Appendix E — Time reversal
Although it isn’t an approximation method in and of itself, time reversal symmetry can be a powerful tool in constraining what can happen when a system evolves in time, so it’s worth considering here. Time reversal is similar to parity, in that it’s just an inversion of one of our coordinates, mapping t \rightarrow -t. But because time plays a special role in quantum mechanics (with time evolution being included in the postulates explicitly), the consequences are quite different.
As a warm-up, let’s remind ourselves that time reversal is a good symmetry of (conservative) classical mechanics. Given a classical Hamiltonian of the form H = p^2/(2m) + V(x), the equation of motion is m \ddot{x} = -\frac{\partial V}{\partial x}. Because only the second derivative appears, inverting t \rightarrow -t leaves this equation invariant. Another way to state this is that if x(t) solves the equations of motion, so does x(-t). This means that if we record a movie of a classical system evolving in time, and then play the movie backwards, either trajectory that we observe will be consistent with Newton’s laws.
To transfer this idea into quantum mechanics, following our earlier treatment of parity, we have to define an operator that will take t \rightarrow -t. However, our first problem is that the Schrodinger equation will not be invariant under just reversing time, since it depends on the first time derivative of the wavefunction and not the second: i\hbar \frac{\partial \psi}{\partial t} = \hat{H} \psi. But now we notice the i in front of the time derivative; if we both reverse time and complex conjugate the equation, then the result is unchanged. In other words, \psi^\star(x,-t) and \psi(x,t) obey the same equation of motion. This suggests that the time-reversal operator has to return the conjugate of the wavefunction, in addition to flipping the sign of time.
We can proceed by trying to expand \ket{\psi(t)} out and consider what the time-reversal operator has to look like to give this behavior, but because \psi^\star(x,t) is most conveniently thought of as a bra \bra{\psi(t)}, things will quickly get confusing. Instead, as with parity, it’s better to start in the operator picture and describe how it acts there. As a reminder, parity acts on some familiar vectors as follows: \hat{P}^{-1} \hat{\vec{x}} \hat{P} = -\hat{\vec{x}},\ \ \hat{P}^{-1} \hat{\vec{p}} \hat{P} = -\hat{\vec{p}},\ \ \hat{P}^{-1} \hat{\vec{L}} \hat{P} = +\hat{\vec{L}}. (note that \hat{\vec{L}} is an axial vector, i.e. it’s a cross product of vectors, so doesn’t flip sign under parity.) If we call the time reversal operator \hat{\Theta}, then we have the following action for the same vectors: \hat{\Theta}^{-1} \hat{\vec{x}} \hat{\Theta} = +\hat{\vec{x}},\ \ \hat{\Theta}^{-1} \hat{\vec{p}} \hat{\Theta} = -\hat{\vec{p}},\ \ \hat{\Theta}^{-1} \hat{\vec{L}} \hat{\Theta} = -\hat{\vec{L}}. Briefly, the position operator doesn’t depend on time, so \Theta doesn’t affect it; but both linear and angular momentum have single time derivatives in their definitions, so they are reversed by \Theta.
Now, there is an important relation between position and momentum; as we discussed in Section 10.4, identifying momentum as the generator of translation symmetry implies the canonical commutation relation [\hat{x}, \hat{p}] = i\hbar. Let’s try conjugating with \hat{\Theta} on this equation: \hat{\Theta}^{-1} (\hat{x} \hat{p} - \hat{p} \hat{x}) \hat{\Theta} = \hat{\Theta}^{-1} i\hbar \hat{\Theta} \\ \hat{\Theta}^{-1} (\hat{x} \hat{\Theta} \hat{\Theta}^{-1} \hat{p} - \hat{p} \hat{\Theta} \hat{\Theta}^{-1} \hat{x})\hat{\Theta} = \hat{\Theta}^{-1} i\hbar \hat{\Theta} \\ \hat{x} (-\hat{p}) - (-\hat{p}) \hat{x} = \hat{\Theta}^{-1} i\hbar \hat{\Theta} \\ -[\hat{x}, \hat{p}] = \hat{\Theta}^{-1} i\hbar \hat{\Theta}. Here I’ve deliberately not simplified the right-hand side yet. If we go too quickly and assume we can just collapse \hat{\Theta} \hat{\Theta} = \hat{1} on the right-hand side, then we end up with a contradiction; -i\hbar = i\hbar. What is going on?
