Quantum Mechanics (Physics 212A) Fall 2015 Assignment 4 – Solutions

```University of California at San Diego – Department of Physics – Prof. John McGreevy TA:
Shauna Kravec
Quantum Mechanics (Physics 212A) Fall 2015
Assignment 4 –
Solutions
Due 12:30pm Wednesday, October 28, 2015
1. Minimum uncertainty state. [from Commins] The Heisenberg uncertainty relation
for a coordinate and its conjugate momentum
∆x∆p ≥
~
2
is derived from the commutation relation [x, p] = i~ with the aid of the CauchySchwarz inequality. From consideration of the conditions that must be satisfied for that
inequality to become an equality, show that the minimum uncertainty wavefunction
must be a Gaussian.
Consider A0 ≡ A − hAiψ and B0 ≡ B − hBiψ so that hA20 i = hA2 i − hAi2 = ∆A
CS implies hA20 ihB02 i ≥ |hA0 B0 i|2 where hA0 B0 i = 21 h[A, B]i + 12 h{A0 , B0 }i
Dropping the anti-commutator piece gives (∆A)2 (∆B)2 ≥ 41 |h[A, B]i|2 which produces
the above using [x, p] = i~
CS becomes an equality when the vectors are colinear. So we’d demand A0 |ψi ∝ B0 |ψi
Additionally, to minimize this inequality, the anti-commutator term should vanish.
Assume for argument hxi = 0 = hpi and plug in these operators above. Linearity
demands: (p̂ − λx̂)|ψi = 0 or in position space (−i∂x − λx)ψ(x) = 0
2
One can check ψ(x) = Ae−αx for α &gt; 0 solves for λ = 2iα. The anti-commutator
further constrains α
Relaxing the assumption simply shifts the Gaussian by hxi and adds a phase
2. Time-energy uncertainty. Consider an observable A which has no explicit dependence on time. Define the variance of A in a state ψ to be
q
∆A ≡ h(A − hAiψ )2 iψ
as usual, with hAiψ ≡ hψ|A|ψi. Let
∆T ≡
1
∆A
.
|∂t hAiψ |
This is a measure of the time required for hAi to change significantly, i.e. by an amount
comparable to its variance ∆A. Show that
~
,
2
where ∆E is the variance of H, the Hamiltonian.
∆E∆T ≥
Let H0 ≡ H − hHiψ ≡ H − E and A0 ≡ A − hAiψ . Apply Cauchy-Schwarz to
|αi ≡ A0 |ψi and |βi ≡ H0 |ψi. This gives
1
1
(∆A)2 (∆E)2 ≥ |hψ|[A, H]|ψi|2 = |i~∂t hAiψ |2 .
4
4
where in the last step we’ve used the Heisenberg equation of motion.
3. Wigner distribution function. [from Commins]
Let ψ(x, t) = hx|ψ(t)i be a position-space wavefunction for a particle in one dimension,
and let φ(p, t) = hp|ψ(t)i be its momentum-space counterpart. Define the associated
Wigner (or phase-space) distribution function by
Z
1
dye2ipy ψ ? (x + y, t)ψ(x − y, t).
W (x, p, t) ≡
π
The arguments of W , x and p are c-numbers. (The dependence on time will not play
an important role in our discussion.)
(a) Show that the probability density in position space is
Z
2
ρ(x, t) ≡ |ψ(x, t)| = dpW (x, p, t).
R
dp W = π1 dpdy e2ipy ψ ? (x + y, t)ψ(x − y, t)
R
Doing the p-integral produces dp e2ipy = 2πδ(2y) = πδ(y) and then doing the
y-integral yields: ψ ∗ (x, t)ψ(x, t) = ρ(x, t)
R
2
(b) Evaluate the Wigner function for a gaussian wavefunction, ψ(x) =
R∞
W =
−∞
−
e2ipy e
(x+y)2
2σ 2
−
e
π(2πσ 2 )
(x−y)2
2σ 2
dy
q
=
1 −
e
σ2
x
√1 e− 2σ2 .
