2 c_{1}) \chi_{1}=(c_{0}+a+b) \chi_{1} \tag{4}\\
& H \chi_{2}=(c_{0}-2 c_{1}) \chi_{2}=(c_{0}-a-b) \chi_{2} \tag{5}
\end{align*}
Thus, we obtain two energy levels of the system: $E=c_{0} \pm 2 c_{1}$.
It is easy to calculate the action of $(\sigma_{1 z}-\sigma_{2 z})$ on the basis vectors, which is
\begin{array}{cl}
(\sigma_{1 z}-\sigma_{2 z}) \chi_{1}=0, & (\sigma_{1 z}-\sigma_{2 z}) \chi_{2}=0 \\
(\sigma_{1 z}-\sigma_{2 z}) \chi_{3}=2 \chi_{4}, & (\sigma_{1 z}-\sigma_{2 z}) \chi_{4}=2 \chi_{3} \tag{7}
\end{array}
Therefore, in the subspace ${\chi_{3}, \chi_{4}}$, the matrix elements of $(\sigma_{1 z}-\sigma_{2 z})$ are
\begin{array}{l}
(\sigma_{1 z}-\sigma_{2 z})_{33}=(\sigma_{1 z}-\sigma_{2 z})_{44}=0 \tag{7'}\\
(\sigma_{1 z}-\sigma_{2 z})_{34}=(\sigma_{1 z}-\sigma_{2 z})_{43}=2
\end{array}
All matrix elements of $(\sigma_{1 z}+\sigma_{2 z})$ are zero. The matrix representation of $H$ is
\begin{equation}
H=\begin{pmatrix}
c_{0} & 2 c_{2} \tag{8}\\
2 c_{2} & -3 c_{0}
\end{pmatrix}
\end{equation}
Assume the energy eigenstate is
\begin{equation*}
\chi=f_{3} \chi_{3}+f_{4} \chi_{4} \tag{9}
\end{equation*}
Substitute into the energy eigenvalue equation
\begin{equation*}
H \chi=E \chi \tag{10}
\end{equation*}
resulting in
[\begin{array}{cc}
c_{0}-E & 2 c_{2} \tag{$\prime$}\\
2 c_{2} & -3 c_{0}-E
\end{array}][\begin{array}{l}
f_{1} \\
f_{2}
\end{array}]=0
The energy level $E$ is determined by:
\begin{equation*}
\operatorname{det}(H-E)=0 \tag{11}
\end{equation*}
namely
|\begin{array}{cc}
c_{0}-E & 2 c_{2} \tag{11'}\\
2 c_{2} & -3 c_{0}-E
\end{array}|=(E-c_{0})(E+3 c_{0})-4 c_{2}^{2}=0
Solving gives
E=-c_{0} \pm 2 \sqrt{c_{0}^{2}+c_{2}^{2}}
Conclusion: This problem has four energy levels (excluding accidental degeneracy), which are
\begin{equation*}
E=c_{0} \pm 2 c_{1}, \quad-c_{0} \pm 2 \sqrt{c_{0}^{2}+c_{2}^{2}} \tag{12}
\end{equation*}
The energy eigenstates of the first two energy levels are $\chi_{1}$ and $\chi_{2}$, respectively, and the energy eigenstates of the last two energy levels are linear combinations of $\chi_{3}$ and $\chi_{4}$.
|
[
"c_{0} + a + b"
] |
Expression
|
Theoretical Foundations
|
$c_{0}$: Real constant arising from the interaction between the two particles
$a$: Real constant arising from the interaction between the magnetic field and particle 1's intrinsic magnetic moment
$b$: Real constant arising from the interaction between the magnetic field and particle 2's intrinsic magnetic moment
|
31
|
Consider a system composed of three non-identical spin $1/2$ particles, with the Hamiltonian given by
\begin{equation*}
H=A \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}+B(\boldsymbol{s}_{1}+\boldsymbol{s}_{2}) \cdot \boldsymbol{s}_{3}, \tag{1}
\end{equation*}
where $A, ~ B$ are real constants. Let $\boldsymbol{S}_{12}=\boldsymbol{s}_{1}+\boldsymbol{s}_{2}$ and $\boldsymbol{S}_{123}=\boldsymbol{S}_{12}+\boldsymbol{s}_{3}$. Try to express the Hamiltonian $H$ as a function of $\boldsymbol{S}_{12}^{2}$ and $\boldsymbol{S}_{123}^{2}$. (Take $\hbar=1$ )
|
The sum of the spins of particles 1 and 2 is denoted as $\boldsymbol{S}_{12}$, and the total spin is denoted as $\boldsymbol{S}_{123}$, that is
\begin{equation*}
\boldsymbol{S}_{12}=\boldsymbol{s}_{1}+\boldsymbol{s}_{2}, \quad \boldsymbol{S}_{123}=\boldsymbol{s}_{1}+\boldsymbol{s}_{2}+\boldsymbol{s}_{3}=\boldsymbol{S}_{12}+\boldsymbol{s}_{3} \tag{2}
\end{equation*}
Evidently, $\boldsymbol{S}_{12}$ and $\boldsymbol{S}_{123}$ both possess the properties of angular momentum, satisfying the commutation relation
\begin{array}{ll}
\boldsymbol{S}_{12} \times \boldsymbol{S}_{12}=\mathrm{i} \boldsymbol{S}_{12}, & {[\boldsymbol{S}_{12}^{2}, \boldsymbol{S}_{12}]=0} \\
\boldsymbol{S}_{123} \times \boldsymbol{S}_{123}=\mathrm{i} \boldsymbol{S}_{123}, & {[\boldsymbol{S}_{123}^{2}, \boldsymbol{S}_{123}]=0} \tag{4}
\end{array}
$\boldsymbol{s}_{1}, ~ \boldsymbol{s}_{2}, ~ \boldsymbol{s}_{3}$ commute with each other, and
\begin{equation*}
s_{1}^{2}=s_{2}^{2}=s_{3}^{2}=\frac{3}{4} \tag{5}
\end{equation*}
Therefore
\begin{align*}
& \boldsymbol{S}_{12}^{2}=\boldsymbol{s}_{1}^{2}+\boldsymbol{s}_{2}^{2}+2 \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}=\frac{3}{2}+2 \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2} \tag{6}\\
& \boldsymbol{S}_{123}^{2}=\frac{9}{4}+2(\boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}+\boldsymbol{s}_{2} \cdot \boldsymbol{s}_{3}+\boldsymbol{s}_{3} \cdot \boldsymbol{s}_{1}) \tag{7}
\end{align*}
Based on this, $H$ can be written as
\begin{align*}
H & =(A-B) \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}+B(\boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}+\boldsymbol{s}_{2} \cdot \boldsymbol{s}_{3}+\boldsymbol{s}_{3} \cdot \boldsymbol{s}_{1}) \\
& =\frac{1}{2}(A-B) \boldsymbol{S}_{12}^{2}+\frac{B}{2} \boldsymbol{S}_{123}^{2}-\frac{3}{8}(2 A+B) \tag{$\prime$}
\end{align*}
|
[
"H = \\frac{1}{2}(A-B) \\boldsymbol{S}_{12}^{2}+\\frac{B}{2} \\boldsymbol{S}_{123}^{2}-\\frac{3}{8}(2 A+B)"
] |
Expression
|
Theoretical Foundations
|
$H$: Hamiltonian of the system.
$A$: Real constant in the Hamiltonian.
$B$: Real constant in the Hamiltonian.
$\boldsymbol{S}_{12}^{2}$: Square of the magnitude of the sum of spins of particles 1 and 2.
$\boldsymbol{S}_{123}^{2}$: Square of the magnitude of the total spin of particles 1, 2, and 3.
|
32
|
Same as the previous question, for any value, find
\begin{equation*}
d_{j m}^{j}(\lambda)=\langle j j| \mathrm{e}^{-\mathrm{i} \lambda J_{y}}|j m\rangle \tag{1}
\end{equation*}
|
According to the general theory of angular momentum,
\begin{array}{rl}
J_{+}|j m\rangle & =(J_{x}+i J_{y})|j m\rangle \tag{2}\\
J_{-}|j m\rangle & =a_{j m}|j m+1\rangle \\
J_{x}-i J_{y})|j m\rangle & =a_{j,-m}|j m-1\rangle
\end{array}}
where
\begin{equation*}
a_{j m}=\sqrt{(j-m)(j+m+1)} \tag{3}
\end{equation*}
When $m=j$
\begin{equation*}
J_{+}|j j\rangle=0, \quad\langle j j| J_{-}=0 \tag{4}
\end{equation*}
(The second equation is the conjugate of the first equation.) Therefore
\begin{equation*}
\langle j j| J_{-} \mathrm{e}^{-\mathrm{i} \lambda J_{y}}|j m\rangle=0. \tag{5}
\end{equation*}
We also have:
\begin{align*}
J_{-} \mathrm{e}^{-\mathrm{i} \mathrm{i} J_{y}} & =\mathrm{e}^{-\mathrm{i} \lambda J_{y}}(J_{z} \sin \lambda+J_{x} \cos \lambda-\mathrm{i} J_{y}) \\
& =\mathrm{e}^{-\mathrm{i} \lambda J_{y}}[J_{z} \sin \lambda+\frac{1}{2}(\cos \lambda-1) J_{+}+\frac{1}{2}(\cos \lambda+1) J_{-}] \tag{6}
\end{align*}
Substitute into equation (5), and use equation (2) to obtain
\begin{gather*}
m \sin \lambda d_{j m}^{j}(\lambda)+\frac{1}{2}(\cos \lambda-1) a_{j m} d_{j m+1}^{j}(\lambda) \\
\quad+\frac{1}{2}(\cos \lambda+1) a_{j,-m} d_{j m-1}^{j}(\lambda)=0 \tag{7}
\end{gather*}
Additionally, we can also obtain
\begin{equation*}
J_{z} \mathrm{e}^{-\mathrm{i} \lambda J_{y}}=\mathrm{e}^{-\mathrm{i} \lambda J_{y}}(J_{z} \cos \lambda-J_{x} \sin \lambda)=\mathrm{e}^{-\mathrm{i} \lambda J_{y}}[J_{z} \cos \lambda-\frac{1}{2} \sin \lambda(J_{+}+J_{-})] \tag{8}
\end{equation*}
Multiply the above equation from the left with $\langle j|$ and from the right with $|j m\rangle$, yielding
\begin{equation*}
(j-m \cos \lambda) d_{j m}^{j}(\lambda)+\frac{1}{2} \sin \lambda[a_{j m} d_{j m+1}^{j}(\lambda)+a_{j,-m} d_{j m-1}^{j}(\lambda)]=0 \tag{9}
\end{equation*}
Combine equations (7) and (9), eliminate $d_{j m-1}^{j}(\lambda)$, and obtain a simpler recursive relation:
\begin{equation*}
\sin \lambda \cdot a_{j m} d_{j m+1}^{j}(\lambda)=-(j-m)(1+\cos \lambda) d_{j m}^{j}(\lambda) \tag{10}
\end{equation*}
That is
\begin{equation*}
d_{j m}^{j}(\lambda)=-\frac{\sin \frac{\lambda}{2}}{\cos \frac{\lambda}{2}}(\frac{j+m+1}{j-m})^{\frac{1}{2}} d_{j m+1}^{j}(\lambda) \tag{$10^\prime$}
\end{equation*}
Thus, once $d_{j j}^{j}$ is found, all $d_{j m}^{j}$ can be recursively derived. Let's first find $d_{j j}^{j}$.
