text stringlengths 30 4k | source stringlengths 60 201 |
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3.3 Neutrino oscillations and the MSW effect
3.4 Born-Oppenheimer approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
4 Scattering
24
. . . . . . . . . . . . . . . . . . . . . . . . . . . . ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
0. This is a
bit abstract, so here is a more concrete version of the example. H0 is the natural Hamiltonian of
1
the hydrogen atom and δH(t) comes from electric and/or magnetic fields that we temporarily turn
on. If we start in the 1s state, then what is the probability that after some time we will be in the
2p state? ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
8.05 we saw the Schr¨odinger
Aside: comparison to Schr¨odinger and Heisenberg pictures.
picture and the Heisenberg picture. In the former, states evolve according to H and operators re-
main the same; in the latter, states stay the same and operators evolve according to H. The
interaction picture can be thought of as i... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
¨odinger
equation in the rotating frame
where we have defined
i
d(cid:126)
dt
| ˜
ψ(t)(cid:105) = δ(cid:102)H(t)
˜
|ψ(t)(cid:105)
δH(t) = e (cid:126) δH(t)e−
(cid:102)
iH t
0
iH
t0
(cid:126)
.
(3)
This has a simple interpretation as a matrix. Suppose that the eigenvalues and eigenvectors of H0
(reminder: we work with th... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
H mn(t)cn(t) =
eiωmntδHmncn(t).
1.2 Perturbation expansion
n
n
So far everything has been exact, although sometimes this is already enough to solve interesting
problems. But often we will need approximate solutions. So assume that δH(t) = O(λ) and expand
the wavefunction in powers of λ, i.e.
cm(t) = cm (t) + cm (t) + c... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
(cid:125)
(cid:125)
(cid:124)
(cid:125)
(4)
The solution is much simp
the RHS, so the zeroth order approximation is simply that nothing happens:
time-dep
ler than
t case.
There
enden
the
in
is
no zeroth
| ˜ψ(0)(t)(cid:105) = | ˜ψ(0)(0)(cid:105) = |ψ(0)(cid:105)
i(cid:126)∂t| ˜ψ(1)(t)(cid:105)
= δ(cid:102)H(t)|ψ(0)(t)(c... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
)
(cid:12)
second-order solution
is
| ˜ψ(2)(t)(cid:105) =
(cid:90) t(cid:48)
(cid:90) t
dt(cid:48)
0
0
dt(cid:48)(cid:48)
(cid:102)H(t(cid:48)) δ
δ
i(cid:126)
(cid:102)H(t(cid:48)(cid:48))
i(cid:126)
|ψ(0)(cid:105) .
(8)
3
1.3 NMR
In some cases the rotating frame already helps us solve nontrivial problems exactly with... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
)ω2
the zˆ axis.
2
0 + Ω .
If Ω (cid:28) ω0 then this is very close to precession around
Now let’s look at this problem using first-order perturbation theory.
ω t
0
H(t) = ei
δ(cid:102)
(cid:90) t
| ˜ψ(1)(t)(cid:105) =
= Ω (cos(ω0t)Sx − sin(ω0t)Sy)
ω t
0
2
σz
σz
dt(cid:48)
2 ΩSxe−i
δH(t(cid:48))
(cid:102)
i(cid:126) |ψ(... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
= Ω (cos(ω0t)Sx + sin(ω0t)Sy) .
4
We have already computed Sx above. In the rotating frame we have
˜
Thus
˜Sx = (cos(ω0t)Sx − sin(ω0t)Sy)
˜Sy = (cos(ω0t)Sy + sin(ω0t)Sx)
δ(cid:102)H(t) = ΩSx.
The rotating-frame solution is now very simple:
| ˜ψ(t)(cid:105)
i
= e− Ωt σx
2
|ψ(0)(cid:105).
This can be easily translated b... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
) mn
i(cid:126)
dt(cid:48) (cid:16)
0
(cid:90) t
Vmn
2i(cid:126) 0
(cid:34)
V
mn
2i(cid:126)
ei(ωmn+ω)t(cid:48)
+ ei(ωmn−ω)t(cid:48)
ei(ωmn+ω)t
−
ωmn + ω
1
+
ei(ωmn−ω)t
−
ωmn − ω
(cid:17)
(cid:35)
1
The ωmn ± ω terms in the denominator mean that we will get the largest contribution when
ω ≈ |ωmn|. (A word about signs. ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
t
2
α2
(9)
For fixed α, f (t, α) is periodic in t.
It is more interesting to consider the case of fixed t, for which f has a sinc-like appearance (see
Fig. 1).
Figure 1: The function f (α, t) from (9), representing the amount of amplitude transfered at a fixed
time t as we vary the detuning α ≡ ωmn − ω.
t
It has zeroes as... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
. So as t → ∞
On the other hand,
is always
nonnegative, always has total mass roughly independent of t, but approaches zero for all nonzero
α. This means that it approaches a delta function. A more detailed calculation (omitted, but it
uses complex analysis) shows that
we see that
f (t,α)
t
f (t, α)
lim
t→∞ t
= δ(α).
π... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
The driving frequency
might come from a thermal distribution or a laser, both of which output a distribution of frequen-
cies. The linewidth of a laser is much lower but still nonzero. The energy difference (cid:126)ωmn seems
like a universal constant, can also be replaced by a distribution by phenomena such as Doppler
... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
”). In those cases transition rates are much lower. One dramatic example is that 2p → 1s
transition in hydrogen takes 1.6ns because it occurs at first order while the 2s → 1s transition takes
0.12 seconds. For this reason states such as the 2s states are called “metastable.”
We now consider the most important special ca... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
treatment we’ve described here. Nor do the photons. “Absorption” means jumping
from a low-energy state to a higher-energy state, and “stimulated emission” means jumping from
high energy to low energy. In the former case, we reason from energy conservation that a photon
must have been absorbed, and in the latter, a phot... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
. These vacuum fluctuations lead to a small separation in energy between the 2s
and 2p levels of hydrogen.
(cid:104)E
2
2
Dipole moment
In the Stark effect we looked at the interaction of the hydrogen atom with an
electric field. This was a special case of the interaction between a E field and the dipole moment
of a collec... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
:126)x · E for any constant C. However, this would only have added
an overall constant to the Hamiltonian, which would have no physical effect.
N
i=1 i
(cid:126)
What if the electric field is spatially varying? If this spatial variation is small and we are near
the origin, we use the first few terms of the Taylor expansio... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
. The only operator here is the dipole
moment d = (d1, d2, d3), so the matrix elements of V are given by
ˆ (cid:126)
(cid:126)
ˆ
ˆ
Vmn = − ˆ (cid:126)
E0P · dmn = −E0
3
(cid:88)
i=1
Pi(cid:104)m|di
|n(cid:105).
Since the transition rate depends on |V
mn , we will average this quantity over the choice of polar-
|2
9
iz... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
≡ 0
3
|(cid:126)dmn|2
P
How did we calculate (cid:104)PiPj(cid:105) ˆ? This can be done by explicit calculation, but it is easier to use sym-
metry. First, observe that the uniform distribution over unit vectors is invariant under reflection.
Thus, if i = j, then (cid:104)PiPj(cid:105) = (cid:104)(−Pi)Pj(cid:105) = 0. O... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
6)2 |
d
2
|
This last expression is known as Fermi’s Golden Rule. (It was discovered by Dirac, but Fermi called
it “Goldren Rule #2”.)
2.2 Spontaneous emission
The modern description of spontaneous emission requires QED, but the first derivation of it predates
even modern quantum mechanics! In a simple and elegant argum... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
changes from state b to state a.
ANb
Stimulated
emission
A photon of frequency ωba is emitted and an atom
changes from state b to state a.
BbaNbU (ωba)
These processes depend on the Einstein coefficients A, Bab and Bba for spontaneous emission,
absorption and stimulated emission respectively. They also depend on the popu... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
relation
between spontaneous and stimulated emission can be seen in the fact that both involve an a†
operator acting on the photon field. If there are no photons, then the field is in state |0(cid:105) and we
get the term a†|0(cid:105) = |1(cid:105), corresponding to spontaneous emission. If there are already n photon in... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
126)
Suppose our initial state is our old friend, the ground state of the hydrogen atom, and the
final state is a plane wave.
If the final state is unnormalized, then matrix elements such as
(cid:104)ψfinal|V |ψinitial(cid:105) become less meaningful. One way to fix this is to put the system in a box of
size L × L × L with... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
0
(cid:123)(cid:122)
A
− (cid:126)k · (cid:126)x
i
≡
(cid:112)
eE
0
πa3
0L3
A.
(cid:19)
(cid:125)
We have defined A to equal the difficult-to-ev
aluate in
tegral. The factor
of x3 can be removed by
12
writing A = i ∂ B, where
∂k3
(cid:90)
B =
d3x exp
(cid:18) r
a0
− − (cid:126)ik (cid:126)
· x
(cid:19)
defining µ =
(cid:1... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
=
2πi
k3
1 + 1
ika
0
(cid:16)
− 1
ika0
(cid:17)
2
(cid:17)
2
(cid:17)
2
(cid:16)
− 1
1 + 1
k2a2
0
1
1
− i
ka
0
8
π
=
8πi
k3
(cid:16)
1
a
ik 0
1 + 1
2 2
k a0
=
(cid:17)
2
(cid:16)
0 1
a
k4
1
+ 2a2
k
0
=
(cid:17)
2
8π
(cid:0)
a
0
−
k2 + a
0
2
(cid:1)
2
To compute A, we use the fact that
∂ k2 = 2
3. Thus
k
∂k3
A... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
where θ is the angle between k and the z-axis.
(cid:126)
We can now compute the squared matrix element to be (canceling a factor of π from numerator
and denominator)
|(cid:104)(cid:126)k|V |1, 0, 0(cid:105)|2 =
e2E2 1024π cos2(θ)
0
a3L3
0
a2 10
0k
.
To compute the average rate over some time t, we will multiply this by... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
7)
− ω
.
Let’s pause for a minute to look at what we’ve derived. One strange feature is the 1/L3 term,
because the rate of ionization should not depend on how much empty space surrounds the atom.
