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If the two particles are distinguishable, and number 1 is the one in state $\psi_a$ , then the combined wave function is
$$\psi(x_1, x_2) = \psi_a(x_1)\psi_b(x_2); \qquad [5.15]$$
if they are identical bosons, the composite wave function is (see Problem 5.3 for the normalization)
$$\psi_{+}(x_{1}, x_{2}) = \frac... | {
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between the two protons (Figure 5.1a), and the resulting accumulation of negative charge would attract the protons inward, accounting for the **covalent bond** that holds
**Figure 5.1:** Schematic picture of the covalent bond: (a) Symmetric configuration produces attractive force; (b) ... | {
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The Hamiltonian,
$$H = \left\{ -\frac{\hbar^2}{2m} \nabla_1^2 - \frac{1}{4\pi\epsilon_0} \frac{2e^2}{r_1} \right\} + \left\{ -\frac{\hbar^2}{2m} \nabla_2^2 - \frac{1}{4\pi\epsilon_0} \frac{2e^2}{r_2} \right\} + \frac{1}{4\pi\epsilon_0} \frac{e^2}{|\mathbf{r}_1 - \mathbf{r}_2|}, [5.27]$$
consists of two *hydrogenic*... | {
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But the effect of electron repulsion is to favor the lowest value of l, for the following reason: Angular momentum tends to throw the electron outward (more formally, the expectation value of r increases with increasing l, for a given n), and the farther out an electron gets, the more effectively the inner electronscre... | {
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| $\overline{z}$ | Element | Configuration | |
|----------------|---------|-----------------------------|---------------|
| 1 | Н | (1s) | $^{2}S_{1/2}$ |
| 2 | He | $(1s)^2$ | $^{1}S_{0}$ |
| 3 |... | {
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Letting
$$k_x \equiv \frac{\sqrt{2mE_x}}{\hbar}, \ k_y \equiv \frac{\sqrt{2mE_y}}{\hbar}, \ k_z \equiv \frac{\sqrt{2mE_z}}{\hbar},$$
we obtain the general solutions
$$X(x) = A_x \sin(k_x x) + B_x \cos(k_x x), \ Y(y) = A_y \sin(k_y y) + B_y \cos(k_y y),$$
$$Z(z) = A_z \sin(k_z z) + B_z \cos(k_z z).$$
The bound... | {
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It is sometimes called **degeneracy pressure**, although "exclusion pressure" might be a better term.<sup>12</sup>
**Problem 5.13** The density of copper is 8.96 gm/cm<sup>3</sup>, and its atomic weight is 63.5 gm/mole.
- (a) Calculate the Fermi energy for copper (Equation 5.43). Assume $q_1 = 1$ , and give your a... | {
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<sup>&</sup>lt;sup>15</sup>See, for instance, D. Park, *Introduction to the Quantum Theory*, 3rd ed., (New York: McGraw-Hill, 1992).
According to Bloch's theorem, the wave function in the cell immediately to the *left* of the origin is
$$\psi(x) = e^{-iKa} [A\sin k(x+a) + B\cos k(x+a)], \quad (-a < x < 0). \quad ... | {
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The arguments are relatively straightforward, but the arithmetic gets pretty dense, so I'm going to begin with an absurdly simple example, so you'll have a clear sense of what is at issue when we come to the general case.
#### 5.4.1 Example
Suppose we have just *three* noninteracting particles (all of mass m) in th... | {
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#### \*Problem 5.19
- (a) Construct the completely antisymmetric wave function $\psi(x_A, x_B, x_C)$ for three identical fermions, one in the state $\psi_5$ , one in the state $\psi_7$ , and one in the state $\psi_{13}$ .