\begin{equation*}
d_{j j}^{j}(\lambda)=\langle j j| \mathrm{e}^{-\mathrm{i} \lambda J_{y}}|j j\rangle \tag{11}
\end{equation*}
Note that $d_{j j}^{j}(0)=1$, differentiating the above equation with respect to $\lambda$, we get
\begin{align*}
\frac{\mathrm{d}}{\mathrm{~d} \lambda d_{j j}^{j}(\lambda)} & =\langle j j| \mathrm{e}^{-\mathrm{i} \lambda J_{y}}(-\mathrm{i} J_{y})|j j\rangle \\
& =\frac{1}{2}\langle j j| \mathrm{e}^{-\mathrm{i} \lambda J_{y}}(J_{-}-J_{+})|j j\rangle \\
& =\frac{1}{2} a_{j-j} d_{j j-1}^{j}(\lambda) \tag{12}
\end{align*}
While equation (7) takes $m=j$, noticing $a_{j j}=0$, we get
\begin{equation*}
\frac{1}{2}(\cos \lambda+1) a_{j,-j} d_{j j-1}^{j}(\lambda)=-j \sin \lambda \cdot d_{j j}^{j}(\lambda) \tag{13}
\end{equation*}
Substituting into equation (12), we get
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} \lambda} d_{j j}^{j}(\lambda)=-j \frac{\sin \lambda}{1+\cos \lambda} d_{j j}^{j}(\lambda) \tag{14}
\end{equation*}
Integrate, and using $d_{j j}^{j}(0)=1$, we obtain
\begin{equation*}
d_{j j}^{j}(\lambda)=(\cos \frac{\lambda}{2})^{2 j} \tag{15}
\end{equation*}
This formula can also be derived using the model in another format. Using equation (15) and the recursive relation ( $10^{\prime}$), we can sequentially obtain
\begin{aligned}
& d_{j j-1}^{j}(\lambda)=-(2 j)^{1 / 2}(\cos \frac{\lambda}{2})^{2 j-1} \sin \frac{\lambda}{2} \\
& d_{j j-2}^{j}(\lambda)=[\frac{2 j(2 j-1)}{2!}]^{\frac{1}{2}}(\cos \frac{\lambda}{2})^{2 j-2}(\sin \frac{\lambda}{2})^{2} \\
& \vdots
\end{aligned}
\begin{align*}
& d_{j m}^{j}(\lambda)=(-1)^{j-m}[\frac{(2 j)!}{(j+m)!(j-m)!}]^{\frac{1}{2}}(\cos \frac{\lambda}{2})^{j+m}(\sin \frac{\lambda}{2})^{j-m} \tag{16}\\
& \vdots \\
& d_{j-j}^{j}(\lambda)=(-1)^{2 j}(\sin \frac{\lambda}{2})^{2 j}
\end{align*}
Discussion $1^{\circ}$ When $\lambda \rightarrow-\lambda$, clearly
d_{j m}^{j_{m}}(-\lambda)=(-1)^{\jmath^{-m}} d_{j m}^{\prime}(\lambda)
This result matches the general property of $d_{m^{\prime} m}^{\prime}$.
$\mathbf{2}^{\circ}$ When $\lambda=\pi$, the only non-zero matrix element is clearly
\begin{equation*}
d_{j-\jmath}^{j}(\pi)=(-1)^{2 j} \tag{18}
\end{equation*}
This is because
\begin{align*}
& \mathrm{e}^{-i \pi J_{y}}|j m\rangle=(-1)^{j-m}|j-m\rangle \tag{19}\\
& \mathrm{e}^{-i \pi J_{y}}|j-j\rangle=(-1)^{2 j}|j j\rangle
\end{align*}
[The meaning of operator $\mathrm{e}^{-\mathrm{i} \pi_{y}}$ is to rotate the system by $180^{\circ}$ about the y-axis, so the state $|j m\rangle$ becomes the state $|j-m\rangle$, and equation (19) specifies the relative phase factor for the relationship between $|j m\rangle$ and $|j-m\rangle$.]
|
[
"d_{j m}^{j}(\\lambda)=(-1)^{j-m}[\\frac{(2 j)!}{(j+m)!(j-m)!}]^{\\frac{1}{2}}(\\cos \\frac{\\lambda}{2})^{j+m}(\\sin \\frac{\\lambda}{2})^{j-m}"
] |
Expression
|
Theoretical Foundations
|
$d_{j m}^{j}(\lambda)$: Wigner d-matrix element, defined as $\langle j j| \mathrm{e}^{-\mathrm{i} \lambda J_{y}}|j m\rangle$
$j$: Total angular momentum quantum number
$m$: Magnetic quantum number
$\lambda$: Angle of rotation about the y-axis
|
33
|
Let $\boldsymbol{J}_{1}$ and $\boldsymbol{J}_{2}$ be angular momenta corresponding to different degrees of freedom, then their sum $\boldsymbol{J}=\boldsymbol{J}_{1}+\boldsymbol{J}_{2}$ is also an angular momentum. Try to compute the expectation values of $J_{1 z}$ for the common eigenstate $|j_{1} j_{2} j m\rangle$ of $(\boldsymbol{J}_{1}^{2}, \boldsymbol{J}_{2}^{2}, \boldsymbol{J}^{2}, J_{z})$. (Take $\hbar=1$)
|
$J_{1}, ~ J_{2}$ satisfy the fundamental commutation relations of angular momentum operators
\begin{equation*}
\boldsymbol{J}_{1} \times \boldsymbol{J}_{1}=\mathrm{i} \boldsymbol{J}_{1}, \quad \boldsymbol{J}_{2} \times \boldsymbol{J}_{2}=\mathrm{i} \boldsymbol{J}_{2} \tag{1}
\end{equation*}
$\boldsymbol{J}_{1}, ~ J_{2}$ belong to different degrees of freedom and commute with each other, so
\begin{align*}
& {[J_{x}, J_{1 x}]=[J_{1 x}+J_{2 x}, J_{1 x}]=0} \tag{2}\\
& {[J_{x}, J_{1 y}]=[J_{1 x}+J_{2 x}, J_{1 y}]=\mathrm{i} J_{1 z}}
\end{align*}
$\boldsymbol{J}$ and $\boldsymbol{J}_{2}$ have similar relationships. In summary, $\boldsymbol{J}_{1}$ or $\boldsymbol{J}_{2}$ and $\boldsymbol{J}$ satisfy all the relations between the vector operator $\boldsymbol{A}$ and $\boldsymbol{J}$ from the previous problem. Moreover,
\begin{align*}
& \boldsymbol{J} \cdot \boldsymbol{J}_{1}=\boldsymbol{J}_{1}^{2}+\boldsymbol{J}_{2} \cdot \boldsymbol{J}_{1}=\frac{1}{2}(\boldsymbol{J}^{2}+\boldsymbol{J}_{1}^{2}-\boldsymbol{J}_{2}^{2}) \tag{3}\\
& \boldsymbol{J} \cdot \boldsymbol{J}_{2}=\boldsymbol{J}_{2}^{2}+\boldsymbol{J}_{1} \cdot \boldsymbol{J}_{2}=\frac{1}{2}(\boldsymbol{J}^{2}+\boldsymbol{J}_{2}^{2}-\boldsymbol{J}_{1}^{2}) \tag{4}
\end{align*}
Using equation (5) from the previous problem, we have
\begin{align*}
& j(j+1)\langle\boldsymbol{J}_{1}\rangle_{j_{1} j_{2} j m}=\frac{1}{2}[j(j+1)+j_{1}(j_{1}+1)-j_{2}(j_{2}+1)]\langle\boldsymbol{J}\rangle_{j_{1} j_{2} j m} \tag{5}\\
& j(j+1)\langle\boldsymbol{J}_{2}\rangle_{j_{1} j_{2} j m}=\frac{1}{2}[j(j+1)+j_{2}(j_{2}+1)-j_{1}(j_{1}+1)]\langle\boldsymbol{J}\rangle_{j_{1} j_{2} j m} \tag{6}
\end{align*}
Since in the state $|J_{z}=m\rangle$
$$ \langle J_{x}\rangle=\langle J_{y}\rangle=0, \quad\langle J_{z}\rangle=m $$
Therefore, equations (5) and (6) yield
$$ \langle J_{1 x}\rangle_{j_{1} j_{2} j m}=\langle J_{1 y}\rangle_{j_{1} j_{2} j m}=\langle J_{2 x}\rangle_{j_{1} j_{2} j m}=\langle J_{2 y}\rangle_{j_{1} j_{2} j m}=0 $$
\begin{align*}
\langle J_{1 z}\rangle_{j_{1} j_{2} j m} & =m \frac{j(j+1)+j_{1}(j_{1}+1)-j_{2}(j_{2}+1)}{2 j(j+1)} \tag{7}\\
\langle J_{2 z}\rangle_{j_{1} j_{2} j m} & =m \frac{j(j+1)+j_{2}(j_{2}+1)-j_{1}(j_{1}+1)}{2 j(j+1)} \\
& =m-\langle J_{1 z}\rangle_{j_{1} j_{2}>m}
\end{align*}
|
[
"m \\frac{j(j+1)+j_{1}(j_{1}+1)-j_{2}(j_{2}+1)}{2 j(j+1)}"
] |
Expression
|
Theoretical Foundations
|
$m$: Magnetic quantum number, representing the eigenvalue of $J_z$.