Another strange thing appears to happen when we take t large, so that f (t, α)/t will approach
π δ(α). This would cause the ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
:12)
=
=
=
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
0
(cid:90) 1/t
0
(cid:90)
1/t
t
t
(cid:90)
1/t
(cid:32)
t
(cid:33)
dα g(α)
−
g(0)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
dα (g(α)
(cid:12)
(cid:12)
(cid:12)
− g(0))
(ci... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
(cid:126)
1
t
1 +
(cid:19)
1
tω
(cid:114) 2mω
(cid:126) ≤ ≤
k
(cid:115)
(cid:18)
1 +
2mω
(cid:126)
(cid:19)
1
tω
(cid:114) 2mω
(cid:126)
≈
(cid:19)
(cid:18) 1
+
2tω
(17)
14
(cid:126)
(cid:126)
How many k satisfy (17)? Valid k live on a cubic lattice with spacing 2π/L, and thus have density
(cid:126)
(L/2π)3. Thus we c... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
k
π
= t
2
· # valid k =
(cid:126)
L3mk
(cid:126)
4π
.
We have obtained our factor of L3 that removes the unphysical dependence on the boundary
conditions. Putting everything together we get
R
1,0,0→all k
(cid:126) =
256me2E2
0 .
3(cid:126)3a5
0k9
3 Adiabatic evolution
3.1 The adiabatic approximation
We now turn to a di... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
some n. Then if H is
changed slowly for 0 ≤ t ≤ T , then at time T we will have |Ψ(T )(cid:105) ≈ |ψn(T )(cid:105).
˙
This theorem is stated in somewhat vague terms, e.g. what does “changed slowly” mean? H
should be small, but relative to what? One clue is the reference to the nth eigenstate |ψn(t)(cid:105).
This is on... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
obtain
n
i(cid:126)
d
dt
|Ψ(t)(cid:105) = i(cid:126) (cid:88)
c˙n(t)|ψn(t)(cid:105) + cn(t) ψn(t)(cid:105) =
| ˙
n
(cid:88)
n
cn(t)En(t)|ψn(t)
(cid:105).
Multiply both sides by (cid:104)ψk(t)| and we obtain
i(cid:126)
c˙k = Ekck − i(cid:126) (cid:88)
˙
(cid:104)ψk|ψn(cid:105)cn.
n
(20)
˙
Now we need a way to evaluate (... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
˙H|ψn(cid:105)
˙
(cid:104)ψ | ˙H|ψ (cid:105)
n
k
− Ek
En
=
ψ | ˙
k
ψn(cid:105)
≡
˙H
k
En − Ek
n
In the last step, we used Hkn to refer to the matrix elements of H in the {|ψn(cid:105)} basis.
˙
˙
Plugging this into (20) we find
i(cid:126)c˙k =
(cid:104)ψk| ˙ψk(cid:105)
(Ek − i(cid:126)
)ck − i(cid:126)
(cid:125)
(cid:12... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
fferential equation is
not there. i.e. suppose that i(cid:126)
ck(t) = c (0)eiθk(t)eiγk(t)
(cid:90) t
k
1
θk(t) ≡ − (cid:126)
Ek(t(cid:48))dt(cid:48)
0
γk(t) ≡
(cid:90)
t
0
νk(t(cid:48))dt(cid:48)
˙
| k(cid:105)
νk(t) ≡ i(cid:104)ψk ψ
(22a)
(22b)
The θk(t) term is called the “dynamical” phase and corresponds to exactly ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
matrix element of V . This decreases as T and ∆ increase, which is good. But if we
add up this rate of transitions over time T , then the total transition amplitude can be as large as
∼ V /∆. Thus, going more slowly appears not to reduce the total probability of transition!
T ∆
What went wrong? Well, we assumed that am... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
energy levels are roughly constant, so we can replace it with e−iωnkt, where ωnk = (En − Ek)/(cid:126).
Now when we integrate the contribution of this term from t = 0 to t = T we get
(cid:90) T ˙H e−iωnkt
kn
0 En − Ek
dt
∼
V e−iωnkT
(cid:126)ω2
T
nk
− 1
∼
V
T (cid:126)ω2
nk
∼
(cid:126)V
∆2T
Finally we obtain that the p... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
the bottom eigenstate. When t = 0, the eigenstates are
|−(cid:105)
|+
(cid:105)
Suppose that ∆ = 0 and we start in the |−(cid:105) at t = −∞. Then at t = ∞ we will still be in
the |−(cid:105) state, with only the phase having changed. But if ∆ > 0 and we move slowly enough then
the adiabatic approximation says we will ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
))
dt(cid:48)
1
(cid:126)
0
γn(t) ≡
(cid:90)
t
0
n(t(cid:48))dt(cid:48)
ν
˙
νn(t) ≡ i(cid:104)ψn|ψn
(cid:105)
(24a)
(24b)
The phase γn(t) is called the geometric phase, or the Berry phase, after Michael Berry’s 1984
explanation of it.
Do the phases in the adiabatic approximation matter? This is a somewhat subtle questi... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
that is a function of n and (cid:126)r.
In fact, even if the particle were in a superposition of positions as in the two-slit experiment,
then we could still only see interference effects between branches of the wavefunction with the same
value of (cid:126)r. Thus, again we can define an arbitrary (n, (cid:126)r)-depende... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
126) (0)
R
i(cid:104)
ψn
(cid:126)
(cid:126)
|∇ (cid:126)R|ψn(cid:105) · dR
0
The answer is in terms of a line integral, which depends only on the path and not on time (unlike
the dynamical phase).
