Time evolution and anti-unitary operators

Question

Let $\hat{H}$ be a system of lattice fermions with two internal states per site, so they can be described using operators $\hat{c}_{m,\alpha}$, with $m$ denoting the lattice site and $\alpha$ the internal state. An anti-unitary operator $\hat{U}$ acts on these operators as $$\hat{U}\hat c_{m\alpha}\hat{U}^{-1}=\sum_\beta\mathcal U_{\alpha\beta}\hat{c}_{m,\beta}.$$ Since $\hat{U}$ is anti-unitary, it also satisfies that it acts over imaginary scalars as $$\hat{U}i\hat{U}^{-1}=-i.$$

Now suppose that at time $t=0$ I have a state $|\psi(0)\rangle$that is invariant under this transformation, $\hat{U}|\psi(0)\rangle=|\psi(0)\rangle$. Now, I evolve the state with a second-quantized Hamiltonian $\hat{H}=\sum_{m,n,\alpha,\beta}\hat{c}_{m\alpha}^\dagger H_{m\alpha,n\beta}\hat{c}_{n\beta}$ that is symmetric under the transformation $\hat{U}$, i.e. $\left[\hat{H},\,\hat{U}\right]=0$. Then I have $$\hat{U}|\psi(t)\rangle=\hat{U}e^{-i\hat{H}t}|\psi(0)\rangle=\hat{U}e^{-i\hat{H}t}\hat{U}^{-1}\hat{U}|\psi(0)\rangle=e^{+i\hat{H}t}|\psi(0)\rangle\neq|\psi(t)\rangle.$$

Therefore, even though both the initial state and the Hamiltonian are invariant under this transformation, the anti-unitary character makes that any evolved state will break that symmetry. However, this is conceptually a bit strange for me, especially for anti-unitary transformations that have nothing to do with time or time-reversal (for example, sub-lattice symmetry). Since these transformations are used e.g. to classify topological insulators, how can it be that a mere time evolution breaks the symmetry? This has no counterpart for unitary transformations and I don't really get the conceptual meaning of this fact.

Answer 1

In general if $H$ is symmetric under a unitary or anti-unitary transformation $\hat U$, it implies that if $|\psi(t)\rangle$ is a solution to the Schroedinger equation $i\hbar \frac{d}{dt}|\psi(t)\rangle = H|\psi(t)\rangle$, so is $\hat U|\psi(t)\rangle$. For example, the time reversal operator $T$ is anti-unitary: if a Hamiltonian is symmetric under time reversal, then if $|\psi(t)\rangle$ is a solution to $i\frac{d}{dt}|\psi(t)\rangle = H|\psi(t)\rangle$, then so is $|\psi(-t)\rangle = T|\psi(t)\rangle$.

For anti-unitary symmetries, this implies that if $|E\rangle$ is a stationary state of $H$ with energy $E$ (i.e. $e^{-iHt}|E\rangle = e^{-iEt}|E\rangle$), then $|-E\rangle$ is stationary as well (with energy $-E$), so the spectrum is symmetric about $E=0$.

If the initial state $|\psi\rangle$ maps to itself under $\hat U$, $|\psi\rangle$ can be written as $|\psi_+\rangle + |\psi_-\rangle + |\psi_0\rangle$, where $|\psi_+\rangle$ and $|\psi_-\rangle$ are projections of $|\psi\rangle$ to positive and negative energy states respectively (and $|\psi_0\rangle$ is the remainder), and $\hat U|\psi_\pm\rangle = |\psi_\mp\rangle$, $\hat U|\psi_0\rangle = |\psi_0\rangle$. Then (letting $H=H_+ + H_- + H_0$, where $H_+=\sum_{E>0}E|E\rangle\langle E|$ and $H_- = \sum_{E<0} E|E\rangle\langle E|$) \begin{align*} |\psi(t)\rangle = e^{-iHt}|\psi\rangle &= e^{-iHt}\Big(|\psi_+\rangle + |\psi_-\rangle + |\psi_0\rangle\Big)\\ &= \Big(e^{-iH_+t}|\psi_+\rangle + e^{-iH_-t}|\psi_-\rangle + |\psi_0\rangle\Big)\\ &= \Big(\hat U e^{iH_+t}\hat U|\psi_+\rangle + \hat U e^{iH_-t}\hat U|\psi_-\rangle + |\psi_0\rangle\Big)\\ &= \Big(\hat U e^{iH_+t}|\psi_-\rangle + \hat U e^{iH_-t}|\psi_+\rangle + \hat U|\psi_0\rangle\Big)\\ &= \hat U\Big(e^{iH_+t}|\psi_-\rangle + e^{iH_-t}|\psi_+\rangle + |\psi_0\rangle\Big)\\ &= \hat U\Big(e^{-iH_-t}|\psi_-\rangle + e^{-iH_+t}|\psi_+\rangle + |\psi_0\rangle\Big)\\ &= \hat U|\psi(t)\rangle \end{align*}

Answer 2

Ah, I see. It comes down to the fact that $\hat{U}$ is not linear. (It is antilinear, duh!). And it doesn't play well with unitary transformation.

Let's say you define $\hat{U}$ by what it does to the complete set of energy eigenstates. Normally if $\hat{U}$ is unitary, the phases of these states don't matter. For antiunitary and antilinear $\hat{U}$, however, you have to specify a precise gauge choice, because a phase factor $e^{i\theta}$ doesn't commute with $\hat{U}$! Make even a gauge change, and $\hat{U}$ must transform.

More generally, let $\psi_a$ be some second quantized fermion operator with some index $a$, and let the U-transformation be defined by $$ \hat{U} \psi_a \hat{U}^{\dagger} = J_{ab} \psi_b. $$ If you make a unitary transformation $\psi_a \rightarrow \tilde{\psi}_a = V_{ab} \psi_b$, the matrix $J$ has to become $$ J \rightarrow \tilde{J} = V^{*} J V^{\dagger} $$ so that $\hat{U} \tilde{\psi}_a \hat{U}^{\dagger} = \tilde{J}_{ab} \tilde{\psi}_b$. Even a mere phase shift of the basis alters $\hat{U}$.

Time evolution is also a unitary transformation. I guess $\hat{U}$ is, in fact, time dependent.