σ 2π
p2 σ 4 +x2
σ2
2π 3/2
(c) Show that the probability density in momentum space is
Z
2
|φ(p, t)| = dxW (x, p, t).
For this and the next problem I’m going to just drop normalization factors and
write the RHS without them. I’m going to suppress time dependency as well.
R
R
Insert 1 = dp1 |p1 ihp1 | = dp2 |p2 ihp2 | to get:
R
dxdydp1 dp2 e2ipy eip1 (x−y) e−ip2 (x+y) φ∗ (p2 )φ(p1 )
Let’s look at the exponential factors: e2ipy eip1 (x−y) e−ip2 (x+y) = e2ipy ei(p1 −p2 )x e−i(p1 +p2 )y
Do the x-integral to produce δ(p1 −p2 ). Then do the p2 integral to set p1 = p2 ≡ p0
R
0
dydp0 e2i(p−p )y φ∗ (p0 , t)φ(p0 , t) . Do the y-integral to produce δ(p − p0 ) and then
finally the p0 -integral to set p0 = p and finally reach |φ(p, t)|2
2
(d) Let W1 (x, p, t), W2 (x, p, t) be the Wigner functions associated with two states
|ψ1 (t)i, |ψ2 (t)i. Show that their overlap-squared is
Z
2
|hψ1 |ψ2 i| = 2π dxdp W1? (x, p, t)W2 (x, p, t).
dxdpdy1 dy2 e2ip(y2 −y1 ) ψ1∗ (x − y1 )ψ1 (x + y1 )ψ2∗ (x + y2 )ψ2 (x − y2 )
Do the p-integral to produce δ(y2 − y1 ) and then the y2 integal to set y1 = y2 ≡ y
R
→ dxdy ψ1∗ (x − y, t)ψ1 (x + y)ψ2∗ (x + y)ψ2 (x − y)
R
= dxdy hψ2 |x + yihx + y|ψ1 ihψ1 |x − yihx − y|ψ2 i
Change coordinates to z1 ≡ x − y and z2 ≡ x + Ry ; it’s easy to see
R the Jacobian
factor is constant. Then simply remove two 1 = dz1 |z1 ihz1 | = dz2 |z2 ihz2 |
R
4. [optional] Can a general operator acting on the Hilbert space of a particle on a line be
written formally in terms of the position and wavenumber (k = −i∂) operators as
X
A=
anm kn xm ?
nm
If so, find the condition on anm for A to be hermitian.
The above form can define the (formal) Laurent series for any function of x, k, or their
products. That’s a whole lot of things! Even δ(x) has such a representation.
P
To simplify life let’s redefine A0 = nm anm kn xm +bnm xm kn which should be equivalent
to A expression above up to redefinitions of constants due to the commutator [xm , kn ]
P
A0† = nm a∗nm xm kn + b∗nm kn xm so then if b∗nm = anm and a∗nm = bnm for all n, m we
have a hermitian operator.
5. [optional] Is the uncertainty principle still true for mixed states? If so, prove it. I
suggest the form
? 1
∆ρ A∆ρ B ≥ |h[A, B]iρ
2
with
q
hAiρ ≡ tr ρA, ∆ρ A ≡ (A − hAiρ )2 .
See: here . section 2.3.2
The following three problems form a triptych, on the subject of resolving the various infinities
involved in the quantum mechanics of a particle on the real line. There are two such infinities:
one is the fact that the real line goes on forever; this is resolved in problem 6. The other is
the fact that in between any two points there are infinitely many points; this is resolved in
problem 7. In problem 8 we resolve both to get a finite-dimensional Hilbert space.
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6. Particle on a circle.
Consider a particle which lives on a circle:
That is, its coordinate x takes values in [0, 2πR] and we identify x ' x + 2πR.