Let’s be extra careful. Strictly speaking, what we have on the right-hand side is the operator \hat{\Theta}^{-1} times the operator i\hbar \hat{\Theta}. Our usual interpretation of multiplying an operator times a scalar relies on the property of linearity, that \hat{O}(c_\alpha \ket{\alpha} + c_\beta \ket{\beta}) = c_\alpha \hat{O} \ket{\alpha} + c_\beta \hat{O} \ket{\beta}. Linearity tells us that i\hbar \hat{\Theta} \ket{\alpha} should be the same thing as \hat{\Theta} (i\hbar \ket{\alpha}), which would let us cancel against its own inverse. So in order for a time reversal operator to exist, we can only conclude that it is not linear!
In fact, what we want to be true in order to preserve the canonical commutation relation is that i\hbar \hat{\Theta} \ket{\alpha} = -\hat{\Theta} (i\hbar \ket{\alpha}), which will fix the sign for us. This is the behavior of an anti-linear operator:
An anti-linear operator on a Hilbert space is an operator \hat{A}: \mathcal{H} \rightarrow \mathcal{H} which satisfies the condition \hat{A}(c_\alpha \ket{\alpha} + c_\beta \ket{\beta}) = c_\alpha^\star \hat{A} \ket{\alpha} + c_\beta^\star \hat{A} \ket{\beta}, i.e. it can still be distributed linearly over sums of kets, but it has the additional effect of complex conjugating their coefficients.
Anti-linear operators have some rather unintuitve properties, especially since we have become used to working in Dirac bra-ket notation - which, unfortunately, can be extremely misleading for some specific properties of anti-linear operators! Most importantly, for the linear operators we’ve encountered so far, the idea of “applying the operator to a bra” is well-defined through the inner product: \hat{O} \ket{\alpha} = \ket{\beta} \Rightarrow \bra{\alpha} \hat{O}^\dagger = \bra{\beta}, so if we’re handed a matrix element \bra{\alpha} \hat{O} \ket{\beta} we can always apply the operator to the left by conjugating it (we’ve done this frequently, for example when using SHO ladder operators.) In other words, (\bra{\alpha} \hat{O}) \ket{\beta} = \bra{\alpha} (\hat{O} \ket{\beta}).
This fails spectacularly for anti-linear operators! If we want to define \hat{A}^{\dagger}, or equivalently if we want to act on a bra to the left, we have to be extra careful and include some additional complex conjugation operations.
For these notes, I am not going to be so careful or general about anti-linear operators; see David Tong’s notes for a more careful treatment, or this paper if you really want all of the rigorous mathematical details. For time reversal, we can see the basic physical effects by always acting to the right on kets and not worrying about the adjoint (just keep in mind that you can’t manipulate \hat{\Theta} in all the same ways if matrix elements are involved!)
The anti-linearity of time reversal gives us exactly the behavior that we want in terms of how it interacts with the time evolution operator. If we start with a state \ket{\psi(0)}, its time evolution is given by the usual exponential of the Hamiltonian, \ket{\psi(t)} = e^{-i\hat{H} t/\hbar} \ket{\psi(0)}. If we act with \hat{\Theta}, the resulting state \hat{\Theta} \ket{\psi} = \ket{\psi_\Theta} should evolve backwards instead of forwards, meaning that: \ket{\psi_\Theta(-t)} = \hat{\Theta} \ket{\psi(t)} \\ e^{+i\hat{H} t/\hbar} \hat{\Theta} \ket{\psi(0)} = \hat{\Theta} e^{-i\hat{H} t/\hbar} \ket{\psi(0)} \\ \Rightarrow e^{i\hat{H} t/\hbar} \hat{\Theta} = \hat{\Theta} e^{-i\hat{H} t/\hbar}. This will be true thanks to the complex-conjugation that happens when we push \hat{\Theta} through the time evolution, so long as it commutes with the Hamiltonian, [\hat{H}, \hat{\Theta}] = 0.