(cid:126)
(cid:126)
) ? Suppose we replace |ψn(R)
(cid:126)
(cid:105) with |ψn(R)
˜
(cid:105) =
How does this change if we... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
the endpoints of the path. Thus, we can eliminate
the phase for any fixed path with R(t) = R(0), but not simultaneously for all paths. In particular,
if a particle takes two different paths to the same point, the difference in their phases cannot be
redefined away. More simply, suppose the path is a loop, so that R(0) = R(... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
-d, because it contains
2-d as a special case. Then
if S denotes the surface enclosed by curve C, we have
(cid:73)
C
A(cid:126)
(cid:126)
n(R)
(cid:126)
· dR =
(cid:90) (cid:90)
S
(
(cid:126)
(cid:126)
∇ (cid:126)R × An) · d(cid:126)a ≡
(cid:90) (cid:90)
S
(cid:126)
D
n
· d(cid:126)a.
19
(cid:54)
Here we define the Ber... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
(cid:105)).
(25)
j,k
d
Example: electron spin in a magnetic field. The Hamiltonian is
(cid:126)
H = µ(cid:126)σ · B
µ =
(cid:126)
e
mc
.
Suppose that B = B(cid:126)r where B is fixed and we slowly trace out a closed path in the unit sphere
with (cid:126)r. Suppose that we start in the state
(cid:126)
|(cid:126)r; +(cid:1... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
(cid:126)r
(cid:125)
0
ieiφ sin(θ/2)
ˆ
φ.
This first term will not contribute to the Berry connection, and so we obtain
A(cid:126)
+((cid:126)r) = i(cid:104)(cid:126)r|∇|(cid:126)r(cid:105) = −
(cid:126)
1 sin2(θ/2)
sin(θ)
r
ˆ
φ.
Finally the Berry curvature is
(cid:126)
(cid:126)
D+ = ∇ × A+ =
(cid:126)
1
d
r si... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
the same computation we find that no
w
A(cid:126)
+ =
1 cos2(θ/2)
sin(θ)
r
ˆ
φ
and
(cid:126)
D+ =
−
1
2r2
.
rˆ
dθ
We see that the d sin2(θ/2) was replaced by a d (
dθ − cos2(θ/2)) which gives the same answer. This
is an example of the general principle that the Berry connection is sensitive to our choice of phase
conven... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
name neutrino means “little neutral one” and neutrinos are spin-1/2, electri-
cally neutral, almost massless and very weakly interacting particles. Neutrinos were first proposed
by Pauli in 1930 to explain the apparent violation of energy, momentum and angular momentum
conservation in beta decay. (Since beta decay invol... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
the range
produced in the sun are electron neutrinos.
0.5-20MeV. Almost all of neutrinos
Detection Neutrinos can be detected via inverse beta decay, corresponding to the reaction
A + νe (cid:55)→ A(cid:48) + e−,
where A, A(cid:48) are different atomic nuclei. For solar neutrinos this will only happen for electron
neutri... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
neutrino, in its rest frame, has a Hamiltonian with eigenstates |ν1(cid:105), |ν2(cid:105), |ν3(cid:105)
that in general will be different from the flavor eigenstates |νe(cid:105), |νµ(cid:105), |ντ (cid:105) that participate in weak-
interaction processes such as beta decay.
We will treat this in a simplified way by negl... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
θ come from (27) (θ ≈ π/6 is the “mixing angle” that measures how far the flavor states
are from being eigenstates), C is a constant and Ne = Ne((cid:126)r) is the local electron density. If the
neutrino is traveling at speed ≈ c in direction xˆ, then (cid:126)r ≈ ctxˆ. Thus we can think of Ne as
time-dependent. We then... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
H = −
N
(cid:88) (cid:126)2
j=1
∇(cid:126) 2
2M R
j
j
(cid:126) + Hel(R).
(30)
(cid:126)
(cid:126)
Here R = (R1, . . . , RN ) denotes the positions of the N nuclei and Hel(R) includes all the other
terms, i.e. kinetic energy of the electrons as well as the potential energy terms which include
electron-electron, nuclei-... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
:126) Ψ(R, r) = (
Rj
(cid:126)
∇Rj
(cid:126) γ(R))ΦR(r) + γ(R)
(cid:126)
∇Rj
(cid:126) ΦR(r).
Using the adiabatic approximation we neglect the overlap of ∇ (cid:126) Ψ(R, r) with all states to
Rj
Equivalently we can multiply on the left by (cid:104)ΦR|. This results in
(cid:126)
(cid:90)
d3nrΦR(r)∗∇(cid:126)
(cid:126)
... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
before the Aj terms can have an effect,
for the same reason that we do not see a Berry phase unless we trace out a loop in a parameter
space of dimension ≥ 2.
(cid:126)
The Born-Oppenheimer applies not just to nuclei and electrons but whenever we can divide a
system into fast and slow-moving degrees of freedom; e.g. we ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
led to a model of atoms in which electrons orbit nuclei like planets, and resolving the problems of
this model in turn was one of the early successes of quantum mechanics.
