University of California at San Diego – Department of Physics – Prof. John McGreevy TA:
Shauna Kravec
Quantum Mechanics (Physics 212A) Fall 2015
Assignment 6 –
Solutions
Due 12:30pm Wednesday, November 11, 2015
1. Coherent states.
Consider a quantum harmonic oscillator with frequency ω. The creation and annihilation operators a† and a satisfy the algebra
[a, a† ] = 1
and the vacuum state |0i satisfies a|0i = 0. Coherent states are eigenstates of the
annihilation operator:
a|αi = α|αi.
(a) Show that
−|α|2 /2 αa†
|αi = e
e
−|α|2 /2
|0i = e
∞
X
αn
√ |ni
n!
n=0
is an eigenstate of a with eigenvalue α. (a is not hermitian, so its eigenvalues
need not be real.)
P
P∞ α n √
√
2
αn
−|α|2 /2
√
√
â|ni = n|n − 1i → â|αi = e−|α| /2 ∞
n=0 n! â|ni = e
n=1 n! n|n − 1i
Where we’ve used the fact â annihilates the vacuum. Reshuffling the summand:
P
n+1
2
α√
â|ni = e−|α| /2 ∞
|ni = α|αi
n=0
n!
(b) Coherent states with different α are not orthogonal. (a is not hermitian, so its
2
eigenstates need not be orthogonal.) Show that |hα1 |α2 i|2 = e−|α1 −α2 | .
n
P
P∞ α∗n
P∞ α∗n
2
2
2
2
αm
1 α2
√1 √ 2 hn|mi = e−|α1 | /2 e−|α2 | /2
hα1 |α2 i = e−|α1 | /2 e−|α2 | /2 ∞
n=0
m=0 n! m!
n=0
n!
Where in the last step we have used the orthogonality of {|ni}. We recognize this
sum as an exponential:
2
2
∗
2
hα1 |α2 i = e−|α1 | /2 e−|α2 | /2 eα1 α2 → |hα1 |α2 i|2 = e−|α1 −α2 |
(c) Compute the expectation value of the number operator n = a† a in the coherent
state |αi.
hα|↠â|αi = |α|2 hα|αi = |α|2
(d) Time evolution acts nicely on coherent states. The hamiltonian is H = ~ω a† a + 21 .
Show that a coherent state evolves into a coherent state with an eigenvalue α(t):
e−iHt |αi = e−iωt/2 |α(t)i
1
where α(t) = e−iωt α.
|α(t)i = e−iĤt |α0 i = e−|α0 |
2 /2
P∞
n=0
1
αn
√ 0 e−iω(n+ 2 )t |ni
n!
ωt
We pull out the ground state contribution: = e−i 2 e−|α0 |
2 /2
P∞
n=0
(α0 e−iωt )n
√
|ni
n!
Which by looking at the definition |α(t)i = |e−iωt α0 i we have shown the result.
(e) Show that the coherent states can be used to resolve the identity in the form
Z 2
dα
1=
|αihα|
π
where d2 α ≡ dα1 dα2 in terms of the real and imaginary
P∞parts of α = α1 + iα2 .
One way to do this is to relate this expression to 1 = n=0 |nihn|.
R d2 α
R 2 −|α|2 m ∗ n
P
|α|2 P
1
αn
√
Recall |αi = e− 2
|αihα| = π1 n,m √n!m!
d αe
α (α ) |mihn|
n n! |ni so that
π
One can evaluate the α integral now. Switch to polar coordinates α = reiθ
R
R
2
2
drdθ e−r rn+m+1 eiθ(m−n) = 2πδmn dre−r rn+m+1 = πδmn Γ(1 + m+n
)
2
R d2 α
P 1
|αihα| = n n! Γ(1 + n)|nihn| = 1
So
π
2. Oscillation of excited oscillator states.
Consider a 1d harmonic oscillator of frequency ω. Consider the initial state
|ψn,s (0)i ≡ T(s)|ni
where |ni ≡
operator.
√1
n!
a†
n
|0i is the nth excited state and T(s) ≡ e−iPs is the displacement
Describe (plot it as a function of q for some n, t, s > 0) the time evolution of the
probability distribution: ρ(q, t) = |ψn,s (q, t)|2 where ψn,s (q, t) ≡ hq|e−iHt |ψn,s (0)i. Does
it keep its shape like it does for n = 0?