- (b) Construct the completely symmetric wave function $\psi(x_A, x_B, x_C)$ for three ... | {
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So there are in fact
$$\frac{(N_n + d_n - 1)!}{N_n!(d_n - 1)!} = \binom{N_n + d_n - 1}{N_n}$$
[5.75]
distinct ways of assigning the $N_n$ particles to the $d_n$ one-particle states in the nth bin, and we conclude that
$$Q(N_1, N_2, N_3, \ldots) = \prod_{n=1}^{\infty} \frac{(N_n + d_n - 1)!}{N_n!(d_n - 1)!}.$$... | {
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In that case
$$G \approx \sum_{n=1}^{\infty} \left[ \ln(d_n!) - N_n \ln(N_n) + N_n - (d_n - N_n) \ln(d_n - N_n) + (d_n - N_n) - \alpha N_n - \beta E_n N_n \right] + \alpha N + \beta E,$$
[5.88]
so
$$\frac{\partial G}{\partial N_n} = -\ln(N_n) + \ln(d_n - N_n) - \alpha - \beta E_n.$$
[5.89]
Setting this equal to... | {
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It is customary to replace $\alpha$ (which, as is clear from the special case of Equation 5.97, is a function of T) by the so-called **chemical potential**,
$$\mu(T) \equiv -\alpha k_B T, \tag{5.10}$$
and rewrite Equations 5.86, 5.90, and 5.94 as formulas for the most probable number of particles in a particula... | {
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Thus the most probable occupation number for photons is (Equation 5.94)
$$N_{\omega} = \frac{d_k}{e^{\hbar\omega/k_BT} - 1}.$$
[5.110]
For free photons in a box of volume V, $d_k$ is given by Equation 5.96,<sup>24</sup> multiplied by 2 for spin (item 3), and expressed in terms of $\omega$ instead of k (item 2):... | {
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Note that by differentiating the geometric series,
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n,$$
[5.115]
you can get
$$\frac{d}{dx}\left(\frac{x}{1-x}\right) = \sum_{n=0}^{\infty} (n+1)x^n$$
and similar results for higher derivatives. Answer:
$$E = \frac{3}{2} N\hbar\omega \left( \frac{1 + e^{-\hbar\omega/k_B ... | {
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#### 6.1 NONDEGENERATE PERTURBATION THEORY
#### 6.1.1 General Formulation
Suppose we have solved the (time-independent) Schrödinger equation for some potential (say, the one-dimensional infinite square well):
$$H^0 \psi_n^0 = E_n^0 \psi_n^0, ag{6.1}$$
obtaining a complete set of orthonormal eigenfunctions, $\p... | {
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Evidently, for a *constant* perturbation all the higher corrections vanish.<sup>2</sup> If, on the other hand, the perturbation extends only halfway across the well (Figure 6.3), then
$$E_n^1 = \frac{2V_0}{a} \int_0^{a/2} \sin^2\left(\frac{n\pi}{a}x\right) dx = \frac{V_0}{2}.$$
In this case every energy level is li... | {
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and we are left with a formula for $E_n^2$ :
$$E_n^2 = \langle \psi_n^0 | H' | \psi_n^1 \rangle - E_n^1 \langle \psi_n^0 | \psi_n^1 \rangle.$$
But
$$\langle \psi_n^0 | \psi_n^1 \rangle = \sum_{m \neq n} c_m^{(n)} \langle \psi_n^0 | \psi_m^0 \rangle = 0,$$
so
$$E_n^2 = \langle \psi_n^0 | H' | \psi_n^1 \rangle... | {
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Multiplying Equation 6.23 by $W_{ab}$ , and using Equation 6.21 to eliminate $\beta W_{ab}$ , we find
$$\alpha[W_{ab}W_{ba} - (E^1 - W_{aa})(E^1 - W_{bb})] = 0.$$
[6.24]
If $\alpha$ is not zero, Equation 6.24 yields an equation for $E^1$ :
$$(E^{1})^{2} - E^{1}(W_{aa} + W_{bb}) + (W_{aa}W_{bb} - W_{ab}W_{ba}... | {
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**Example.** Consider the three-dimensional infinite cubical well (Problem 4.2):
$$V(x, y, z) = \begin{cases} 0, & \text{if } 0 < x < a, 0 < y < a, \text{ and } 0 < z < a; \\ \infty & \text{otherwise.} \end{cases}$$
[6.29]
The stationary states are
$$\psi_{n_x n_y n_z}^0(x, y, z) = \left(\frac{2}{a}\right)^{3/2... | {
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\end{cases}$$
[6.40]
**Problem 6.8** Suppose we perturb the infinite cubical well (Equation 6.29) by putting a delta-function "bump" at the point (a/4, a/2, 3a/4):
$$H' = a^3 V_0 \delta(x - a/4) \delta(y - a/2) \delta(z - 3a/4).$$
Find the first-order corrections to the energy of the ground state and the (triply ... | {
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ag{6.49}$$
In first-order perturbation theory, the correction to $E_n$ is given by the expectation value of H' in the unperturbed state (Equation 6.9):
$$E_r^1 = \langle H_r' \rangle = -\frac{1}{8m^3c^2} \langle \psi | \hat{p}^4 \psi \rangle = -\frac{1}{8m^3c^2} \langle \hat{p}^2 \psi | \hat{p}^2 \psi \rangle.$$
... | {