Scattering cross section:
In scattering problems it is important to think about which physical
quantities can be observed. The incoming particles ha... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
at . The resulting differential cross-section dσ
dΩ
dΩ dt
is defined to be
dσ
dΩ
(θ, φ)
scat
d2N
≡ dΩ dt
d2
Nin
dAdt
(38)
Here the spherical coordinates (θ, φ) denote the direction of the outgoing particles. It is conventional
to define the axes so that the incoming particles have momentum in the zˆ direction, so θ is the... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
41)
(42)
A general solution can be written as a superposition of separable solutions. Separable solutions to
(42) can in turn be written as
in terms of some functions u(r), f (θ, φ). In terms of these (42) becomes
rψ(r, θ, φ) = u(r)f (θ, φ),
u(cid:48)(cid:48)f +
ˆ
1
uL2f +k2uf = 0.
r2
(cid:123)(cid:122)
(cid:124)
0 as
... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
be a combination of an un-scattered part, which looks like the original wave packet continuing
forward in the zˆ direction, and a scattered part, which is a spherical outgoing wavepacket with a
f (θ, φ) angular dependence. However, we can treat the incoming wave instead as the static plane
wave eikz
eikr. (Both of thes... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
8)
In the last step we have used v = (cid:126)k/m. The units here are off because technically the wavefunction
√
should be not eikz but something more like eikz/
V , where V has units of volume. But neglecting
this factor in both the numerator and denominator of (38) will cause this to cancel out. Keeping
that in mind, ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
.
(52) is valid everywhere except at θ = 0. There we have to also
consider interference between the scattered and the unscattered wave. (Unlike the incoming wave,
the outgoing unscattered does coexist with the scattered wave.) The resulting flux is
(cid:126)
S = Im
out
(cid:126)
m
(cid:18)
e−ikz
f ∗
+ e−ikr
r
(cid:19) (... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
−ik
2
2
ρ
ρ
2z + f e 2z
ik
(cid:19)
(cid:18)
ˆ
z
z
0
(cid:90) ∞ dρ2
2
0
(cid:90) ∞
0
v
= 4π Re
z
2πv
z
Re
=
2
ikρ
f e 2z
iky
dye 2z f (0) =
−
4πv
k
Im f (0)
0
(54a)
(54b)
(54c)
27
Since the outgoing flux should equal the incoming flux, we can define A to be the beam area and
find
Av = Av + v
dΩ|f (θ, φ)|2 −
Im f (0).
(cid... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
5) = U |ψ(cid:105)
where
U ≡
2m
(cid:126)2
V.
(57)
This looks like a basic linear algebra question. Can we solve it by inverting (∇2 + k2) to obtain
(cid:126)
? (cid:126)
ψ = (∇2 + k2)−1U |ψ(cid:105)?
(58)
To answer this, we first review some basic linear algebra. Suppose we want to solve the equation
(cid:126)
A(cid:12... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
operator G to be (∇2 + k2)|−1
. The calculation of G is rather subtle and the details can be
found in Griffiths. However, on general principles we can make a fair amount of progress. Since
(cid:126)(∇2 + k2) is diagonal in the momentum basis, then G should be as well. Thus G should be written
as an integral over |p(cid:1... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
p = (cid:126)k. To handle the diverging denominator, one method is
(cid:48) denotes the integral
to write
(cid:90)
G = lim
(cid:15)→0
d3p(cid:126)(−(cid:126)2p2 + k2 + i(cid:15))−1|p(cid:126) p(cid:126)
(cid:105)(cid:104) |.
(64)
Finally we can write this in the position basis according to (62) and obtain the position-... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
(cid:105) = |ψ0(cid:105) + GU |ψ0(cid:105).
(68)
The second Born approximation consists of using (68) to approximate |ψ(cid:105) in the RHS of (67), which
yields
|ψ(cid:105) = |ψ0(cid:105) + GU (|ψ0(cid:105) + GU |ψ0(cid:105)) = |ψ0(cid:105) + GU |ψ0(cid:105) + GU GU |ψ0(cid:105).
Of course we could also rewrite (67) a... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
8)|
U ((cid:126)r(cid:48))
(70a)
(70b)
If we assume that the potential is short range and we evaluate this quantity for r far outside the
range of U , then we will have r (cid:29) r(cid:48) for all the points where the integral has a nonzero contribution.
In this case |(cid:126)r − (cid:126)r(cid:48)| ≈ r − rˆ · (cid:1... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
2b)
(72c)
The quantity in parentheses is then f (θ, φ). If we define V ((cid:126)q) to be the Fourier transform of V ((cid:126)r)
then we obtain
m ˜ (cid:126)
V (k
(cid:126)
2π 2
where f1 refers to the first Born approximation. One very simple example is when V ((cid:126)r) = V0δ((cid:126)r).
Then we simply have f1 = − m... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
θ) =
2meQ
(cid:126)2q2
=
meQ
2(cid:126)2k2 sin2(θ/2)
.
A good exercise is to rederive the Born approximation by using Fermi’s Golden Rule and
counting the number of outgoing states in the vicinity of a given k. See section 7.11 of Sakurai for
details. Another version is in Merzbacher, section 20.1.
(cid:126)
30
Rigoro... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
) =
∞
(cid:88)
l=0
Rl(r)Pl(cos θ).
If we define ul(r) = rRl(r), then the eigenvalue equation becomes (for each l)
− u(cid:48)(cid:48)
l + Vefful = k2ul
where
Veff ≡
2mV
(cid:126)2
+
l(l + 1)
r2
.