$j$: Quantum number associated with the magnitude of the total angular momentum $\boldsymbol{J}$, specifically for the eigenvalue of $\boldsymbol{J}^{2}$.
$j_{1}$: Quantum number associated with the magnitude of $\boldsymbol{J}_{1}$, specifically for the eigenvalue of $\boldsymbol{J}_{1}^{2}$.
$j_{2}$: Quantum number associated with the magnitude of $\boldsymbol{J}_{2}$, specifically for the eigenvalue of $\boldsymbol{J}_{2}^{2}$.
|
34
|
Two angular momenta $\boldsymbol{J}_{1}$ and $\boldsymbol{J}_{2}$, of equal magnitude but belonging to different degrees of freedom, couple to form a total angular momentum $\boldsymbol{J}=\boldsymbol{J}_{1}+\boldsymbol{J}_{2}$, with $\hbar=1$, and assume $\boldsymbol{J}_{1}^{2}=\boldsymbol{J}_{2}^{2}=j(j+1)$. In the state where the total angular momentum quantum number $J=0$, what is the probability when $J_{1z}$ takes the value $m$ (with $J_{2z}$ simultaneously taking the value $-m$)?
|
The eigenstate of $(\boldsymbol{J}_{1}^{2}, J_{1 z})$ is denoted by $|j m_{1}\rangle_{1}$, and the eigenstate of $(\boldsymbol{J}_{2}^{2}, J_{2 z})$ is denoted by $|j m_{2}\rangle_{2}$. The common eigenstate of $(\boldsymbol{J}_{1}^{2}, \boldsymbol{J}_{2}^{2}$, $\mathbf{J}^{2}, J_{z})$ is denoted by $|j j J M\rangle$, where $M$ is the eigenvalue of $J_{z}$.
$|j j J M\rangle$ can also be abbreviated as $|J M\rangle$.
When $J=0$, $M=0, m_{1}=-m_{2}=m$, so the state under discussion can be expressed as
\begin{equation*}
|j j 00\rangle=\sum_{m} C_{m}|j m\rangle_{1}|j-m\rangle_{2} \tag{3}
\end{equation*}
$C_{m}$ is the C.G. coefficient $\langle j_{1} m_{1} j_{2} m_{2} \mid J M\rangle$ when $j_{1}=j_{2}=j, J=M=0, m_{1}=-m_{2}=m$. $|C_{m}|^{2}$ is the probability that $J_{1}$ takes the value $m$ (with $J_{2 z}$ taking the value $-m$ at the same time). Below, we solve for $C_{m}$.
Since $\boldsymbol{J}^{2}|j j 00\rangle=0$, and $\boldsymbol{J}^{2}$ is positive definite, it must be
\begin{equation*}
\boldsymbol{J}|j j 00\rangle=0 \tag{4}
\end{equation*}
Thus,
\begin{equation*}
(J_{x}+\mathrm{i} J_{y})|j j 00\rangle=0 \tag{5}
\end{equation*}
Where
\begin{equation*}
J_{x}+\mathrm{i} J_{y}=(J_{1 x}+\mathrm{i} J_{1 y})+(J_{2 x}+\mathrm{i} J_{2 y})=J_{1+}+J_{2+} \tag{6}
\end{equation*}
According to the basic formula for angular momentum ladder operators,
\begin{align*}
J_{1+}|j m\rangle_{1} & =(J_{1 x}+\mathrm{i} J_{1 y})|j m\rangle_{1}=a_{j m}|j m+1\rangle_{1} \\
a_{j m} & =\sqrt{(j-m)(j+m+1)} \tag{7a}\\
J_{2+}|j,-m\rangle_{2} & =(J_{2 x}+\mathrm{i} J_{2 y})|j,-m\rangle_{2}=a_{j,-m}|j, 1-m\rangle_{2} \\
a_{j,-m} & =\sqrt{(j+m)(j-m+1)}=a_{,, m-1} \tag{7b}
\end{align*}
Substituting expressions (5) to (7) into (3), we obtain
\begin{equation*}
\sum_{m} C_{m}[a_{j m}|j, m+1\rangle_{1}|j,-m\rangle_{2}+a_{J, m-1}|j m\rangle_{1}|j, 1-m\rangle_{2}]=0 \tag{8}
\end{equation*}
Since
\sum_{m} C_{m} a_{j, m-1}|j m\rangle_{1}|j, 1-m\rangle_{2} \xrightarrow{m \rightarrow m+1} \sum_{m} C_{m+1} a_{j m}|j m+1\rangle_{1}|j,-m\rangle_{2}
Therefore, equation (8) becomes
\begin{equation*}
\sum_{m}(C_{m}+C_{m+1}) a_{j m}|j m+1\rangle_{1}|j,-m\rangle_{2}=0 \tag{$\prime$}
\end{equation*}
Since each basis vector is linearly independent, it must be
\begin{equation*}
(C_{m}+C_{m+1}) a_{j m}=0 \tag{9}
\end{equation*}
Which implies
\begin{equation*}
C_{m}=-C_{m+1}, \quad m=j-1, j-2, \cdots,(-j) \tag{\prime}
\end{equation*}
In equation (3), $|j j 00\rangle, ~|j m\rangle_{1}, ~|j,-m\rangle_{2}$ etc. are all orthonormalized, and $m$ has a total of $(2 j+1)$ possible values. According to the normalization condition
\begin{equation*}
\sum_{m}|C_{m}|^{2}=1 \tag{10}
\end{equation*}
And equation ($9^{\prime}$), we immediately have $|C_{m}|^{2}=1 /(2 j+1)$. If we take
C_{j}=1 / \sqrt{2 j+1}
We get
\begin{equation*}
C_{m}=(-1)^{j-m} C_{j}=(-1)^{j-m} \frac{1}{\sqrt{2 j+1}} \tag{11}
\end{equation*}
Substituting into equation (3), we obtain
\begin{equation*}
|j j 00\rangle=\frac{1}{\sqrt{2 j+1}} \sum_{m}(-1)^{1^{-m}}|j m\rangle_{1}|j,-m\rangle_{2} \tag{12}
\end{equation*}
Clearly, under the premise $J_{1 z}=-J_{2 z}$, the probabilities that $J_{1 z}$ and $J_{2 z}$ take each eigenvalue $(j, j-1, \cdots,-j)$ are equal, both being $1 /(2 j+1)$.
|
[
"1/(2j+1)"
] |
Expression
|
Theoretical Foundations
|
$j$: Angular momentum quantum number for $\boldsymbol{J}_{1}$ and $\boldsymbol{J}_{2}$.
|
35
|
A particle with mass $\mu$ and charge $q$ moves in a magnetic field $\boldsymbol{B}=\nabla \times \boldsymbol{A}$, where the Hamiltonian is $H = \frac{1}{2} \mu \boldsymbol{v}^{2}$, with $\boldsymbol{v}$ as the velocity operator. Calculate $\mathrm{d} \boldsymbol{v} / \mathrm{d} t$.
|
The Hamiltonian operator can be expressed as
\begin{equation*}
H=\frac{1}{2 \mu}(\boldsymbol{p}-\frac{q}{c} \boldsymbol{A})^{2}=\frac{1}{2} \mu \boldsymbol{v}^{2} \tag{2}
\end{equation*}
Using the commutation relations of $\boldsymbol{v}$ and $v^{2}$, it can be easily demonstrated that
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} t} \boldsymbol{v}=\frac{1}{\mathrm{i} \hbar}[\boldsymbol{v}, H]=\frac{q}{2 \mu c}(\boldsymbol{v} \times \boldsymbol{B}-\boldsymbol{B} \times \boldsymbol{v}) \tag{3}
\end{equation*}
|
[
"\\frac{q}{2 \\mu c}(\\boldsymbol{v} \\times \\boldsymbol{B}-\\boldsymbol{B} \\times \\boldsymbol{v})"
] |
Expression
|
Theoretical Foundations
|
$q$: Charge of the particle
$\mu$: Mass of the particle
$c$: Speed of light
$\boldsymbol{v}$: Velocity operator
$\boldsymbol{B}$: Magnetic field, defined as $\boldsymbol{B}=\nabla \times \boldsymbol{A}$
|
36
|
A particle with mass $\mu$ and charge $q$ moves in a magnetic field $\boldsymbol{B}=\nabla \times \boldsymbol{A}$, where the Hamiltonian is $H = \frac{1}{2} \mu \boldsymbol{v}^{2}$, with $\boldsymbol{v}$ as the velocity operator. Let $\boldsymbol{L}$ be the the angular momentum operator. Calculate $\mathrm{d} \boldsymbol{L} / \mathrm{d} t$.