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The magnetic dipole moment of a spinning charge is related to its (spin) angular momentum; the proportionality factor is the gyromagnetic ratio (which we already encountered in Section 4.4.2 Let's derive it, using classical electrodynamics. Consider first a charge q smeared out around a ring of radius r, which rotates ... | {
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The azimuthal eigenvalues for orbital and spin angular momentum $(m_l \text{ and } m_s)$ are no longer "good" quantum numbers—the stationary states are linear combinations of states with different values of these quantities; the "good" quantum numbers are n, l, s, j, and $m_j$ . <sup>10</sup>
<sup>&</sup>lt;sup>10... | {
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In particular, the (time) *average* value of S is just its projection along J:
$$\mathbf{S}_{\text{ave}} = \frac{(\mathbf{S} \cdot \mathbf{J})}{I^2} \,\mathbf{J}.\tag{6.72}$$
But $\mathbf{L} = \mathbf{J} - \mathbf{S}$ , so $L^2 = J^2 + S^2 - 2\mathbf{J} \cdot \mathbf{S}$ , and hence
$$\mathbf{S} \cdot \mathbf{J... | {
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Using the Clebsch-Gordan coefficients (Problem 4.45 or Table 4.7) to express $|jm_i\rangle$ as a linear combination of $|l m_l\rangle |s m_s\rangle$ , we have
$$l = 0 \begin{cases} \psi_1 & \equiv |\frac{1}{2}\frac{1}{2}\rangle = |00\rangle|\frac{1}{2}\frac{1}{2}\rangle, \\ \psi_2 & \equiv |\frac{1}{2}\frac{-1}{2}... | {
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#### 6.5 HYPERFINE SPLITTING
The proton itself constitutes a magnetic dipole, though its dipole moment is much smaller than the electron's because of the mass in the denominator (Equation 6.59):
$$\boldsymbol{\mu}_p = \frac{ge}{2m_p} \mathbf{S}_p, \quad \boldsymbol{\mu}_e = -\frac{e}{m_e} \mathbf{S}_e.$$
[6.84] ... | {
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Incidentally, the answer you get for positronium $(4.85 \times 10^{-4} \text{ eV})$ is quite far from the experimental value $(8.41 \times 10^{-4} \text{ eV})$ ; the large discrepancy is due to pair annihilation $(e^+ + e^- \rightarrow \gamma + \gamma)$ , which contributes an extra $(3/4)\Delta E$ and does not oc... | {
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There are initially nine degenerate states, $\psi_{3lm}$ (neglecting spin, of course), and we turn on an electric field in the z direction.
- (a) Construct the $9 \times 9$ matrix representing the perturbing Hamiltonian. Partial answer: $\langle 3\ 0\ 0|z|3\ 1\ 0\rangle = -3\sqrt{6}a$ , $\langle 3\ 1\ 0|z|3\ 2\ 0... | {
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#### 7.1 THEORY
Suppose you want to calculate the ground-state energy $E_g$ for a system described by the Hamiltonian H, but you are unable to solve the (time-independent) Schrödinger equation. Pick any normalized function $\psi$ whatsoever.
Theorem:
$$E_g \le \langle \psi | H | \psi \rangle \equiv \langle H ... | {
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**Example 3.** Find an upper bound on the ground-state energy of the one-dimensional infinite square well (Equation 2.15), using the "triangular" trial wave function (Figure 7.1)<sup>2</sup>:
$$\psi(x) = \begin{cases} Ax, & \text{if } 0 \le x \le a/2, \\ A(a-x), & \text{if } a/2 \le x \le a, \\ 0, & \text{otherwise... | {
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To get a better approximation for $E_g$ , we'll apply the variational principle, using $\psi_0$ as the trial wave function. This is a particularly convenient choice because it's an eigenfunction of *most* of the Hamiltonian:
$$H\psi_0 = (8E_1 + V_{ee})\psi_0. ag{7.18}$$
Thus
$$\langle H \rangle = 8E_1 + \lan... | {
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With this in mind, we rewrite H (Equation 7.14) as follows:
$$H = -\frac{\hbar^2}{2m} (\nabla_1^2 + \nabla_2^2) - \frac{e^2}{4\pi\epsilon_0} \left( \frac{Z}{r_1} + \frac{Z}{r_2} \right) + \frac{e^2}{4\pi\epsilon_0} \left( \frac{(Z-2)}{r_1} + \frac{(Z-2)}{r_2} + \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} \right).$$
[7.28... | {
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\tag{7.37}$$
(Quantum chemists call this the LCAO technique, because we are expressing the *molecular* wave function as a *l*inear combination of atomic orbitals.)
Our first task is to normalize the trial function:
$$1 = \int |\psi|^2 d^3 \mathbf{r} = |A|^2 \Big[ \int |\psi_g(r_1)|^2 d^3 \mathbf{r} + \int |\psi_g... | {
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\*\*Problem 7.9 Suppose we used a *minus* sign in our trial wave function (Equation 7.37):