(a) Let’s assume that the wavefunction of the particle is periodic in x:
ψ(x + 2πR) = ψ(x) .
What set of values can its momentum (that is, eigenvalues of the operator p =
−i~∂x ) take?
hx + 2πR|ψi = hx|ψi
R dp
R dp
hx + 2πR|pihp|ψi = 2π
hx|pihp|ψi
2π
R dp i(x+2πR)p
R dp ixp
e
hp|ψi = 2π e hp|ψi
2π
For this to be true e2πiRp = 1 thus quantizing p = Rn for n ∈ Z
To emphasize: x ∈ S 1 =⇒ p ∈ Z
(b) Recall that the overall phase of the state vector is not physical data. This suggests
the possibility that the wavefunction might not be periodic, but instead might
acquire a phase when we go around the circle:
ψ(x + 2πR) = eiϕ ψ(x)
for some fixed ϕ. In this case what values does the momentum take?
The same logic of the above holds only now e2πiRp = eiφ implying p =
n
R
+
φ
2πR
7. Particle on a lattice.
Now consider a particle which lives on a lattice: its position can take only the discrete
values x = na, n ∈ Z where a is some unit of length and n is an integer. We’ll call the
corresponding position eigenstates |ni. The Hilbert space is still infinite-dimensional,
but at least we have in our hands a countably infinite basis.
In this problem we will determine: what is the spectrum of the momentum operator
p in this system?
(a) Consider the state
1 X inθ
|θi = √
e |ni.
N n∈Z
4
Show that |θi is an eigenstate of the translation operator T̂ , defined by
X
T̂ =
|n + 1ihn|.
n∈Z
Why do I want to call θ momentum?
P
T |θi = n∈Z einθ |n + 1i = e−iθ |θi. The values of n shift along Z.
Recall that T = e−ip̂a so e−ipa |θi = e−iθ |θi implying θ = pa.
(b) What range of values of θ give different states |θi? [Recall that n is an integer.]
Since n is an integer |θi = |θ + 2πi. We’ve found that for x ∈ Z =⇒ p ∈ S 1
(this circle is called the Brillioun zone)!
8. Discrete Laplacian.
Consider again a particle which lives on a lattice, but now we’ll wrap the lattice
around a circle, in the following sense. Its position can take only the discrete values x = a, 2a, 3a, ..., N a (where, again, a is some unit of length and again we’ll call
the corresponding position eigenstates |ni). Suppose further that the particle lives on
a circle, so that the site labelled x = (N + 1)a is the same as the site labelled x = a.
We can visualize this as in the figure:
In this case, the Hilbert space has finite dimension N .
Consider the following N &times; N matrix representation of a Hamiltonian operator (a is a
5
constant):

 2 −1 0
0
0 &middot;&middot;&middot;

 −1 2 −1 0
0 &middot;&middot;&middot;

 0 −1 2 −1 0 &middot; &middot; &middot;
1 

0 −1 2 −1 &middot; &middot; &middot;
H= 2 0
a  ..
..
..
..
.. . .
 .
.
.
.
.
.

 0
0
0
0
0 &middot;&middot;&middot;

 −1 0
0
0
0 &middot;&middot;&middot;
|
{z
N
 


 

0 −1

 


0
0





0
0
 
0
0 N


..
.. 


.
. 




 
2 −1




−1 2 


}
(a) Convince yourself that this is equivalent to the following: Acting on an N dimensional Hilbert space with orthonormal basis {|ni, n = 1, . . . , N }, Ĥ acts
by
a2 Ĥ|ni = 2|ni − |n + 1i − |n − 1i, with |N + 1i ' |1i
that is, we consider the arguments of the ket to be integers modulo N .
I will set a = 1 until needed. Recall that Hnm ≡ hn|H|mi. Our claim above
is compatible with Hnn = 2 by orthogonality as well as off diagonals Hn+1,n =
Hn−1,n = −1.