|
The Hamiltonian operator can be expressed as
\begin{equation*}
H=\frac{1}{2 \mu}(\boldsymbol{p}-\frac{q}{c} \boldsymbol{A})^{2}=\frac{1}{2} \mu \boldsymbol{v}^{2} \tag{2}
\end{equation*}
Using the commutation relations of $\boldsymbol{v}$ and $v^{2}$, it can be easily demonstrated that
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} t} \boldsymbol{v}=\frac{1}{\mathrm{i} \hbar}[\boldsymbol{v}, H]=\frac{q}{2 \mu c}(\boldsymbol{v} \times \boldsymbol{B}-\boldsymbol{B} \times \boldsymbol{v}) \tag{3}
\end{equation*}
In classical electrodynamics, the Lorentz force is
$$ \boldsymbol{f}=\frac{q}{c} \boldsymbol{v} \times \boldsymbol{B}. $$
The equation of motion is
\begin{equation*}
\mu \frac{\mathrm{d}}{\mathrm{~d} t} \boldsymbol{v}=\boldsymbol{f}=\frac{q}{c} \boldsymbol{v} \times \boldsymbol{B} \tag{$\prime$}
\end{equation*}
Equation (3) is the quantum mechanical extension of the classical equation of motion ($3^{\prime}$). In equation (3),
\begin{equation*}
\boldsymbol{v} \times \boldsymbol{B}=\frac{1}{\mu}(\boldsymbol{p}-\frac{q}{c} \boldsymbol{A}) \times \boldsymbol{B}=-\boldsymbol{B} \times \boldsymbol{v}-\frac{\mathrm{i} \hbar}{\mu} \nabla \times \boldsymbol{B} \tag{4}
\end{equation*}
From electrodynamics,
$$\nabla \times \boldsymbol{B}=\frac{4 \pi}{c} \boldsymbol{j}$$
$\boldsymbol{j}$ is the source current that generates the magnetic field. Thus, if in the space where the particle moves, the source current $\boldsymbol{j}=0$, then
\begin{equation*}
\frac{q^{v}}{c} \boldsymbol{v} \times \boldsymbol{B}=-\frac{q}{c} \boldsymbol{B} \times \boldsymbol{v}=\boldsymbol{f} \tag{5}
\end{equation*}
Equation (3) can still be rewritten in the form of equation ($3^{\prime}$). For uniform fields, it is obvious that equations (3) and ($3^{\prime}$) are equivalent.
The mechanical angular momentum can also be expressed as
\begin{equation*}
\boldsymbol{L}=\mu \boldsymbol{r} \times \boldsymbol{v}=\frac{1}{2} \mu(\boldsymbol{r} \times \boldsymbol{v}-\boldsymbol{v} \times \boldsymbol{r}) \tag{$\prime$}
\end{equation*}
(Note that in $\boldsymbol{r} \times \boldsymbol{v}$, the relevant components of $\boldsymbol{r}$ and $\boldsymbol{v}$ are commutative.) The time derivative of $\boldsymbol{L}$ is
$$\frac{\mathrm{d} \boldsymbol{L}}{\mathrm{~d} t}=\frac{\mu}{2}(\frac{\mathrm{~d} \boldsymbol{r}}{\mathrm{~d} t} \times \boldsymbol{v}+\boldsymbol{r} \times \frac{\mathrm{d} \boldsymbol{v}}{\mathrm{~d} t}-\frac{\mathrm{d} \boldsymbol{v}}{\mathrm{~d} t} \times \boldsymbol{r}-\boldsymbol{v} \times \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{~d} t}).$$
Using equation (3) and $\mathrm{d} \boldsymbol{r} / \mathrm{d} t=\boldsymbol{v}$, we obtain
\begin{equation*}
\frac{\mathrm{d} \boldsymbol{L}}{\mathrm{~d} t}=\frac{q}{4 c}[\boldsymbol{r} \times(\boldsymbol{v} \times \boldsymbol{B})+(\boldsymbol{B} \times \boldsymbol{v}) \times \boldsymbol{r}-\boldsymbol{r} \times(\boldsymbol{B} \times \boldsymbol{v})-(\boldsymbol{v} \times \boldsymbol{B}) \times \boldsymbol{r}] \tag{6}
\end{equation*}
Under the conditions where equation (5) holds, equation (6) can be simplified to
\begin{equation*}
\frac{\mathrm{d} \boldsymbol{L}}{\mathrm{~d} t}=\frac{q}{2 c}[\boldsymbol{r} \times(\boldsymbol{v} \times \boldsymbol{B})+(\boldsymbol{B} \times \boldsymbol{v}) \times \boldsymbol{r}]=\frac{1}{2}(\boldsymbol{r} \times \boldsymbol{f}-\boldsymbol{f} \times \boldsymbol{r}) \tag{7}
\end{equation*}
Note that in $\boldsymbol{r} \times \boldsymbol{f}$, the relevant components of $\boldsymbol{r}$ and $\boldsymbol{f}$ are non-commutative, and $\boldsymbol{r} \times \boldsymbol{f} \neq-\boldsymbol{f} \times \boldsymbol{r}$, hence the right side of equation (7) is not equivalent to $\boldsymbol{r} \times \boldsymbol{f}$.
|
[
"\\frac{q}{2 c}[\\boldsymbol{r} \\times(\\boldsymbol{v} \\times \\boldsymbol{B})+(\\boldsymbol{B} \\times \\boldsymbol{v}) \\times \\boldsymbol{r}]"
] |
Expression
|
Theoretical Foundations
|
$q$: Charge of the particle
$c$: Speed of light
$\boldsymbol{r}$: Position vector
$\boldsymbol{v}$: Velocity operator
$\boldsymbol{B}$: Magnetic field
|
37
|
A three-dimensional isotropic oscillator with mass $\mu$, charge $q$, and natural frequency $\omega_{0}$ is placed in a uniform external magnetic field $\boldsymbol{B}$. Find the formula for energy levels.
|
Compared to the previous two questions, an additional harmonic oscillator potential $\frac{1}{2} \mu \omega_{0}^{2}(x^{2}+y^{2}+z^{2})$ should be added to the Hamiltonian in this question, thus
\begin{align*}
H & =\frac{1}{2 \mu} \boldsymbol{p}^{2}+\frac{1}{2} \mu \omega_{0}^{2}(x^{2}+y^{2}+z^{2})+\frac{q^{2} B^{2}}{8 \mu c^{2}}(x^{2}+y^{2})-\frac{q B}{2 \mu c} l_{z} \\
& =H_{1}+H_{2}-\omega_{\mathrm{L}} l_{z} \tag{1}
\end{align*}
where
\begin{align*}
& H_{1}=\frac{1}{2 \mu} p_{z}^{2}+\frac{1}{2} \mu \omega_{0}^{2} z^{2} \\
& H_{2}=\frac{1}{2 \mu}(p_{x}^{2}+p_{y}^{2})+\frac{1}{2} \mu(\omega_{0}^{2}+\omega_{\mathrm{L}}^{2})(x^{2}+y^{2}) \tag{2}\\
& \omega_{\mathrm{L}}=\frac{q B}{2 \mu c}
\end{align*}
$H_{1}, ~ H_{2}, ~ l_{z}$ are mutually commuting conserved quantities and can be chosen as a complete set of commuting observables. $H_{1}$ corresponds to a one-dimensional harmonic oscillator energy operator, with eigenvalues given by
\begin{equation*}
E_{n_{1}}=(n_{1}+\frac{1}{2}) \hbar \omega_{0}, \quad n_{1}=0,1,2, \cdots \tag{3}
\end{equation*}
$\mathrm{H}_{2}$ corresponds to a two-dimensional isotropic oscillator total energy operator, with eigenvalues given by
\begin{align*}
& E_{n_{2} m}=(2 n_{2}+1+|m|) \hbar \omega \tag{4}\\
& n_{2}=0,1,2, \cdots, \quad m=0, \pm 1, \pm 2, \cdots
\end{align*}
where
\begin{equation*}
\omega=\sqrt{\omega_{0}^{2}+\omega_{L}^{2}} \tag{5}
\end{equation*}
The eigenvalue of $l_{z}$ is $m \hbar$. Therefore, the energy levels for this problem are
\begin{equation*}
E_{n_{1} n_{2} m}=(n_{1}+\frac{1}{2}) \hbar \omega_{0}+(2 n_{2}+1+|m|) \hbar \omega-m \hbar \omega_{\mathrm{L}} \tag{6}
\end{equation*}
Note: $\omega_{\mathrm{L}}$ can be positive or negative depending on the sign of charge $q$; the magnetic quantum number $m$ can also be positive or negative. Therefore, the sign of $q$ does not affect the overall energy spectrum. For precision, only the case of $q>0(\omega_{\mathrm{L}}>0)$ is discussed below.