$$\psi = A[\psi_g(r_1) - \psi_g(r_2)]. \tag{7.52}$$
**Figure 7.7:** Plot of the function F(x), Equation 7.51, showing existence of a bound state.
Without doing any new integrals, find F(x) (... | {
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Chandrasekhar<sup>10</sup> used a trial wave function of the form
$$\psi(\mathbf{r}_1, \mathbf{r}_2) \equiv A[\psi_1(r_1)\psi_2(r_2) + \psi_2(r_1)\psi_1(r_2)], \tag{7.62}$$
where
$$\psi_1(r) \equiv \sqrt{\frac{Z_1^3}{\pi a^3}} e^{-Z_1 r/a}$$
, and $\psi_2(r) \equiv \sqrt{\frac{Z_2^3}{\pi a^3}} e^{-Z_2 r/a}$ . [7... | {
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The WKB (Wentzel, Kramers, Brillouin)<sup>1</sup> method is a technique for obtaining approximate solutions to the time-independent Schrödinger equation in one dimension (the same basic idea can be applied to many other differential equations, and to the radial part of the Schrödinger equation in three dimensions). It ... | {
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Inside the well [assuming E > V(x) throughout] we have
$$\psi(x) \cong \frac{1}{\sqrt{p(x)}} \left[ C_+ e^{i\phi(x)} + C_- e^{-i\phi(x)} \right],$$
or, more conveniently,
$$\psi(x) \cong \frac{1}{\sqrt{p(x)}} \left[ C_1 \sin \phi(x) + C_2 \cos \phi(x) \right], \qquad [8.13]$$
where (exploiting the freedom not... | {
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Gamow pictured the potential energy curve for the alpha particle as a finite square well (representing the attractive nuclear force), extending out to $r_1$ (the radius of the nucleus), joined to a repulsive Coulombic tail (Figure 8.5). If E is the energy of the emitted alpha particle, the outer turning point $(r_2)... | {
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In the WKB approximation, we have
$$\psi(x) \cong \begin{cases} \frac{1}{\sqrt{p(x)}} \left[ Be^{\frac{i}{\hbar} \int_{x}^{0} p(x') dx'} + Ce^{-\frac{i}{\hbar} \int_{x}^{0} p(x') dx'} \right], & \text{if } x < 0, \\ \frac{1}{\sqrt{|p(x)|}} De^{-\frac{1}{\hbar} \int_{0}^{x} |p(x')| dx'}, & \text{if } x > 0. \end{cases... | {
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But notice that the argument here is $\alpha x$ , and if you study the matter carefully (see Problem 8.8) you will find that there *is* (typically) a region in which $\alpha x$ is large, but at the same time it is reasonable to approximate V(x) by a straight line.
$$\psi(x) \cong \frac{1}{\sqrt{\hbar}\alpha^{3/4}(... | {
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In any event, the result is extraordinarily powerful, for it enables us to calculate (approximate) allowed energies without ever solving the Schrödinger equation, by simply evaluating one integral. The wave function itself has dropped out of sight.
- \*\*Problem 8.5 Consider the quantum mechanical analog to the class... | {
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(a) Write down the WKB wave functions in regions (i) $x > x_2$ , (ii) $x_1 < x < x_2$ , and (iii) $0 < x < x_1$ . Impose the appropriate connection formulas at $x_1$ and $x_2$ (this has already been done, in Equation 8.46, for $x_2$ ; you will have to work out $x_1$ for yourself), to show that
$$\psi(x) \... | {
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Up to this point, practically everything we have done belongs to the subject that might properly be called **quantum statics**, in which the potential energy function is independent of time: $V(\mathbf{r}, t) = V(\mathbf{r})$ . In that case the (time-dependent) Schrödinger equation,
$$H\Psi = i\hbar \frac{\partial \... | {
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\tag{9.11}$$
Equations 9.10 and 9.11 determine $c_a(t)$ and $c_b(t)$ ; taken together, they are completely equivalent to the (time-dependent) Schrödinger equation, for a two-level system. Typically, the diagonal matrix elements of H' vanish (see Problem 9.4 for the more general case in which the diagonal terms are... | {
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Let
$$d_a \equiv e^{\frac{i}{\hbar} \int_0^t H'_{aa}(t')dt'} c_a, \quad d_b \equiv e^{\frac{i}{\hbar} \int_0^t H'_{bb}(t')dt'} c_b.$$
[9.19]
Show that
$$\dot{d}_{a} = -\frac{i}{\hbar} e^{i\phi} H'_{ab} e^{-i\omega_{0}t} d_{b}; \quad \dot{d}_{b} = -\frac{i}{\hbar} e^{-i\phi} H'_{ba} e^{i\omega_{0}t} d_{a}, \quad [... | {
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Express your results $[c_a(t)]$ and
$c_b(t)$ ] in terms of the Rabi flopping frequency,
$$\omega_r \equiv \frac{1}{2} \sqrt{(\omega - \omega_0)^2 + (|V_{ab}|/\hbar)^2}.$$
[9.30]
- **(b)** Determine the transition probability, $P_{a\to b}(t)$ , and show that it never exceeds 1. Confirm that $|c_a(t)|^2 + |c_b(... | {
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And so it would, if it were really free of all external perturbations. However, in quantum electrodynamics the fields are nonzero even in the ground state—just as the harmonic oscillator (for example) has nonzero energy (to wit, $\hbar\omega/2$ ) in its ground state. You can turn out all the lights, and cool the room ... | {
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**Propagation direction:** Now let's set the polar axis along $\varphi$ and integrate over all propagation directions to get the polarization–propagation average (subscript pp):
$$(\hat{n} \cdot \boldsymbol{\wp})_{pp}^2 = \frac{1}{4\pi} \int \left[ \frac{1}{2} \wp^2 \sin^2 \theta \right] \sin \theta \, d\theta \,... | {
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From Equation 9.44, we have
$$\boldsymbol{\wp} = q \langle n | x | n' \rangle \hat{\iota}.$$
<sup>&</sup>lt;sup>14</sup>This situation is not to be confused with the case of thermal equilibrium, which we considered in the previous section. We assume here that the atoms have been lifted *out* of equilibrium, and are... | {
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#### **Conclusion:**
$$(m'-m)\langle n'l'm'|x|nlm\rangle = i\langle n'l'm'|y|nlm\rangle$$
[9.70]
So you never have to compute matrix elements of y—you can always get them from the corresponding matrix elements of x.
