University of California at San Diego – Department of Physics – TA: Shauna Kravec
Quantum Mechanics C (Physics 212A) Fall 2015
Worksheet 4 –
Solutions
Announcements
• The 212A web site is:
http://physics.ucsd.edu/∼mcgreevy/f15/ .
Please check it regularly! It contains relevant course information!
Problems
1. Talking ’bout My Generation Let’s think about translations.
(a) Define the operator T (x) by the following:
T (x)|yi = |y + xi
(1)
Show that T (x) is unitary by considering the action of T −1 (x) = T (−x)
R
R
R
T −1 (x) = T (−x) = dy T (−x)|yihy| = dy |y − xihy| = dy |yihy + x|
R
R
dy |yihy + x| = dy |yihy|T † (x) = T † (x)
(b) Given the above we can express 1 by T (x) = e−iKx where K is hermitian.
Let |ki be an eigenbasis of K; K|ki = k|ki. What is the action of T (x)|ki?
By Taylor expansion: T (x)|ki = e−iKx |ki = e−ikx |ki
Denote hy|ki = φk (y). Note that by unitarity:
hy|T (x)|ki = hy|T † (−x)|ki
(2)
What does 2 imply about φk (y)?
LHS: hy|T (x)|ki = e−ikx φk (y) RHS: hy|T † (−x)|ki = hx + y|ki = φk (x + y)
Therefore φk (x + y) = e−ikx φk (y)
φk is a plane wave if this is a condition satisfied for all x.
(c) Recall plane waves have momentum related to their de Broglie wavelength: p = ~k
Rewrite T (x) with the implied expression for the operator P . This is why we say
momentum generates translations!
P
T (x) = e−i ~ x
∂
We can use a clever trick to get P on its own: ∂x
T (x) = − ~i P T (x)
1
Write an expression for hx0 |P |xi Hint: Derivative of a delta function
∂
hx0 |P |xi = hx0 |P T (x)|0i = hx0 |i~ ∂x
T (x)|0i = i~∂x hx0 |xi = i~∂x δ(x − x0 )
This is valid as x in the derivative and T (x) is just a parameter.
Recall that by some integration by parts:
Z
Z
0
dx δ (x − y)f (x) = 0 − dx δ(x − y)f 0 (x) = −f 0 (y)
(3)
For f which vanishes at ±∞
Use (3) to derive the familiar expression for P by considering:
P ψ(x) = hx|P |ψi
(4)
R
R
P ψ(x) = hx|P |ψi = dx0 hx|P |x0 ihx0 |ψi = dx0 i~δ 0 (x0 − x)ψ(x0 )
R
∂
ψ(x)
Now using IBP: = −i~ dx0 δ(x0 − x) ∂x∂ 0 ψ(x0 ) = −i~ ∂x
2. Building Bloch’s Theorem
Consider a 1D Hamiltonian with a periodic potential V (x) = V (x + na) for n ∈ Z and
a the lattice spacing.
(a) Define the operator T n by T n |xi = |x + nai. Show this is a symmetry.
We would need that [H, T ] = 0 or equivalently H 0 = T † HT = H. Computing
H 0 ψ(x) = hx|H 0 |ψi. Writing H 0 in the position basis: H 0 = − 12 ∂x20 + V (x0 ).
0
V (x0 ) = V (x + a) = V (x) by assumption. One computes the Jacobian: ∂x
=
∂x
∂x+a
∂x
0
= ∂x = 1 to see the kinetic part is also unchanged. Thus H is invariant and
∂x
the result is shown!
(b) Assuming H has no shared degeneracy with T , show that any eigenfunctions of
this system can be chosen to obey
ψk (x − a) = e−ika ψk (x)
(5)
Recall that T |ki = e−ika |ki and hx|ki ≡ ψk (x).
Since T is a symmetry, and there is no degeneracy, I can diagonalize H in the
basis of T : |ki. The above implies that T |ki = e−ika |ki and that H|ki = Ek |ki.
If there were degeneracy then not every eigenvector of H need be an eigenvector
of T . This is called spontaneous symmetry breaking
The proof then is very easy: hx − a|ki = hx|T |ki = e−ika ψk (x)
(c) Infer from (5) that one can then write ψk (x) = eikx uk (x) where uk (x) = uk (x + a)
This amounts to showing the claimed form for ψk satisfies (5)
ψk (x − a) = eik(x−a) uk (x − a) = e−ika eikx uk (x) done.
Note that because not every a is a valid symmetry transformation the function
uk is not generically a constant.
Note that k is different from our usual momentum. It’s a crystal momentum!
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(d) Show explicitly that for P = −i∂x that P ψk (x) 6= kψk (x)
−i∂x (ψk (x)) = keikx uk (x) − ieikx u0k (x)
(e) Show that
−π
a
≤ k ≤ πa . What is k +
2π
?
a
2π
2π
Consider ψk+ 2π (x) = eikx ei a x uk+ 2π (x) and notice that the product ei a x uk+ 2π (x)
a
a
a
is still a periodic function of x with period a.
2π
Therefore this is just a relabelling of the function uk . Call vk (x) ≡ ei a x uk+ 2π (x)
a
Then ψk+ 2π = eikx vk (x) ; this transformation on k did nothing! Therefore k+ 2π
≡
a
a
k and that’s why we only need the finite region.
3. A Theorem of Kramer
Most symmetries are unitary. Some are anti -unitary. Time reversal is one of the latter.
Denote this operator with T .
Something one might expect classically is that T xT −1 = x but T pT −1 = −p. It makes
things run backwards.
A similar story is true for angular and spin momentum. T ST −1 = −S
(a) Consider the action of T on a spin- 21 : T |0i where Sz |0i = 12 |0i.
Sz T |0i = −T Sz |0i = − 12 T |0i
This implies T |0i is also an eigenstate of Sz with eigenvalue − 21 . Thus T |0i = |1i
Show that T = −iY K is a suitable representation
for T where K implements
0 −i
complex conjugation 1 and Y is the Pauli matrix
.
i 0
This follows from Y |0i = i|1i and Y |1i = −i|0i
What is T 2 in this case? T 2 = (−iY K)2 = (−i)2 1 = −1
(b) Consider a system whose Hamiltonian H is time reversal symmetric. Show that
if |ψi is an eigenstate then T |ψi is as well.
Does this change the energy of the state?
If HT |ψi = T H|ψi = T E|ψi which since E is a real number it’s unchanged; the
states are degenerate in energy!
(c) Imagine this is a spin- 21 system such that |ψi is an eigenstate of Sz as well.
How are |ψi and T |ψi related? Can they be the same?
If it’s a spin- 21 system then T |ψi is the other eigenvector of Sz which is orthogonal
to |ψi. Therefore in a time-reversal symmetric system with a net spin- 12 there
must be two degenerate, in energy, states! In other spin systems it’s possible that
T |ψi = |ψi
1
This is what makes it anti-unitary.
3