The relationship among the three frequencies $\omega_{0}, ~ \omega_{L}, ~ \omega$ is given by:
\begin{equation*}
0<(\omega-\omega_{\mathrm{L}})<\omega_{0} \tag{7}
\end{equation*}
The dependence of the energy levels on the quantum numbers (listed in the order of their magnitude of change) is
\begin{array}{lll}
m \geqslant 0, & m \text { increases by } 1, & E \text { increases by } \hbar(\omega-\omega_{\mathrm{L}}) \\
& n_{1} \text { increases by } 1, & E \text { increases by } \hbar \omega_{0} \\
m \leqslant 0, & |m| \text { increases by } 1, & E \text { increases by } \hbar(\omega+\omega_{\mathrm{L}}) \\
& n_{2} \text { increases by } 1, & E \text { increases by } 2 \hbar \omega
\end{array}
For the ground state $E_{0}$, obviously, $n_{1}, ~ n_{2}, ~|m|$ all take their minimum values,
\begin{equation*}
E_{0}=E_{000}=\frac{1}{2} \hbar \omega_{0}+\hbar \omega \tag{8}
\end{equation*}
For the first excited state $E_{1}, n_{1}=n_{2}=0, m=1$,
\begin{equation*}
E_{1}=E_{001}=\frac{1}{2} \hbar \omega_{0}+2 \hbar \omega-\hbar \omega_{\mathrm{L}}=E_{0}+\hbar(\omega-\omega_{\mathrm{L}}) \tag{9}
\end{equation*}
|
[
"E_{n_{1} n_{2} m}=(n_{1}+\\frac{1}{2}) \\hbar \\omega_{0}+(2 n_{2}+1+|m|) \\hbar \\omega-m \\hbar \\omega_{\\mathrm{L}}"
] |
Expression
|
Theoretical Foundations
|
$E_{n_{1} n_{2} m}$: Total energy level for the three-dimensional isotropic oscillator in a magnetic field.
$n_{1}$: Quantum number for the one-dimensional harmonic oscillator along the z-axis.
$\hbar$: Reduced Planck's constant.
$\omega_{0}$: Natural frequency of the three-dimensional isotropic oscillator.
$n_{2}$: Radial quantum number for the two-dimensional isotropic oscillator in the xy-plane.
$m$: Magnetic quantum number.
$\omega$: Effective frequency for the two-dimensional isotropic oscillator, defined as $\omega=\sqrt{\omega_{0}^{2}+\omega_{L}^{2}}$.
$\omega_{\mathrm{L}}$: Larmor frequency, defined as $\omega_{\mathrm{L}}=\frac{q B}{2 \mu c}$.
|
38
|
A particle with mass $\mu$ and charge $q$ moves in a uniform electric field $\mathscr{E}$ (along the x-axis) and a uniform magnetic field $\boldsymbol{B}$ (along the z-axis) that are perpendicular to each other. If the momentum of the particle in the y-direction is $p_y$ and in the z-direction is $p_z$, find the energy level expression of the system.
|
With the electric field direction as the $x$ axis and the magnetic field direction as the $z$ axis, then
\begin{equation*}
\mathscr{L}=(\mathscr{E}, 0,0), \quad \boldsymbol{B}=(0,0, B) \tag{1}
\end{equation*}
Taking the scalar and vector potentials of the electromagnetic field as
\begin{equation*}
\phi=-\mathscr{E} x, \quad \boldsymbol{A}=(0, B x, 0) \tag{2}
\end{equation*}
Satisfying the relation
$$\mathscr{E}=-\nabla \phi, \quad \boldsymbol{B}=\nabla \times \boldsymbol{A}.$$
The Hamiltonian of the particle is
\begin{equation*}
H=\frac{1}{2 \mu}[p_{x}^{2}+(p_{y}-\frac{q B}{c} x)^{2}+p_{z}^{2}]-q \mathscr{E} x \tag{3}
\end{equation*}
With the constants of motion set as $(H, p_{y}, p_{z})$, their common eigenfunction can be written as
\begin{equation*}
\psi(x, y, z)=\psi(x) \mathrm{e}^{\mathrm{i}(p_{y} y+p_{z} z) / \hbar} \tag{4}
\end{equation*}
where $p_{y}$ and $p_{z}$ are eigenvalues and can be any real numbers.
$\psi(x, y, z)$ satisfies the energy eigen-equation
$$ H \psi(x, y, z)=E \psi(x, y, z) $$
Thus, $\psi(x)$ satisfies the equation
\begin{equation*}
\frac{1}{2 \mu}[p_{x}^{2}+(p_{y}-\frac{q B}{c} x)^{2}+p_{z}^{2}] \psi-q \mathscr{E} x \psi=E \psi \tag{5}
\end{equation*}
That is, for $\psi(x)$, $H$ is equivalent to the following:
\begin{align*}
H & \Rightarrow-\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2}}{\partial x^{2}}+\frac{q^{2} B^{2}}{2 \mu c^{2}} x^{2}-(q \mathscr{E}+\frac{q B}{\mu c} p_{y}) x+\frac{1}{2 \mu}(p_{y}^{2}+p_{x}^{2}) \\
& =-\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2}}{\partial x^{2}}+\frac{q^{2} B^{2}}{2 \mu c^{2}}(x-x_{0})^{2}-\frac{q^{2} B^{2}}{2 \mu c^{2}} x_{0}^{2}+\frac{1}{2 \mu}(p_{y}^{2}+p_{z}^{2}) \tag{6}
\end{align*}
where
\begin{equation*}
x_{0}=\frac{\mu c^{2}}{q^{2} B^{2}}(q \mathscr{E}+\frac{q B}{\mu c} p_{y})=\frac{\mu c}{q B}(\frac{c \mathscr{E}}{B}+\frac{p_{y}}{\mu}) \tag{7}
\end{equation*}
Equation (6) corresponds to a one-dimensional harmonic oscillator energy operator
$$-\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2}}{\partial x^{2}}+\frac{1}{2} \mu \omega^{2}(x-x_{0})^{2}, \quad \omega=\frac{|q| B}{\mu c} $$
Plus two constant terms. Therefore, the energy level of this problem is
\begin{align*}
E & =(n+\frac{1}{2}) \hbar \omega-\frac{q^{2} B^{2}}{2 \mu c^{2}} x_{0}^{2}+\frac{1}{2 \mu}(p_{y}^{2}+p_{z}^{2}) \\
& =(n+\frac{1}{2}) \frac{\hbar B|q|}{\mu c}-\frac{c^{2} \mathscr{E}^{2} \mu}{2 B^{2}}-\frac{c \mathscr{E}}{B} p_{y}+\frac{1}{2 \mu} p_{z}^{2} \tag{8}
\end{align*}
where $p_{y}, ~ p_{z}$ are any real numbers, $n=0,1,2, \cdots$.
In equation (4), $\psi(x)$ is the one-dimensional harmonic oscillator energy eigenfunction with $(x-x_{0})$ as the variable, i.e.,
\begin{equation*}
\psi(x)=\psi_{n}(x-x_{0})=H_{n}(\xi) \mathrm{e}^{-\frac{1}{2} \xi^{2}} \tag{9}
\end{equation*}
$H_{n}(\xi)$ is the Hermite polynomial, $\xi=(\frac{|q| B}{\hbar c})^{\frac{1}{2}}(x-x_{0})$.
|
[
"E = (n+\\frac{1}{2}) \\frac{\\hbar B|q|}{\\mu c}-\\frac{c^{2} \\mathscr{E}^{2} \\mu}{2 B^{2}}-\\frac{c \\mathscr{E}}{B} p_{y}+\\frac{1}{2 \\mu} p_{z}^{2}"
] |
Expression
|
Theoretical Foundations
|
$E$: Energy eigenvalue of the system.
$n$: Quantum number for the energy levels, $n=0,1,2, \cdots$.
$\hbar$: Reduced Planck's constant.
$B$: Magnitude of the uniform magnetic field.
$|q|$: Absolute value of the charge of the particle.
$\mu$: Mass of the particle.
$c$: Speed of light.
$\mathscr{E}$: Magnitude of the uniform electric field, directed along the x-axis.
$p_y$: Momentum of the particle in the y-direction.
$p_z$: Momentum of the particle in the z-direction.
|
39
|
A particle moves in one dimension. When the total energy operator is
\begin{equation*}
H_{0}=\frac{p^{2}}{2 m}+V(x) \tag{1}
\end{equation*}
the energy level is $E_{n}^{(0)}$. If the total energy operator becomes
\begin{equation*}
H=H_{0}+\frac{\lambda p}{m} \tag{2}
\end{equation*}
find the energy level $E_{n}$.
|
First, treat $\lambda$ as a parameter, then
\begin{equation*}
\frac{\partial H}{\partial \lambda}=\frac{p}{m} \tag{3}
\end{equation*}
According to the Hellmann theorem, we have
\begin{equation*}
\frac{\partial E_{n}}{\partial \lambda}=\langle\frac{\partial H}{\partial \lambda}\rangle_{n}=\frac{1}{m}\langle p\rangle_{n} \tag{4}
\end{equation*}
However, since
\begin{equation*}
\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{1}{\mathrm{i} \hbar}[x, H]=\frac{1}{\mathrm{i} \hbar m}[x, \frac{p^{2}}{2}+\lambda p]=\frac{1}{m}(p+\lambda) \tag{5}
\end{equation*}
For any bound state,
$$ \langle\frac{\mathrm{d} x}{\mathrm{~d} t}\rangle_{n}=\frac{1}{\mathrm{i} \hbar}\langle x H-H x\rangle_{n}=0, $$
therefore
\begin{equation*}
\langle p\rangle_{n}=-\lambda \tag{6}
\end{equation*}
Substitute into equation (4), obtaining
\begin{equation*}
\frac{\partial E_{n}}{\partial \lambda}=-\frac{\lambda}{m} \tag{7}
\end{equation*}
Integrate to get
\begin{equation*}
E_{n}=-\frac{\lambda^{2}}{2 m}+C \tag{8}
\end{equation*}
$C$ is the integration constant. Since when $\lambda=0$, $H=H_{0}, E_{n}=E_{n}^{(0)}$, so $C=E_{n}^{(0)}$. Substitute into equation (8), obtaining
\begin{equation*}
E_{n}=E_{n}^{(0)}-\frac{\lambda^{2}}{2 m} \tag{9}
\end{equation*}
Solution Two: Write $H$ as
\begin{equation*}
H=\frac{p^{2}}{2 m}+\frac{\lambda p}{m}+V(x)=\frac{P^{2}}{2 m}+V(x)-\frac{\lambda^{2}}{2 m} \tag{10}
\end{equation*}
where
\begin{equation*}
P=p+\lambda \tag{11}
\end{equation*}
In the momentum representation,
\begin{equation*}
x=\mathrm{i} \hbar \frac{\partial}{\partial p}=\mathrm{i} \hbar \frac{\partial}{\partial P} \tag{12}
\end{equation*}
Therefore,
\begin{align*}
H_{0} & =\frac{p^{2}}{2 m}+V(i \hbar \frac{\partial}{\partial p}) \tag{13}\\
H & =\frac{P^{2}}{2 m}+V(i \hbar \frac{\partial}{\partial P})-\frac{\lambda^{2}}{2 m} \tag{14}
\end{align*}
The difference between $H$ and $H_{0}$, aside from the constant term, is just replacing $p$ with $P$, which does not affect the energy levels. Therefore,
\begin{equation*}
E_{n}=E_{n}^{(0)}-\frac{\lambda^{2}}{2 m} \tag{15}
\end{equation*}
|
[
"E_{n}=E_{n}^{(0)}-\\frac{\\lambda^{2}}{2 m}"
] |
Expression
|
Theoretical Foundations
|
$E_n$: Energy level corresponding to the modified total energy operator $H$.