Finally, the commutator of $L_z$ with y yields
$$\langle n'l'm'|[L_z, y]|nlm\rangle = \langl... | {
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#### FURTHER PROBLEMS FOR CHAPTER 9
\*\* **Problem 9.14** Develop time-dependent perturbation theory for a multilevel system, starting with the generalization of Equations 9.1 and 9.2:
$$H_0\psi_n = E_n\psi_n, \quad \langle \psi_n | \psi_m \rangle = \delta_{nm}.$$
[9.79]
At time t = 0 we turn on a perturbation ... | {
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Suppose we keep the first-order correction:
$$\mathbf{E}(\mathbf{r},t) = \mathbf{E}_0[\cos(\omega t) + (\mathbf{k} \cdot \mathbf{r})\sin(\omega t)].$$
[9.93]
The first term gives rise to the **allowed** (**electric dipole**) transitions we considered in the text; the second gives rise to so-called **forbidden** (**... | {
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#### 10.1 THE ADIABATIC THEOREM
#### 10.1.1 Adiabatic Processes
Imagine a perfect pendulum, with no friction or air resistance, oscillating back and forth in a vertical plane. If I grab the support and shake it in a jerky manner, the bob will swing around in a wild chaotic fashion. But if I very gently and steadily... | {
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#### 10.1.2 Proof of the Adiabatic Theorem
The adiabatic theorem is simple to state, and it *sounds* plausible, but it is not easy to prove.<sup>3</sup> Suppose the time-dependent part of the Hamiltonian can be written in the form<sup>4</sup>
$$H'(t) = V f(t), ag{10.7}$$
where f(t) is a function that starts out... | {
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\end{cases}$$
[10.19]
It follows that
$$|\langle \Psi(T)|\psi_n^f \rangle|^2 = 1, \qquad [10.20]$$
while (for $m \neq n$ )
$$|\langle \Psi(T)|\psi_m^f\rangle|^2 = 0.$$
[10.21]
Ostensibly, either of these would suffice to establish the desired result (Equation 10.9). However, Equation 10.20 is only accurate t... | {
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The corresponding eigenvalues are
$$E_{\pm} = \mp \frac{\hbar \omega_1}{2}.\tag{10.29}$$
Suppose the electron starts out with spin up, along $\mathbf{B}(0)$ :
$$\chi(0) = \begin{pmatrix} \cos(\alpha/2) \\ \sin(\alpha/2) \end{pmatrix}.$$
[10.30]
The exact solution to the time-dependent Schrödinger equation is (... | {
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<sup>&</sup>lt;sup>8</sup>See, for example, Jerry B. Marion, *Classical Dynamics*, 2nd ed. (New York: Academic Press, 1970), Section 11.4. Geographers measure latitude ( $\lambda$ ) up from the equator, rather than down from the pole, so $\cos \theta_0 = \sin \lambda$ .