(76)
(77)
If (75) holds, then for sufficiently large r, we can approximate Veff ≈ l(l + 1)/r2. In this region
the solutions to (77... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
starts with the orthogonality condition
2 δmn. (The δmn is the important term here;
2
2n+1
2n+1
Gram-Schmidt procedure to 1, x, x2, . . . , we obtain the Legendre polynomials P0 = 1, P1 = 1,
Thus any degree-n polynomial can be written as a linear combination of P0, . . . , Pn and vice-versa.
1
P− m(x)Pn(x)dx =
1
is a s... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
of the first and second
kind:
h = j
l + inl
and
(1)
l
h = j
l − inl
(2)
l
(1)
l
For large r, h (kr)
precisely, Rl(r) (i.e.
Putting this together we get
→ − l+1 eikr
i)
kr
(
, and so our scattered wave should be proportional to h . More
the angular-momentum-l component) should be proportional to h (kr).
ψ(r, θ)
when V 0
... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
θ))
(cid:12)
(cid:12)
.
Recall that the Legendre polynomials satify the orthogonality relation
Thus we can calculate
(cid:90) 1
−
1
dzPl(z)Pl(cid:48)(z) = δl,l(cid:48)
2
2l + 1
.
(cid:90)
σ =
dΩ
dσ
dΩ
(cid:88)
= 4π
(2l + 1)|a |2
l
.
l
(85)
(86)
(87)
(88)
(89)
While the differential cross section involves interference be... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
�)
l≥
0
l(kr)
j
(cid:123)(cid:122)
(cid:125)
(cid:124)
wa
plane
ve
(1)
+ ikalh (kr)
l
(cid:123)(cid:122)
(cid:125)
scattered
(cid:124)
=
1
2
(cid:88)
l≥0
(1)
il(2l + 1)Pl(cos θ)
)(1 + 2ika +
(kr
h
l
(cid:124)
(cid:125)
(cid:123)(cid:122)
outgoing
l)
(2)
(kr
h
)
l
(cid:123)(cid:122)
(cid:125)
(cid:124)... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
=
1
k
(cid:88)
(2l + 1)Pl(cos θ)eiδl sin(δl)
σ =
4π
k2
l
(cid:88)
l
(2l + 1) sin2(δl).
(94)
(95)
(96)
(97)
As an application we can verify that this satisfies the optical theorem. Observe that Pl(1) = 1 for
all l. Thus
Im f (0) =
(2l + 1)P (1) sin2
l
(δl) = σ.
(98)
4π
k
4π 1
k k
(cid:88)
l
Another easy application is pa... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
can compute the phase shift by matching
for r = b
Here is a simple example. Consider a hard sphere of radius b. Then Rl(r) = 0 for r ≤ b and we
R(cid:48)(r)
l
Rl(r)
±
(cid:15).
have
jl(kb) −
tan(δl)nl(kb) = 0 =⇒ δl = tan−1
(cid:18)
(cid:19)
.
j
l(kb
)
nl(kb)
(101)
One particularly simple case is when l = 0. Then
δ0 = t... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
(assuming kb (cid:28) 1). Even if kb > 1 this drops exponentially once l (cid:29) kb, which
confirms our intuition that the angular momentum should be on the order of (cid:126)kb.
Another way to think about this reason to favor low values of l is because the l(l + 1)/r2
term in Veff forms an angular momentum “barrier” th... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
,
look up “Arago spot” on wikipedia.
4π
Phase shifts. As we have seen, scattering can be understood in terms of phase shifts. Now we
describe a simple physical way of seeing this. If V = 0 then a plane wave has u0(r) = sin(kr),
due to the u0(0) = 0 boundary condition. When there is scattering, the phase shift δ0 will b... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
7)
Rearranging we have tan(δ0) = −ka, and in the ka (cid:28) 1 limit this yields
σ
0 =
4π
k2
2
2
sin (δ0) ≈ 4πa ,
which is again the hard sphere result. Similar results hold for larger value of l. Thus the scattering
length can be thought of as an effective size of a scattering target.
36
MIT OpenCourseWare
http://ocw.... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/0c27511c09675d8d385577023328248b_MIT8_06S16_chap2.pdf |
18.725 Algebraic Geometry I
Lecture
9
Lecture 9: Chow’s Lemma, Blowups
Last time we showed that projective varieties are complete. The following result from Wei-Liang Chow gives
a partial converse. Recall that a birational morphism between two varieties is an isomorphism on some pair
of open subsets.
Lemma 1 (Chow’s Le... | https://ocw.mit.edu/courses/18-725-algebraic-geometry-fall-2015/0c3f5291cb00434b36fd840f150e5143_MIT18_725F15_lec09.pdf |
π−1
2 (Vi) = f −1(Ui), since Ui cover X. Consider W = f −1(U ) = φ(U ) as an open subset in f −1(Ui): on W
we have f = pig, so the same holds on f −1(Ui) and the covering property follows.
˜
i
˜
It remains to show that X ∩ Vi → Ui are closed embeddings. Noting that Vi = Y1 × . . .