$E_n^{(0)}$: Initial energy level when the total energy operator is $H_0$.
$\lambda$: Parameter representing the strength of the perturbation.
$m$: Mass of the particle.
|
40
|
A particle with mass $\mu$ moves in a central force field,
\begin{equation*}
V(r)=\lambda r^{\nu}, \quad-2<\nu, \quad \nu / \lambda>0 \tag{1}
\end{equation*}
Use the Hellmann theorem and the virial theorem to analyze the dependence of the energy level structure on $\hbar, ~ \lambda, ~ \mu$.
|
The energy operator is
\begin{equation*}
H=T+V=-\frac{\hbar^{2}}{2 \mu} \nabla^{2}+\lambda r^{\nu} \tag{2}
\end{equation*}
Let $\beta=\hbar^{2} / 2 \mu, \lambda$ and $\beta$ be independent parameters. It is evident that
\begin{equation*}
\beta \frac{\partial H}{\partial \beta}=T, \quad \lambda \frac{\partial H}{\partial \lambda}=V \tag{3}
\end{equation*}
According to the Hellmann theorem, for any bound state,
\begin{align*}
& \langle T\rangle=\frac{\beta \partial E}{\partial \beta} \tag{4}\\
& \langle V\rangle=\frac{\lambda \partial E}{\partial \lambda} \tag{5}
\end{align*}
Adding the two equations yields
\begin{equation*}
\beta \frac{\partial E}{\partial \beta}+\lambda \frac{\partial E}{\partial \lambda}=\langle T+V\rangle=E \tag{6}
\end{equation*}
And from the virial theorem, we have
$$ \langle T\rangle=\frac{\nu}{2}\langle V\rangle$$
That is
\begin{equation*}
\beta \frac{\partial E}{\partial \beta}=\frac{\nu}{2} \lambda \frac{\partial E}{\partial \lambda} \tag{7}
\end{equation*}
Substituting equation (7) into equation (6), we get
\begin{align*}
& (1+\frac{\nu}{2}) \lambda \frac{\partial E}{\partial \lambda}=E \tag{8}\\
& (1+\frac{2}{\nu}) \beta \frac{\partial E}{\partial \beta}=E \tag{9}
\end{align*}
Integrating equation (8), we obtain the construction relationship between $E$ and $\lambda$
\begin{equation*}
E=C_{1} \lambda^{2 /(2+\iota)} \tag{10}
\end{equation*}
$C_{1}$ is the "integration constant" and is independent of $\lambda$. Integrating equation (9), we obtain the construction relationship between $E$ and $\beta$
\begin{equation*}
E=C_{2} \beta^{\nu /(2 \downarrow \imath)} \tag{11}
\end{equation*}
$C_{2}$ is independent of $\nu$. Comparing equations (10) and (11), it follows that
\begin{equation*}
E=C \lambda^{2 /(2+\tau)} \beta^{\nu /(2+\nu)}=C \lambda^{2 /(2+\tau)}(\frac{\hbar^{2}}{2 \mu})^{\nu /(2+\nu)} \tag{12}
\end{equation*}
$C$ is independent of $\lambda, ~ \beta$, and is a dimensionless pure number (related to $\nu$ and quantum numbers). It is easy to verify that the above expression is the only possible energy construction that is dimensionally correct.
From equation (12), it is evident (note $\nu>-2$) that as the interaction strength $|\lambda|$ increases, $|E|$ increases, and the energy level spacing increases. If $\lambda$ is independent of the particle's mass, then when $\nu>0, ~ \mu$ increases, $|E|$ decreases; when $\nu<0, ~ \mu$ increases, $|E|$ increases.
If $\lambda$ is independent of $\mu$, from equations (2) and (4) it can also be seen that
\begin{equation*}
\mu \frac{\partial E}{\partial \mu}=-\langle T\rangle<0 \tag{13}
\end{equation*}
That is, an increase in the particle's mass always leads to a decrease in the algebraic value of the energy levels.
From equation (12) it can also be seen that if $\lambda \propto \mu^{\nu / 2}$, then $E$ is independent of $\mu$. A famous example of this situation is the harmonic oscillator, i.e.,
\begin{align*}
& V(\boldsymbol{r})=\lambda r^{2}=\frac{1}{2} \mu \omega^{2} r^{2} \\
& (\nu=2, \quad \lambda=\frac{1}{2} \mu \omega^{2})
\end{align*}
The energy levels are
$$ E_{N}=(N+\frac{3}{2}) \hbar \omega, \quad N=0,1,2, \cdots $$
If $\omega$ remains constant, $E_{N}$ is independent of $\mu$.
|
[
"E=C \\lambda^{2 /(2+\\nu)}(\\frac{\\hbar^{2}}{2 \\mu})^{\\nu /(2+\\nu)}"
] |
Expression
|
Theoretical Foundations
|
$E$: Energy of a bound state, an eigenvalue of the Hamiltonian.
$C$: Dimensionless pure number constant, independent of $\lambda$ and $\beta$, related to $\nu$ and quantum numbers.
$\lambda$: Constant parameter defining the strength and sign of the central force potential.
$\nu$: Exponent parameter defining the radial dependence of the central force potential.
$\hbar$: Reduced Planck's constant.
$\mu$: Mass of the particle.
$E_n$: Energy level associated with quantum number $n$.
|
41
|
A particle moves in a potential field
\begin{equation*}
V(x)=V_{0}|x / a|^{\nu}, \quad V_{0}, a>0 \tag{1}.
\end{equation*}
Find the dependence of energy levels on parameters as $\nu \rightarrow \infty$.
|
The total energy operator is
\begin{equation*}
H=T+V=-\frac{\hbar^{2}}{2 \mu} \frac{\mathrm{~d}^{2}}{\mathrm{~d} x^{2}}+V_{0}|x / a|^{\nu} \tag{2}
\end{equation*}
From dimensional analysis, if $x_{0}$ represents the characteristic length, we have
\begin{equation*}
E \sim \frac{\hbar^{2}}{\mu x_{0}^{2}} \sim \frac{V_{0} x_{0}^{\nu}}{a^{\nu}} \tag{3}
\end{equation*}
It is not difficult to solve for
\begin{equation*}
x_{0} \sim(\frac{\hbar^{2} a^{\nu}}{\mu V_{0}})^{\frac{1}{\nu+2}} \xrightarrow{\nu \rightarrow \infty} a \tag{4}
\end{equation*}
\begin{equation*}
E \sim(\frac{\hbar^{2}}{\mu})^{\frac{1}{\nu+2}} V_{0}^{2 /(\nu+2)} a^{-2 \nu /(\nu+2)} \xrightarrow{\nu \rightarrow \infty} \frac{\hbar^{2}}{\mu a^{2}} \tag{5}
\end{equation*}
The construction of the characteristic length and energy levels is independent of $V_{0}$,
In fact, it is easy to see from Equation (1)
$$ \nu \rightarrow \infty, \quad V(x) \rightarrow \begin{cases}0, & |x|<a \tag{6}\\ \infty, & |x|>a\end{cases} $$
This is precisely an infinitely deep potential well of width $2 a$, with energy levels
\begin{equation*}
E_{n}=\frac{n^{2} \pi^{2} \hbar^{2}}{8 \mu a^{2}} . \quad n=1,2,3, \cdots \tag{7}
\end{equation*}
When $\nu$ is finite, according to the virial theorem, for any bound state, there is a relationship between the average kinetic energy and the average potential energy
\begin{equation*}
\frac{2}{\nu}\langle T\rangle=\langle V\rangle \tag{8}
\end{equation*}
Thus, as $u \rightarrow \infty$, it follows
\begin{equation*}
\langle V\rangle \rightarrow 0, \quad\langle T\rangle \rightarrow E \tag{9}
\end{equation*}
This conclusion is also consistent with the infinite potential well problem.
|
[
"E_{n}=\\frac{n^{2} \\pi^{2} \\hbar^{2}}{8 \\mu a^{2}}"
] |
Expression
|
Theoretical Foundations
|
$E_n$: Energy levels for an infinite potential well.
$n$: Principal quantum number, an integer ($n=1,2,3,...$).
$\pi$: Mathematical constant pi.
$\hbar$: Reduced Planck's constant.
$\mu$: Mass of the particle.