the system are changed in some fashion that ... | {
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This time we have
$$\gamma_n(t) = i \int_{\mathbf{R}_i}^{\mathbf{R}_f} \langle \psi_n | \nabla_R \psi_n \rangle \cdot d\mathbf{R}, \qquad [10.48]$$
and if the Hamiltonian returns to its original form after a time T, the net geometric phase change is
$$\gamma_n(T) = i \oint \langle \psi_n | \nabla_R \psi_n \rangle... | {
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For consistency (noting, while I'm at it, that $\gamma_n$ is already first order, so $e^{i\gamma_n} \cong 1$ on the right), I should have written
$$\frac{\partial \psi_n}{\partial t} + i \psi_n \frac{d \gamma_n}{dt} = -e^{-i\theta_n} \epsilon \sum_{m \neq n} \left( \frac{i}{\hbar} c_m E_m + \frac{d c_m}{dt} \righ... | {
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Looking up the gradient in spherical coordinates, we find
$$\nabla \chi_{+} = \frac{\partial \chi_{+}}{\partial r} \hat{r} + \frac{1}{r} \frac{\partial \chi_{+}}{\partial \theta} \hat{\theta} + \frac{1}{r \sin \theta} \frac{\partial \chi_{+}}{\partial \phi} \hat{\phi}$$
$$= \frac{1}{r} \left( \frac{-(1/2) \sin(\the... | {
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In this case the Hamiltonian (Equation 10.74) becomes
$$H = \frac{1}{2m} \left[ -\hbar^2 \nabla^2 + q^2 A^2 + 2i\hbar q \mathbf{A} \cdot \nabla \right].$$
[10.76]
But the wave function depends only on the azimuthal angle $\phi$ , $(\theta = \pi/2 \text{ and } r = b)$ so $\nabla \to (\hat{\phi}/b)(d/d\phi)$ , an... | {
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The beams arrive *out of phase* by an amount proportional to the magnetic flux their paths encircle:
phase difference =
$$\frac{q \Phi}{\hbar}$$
. [10.91]
This phase shift leads to measurable interference (as in Equation 10.53), and has been confirmed experimentally by Chambers and others.<sup>17</sup>
The Aharon... | {
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Answer:
$$x_c(t) = \omega \int_0^t f(t') \sin[\omega(t - t')] dt'.$$
[10.100]
(b) Show that the solution to the (time-dependent) Schrödinger equation for this oscillator, assuming it started out in the *n*th state of the *undriven* oscillator $[\Psi(x,0) = \psi_n(x)]$ , where $\psi_n(x)$ is given by Equation 2.5... | {
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#### 11.1 INTRODUCTION
#### 11.1.1 Classical Scattering Theory
Imagine a particle incident on some scattering center (say, a proton fired at a heavy nucleus). It comes in with an energy E and an **impact parameter** b, and it emerges at some **scattering angle** $\theta$ —see Figure 11.1. (I'll assume for simplici... | {
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Indeed, the probability that the incident particle, traveling at speed v, passes through the infinitesimal area $d\sigma$ , in time dt, is (see Figure 11.5)
$$dP = |\psi_{\text{incident}}|^2 dV = |A|^2 (v dt) d\sigma.$$
**Figure 11.4:** Scattering of waves; incoming plane wave gener... | {
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$$h_0^{(1)} = -i\frac{e^{iz}}{z} \qquad h_0^{(2)} = i\frac{e^{-iz}}{z}$$
$$h_1^{(1)} = \left(-\frac{i}{z^2} - \frac{1}{z}\right)e^{iz} \qquad h_1^{(2)} = \left(\frac{i}{z^2} - \frac{1}{z}\right)e^{-iz}$$
$$h_2^{(1)} = \left(-\frac{3i}{z^3} - \frac{3}{z^2} + \frac{i}{z}\right)e^{iz} \qquad h_2^{(2)} = \left(\frac{... | {
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(Since $k = 2\pi/\lambda$ , this amounts to saying that the wavelength is much greater than the radius of the sphere.) Referring to Table 4.3, we note that $n_l(z)$ is much larger than $j_l(z)$ , for small z, so
$$\frac{j_l(z)}{h_l^{(1)}(z)} = \frac{j_l(z)}{j_l(z) + in_l(z)} \approx -i \frac{j_l(z)}{n_l(z)}$$
$... | {
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\*\*Problem 11.4 Consider the case of low-energy scattering from a spherical deltafunction shell:
$$V(r) = \alpha \delta(r - a),$$
where $\alpha$ and a are constants. Calculate the scattering amplitude $f(\theta)$ , the differential cross-section $D(\theta)$ , and the total cross-section $\sigma$ . Assume $ka... | {
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We have to decide how to skirt the poles—I'll go *over* the one at -k and *under* the one at +k (Figure 11.8). (You're welcome to choose some *other* convention if you like—even winding seven times around each pole; you'll get a different Green's function, but, as I'll show you in a minute, they're all equally acceptab... | {
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This is in the standard form (Equation 11.12), and we can read off the scattering amplitude:
$$f(\theta, \phi) = -\frac{m}{2\pi\hbar^2 A} \int e^{-i\mathbf{k}\cdot\mathbf{r}_0} V(\mathbf{r}_0) \psi(\mathbf{r}_0) d^3\mathbf{r}_0.$$
[11.66]
So far, this is *exact*. Now we invoke the **Born approximation**: Suppose ... | {
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#### 11.3.3 The Born Series
The Born approximation is similar in spirit to the **impulse approximation** in classical scattering theory. In the impulse approximation we begin by pretending that the particle keeps going in a straight line (Figure 11.11), and compute the transverse impulse that would be delivered to ... | {
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Now that you have (I hope) a sound understanding of what quantum mechanics says, I should like to return to the question of what it means—continuing the story begun in Section 1.2. The source of the problem is the indeterminacy associated with the statistical interpretation of the wave function. For $\Psi$ (or, more ... | {
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**Figure A.2:** Bell's version of the EPR-Bohm experiment: detectors independently oriented in directions **a** and **b**.