So although the operator itself is strange, we recover the familiar statement that if \hat{\Theta} commutes with the Hamiltonian, then it is a dynamical symmetry of the system. But keep in mind that this is a very different kind of dynamical symmetry, in particular it doesn’t tell us that anything is conserved in time, it just relates forwards and backwards time evolution.
On top of being anti-linear, the time reversal operator \hat{\Theta} turns out to be anti-unitary, which means that it satisfies the property \bra{\chi} \hat{\Theta}^\dagger \hat{\Theta} \ket{\psi} = \left\langle \chi | \psi \right\rangle^\star = \left\langle \psi | \chi \right\rangle.
This is an example of the weirdness (in Dirac notation) that we have to deal with with anti-linear operators; it looks like the unitary condition \hat{U}^\dagger \hat{U} = \hat{1}, except we have to tack on a complex conjugation of the product.
Note that anti-unitary lets us connect back to the story we started with in terms of the Schrödinger equation, since using the fact that \hat{\Theta} \ket{\vec{x}} = \ket{\vec{x}} (since time reversal doesn’t affect the position vector at all), we see that \left\langle \vec{x} | \hat{\Theta} \psi \right\rangle = \left\langle \hat{\Theta} \vec{x} | \hat{\Theta} \psi \right\rangle = \left\langle \psi | \vec{x} \right\rangle = \psi^\star(x).
For time reversal, we should impose one more important physical constraint, which is that applying time reversal twice has to leave the physics of our system unchanged. This means that for any state \ket{\psi} in our Hilbert space, we must have \hat{\Theta}^2 \ket{\psi} = e^{i\alpha} \ket{\psi} where \alpha is a global phase, i.e. it is the same for all states in our space so that all physical observations will be unchanged. This ensures that time reversal squared is a symmetry in the sense of Wigner’s theorem. But now, we see that the following is true: e^{i\alpha} \hat{\Theta} \ket{\psi} = \hat{\Theta}^3 \ket{\psi} = \hat{\Theta} e^{i\alpha} \ket{\psi}, but at the same time by the anti-linearity of \hat{\Theta}, e^{i\alpha} \hat{\Theta} \ket{\psi} = \hat{\Theta} e^{-i\alpha} \ket{\psi}. So the only possible phases satisfy e^{i\alpha} = e^{-i\alpha}, which means \alpha = 0 or \alpha = \pi. In other words, the eigenvalues of \hat{\Theta}^2 are either \pm 1.
Once again, this should remind you of what we found previously with parity, except that it’s a bit stranger. For parity, we proved that the eigenvalues of \hat{P} itself were \pm 1, which means that applying it twice always gives the identity, \hat{P}^2 \ket{\psi} = \ket{\psi}. Here, we see that squaring time reversal (which should take us back to where we started) gives \pm 1 as the result. This still leaves the physics invariant since the -1 is a global phase, but it does actually have an important physical consequence.