√
√
2
s mω
a† −s mω
a
− mωs
−iP s
2
2
4
e
e
|ni for U ≡ e−iHt
|ψ(t)i = U e
|ni = U e
I will use my power of adding/ignoring global phases liberally along with BCH
√ mω †
√ mω †
√ mω
√ mω
|ψ(t)i = U es 2 a U † U e−s 2 a U † U |ni = es 2 a (t) e−s 2 a(t) |ni
Where a† (t) = e−iωt a† using the Heisenberg equation of motion.
√ mω † −iωt iωt
√ mω †
†
|ψ(t)i = es 2 (a e −ae ) |ni = es 2 ((a −a) cos(ωt)−(a+a )i sin(ωt)) |ni = e−is(P cos(ωt)+mωX sin(ωt)) |ni
But we’re interested in |hx|ψi|2 = |hx|e−isP cos(ωt) |ni|2 = |hx − s cos(ωt)|ni|2
So it has the same sloshing for all n
3. Locally linear confining potential.
Determine the spectrum of the hamiltonian
H = p2x + V (x)
with V (x) = a|x|. (You can give your answer in terms of the (not analyticallydeterminable) locations of zeros of a special function.)
2
The potential is symmetric under x → −x so we can focus on eigenstates of parity:
ψ(−x) = ±ψ(x) and the region of x ≥ 0
Notice that even solutions must have ψ 0 (0) = 0 and odd solutions have ψ(0) = 0
(−∂x2 + ax − E)ψ = 0 which behooves a change of variables to u ≡ C(ax − E) for some
constant C which we shall determine.
(−C 2 a2 ∂u2 + C1 u)ψ = 0 so C 3 a2 = 1 makes (∂u2 − u)ψ(u) = 0 which is the Airy equation
ψ(u) = αAi(u) + βBi(u) where β = 0 by normalizability.
To satify the even condition Ai0 (−CE) = 0 and for odd Ai(−CE) = 0 so if you knew
the zeros of the first Airy function (and it’s derivative) that determines the energies.
4. Particle in a uniform force field.
(a) Find the propagator K(x0 , x; t) ≡ hx0 |e−iHt |xi for a particle in a uniform force
field, that is, H = p2 + gx.
To use BCH you need to go out to terms like [x, [p2 , x]]
R
i 2 3
i 2 3
2
2
0
0
2
2
1
e−iHt = e−igxt e−ip t eit gp e− 3 g t so K = 2π
dp ei(x −x)p e−igx t e−ip t eit gp e− 3 g t
p π i (gt2 +(x0 −x))2 −igx0 t − i g2 t3
1
= 2π
e 4t
e
e 3
which after rearrangement
it
q
0
2
1
K(x0 , x; t) = 4πit
exp[ 4i (x −x)
− 2i (x0 + x)gt − 12i g 2 t3 ]
t
(b) Use the result of the first part to find the time evolution of an initial minimaluncertainty wavepacket.
1 x2
The wavefunction of minimal uncertainty is ψ(x, t = 0) = √ 1√ e− 4 σ2
σ 2π
q q
R
i
i 02
i
2
3
0
∞
1
√1 e− 12 g t e− 2 gtx e 4t x I
ψ(x0 , t) = −∞ dx K(x0 , x; t)ψ(x, t = 0) = 4πit
σ 2π
σ 2 2gt2 +x0
(
For I =
R
i
1
i
2
dx e 4 ( t + σ2 )x e
0
)x
− 2i (gt+ x
2t
=
)
2
√
2
2 πe 16t(t−iσ )
q
−i( 1t + i2 )
σ
After combining and throwing away global phases that didn’t look nice.
2 2
x0
1 ( 2 +gt )
p σ
t(σ 2 +it)
− 41 − 4 f (σ,t)
0
with f (σ, t) = σ2
ψ(x , t) = σ2 +it (2π) e
(c) Show that in the accelerating frame in which there is no field, the resulting
wavepacket does not accelerate, but does disperse.
0
Changing frame makes the center x2 + gt2 become stationary but the variance
doesn’t depend on x0 and still disperses in time.
The first two steps of this problem can be done using Gaussian integrals. The first
step will require the BCH formula.
3