Bell proposed to calculate the *average* value of the *product* of the spins, for a given set of detector orientations. Call this average $P(\mathbf{a}, \mathbf... | {
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See D. Greenberger, M. Horne, A. Shimony, and A. Zeilinger, *Am. J. Phys.* **58**, 1131, (1990) and N. David Mermin, *Am. J. Phys.* **58**, 731, (1990).
**Figure A.4:** The shadow of the bug moves across the screen at a velocity v' greater than c, provided that the screen is far enough... | {
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Ballentine, Quantum Mechanics, (Prentice Hall, Englewood Cliffs, NJ, 1990).
<sup>&</sup>lt;sup>11</sup>E. Schrödinger, *Naturwiss.* **48**, 52 (1935); translation by Josef M. Jauch, *Foundations of Quantum Mechanics*, (Reading, MA: Addison-Wesley, 1968), p. 185.
quantum mechanics) interacts with the macroscopic sys... | {
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A
| $\mathbf{A}$ | Alpha: |
|----------------------------------------|-------------------------------------------|
| | decay, 281–84 |
| Absorption, 307–8 | particle, 28... | {
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A. M., 102, 106 Euler's formula, 29 Dirac comb. 198-203 Exchange: Dirac delta function, 52-53, 59, 102 force, 182-85 Dirac equation, 237, 244 operator, 180 Dirac notation, 102, 118-19 Exclusion principle, 180, 189, 193, 195, 203 Direct integral, 269 Expectation value, 7, 9, 16 Dirichlet's theorem, 28-29, 49 rate-of-cha... | {
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046 050 005 | magnetic dipole in magnetic field, 160, 239, 330 |
| g-factor, 246, 250, 225 | Hankel functions, 358–59 |
| deuteron, 255 | Hannay's angle, 335 ... | {
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Department of Physics and Astronomy, University College London
Copyright # 2006 John Wiley & Sons Ltd, The Atrium, Southern Gate, Chichester, West Sussex PO19 8SQ, England
Telephone (+44) 1243 779777
Email (for orders and customer service enquiries): cs-books@wiley.co.uk Visit our Ho... | {
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| | Preface | | xi | | |
|---------------------------------------------------------------------------------------|---------|-------------------------------... | {
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It is common practice to teach nuclear physics and particle physics together in an introductory course and it is for such a course that this book has been written. The material is presented so that different selections can be made for a short course of about 25–30 lectures depending on the lecturer's preferences and th... | {
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Data for particle physics may be obtained from the biannual publications of the Particle Data Group (PDG) and the 2004 edition of the PDG definitive Review of Particle Properties is given in Ei04. The PDG Review is also available at http://pdg.lbl.gov and this site contains links to other sites where compilations of pa... | {
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"Header 2": "Data",
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| Quantity | Symbol | Value |
|-----------------------------|----------------------------------------------------------|-------------------------------------------------------|
| Speed of light in vacuum... | {
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Although this book will not follow a strictly historical development, to 'set the scene' this first chapter will start with a brief review of the most important discoveries that led to the separation of nuclear physics from atomic physics as a subject in its own right and later work that in its turn led to the emergenc... | {
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By the early 1930s, the 19th century view of atoms as indivisible *elementary* particles had been replaced and a larger group of physically smaller entities now enjoyed this status: electrons, protons and neutrons. To these we must add two electrically neutral particles: the photon $(\gamma)$ and the neutrino $(\nu)... | {
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In quantum theory, the interaction is transmitted discontinuously by the exchange of photons, which are members of the family of fundamental spin-1 bosons of the standard model. Photons are referred to as the gauge bosons, or 'force carriers', of the electromagnetic interaction. The use of the word 'gauge' refers to th... | {
"Header 1": "Basic Concepts",
"Header 2": "1.1.2 The emergence of particle physics: the standard model and hadrons",
"token_count": 743,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
Elementary particle physics is also called high-energy physics. One reason for this is that if we wish to produce new particles in a collision between two other particles, then because of the relativistic mass–energy relation E ¼ mc2, energies are needed at least as great as the rest masses of the particles produced. T... | {
"Header 1": "Basic Concepts",
"Header 2": "1.2 Relativity and Antiparticles",
"token_count": 1698,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
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Symmetries and the invariance properties of the underlying interactions play an important role in physics. Some lead to conservation laws that are universal. Familiar examples are translational invariance, leading to the conservation of linear momentum; and rotational invariance, leading to conservation of angular mome... | {
"Header 1": "Basic Concepts",
"Header 2": "1.3 Symmetries and Conservation Laws",
"token_count": 2030,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
From Equation (1.15a), we see that states of type a are eigenstates of $\hat{C}$ with eigenvalues $\pm 1$ , called their *C-parities*. States with distinct antiparticles can only form eigenstates of $\hat{C}$ as linear combinations.
