Quantum Mechanics II

Prof. Boris Altshuler

January 26, 2011

1 Lecture 4

1.1 Time Reversal Continued

Remember last time we were discussing the time reversal symmetry, i.e. the symmetry operation whichreverses the direction of time

t −→ −t, ψ(x, t) −→ ψ∗(x,−t) (1)

We introduced the operator Θ to invert the time and change a state into a time-reversed state

|α〉 −→ Θ |α〉 = |α〉 (2)

Recall that the time-reversal operator differs from usual symmetry operators in a subtle way. Usuallya symmetry operator is a unitary operator

S†S = 1 (3)

which is necessary to preserve inner products between states. However we showed last time that undertime reversal we have instead

〈α|β〉 −→ 〈α|Θ†Θ|β〉 = 〈β|α〉 (4)

An operator satisfying this relation for any states |α〉 and |β〉 is called an antiunitary operator. The Θoperator is an example of an antiunitary operator. It has the property that

Θ (c1 |α1〉+ c2 |α2〉) = c∗1Θ |α1〉+ c∗2Θ |α2〉 (5)

We also call this property antilinearity.We showed last time that {H,Θ} = 0. But that was not quite correct. Now that taking into the effect

of antilinearity, we should write− iHΘ = ΘiH (6)

But because of antilinearity we can’t cancel the i’s from both sides and conclude the anticommutator ofthe two operators is zero. Instead we should take the i on the right hand side and put it in front of the Θoperator with a minus sign and get

[Θ,H] = 0 (7)

Now that we have this relation, we can study the time evolution operator which is just the exponentialof the Hamiltonian

|α, t〉 = U(t) |α〉 , |α, t〉 = U(t)Θ |α〉 = ΘU(−t) |α〉 (8)

1

However the wave function will become

ψα(x, t) = 〈x|α, t〉 = 〈x|ΘU(−t)|α〉 = 〈α,−t|x〉 = ψ∗α(x,−t) (9)

Consider arbitrary linear operator O. Any matrix element will have the following

〈β|O|α〉 = 〈γ|α〉 = 〈α|γ〉 = 〈α|ΘO†|β〉 = 〈α|ΘO†Θ†|β〉 (10)

So for a Hermition operator the O† in the braket will become just O. Let’s define for any operator

A

{is even if ΘAΘ† = A

is odd if ΘAΘ† = −A(11)

And any operator can be written as a linear combination of even and odd operators.Let’s consider some examples. First consider the momentum operator p. We have

〈p〉α = 〈α|p|α〉 = −〈p〉α (12)

So p is an odd operator. The eigenstates of the momentum operator transform as

p |p1〉 = p1 |p1〉 =⇒ pΘ |p1〉 = −ΘpΘ†Θ |p1〉 = −p1Θ |p1〉 (13)

So we have |p1〉 = |−p1〉. In contrast we know the position operator is invariant under time reversal, so xis even and |x〉 = |x〉. Now for a consistency check we can work out the transformation for the commutator

Θ[xi, pj ]Θ† = −[xi, pj ] = −i~δij (14)

The minus sign comes from the fact that p is odd. But this also makes sense because the result is purelyimaginary.

Now let’s consider angular momentum. Because of the definition L = x×p we expect L to be odd. Thisgeneralizes to all angular momenta. It can be also seen from the commutation relation between angularmomenta

[Ji, Jj ] = i~εijkJk (15)

So we should expect that J transforms just as i does, which changes sign under time reversal. For angularwave functions we have the spherical harmonics Y m

l . Because Y ml = eimϕPml , and that Pml = (−1)mP−ml ,

so we expect(Y ml )∗ = Y −ml (−1)m (16)

Now what about spin? Because S is odd, if B were even then B · S is odd, but this is the energy ofa spin coupled to a magnetic field, so we have a contradiction. So we expect B to be odd. We also havethis result in classical physics, as discussed in the last lecture. Now let’s consider the operator S · n. It haseigenstates

|n,+〉 = e−iSzϕ/~e−iSyθ/~ |z,+〉 , |n,−〉 = e−iSzϕ/~e−iSy(θ+π)/~ |z,+〉 (17)

Because we know S is odd, we expect that

Θ |n,+〉 = eiη |n,−〉 (18)

2

where η is an arbitrary phase factor. In light of the above, we can write the time reversal operator as

Θ = eiηe−iSyπ/~K (19)

where K is the complex conjugation operator. Now let’s define this operator. It must be antilinear soK(c |α〉) = c∗K |α〉. If we expand the state in |βi〉 basis, then we will get

K |α〉 = K∑i

〈bi|α〉 |bi〉 =∑i

〈α|bi〉K |bi〉 (20)

We choose a basis that K |bi〉 = |bi〉, then we can define the complex conjugation operator using thisexpression

K |α〉 =∑i

〈α|bi〉 |bi〉 (21)

Note that this definition is basis-dependent. But note that our definition of the time reversal operatoris also basis-dependent, so the dependency actually cancels and there is no problem. In general we canalways write an antiunitary operator into the product of a unitary operator with K:

A = UK (22)

Remember thate−iSyπ/~ = −iσy (23)

So the above representation for the time-reversal operator is actually

Θ = eiη (−iσy)K (24)

This is the correct representation of time reversal operator in the spin-1/2 basis. Now it is interesting toevaluate the square of the time reversal operator

Θ2 =(−ieiησyK

) (−ieiησyK

)= σyKσyK = −σyσyK2 = −1 (25)

In the second last equality we have used the fact that σyK = −Kσy, because σy is a purely imaginarymatrix. Note that this says that for spin 1/2 particle if we reverse the time twice, we don’t come back to thesame state, but pick up a minus sign. This is a profound and counter-intuitive result. The generalizationof this result is that for arbitrary j the action of Θ2 on the eigenstate will be

Θ2 = (−1)2j (26)

This can be seen if we consider the spin as consisting of 2j spin-1/2 particles.There is the so-called Kramers degeneracy which states the follows. Assume that H commutes with

Θ, which means that the system is time-reversal invariant. Let |n〉 be an eigenstate of the Hamiltonian,then Θ |n〉 is an eigenstate with the same energy. This is the analogue of the similar result in classicalmechanics where time-reversed trajectory is also a solution of the equation of motion. Now the questionis: can Θ |n〉 = eiδ |n〉? The answer is no. Assume this statement is true, then we should have

Θ2 |n〉 = Θ(eiδ |n〉

)= e−iδΘ |n〉 = |n〉 (27)

3

But we know this is not true for spin-1/2 particles. So this is a contradiction. For spin-1/2 particles thesetwo states are different states, so there is a 2-fold degeneracy.

Let’s consider the spin-orbit coupling

HSO = λSOL · S (28)

Because both angular momenta are odd, the Hamiltonian is even and symmetric under time-reversal.When there is no spin-orbit coupling, we know that the hydrogen atom has 2s + 1 = 2 degeneracy, butwith spin-orbit coupling there is still 2-fold degeneracy because of the Kramer’s degeneracy.

4