˜
˜
˜
× Yi−1 × Ui ×
i −→ Ui (cid:44)→ ... | https://ocw.mit.edu/courses/18-725-algebraic-geometry-fall-2015/0c3f5291cb00434b36fd840f150e5143_MIT18_725F15_lec09.pdf |
P − (cid:54)⊆ X because dim(Pn−1
irreducible, then X is the irreducible component of π−1
is called the exceptional locus.
n 1
n 1
˜
˜
˜
˜
)
n
1
Next, observe that An is covered by n affine charts. More explicitly, An
(cid:99)
n). On there, the defining equation becomes x
ordinates (ti
A(cid:99)n =∼ An
P (t1xi, . . . , ti−... | https://ocw.mit.edu/courses/18-725-algebraic-geometry-fall-2015/0c3f5291cb00434b36fd840f150e5143_MIT18_725F15_lec09.pdf |
Remark 1. Blx(X) contains X \ x as an open set, so this generalizes to any variety X.
Proposition 1. Suppose X embeds via two embeddings i1, i
exists some x such that i1(x) = i2(x) = 0, then X1 = X2 for two blowups at x.
˜
˜
2 to A and Am respectively, such that there
n
In particular, this tells us that blowup is an in... | https://ocw.mit.edu/courses/18-725-algebraic-geometry-fall-2015/0c3f5291cb00434b36fd840f150e5143_MIT18_725F15_lec09.pdf |
induction on m (or really, just the exactly same argument applied
several times). Now consider the general case of arbitrary i1, i2. First extend the embedding i2 : X → Am
to a map An → Am by lifting each generator (one can switch to the algebraic side, suppose X = Spec A,
then we get two surjective maps ψ1 : k[x1, . .... | https://ocw.mit.edu/courses/18-725-algebraic-geometry-fall-2015/0c3f5291cb00434b36fd840f150e5143_MIT18_725F15_lec09.pdf |
vich and Kurt Augustus Hirsch. Basic algebraic geometry. Vol. 1. Springer,
1977.
2
(cid:54)
MIT OpenCourseWare
http://ocw.mit.edu
18.725 Algebraic Geometry
Fall 2015
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-725-algebraic-geometry-fall-2015/0c3f5291cb00434b36fd840f150e5143_MIT18_725F15_lec09.pdf |
18.175: Lecture 4
Integration
Scott Sheffield
MIT
18.175 Lecture 4
1 Outline
Integration
Expectation
18.175 Lecture 4
2 Outline
Integration
Expectation
18.175 Lecture 4
3 Recall definitions
� Probability space is triple (Ω, F, P) where Ω is sample
space, F is set of events (the σ-algebra) and P : F → [0,... | https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/0c6228b4a4638d5a34b149ec89bc7d19_MIT18_175S14_Lecture4.pdf |
gue: If you can measure, you can integrate.
In more words: if (Ω, F) is a measure space with a measure µ
with µ(Ω) < ∞) and f : Ω → R is F-measurable, then we
can define
fdµ (for non-negative f , also if both f ∨ 0 and
−f ∧ 0 and have finite integrals...)
Idea: define integral, verify linearity and positivity (a.e... | https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/0c6228b4a4638d5a34b149ec89bc7d19_MIT18_175S14_Lecture4.pdf |
f (x)dx = 1E fdλ.
<
E
18.175 Lecture 4
7 Outline
Integration
Expectation
18.175 Lecture 4
8 Outline
Integration
Expectation
18.175 Lecture 4
9 MIT OpenCourseWare
http://ocw.mit.edu
18.175 Theory of Probability
Spring 2014
For information about citing these materials or our Terms of Use, visit: http://ocw.mi... | https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/0c6228b4a4638d5a34b149ec89bc7d19_MIT18_175S14_Lecture4.pdf |
C/C++ empowerment
What is C?
The C memory machine
Logistics
Goodbye
The Adventures of Malloc and New
Lecture 1: The Abstract Memory Machine
Eunsuk Kang and Jean Yang
MIT CSAIL
January 19, 2010
Eunsuk Kang and Jean Yang
The Adventures of Malloc and New
C/C++ empowerment
What is C?
The C memory machine
Logisti... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/0c7443993e151c95e88811f0c5c5bbc5_MIT6_088IAP10_lec01.pdf |
machine.”
5. How to get started with C.
6. Wrap-up and homework.
Eunsuk Kang and Jean Yang
The Adventures of Malloc and New
C/C++ empowerment
What is C?
The C memory machine
Logistics
Goodbye
6.088: a language (rather than programming) course
Images of Wonder Woman and circuit boards removed due to copyright... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/0c7443993e151c95e88811f0c5c5bbc5_MIT6_088IAP10_lec01.pdf |
C/C++ empowerment
What is C?
The C memory machine
Logistics
Goodbye
Administrivia
Homework
• Daily homework to be submitted via the Stellar site.
• Graded �+, �, or �−.
• Homework i will be due 11:59 PM the day after Lecture i;
late submissions up to one day (with deductions).
• Solutions will be released on... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/0c7443993e151c95e88811f0c5c5bbc5_MIT6_088IAP10_lec01.pdf |
OO programming and memory
management.
• Developed at Microsoft, release circa 2001.
Eunsuk Kang and Jean Yang
The Adventures of Malloc and New
C/C++ empowerment
What is C?
The C memory machine
Logistics
Goodbye
Vocabulary check
• Imperative, declarative, functional
• Compiled, interpreted
• Static, dynamic ... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/0c7443993e151c95e88811f0c5c5bbc5_MIT6_088IAP10_lec01.pdf |
Goodbye
It’s a memory world
Controller
ALU
Control/Status
IR
PC
Registers
I/O
Memory
Figure: Processors read from memory, do things, and write to memory.
Figure by MIT OpenCourseWare.