As an example of the latter, consider a $\pi^+\pi^-$ pair with orbital angular... | {
"Header 1": "Basic Concepts",
"Header 2": "1.3 Symmetries and Conservation Laws",
"token_count": 620,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
We now turn to a discussion of particle interactions and how they can be described by the very useful pictorial methods of Feynman diagrams.
#### 1.4.1 Interactions
Interactions involving elementary particles and/or hadrons are conveniently summarized by 'equations' by analogy with chemical reactions, in which the ... | {
"Header 1": "Basic Concepts",
"Header 2": "1.3 Symmetries and Conservation Laws",
"Header 3": "1.4 Interactions and Feynman Diagrams",
"token_count": 2014,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
At each vertex of a Feynman diagram, charge is conserved by construction. We will see later that, depending on the nature of the interaction (strong, weak or electromagnetic), other quantum numbers are also conserved. However, it is easy to show that energy and momentum cannot be conserved simultaneously.
Consider th... | {
"Header 1": "Basic Concepts",
"Header 2": "1.5.1 Range of forces",
"token_count": 1852,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
We have mentioned earlier that Feynman diagrams can be turned into probabilities for a process by a set of mathematical rules (the Feynman Rules) that can be
Although we call g a (point) coupling *constant*, in general it will have a dependence on the momentum carried by the exchanged particle. We ignore this in what... | {
"Header 1": "Basic Concepts",
"Header 2": "Observable Quantities: Cross Sections and Decay Rates",
"token_count": 1525,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
The next step is to relate the amplitude to measurables. For scattering reactions, the appropriate observable is the cross-section. In a typical scattering experiment, a beam of particles is allowed to hit a target and the rates of production of various particles in the final state are counted.36 It is clear that the r... | {
"Header 1": "Basic Concepts",
"Header 2": "1.6.2 Cross-sections",
"token_count": 2044,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
(1.56)
If the initial particles are unpolarized (which is the most common case in practice), then we must average over all possible initial spin configurations (because each is equally likely) and sum over the final configurations. Thus, Equation (1.55) becomes
$$\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \frac{g_... | {
"Header 1": "Basic Concepts",
"Header 2": "1.6.2 Cross-sections",
"token_count": 1263,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
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Most branches of science introduce special units that are convenient for their own purposes. Nuclear and particle physics are no exceptions. Distances tend to be measured in femtometres (fm) or, equivalently *fermis*, with 1 fm $\equiv 10^{-15}$ m. In these units, the radius of the proton is about 0.8 fm. The range o... | {
"Header 1": "Basic Concepts",
"Header 2": "1.7 Units: Length, Mass and Energy",
"token_count": 1905,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
Nuclei are specified by:
Z – atomic number = the number of protons,
N – neutron number = the number of neutrons,
A – mass number = the number of nucleons, so that A = Z + N.
We will also refer to A as the *nucleon number*. The charge on the nucleus is +Ze, where e is the absolute value of the electric charge on... | {
"Header 1": "Basic Concepts",
"Header 2": "2.1 Mass Spectroscopy and Binding Energies",
"token_count": 1867,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
The shape and size of a nucleus may be found from scattering experiments; i.e. a projectile is scattered from the nucleus and the angular distribution of the scattered particles examined, as was done by Rutherford and his collaborators when they deduced the existence of the nucleus. The interpretation is simplest in th... | {
"Header 1": "Basic Concepts",
"Header 2": "2.2 Nuclear Shapes and Sizes",
"token_count": 1975,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
The minima are due to the form factor and we can make this plausible by taking the simple case where the nuclear charge distribution is represented by a hard sphere such that
$$\rho(r) = \text{constant}, \quad r \le a$$
$$= 0 \qquad r > a$$
(2.20)
where a is a constant. In this case, evaluation of Equation (2.1... | {
"Header 1": "Basic Concepts",
"Header 2": "2.2 Nuclear Shapes and Sizes",
"token_count": 1980,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
$$\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} \propto \left[\frac{J_1(qR)}{qR}\right]^2,$$
(2.34)
where $qR \approx pR\theta$ for small $\theta$ and $J_1$ is a first-order Bessel function. For large qR,
$$[J_1(qR)]^2 \approx \left(\frac{2}{\pi qR}\right) \sin^2\left(qR - \frac{\pi}{4}\right),$$
(2.35)
which... | {
"Header 1": "Basic Concepts",
"Header 2": "2.2 Nuclear Shapes and Sizes",
"token_count": 884,
"source_pdf": "datasets/websources/Physics_v1/Physics/Martin - Nuclear and Particle Physics - An Introduction.pdf"
} |
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