$a$: A positive constant parameter defining the characteristic length scale of the potential field.
|
42
|
A particle with mass $m$ moves in a uniform force field $V(x)=F x(F>0)$, with the motion constrained to the range $x \geqslant 0$. Find its ground state energy level.
|
The total energy operator is
\begin{equation*}
H=T+V=\frac{p^{2}}{2 m}+F x \tag{1}
\end{equation*}
In momentum representation, the operator for $x$ is given by
\begin{equation*}
\hat{x}=\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} p} \tag{2}
\end{equation*}
The operator for $H$ is given by
\begin{equation*}
\hat{H}=\frac{p^{2}}{2 m}+\mathrm{i} \hbar F \frac{\mathrm{~d}}{\mathrm{~d} p} \tag{3}
\end{equation*}
The time-independent Schrödinger equation is
\begin{equation*}
\frac{p^{2}}{2 m} \varphi(p)+\mathrm{i} \hbar F \frac{\mathrm{~d}}{\mathrm{~d} p} \varphi(p)=E \varphi(p) \tag{4}
\end{equation*}
Here $\varphi(p)$ is the wave function in the momentum representation. Equation (4) is a very simple first-order differential equation, whose solution is
\begin{equation*}
\varphi(p)=A \exp [\frac{\mathrm{i}}{\hbar F}(\frac{p^{3}}{6 m}-E p)] \tag{5}
\end{equation*}
$A$ is the normalization constant.
Transforming to the $x$ representation, the wave function is
\begin{align*}
\psi(x) & =(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \varphi(p) \mathrm{e}^{\mathrm{i} p x / \hbar} \mathrm{d} p \\
& =A(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \exp \frac{\mathrm{i}}{\hbar}[\frac{p^{3}}{6 m F}+(x-\frac{E}{F}) p] \mathrm{d} p \\
& =2 A(2 \pi \hbar)^{-1 / 2} \int_{0}^{\infty} \cos [\frac{p^{3}}{6 \hbar m F}+\frac{p}{\hbar}(x-\frac{E}{F})] \mathrm{d} p \\
& =\frac{C}{\sqrt{\pi}} \int_{0}^{\infty} \cos (\frac{u^{3}}{3}+u \xi) \mathrm{d} u \tag{6}
\end{align*}
Where
\begin{align*}
& u=p(2 \hbar m F)^{-1 / 3} \tag{7}\\
& \xi=(x-\frac{E}{F})(\frac{2 m F}{\hbar^{2}})^{1 / 3}=\frac{x}{l}-\lambda \tag{8}\\
& l=(\frac{\hbar^{2}}{2 m F})^{1 / 3} \tag{9}\\
& \lambda=(\frac{2 m}{\hbar^{2} F^{2}})^{1 / 3} E=\frac{2 m E}{\hbar^{2}} l^{2} \tag{10}
\end{align*}
$l$ is the characteristic length of this problem.
Except for the normalization constant $C$, the right-hand side of equation (6) resembles the Airy function with $\xi$ as the variable.
When $\xi>0(x>E / F$ , i.e., the classically forbidden region), the Airy function behaves like the modified Bessel function of the second kind, i.e.,
\begin{equation*}
\psi(x)=\sqrt{\xi} K \frac{1}{3}(\frac{2}{3} \xi^{3 / 2}) \xrightarrow{\xi \rightarrow \infty} \sqrt{\xi}(\frac{3 \pi}{4 \xi^{3 / 2}})^{1 / 2} \mathrm{e}^{-\frac{2}{3} \xi^{3 / 2}} \tag{11}
\end{equation*}
When $\xi<0$ ($x<E / F$, i.e., the classically allowed region), the Airy function behaves like the Bessel function of the first kind, i.e.,
\begin{equation*}
\psi(x)=\sqrt{|\xi|}[J_{\frac{1}{3}}(\frac{2}{3}|\xi|^{3 / 2})+J_{-\frac{1}{3}}(\frac{2}{3}|\xi|^{3 / 2})] \tag{12}
\end{equation*}
The energy level $E$ is determined by the boundary condition $\psi(0)=0$, which implies
\begin{equation*}
J_{\frac{1}{3}}(\frac{2}{3} \lambda^{3 / 2})+J_{-\frac{1}{3}}(\frac{2}{3} \lambda^{3 / 2})=0 \tag{13}
\end{equation*}
Solutions can be found in Bessel function tables, yielding
\lambda=2.3381,4.0880,5.5206,6.7867,7.9441, \cdots
The relationship between the energy eigenvalue and $\lambda$ is given by equation (10). The ground state energy level is
\begin{equation*}
E_{1}=1.8558(\frac{\hbar^{2} F^{2}}{m})^{1 / 3} \tag{14}
\end{equation*}
|
[
"E_{1}=1.8558(\\frac{\\hbar^{2} F^{2}}{m})^{1 / 3}"
] |
Expression
|
Theoretical Foundations
|
$E_1$: ground state energy level
$\hbar$: reduced Planck's constant
$F$: magnitude of the uniform force field
$m$: mass of the particle
|
43
|
Calculate the transmission probability of a particle (energy $E>0$) through a $\delta$ potential barrier $V(x)=V_{0} \delta(x)$ in the momentum representation.
|
The stationary Schrödinger equation in the $x$ representation is
\begin{equation*}
\psi^{\prime \prime}+k^{2} \psi-\frac{2 m V_{0}}{\hbar^{2}} \delta(x) \psi=0, \quad k=\sqrt{2 m E} / \hbar \tag{1}
\end{equation*}
Let
\begin{equation*}
\psi(x)=(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \mathrm{d} p \varphi(p) \mathrm{e}^{\mathrm{i} p x / \hbar} \tag{2}
\end{equation*}
where $\varphi(p)$ is the wave function in the momentum representation, which should satisfy the equation
\begin{equation*}
\frac{p^{2}}{2 m} \varphi(p)+\int_{-\infty}^{+\infty} V_{p p^{\prime}} \varphi(p^{\prime}) \mathrm{d} p^{\prime}=E \varphi(p) \tag{3}
\end{equation*}
where
\begin{equation*}
V_{p p^{\prime}}=\frac{1}{2 \pi \hbar} \int_{-\infty}^{+\infty} \mathrm{d} x V_{0} \delta(x) \mathrm{e}^{\mathrm{i}(p^{\prime}-p) x / \hbar}=\frac{V_{0}}{2 \pi \hbar} \tag{4}
\end{equation*}
Thus, using equations (4) and (2), we can solve
\begin{equation*}
\int_{-\infty}^{+\infty} V_{p p^{\prime}} \varphi(p^{\prime}) \mathrm{d} p^{\prime}=\frac{V_{0}}{2 \pi \hbar} \int_{-\infty}^{+\infty} \varphi(p^{\prime}) \mathrm{d} p^{\prime}=\frac{V_{0}}{\sqrt{2 \pi \hbar}} \psi(0) \tag{5}
\end{equation*}
Substituting equations (4) and (5) into equation (3), we obtain
\begin{equation*}
(p^{2}-\hbar^{2} k^{2}) \varphi(p)+\frac{2 m V_{0}}{\sqrt{2 \pi \hbar}} \psi(0)=0 \tag{6}
\end{equation*}
According to the fundamental formula of the $\delta$ function
\begin{equation*}
(\xi-\xi_{0}) \delta(\xi-\xi_{0})=0 \tag{7}
\end{equation*}
The general solution of equation (6) is
\begin{equation*}
\varphi(p)=C_{1} \delta(p-\hbar k)+C_{2} \delta(p+\hbar k)-\frac{2 m V_{0}}{\sqrt{2 \pi \hbar}} \frac{\psi(0)}{(p^{2}-\hbar^{2} k^{2})} \tag{8}
\end{equation*}
where $C_{1}, ~ C_{2}, ~ \psi(0)$ are to be determined. Substituting equation (8) into equation (2), we get the wave function in the $x$ representation
\begin{equation*}
\psi(x)=\frac{C_{1}}{\sqrt{2 \pi \hbar}} \mathrm{e}^{\mathrm{i} k x}+\frac{C_{2}}{\sqrt{2 \pi \hbar}} \mathrm{e}^{-\mathrm{i} k x}-\frac{2 m V_{0}}{2 \pi \hbar} \psi(0) \int_{-\infty}^{+\infty} \frac{\mathrm{d} p}{p^{2}-\hbar^{2} k^{2}} \mathrm{e}^{\mathrm{i} p x / \hbar} \tag{9}
\end{equation*}
The last integral in equation (9) should be taken as the principal value, which can be calculated using the contour integral method in the complex $p$ plane, resulting in
\int_{-\infty}^{+\infty} \frac{\mathrm{d} p}{p^{2}-\hbar^{2} k^{2}} \mathrm{e}^{\mathrm{i} k x / \hbar}= \begin{cases}\frac{\mathrm{i} \pi}{2 \hbar k}(\mathrm{e}^{\mathrm{i} k x}-\mathrm{e}^{-\mathrm{i} k x}), & x>0 \tag{10}\\ \frac{i \pi}{2 \hbar k}(\mathrm{e}^{-\mathrm{i} k x}-\mathrm{e}^{\mathrm{i} k x}), & x<0\end{cases}
When $x \rightarrow 0$, the right side of equation (10) becomes 0, and from equation (9) we get
\begin{equation*}
\psi(0)=(C_{1}+C_{2}) / \sqrt{2 \pi \hbar} \tag{11}
\end{equation*}
Substituting equations (10) and (11) into equation (9), we get
\sqrt{2 \pi \hbar} \psi(x)=C_{1} \mathrm{e}^{\mathrm{i} k x}+C_{2} \mathrm{e}^{-\mathrm{i} k x}-\frac{\mathrm{i} m V_{0}}{2 \hbar^{2} k}(C_{1}+C_{2})(\mathrm{e}^{\mathrm{i} k x}-\mathrm{e}^{-\mathrm{i} k x}), \quad x>0
Given that the incident wave is $\mathrm{e}^{\mathrm{i} k x}$ (i.e., the incident momentum $p=\hbar k$), in the region $x>0$ there should only be the transmitted wave, i.e., the $\mathrm{e}^{\mathrm{i} k x}$ term, and no $\mathrm{e}^{-\mathrm{i} k}$ term. Therefore, $C_{1}, ~ C_{2}$ must satisfy the following relation:
\begin{equation*}
C_{2}=-\mathrm{i} \frac{m V_{0}}{2 \hbar^{2} k}(C_{1}+C_{2}) \tag{12}
\end{equation*}
Thus,
\begin{equation*}
\psi(x)=\frac{C_{1}+C_{2}}{\sqrt{2 \pi \hbar}} \mathrm{e}^{\mathrm{i} k x}, \quad x>0 \tag{13}
\end{equation*}
Similarly, we can get
\begin{equation*}
\psi(x)=\frac{1}{\sqrt{2 \pi \hbar}}[(C_{1}-C_{2}) \mathrm{e}^{i k x}+2 C_{2} \mathrm{e}^{-\mathrm{i} k x}], \quad x<0 \tag{14}
\end{equation*}
where the $\mathrm{e}^{\mathrm{ik} x}$ term is the incident wave, and the $\mathrm{e}^{-\mathrm{i} k x}$ term is the reflected wave. If the incident wave amplitude is set to 1, then equation (14) can be written as
\begin{equation*}
\psi(x)=\mathrm{e}^{\mathrm{i} k \tau}+R \mathrm{e}^{-\mathrm{i} k x} \tag{14'}
\end{equation*}
Then we should take
\begin{equation*}
C_{1}-C_{2}=\sqrt{2 \pi \hbar} \tag{15}
\end{equation*}
From equations (13) and (14), it is easy to see that
$$\text { Transmission coefficient }=|\frac{C_{1}+C_{2}}{C_{1}-C_{2}}|^{2}=\frac{1}{|1-\frac{2 C_{2}}{C_{1}+C_{2}}|^{2}}$$
Using equation (12), we obtain
\begin{equation*}
\text { Transmission coefficient }=\frac{1}{|1+\mathrm{i} m V_{0} / \hbar^{2} k|^{2}}=\frac{1}{1+(m V_{0} / \hbar^{2} k)^{2}} \tag{16}
\end{equation*}
|
[
"\\frac{1}{1+(m V_{0} / \\hbar^{2} k)^{2}}"
] |
Expression
|
Theoretical Foundations
|
$m$: Mass of the particle.