The top right and left corners are compatible by: H0,N = 2h0|N i − h0|N + 1i −
h0|N − 1i = −h0|N + 1 = 0i = −1 making use of the periodicity. The rest are
appropriately 0.
(b) Show that Ĥ and T̂ (where T̂ is the ‘shift operator’ defined by T̂ : |ni 7→ |n + 1i)
can be simultaneously diagonalized.
HT |ni = H|n + 1i = 2|n + 1i − |n + 2i − |ni = T (2|ni − |n + 1i − |n − 1i) = T H|ni
so there is a discrete translation invariance.
Consider again the state
N
1 X inθ
e |ni.
|θi = √
N n=1
(c) Show that |θi is an eigenstate of T̂ , for values of θ that are consistent with the
periodicity n ' n + N .
√
Consider N T|θi = T|0i + Teiθ |1i + &middot; &middot; &middot; + Tei(N −1)θ |N − 1i
= |1i + eiθ |2i + &middot; &middot; &middot; + ei(N −1)θ |0i = e−iθ |θi where θ must be such that eiN θ = 1
The most general solution to eiN θ = 1 is for θ = 2πk
for k ∈ {0, &middot; &middot; &middot; , N − 1}
N
P k∗n
k
This defines a basis of |ω i ≡ n ω |ni where k runs from 0 to N − 1.
(d) What values of θ give different states |θi? [Recall that n is an integer.]
for k ∈ {0, 1, &middot; &middot; &middot; , N − 1}
Specifically θ = 2πk
N
Recalling the relationship between p and θ we arrive at the punchline that for
x ∈ ZN =⇒ p ∈ ZN
6
(e) Find the matrix elements of the unitary operator U which relates position eigenstates |ni to momentum eigenstates |θi: Uθn ≡ hn|θi.
P
0
hn|θi = √1N n0 ein θ hn|n0 i = √1N einθ by orthogonality.
(f) Find the spectrum of Ĥ.
Draw a picture of (θ): plot the energy eigenvalues versus the ‘momentum’ θ.
Because [H, T ] = 0 we can diagonalize them in the same basis.
P
H|θi = √1N n einθ (2|ni − |n + 1i − |n − 1i) = 2|θi − e−iθ |θi − eiθ |θi
Recall that cos θ = 21 (eiθ + e−iθ ) and so consolidating terms: (θ) = 2 + 2 cos θ
(g) Show that the matrix above is an approximation to (minus) the 1-dimensional
Laplacian −∂x2 . That is, show (using Taylor’s theorem) that
a2 ∂x2 f (x) = −2f (x) + (f (x + a) + f (x − a)) + O(a)
(where “O(a)” denotes terms proportional to the small quantity a).
Taylor Expansion: f (x + a) = f (x) + af 0 (x) + 21 a2 f 00 (x) + &middot; &middot; &middot; so we can write that
(x−a)
f 0 (x) = f (x+a)−f
2a
f (x + a) = f (x) + 12 (f (x + a) − f (x − a)) + 21 a2 f 00 (x)
a2 f 00 (x) = f (x + a) + f (x − a) − 2f (x)
Thus our form of H acting on |ni approximates a finite step differentiation.
(h) In the expression for the Hamiltonian, to restore units, I should have written:
~2 1
Ĥ|ni =
(2|ni − |n + 1i − |n − 1i) , with |N + 1i ' |1i
2m a2
where a is the distance between the sites, and m is the mass. Consider the limit
where a → 0, N → ∞ and look at the lowest-energy states (near p = 0); show
p2
that we get the spectrum of a free particle on the line, = 2m
.
Based on the above we have that in the continuum limit H = −∂x2 and thus with
p̂2
the inclusion of the appropriate factors becomes the kinetic energy operator 2m
in position space.
Since H has no potential terms this is the only contribution to the total energy.
The highest momentum states would be the one’s associated with largest values
of θ and to whom the details of our regularization would matter most.
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