E.0.1 Kramers degeneracy
With what we know about \hat{\Theta} so far, we have enough to say something interesting about how it acts on momentum eigenstates. The starting point is that for any angular momentum operator, conjugation by \hat{\Theta} should flip its sign, \hat{\Theta}^{-1} \hat{\vec{J}} \hat{\Theta} = -\hat{\vec{J}}. Now consider \hat{\Theta} acting on an eigenstate \ket{jm}, together with the z component: we have \hat{\Theta} \hat{J}_z \ket{jm} = \hbar m \hat{\Theta} \ket{jm} but also \hat{\Theta} \hat{J}_z \ket{jm} = -\hat{J}_z \hat{\Theta} \ket{jm} which means that \hat{\Theta} acting on \ket{jm} gives us back another \ket{jm} eigenstate, up to a constant. Specifically, \hat{J}_z (\hat{\Theta} \ket{jm}) = -\hbar m (\hat{\Theta} \ket{jm}) \Rightarrow \hat{\Theta} \ket{jm} = \theta_{jm} \ket{j,-m}. We can get more constraints on the phase using the other components of \hat{\vec{J}}. In particular, it’s easy to show that \hat{\Theta} \hat{J}_{\pm} = -\hat{J}_{\mp} \hat{\Theta}, which we can use to act with the lowering operator on the same state: \hat{\Theta} \hat{J}_- \ket{jm} = \hat{\Theta} \hbar \sqrt{(j+m)(j-m+1)} \ket{j,m-1} \\ = \hbar \sqrt{(j+m)(j-m+1)} \theta_{j,m-1} \ket{j, -m+1} \\ = \theta_{j,m-1} \hat{J}_+ \ket{j,-m}. But then commuting the operators, we also have \hat{\Theta} \hat{J}_- \ket{jm} = -\hat{J}_+ \hat{\Theta} \ket{jm} \\ = -\hat{J}_+ \theta_{jm} \ket{j,-m} so comparing, we see the relation \theta_{jm} = -\theta_{j,m-1}. Finally, we get one more condition from applying time reversal twice: \hat{\Theta}^2 \ket{jm} = \pm \ket{jm} \\ \hat{\Theta} \theta_{jm} \ket{j,-m} = \pm \ket{jm} \\ \theta^\star_{jm} \theta_{j,-m} \ket{jm} = \pm \ket{jm} so \theta^\star_{jm} \theta_{j,-m} = \pm 1. But from the alternating sign condition, \theta_{j,-m} = (-1)^{2m} \theta_{j,m}, which means that |\theta_{jm}|^2 = 1, i.e. \theta_{jm} has to be a pure phase. And if we put everything together, we can see that the combined solution to all of the conditions we found is \theta_{jm} = e^{i\pi m} = i^{2m}.
So we see something interesting here: if we consider \hat{\Theta}^2 specifically, we see that \hat{\Theta}^2 \ket{jm} = +\ket{jm} if j is an integer, but \hat{\Theta}^2 \ket{jm} = -\ket{jm} if j is half-integer. In other words, the eigenvalue of \hat{\Theta}^2 is +1 for bosons and -1 for fermions.
Now, suppose that we have a system which preserves time reversal, so that [\hat{H}, \hat{\Theta}] = 0, and consider the energy eigenstates \{\ket{E_n}\}. If time reversal commutes with the Hamiltonian, then \hat{\Theta} \ket{E_n} has the same energy as \ket{E_n}, which means they are equal up to a constant, \hat{\Theta} \ket{E_n} = c_n \ket{E_n}.
But then we apply \hat{\Theta} a second time: \hat{\Theta}^2 \ket{E_n} = \hat{\Theta} (c_n \ket{E_n}) \\ = c_n^\star \hat{\Theta} \ket{E_n} \\ = |c_n|^2 \ket{E_n}. If \hat{\Theta}^2 = +1, then c_n just has to be a pure phase and everything is fine. But if \hat{\Theta}^2 = -1 (i.e. if we are describing a fermion!), then this equation gives us a contradiction. The only way out is if \hat{\Theta} \ket{E_n} gives us a different state with the same energy - a degenerate pair. This is a powerful observation:
For any fermionic system for which time reversal is a dynamical symmetry, all energy eigenstates \ket{E_n} come in degenerate pairs.
For the familiar spin-1/2 system, the degenerate pair is just the two basis states \ket{+} and \ket{-} that we’re familiar with! Anytime they do not have the same energy, we know that we are working with a Hamiltonian that violates time-reversal symmetry. (For example, applying an external magnetic field \vec{B}, which is odd under time reversal - so that the system is not invariant unless we also flip the magnetic field backwards.)
This is a powerful constraint on a wide variety of systems, even systems that we can’t solve analytically, as long as they are time-reversal invariant. This includes arbitrarily complicated condensed matter systems, as long as they are overall fermionic (e.g. systems with an odd number of electrons.)