$V_0$: Strength of the delta potential barrier.
$\hbar$: Reduced Planck's constant.
$k$: Wave number, defined as $k=\sqrt{2 m E} / \hbar$.
|
44
|
The energy operator of a one-dimensional harmonic oscillator is
\begin{equation*}
H=\frac{p_{x}^{2}}{2 \mu}+\frac{1}{2} \mu \omega^{2} x^{2} \tag{1}
\end{equation*}
Try to derive its energy level expression using the Heisenberg equation of motion for operators and the fundamental commutation relation.
|
Using the Heisenberg equation of motion, we obtain
\begin{equation*}
\frac{\mathrm{d} p}{\mathrm{~d} t}=\frac{1}{\mathrm{i} \hbar}[p, H]=\frac{\mu \omega^{2}}{2 \mathrm{i} \hbar}[p, x^{2}]=-\mu \omega^{2} x \tag{2}
\end{equation*}
Take the matrix element in the energy representation, yielding
\begin{equation*}
\mathrm{i} \omega_{k n} p_{k n}=-\mu \omega^{2} x_{k n} \tag{3}
\end{equation*}
From the previous result,
\begin{equation*}
p_{k n}=\mathrm{i} \omega_{k n} \mu x_{k n} \tag{4}
\end{equation*}
Combining equations (3) and (4), we obtain
\begin{equation*}
(\omega^{2}-\omega_{k n}^{2}) x_{k n}=0 \tag{5}
\end{equation*}
where $k, ~ n$ can be understood as quantum state indices. From equation (5) it is evident
\begin{array}{ll}
\text { If } \omega_{k n} \neq \pm \omega, & \text { then } x_{k n}=0 \tag{6}\\
\text { If } x_{k n} \neq 0, \quad \text { then } \omega_{k n}= \pm \omega
\end{array}
Since
\begin{equation*}
(x^{2})_{k k}=\sum_{n} x_{k n} x_{n k}=\sum_{n}|x_{n k}|^{2}>0 \tag{7}
\end{equation*}
For any chosen energy level $E_{k}$, there must exist some $n$ such that $x_{n k} \neq 0$, then from equation (6), the energy level difference $E_{n}-$ $E_{k}= \pm \hbar \omega$, that is, given any energy level, there must exist another energy level differing by $\hbar \omega$. Therefore, all energy levels are
$$ E=E_{0}, \quad E_{0}+\hbar \omega, \quad E_{0}+2 \hbar \omega, \cdots $$
That is
\begin{equation*}
E_{n}=E_{0}+n \hbar \omega, \quad n=0,1,2, \cdots \tag{8}
\end{equation*}
To find the ground state energy $E_{0}$, the equation proven in the previous problem can be used
\begin{equation*}
\sum_{n}(E_{n}-E_{k})|x_{n k}|^{2}=\hbar^{2} / 2 \mu \tag{9}
\end{equation*}
Taking $k$ as the ground state, from equations (6), (7), and (9), we get
$$\frac{\hbar^{2}}{2 \mu}=\hbar \omega \sum_{n}|x_{n 0}|^{2}=\hbar \omega(x^{2})_{00}$$
Thus, the average potential energy of the ground state is
\begin{equation*}
\frac{1}{2} \mu \omega^{2}(x^{2})_{00}=\frac{1}{4} \hbar \omega \tag{10}
\end{equation*}
According to the virial theorem, we have
\langle T\rangle_{0}=\langle V\rangle_{0}=\frac{1}{2} E_{0}
Comparing with equation (10), we obtain
\begin{equation*}
E_{0}=\frac{1}{2} \hbar \omega \tag{11}
\end{equation*}
Substituting into equation (8), we get the energy level formula
\begin{equation*}
E_{n}=(n+\frac{1}{2}) \hbar \omega, \quad n=0,1,2, \cdots \tag{$\prime$}
\end{equation*}
Considering equation (6), equation (9) gives
\begin{equation*}
|x_{k+1, k}|^{2}-|x_{k-1, k}|^{2}=\frac{\hbar}{2 \mu \omega} \tag{12}
\end{equation*}
By appropriately choosing the phase factor $(\mathrm{e}^{\mathrm{i}})$ of each energy eigenfunction, all $x_{n k}$ can be made non-negative real numbers, and equation (12) can be rewritten as (changing $k$ to $n$)
\begin{equation*}
(x_{n+1, n})^{2}-(x_{n, n-1})^{2}=\frac{\hbar}{2 \mu \omega} \tag{$\prime$}
\end{equation*}
When $n=0$, the equation yields
\begin{equation*}
(x_{10})^{2}=\frac{\hbar}{2 \mu \omega}, \quad x_{10}=\sqrt{\frac{\hbar}{2 \mu \omega}} \tag{13}
\end{equation*}
By repeatedly using equation (12＇), we get
\begin{equation*}
x_{n+1, n}=\sqrt{\frac{n+1}{2} \frac{\hbar}{\mu \omega}}, \quad n=0,1,2, \cdots \tag{14}
\end{equation*}
By also using equation (4), we get
\begin{equation*}
p_{n+1, n}=\mathrm{i} \omega \mu x_{n+1, n} \tag{15}
\end{equation*}
Note: The matrix elements of $x$ are real numbers, and the matrix elements of $p$ are purely imaginary numbers. Therefore
\begin{align*}
& x_{n, n+1}=(x_{n+1, n})^{*}=x_{n+1, n} \tag{16}\\
& p_{n, n+1}=(p_{n+1, n})^{*}=-p_{n+1, n} \tag{17}
\end{align*}
|
[
"E_{n}=(n+\\frac{1}{2}) \\hbar \\omega"
] |
Expression
|
Theoretical Foundations
|
$E_n$: Energy of quantum state $n$
$n$: Quantum number for energy levels ($n=0,1,2,...$)
$\hbar$: Reduced Planck's constant
$\omega$: Angular frequency of the harmonic oscillator
|
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in Data Studio
CMPhysBench: A Benchmark for Evaluating Large Language Models in Condensed Matter Physics
🎉🎉🎉 This paper is accpeted by ICLR 2026.
We introduce CMPhysBench, designed to assess the proficiency of Large Language Models (LLMs) in Condensed Matter Physics, as a novel Benchmark. CMPhysBench is composed of more than 520 graduate-level meticulously curated questions covering both representative subfields and foundational theoretical frameworks of condensed matter physics, such as magnetism, superconductivity, strongly correlated systems, etc. To ensure a deep understanding of the problem-solving process,we focus exclusively on calculation problems, requiring LLMs to independently generate comprehensive solutions. Meanwhile, leveraging tree-based representations of expressions, we introduce the Scalable Expression Edit Distance (SEED) score, which provides fine-grained (non-binary) partial credit and yields a more accurate assessment of similarity between prediction and ground-truth. Our results show that even the best models, Grok-4, reach only 36 average SEED score and 28% accuracy on CMPhysBench, underscoring a significant capability gap, especially for this practical and frontier domain relative to traditional physics.
Citations
@article{wang2025cmphysbench,
title={CMPhysBench: A Benchmark for Evaluating Large Language Models in Condensed Matter Physics},
author={Wang, Weida and Huang, Dongchen and Li, Jiatong and Yang, Tengchao and Zheng, Ziyang and Zhang, Di and Han, Dong and Chen, Benteng and Luo, Binzhao and Liu, Zhiyu and others},
journal={arXiv preprint arXiv:2508.18124},
year={2025}
}
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