Eunsuk Kang and Jean Yang
The Adventures of Malloc and New
C/C++ empowerment
What is C?
The C memory machine
Logistics
Goodbye ... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/0c7443993e151c95e88811f0c5c5bbc5_MIT6_088IAP10_lec01.pdf |
can access any memory below the current top of
the stack.
Eunsuk Kang and Jean Yang
The Adventures of Malloc and New
C/C++ empowerment
What is C?
The C memory machine
Logistics
Goodbye
Memory layout: process context
High
Stack
Heap
Bss
Data
Text
0
Uninitialized variables
Initialized variables
Instruction
Figu... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/0c7443993e151c95e88811f0c5c5bbc5_MIT6_088IAP10_lec01.pdf |
s . ∗/
/∗ Arguments
v o i d b a r ( i n t
a r e
x ) {
l o c a l l y
s c o p e d . ∗/
x = 3 ;
}
Eunsuk Kang and Jean Yang
The Adventures of Malloc and New
C/C++ empowerment
What is C?
The C memory machine
Logistics
Goodbye
Conditionals
i n t
f o o ( i n t x ) {
/∗ C h a s t h e u s u a l b o o l e a n ... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/0c7443993e151c95e88811f0c5c5bbc5_MIT6_088IAP10_lec01.pdf |
c v < 1 0 ) {
p r i n t f ( ”%d\n” ,
++l c v ;
l c v ) ;
}
}
Eunsuk Kang and Jean Yang
The Adventures of Malloc and New
C/C++ empowerment
What is C?
The C memory machine
Logistics
Goodbye
When can we call what?
Each function needs to be declared (but not necessarily defined)
before we call it.
/∗ D e c l a... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/0c7443993e151c95e88811f0c5c5bbc5_MIT6_088IAP10_lec01.pdf |
defined somewhere.”
• Function definitions typically go in .c files.
• Angle brackets indicate library header files; quotes indicate
local header files.
#i n c l u d e < s t d i o . h> /∗ L i b r a r y
#i n c l u d e ” m y l i b . h”
/∗ L o c a l
f i l e . ∗/
f i l e . ∗/
• The compiler’s -I flag indicates where to ... | https://ocw.mit.edu/courses/6-088-introduction-to-c-memory-management-and-c-object-oriented-programming-january-iap-2010/0c7443993e151c95e88811f0c5c5bbc5_MIT6_088IAP10_lec01.pdf |
15. Basic Properties of Rings
We first prove some standard results about rings.
Lemma 15.1. Let R be a ring and let a and b be elements of R.
Then
(1) a0 = 0a = 0.
(2) a(−b) = (−a)b = −(ab).
Proof. Let x = a0. We have
x = a0
= a(0 + 0)
= a0 + a0
= x + x.
Adding −x to both sides, we get x = 0, which is (1). ... | https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/0c771e6ff658b800f22aa90016880028_MIT18_703S13_pra_l_15.pdf |
+ a + b + b.
1
MIT OCW: 18.703 Modern AlgebraProf. James McKernan Cancelling an a on the left and a b on the right, we get
b + a = a + b,
which is what we want.
Note the following identity.
D
Lemma 15.3. Let R be a ring and let a and b be any two elements of
R.
Then
(a + b)2 = a 2 + ab + ba + b2 .
Proof. Eas... | https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/0c771e6ff658b800f22aa90016880028_MIT18_703S13_pra_l_15.pdf |
if R − {0} is a group under multiplication. If in addition R is commu
tative, we say that R is a field.
Note that a ring is a division ring iff every non-zero element has a
multiplicative inverse. Similarly for commutative rings and fields.
Example 15.7. The following tower of subsets
Q ⊂ R ⊂ C
is in fact a tower of... | https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/0c771e6ff658b800f22aa90016880028_MIT18_703S13_pra_l_15.pdf |
is a zero-divisor if there is an element b ∈ R, b = 0, such that, either,
ab = 0
or
ba = 0.
If a has a multiplicative inverse in R then a is not a zero divisor.
Proof. Suppose that ba = 0 and that c is the multiplicative inverse of
a. We compute bac, in two different ways.
On the other hand
bac = (ba)c
= 0c
= ... | https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/0c771e6ff658b800f22aa90016880028_MIT18_703S13_pra_l_15.pdf |
2 are disjoint, both non-empty and that their
union is the whole of X). Define f : X −→ R, by
and g : X −→ R, by
f (x) =
0 x ∈ X1
1 x ∈ X2,
g(x) =
1 x ∈ X1
0 x ∈ X2.
Then f g = 0, but neither f not g is zero. Thus f is a zero-divisor.
Now let R be any ring, and suppose that n > 1. I claim that Mn(R)
is not a dom... | https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/0c771e6ff658b800f22aa90016880028_MIT18_703S13_pra_l_15.pdf |
MEASURE AND INTEGRATION: LECTURE 15
Lp spaces. Let 0 < p < ∞ and let f : X
function. We define the Lp norm to be
→
C be a measurable
�f �p =
and the space Lp to be
��
X
�1/p
|f |p dµ
,
Lp(µ) = {f : X → C f is measurable and �f �p
|
< ∞}.
Observe that �f �p = 0 if and only if f = 0 a.e. Thus, if we make
the... | https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/0c85400465928cf42a19e5b87bdc349c_18125_lec15.pdf |
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