# Must a time-reversal symmetric Hamiltonian really have $T^2 = \pm 1$?

It is apparently common knowledge in the condensed matter literature that the time reversal operator $$\hat{T}$$ always squares to $$\pm 1$$ at the first quantised level. When acting on spins the time reversal operator has $$U = \mathrm{e}^{\mathrm{i}\pi S_y}$$ and this is clearly true. However, I have not been able to find any rigorous derivation of this result in the general case. I am sceptical of the `proofs' that I have seen of this result, as Weinberg has shown (see below) that it is in principle possible for the time reversal operator to square to a phase. I am looking for a rigorous (or at least more convincing) proof that the matrix $$U$$ representing time reversal must always satisfy $$U U^*=\pm 1$$ if the time reversal operator commutes with a given Hamiltonian.

Below I give some more detail on the problem, before giving a more precise version of my question.

Consider a Fock space generated by $$N$$ (assumed finite for simplicity) creation operators $$\hat{c}^\dagger_{A=1, \ldots, N}$$. (For clarity, I will use circumflexes to distinguish operators from scalars.) The time reversal operator is an antiunitary operator $$\hat{T}$$ which is defined to act as (see e.g. ref. (2)) \begin{align} \hat{T}\hat{c}^\dagger_A \hat{T}^{-1} &= \sum_B \hat{c}^\dagger_B U_{BA}\\ \hat{T}\hat{c}_A \hat{T}^{-1} &= \sum_B U_{AB}^\dagger\hat{c}_B\\ \hat{T}\mathrm{i}\hat{T}^{-1} &= -\mathrm{i} \end{align} where $$U$$ is some $$N\times N$$ unitary matrix. Under a basis transformation of the creation operators, \begin{align} & \hat{c}^\dagger_A = \sum_B \hat{a}^\dagger_B V_{BA}^\dagger, & \hat{a}^\dagger_A = \sum_B \hat{c}^\dagger_B V_{BA} \end{align} the action of the time reversal operator becomes \begin{align} \hat{T}\hat{a}^\dagger_A \hat{T}^{-1} &= \sum_B \hat{T}\hat{c}^\dagger_B \hat{T}^{-1}\hat{T}V_{BA} \hat{T}^{-1}=\sum_{BC} \hat{c}^\dagger_C U_{CB} V_{BA}^* = \sum_{BCD} \hat{a}^\dagger_D V^\dagger_{DC} U_{CB} V_{BA}^* \end{align} and the new time reversal matrix is $$U' = V^\dagger U V^*.$$ It is shown in Weinberg (ref. (1)) that there does not always exist a basis in which the matrix $$U'$$ is diagonal; in general, the best that can be done is to make $$U'$$ block diagonal, with the blocks either $$1\times 1$$ phases, or $$2\times 2$$ matrices of the form $$\left(\begin{matrix}0 & \mathrm{e}^{\mathrm{i}\phi/2}\\ \mathrm{e}^{-\mathrm{i}\phi/2} & 0\end{matrix}\right), \tag{1}$$ where the phase $$\phi$$ may in general be non-zero. In particular, the action of $$\hat{T}^2$$ on $$\hat{c}^\dagger_A$$ is given by $$\hat{T}^2\hat{c}^\dagger_A \hat{T}^{-2} = \sum_B \hat{T}\hat{c}^\dagger_B U_{BA}\hat{T}^{-1}= \sum_{BC} \hat{c}^\dagger_C U_{CB} U_{BA}^*$$ and so blocks of the form (1) have $$U U^* = \left(\begin{matrix}0 & \mathrm{e}^{\mathrm{i}\phi/2}\\ \mathrm{e}^{-\mathrm{i}\phi/2} & 0\end{matrix}\right)\left(\begin{matrix}0 & \mathrm{e}^{-\mathrm{i}\phi/2}\\ \mathrm{e}^{\mathrm{i}\phi/2} & 0\end{matrix}\right) = \left(\begin{matrix} \mathrm{e}^{\mathrm{i}\phi} & 0\\ 0 & \mathrm{e}^{-\mathrm{i}\phi} \end{matrix}\right) \neq \pm 1,$$ i.e. the time reversal operator need not square to $$\pm 1$$ in general.

While it is therefore clear that $$T^2\neq \pm 1$$ in general, it is argued in e.g. ref. (2) that, for those (quadratic) Hamiltonians $$\hat{H}$$ which commute with $$\hat{T}$$, the matrix $$U U^*$$ must be proportional to the identity. The line of reasoning is as follows (modified from ref. (2)).

Let $$\hat{H} = \sum_{AB}\hat{c}^\dagger_{A} h_{AB} \hat{c}_B$$ be a generic quadratic Hamiltonian. The condition for this Hamiltonian to commute with the time reversal operator is that $$\hat{T}\hat{H}\hat{T}^{-1} = \sum_{AB}\hat{T}\hat{c}^\dagger_{A}\hat{T}^{-1} \hat{T}h_{AB} \hat{T}^{-1}\hat{T}\hat{c}_B\hat{T}^{-1} = \sum_{ABCD}\hat{c}^\dagger_{C}U_{CA} h^*_{AB} U^\dagger_{BD}\hat{c}_D\overset{!}{=}\hat{H}=\sum_{AB}\hat{c}^\dagger_{A} h_{AB} \hat{c}_B$$ which implies that $$[\hat{H}, \hat{T}]=0$$ iff $$U h^* U^\dagger = h$$. Applying this twice gives the condition $$(U U^*) h (U U^*)^\dagger = h$$. The Hamiltonian $$h$$ runs over an irreducible representation space, so by Schur's lemma it follows that $$U U^*$$ is proportional to the identity, $$U U^*= \mathrm{e}^{\mathrm{i}\gamma} 1$$. Then $$U U^* U = \underbrace{U U^*}_{\mathrm{e}^{\mathrm{i}\gamma}} U = U\underbrace{U^* U}_{\mathrm{e}^{-\mathrm{i}\gamma}}$$ so $$U\mathrm{e}^{2\mathrm{i}\gamma} = U$$ and $$\mathrm{e}^{\mathrm{i}\gamma} = \pm 1$$.

I have emphasised the line with which I have difficulty. My first question is:

In what sense is the set of Hermitian matrices $$h$$ that satisfy $$(UU^*) h (UU^*)^\dagger= h$$ an irreducible subspace?

It seems to me that this argument is somewhat cyclic: given a representation $$U$$ of the time reversal operator, we would like to find those Hamiltonians $$\hat{H}$$ which satisfy $$(UU^*) h (UU^*)^\dagger= h$$; if the Hermitian matrices $$h$$ are defined by the fact that they commute with $$UU^*$$, then how could they in any way affect the properties of the matrix $$UU^*$$ itself?

Ultimately, I would like a more rigorous proof that, if a time reversal operator $$\hat{T}$$ represented by a unitary matrix $$U$$ commutes with a (non-trivial) Hamiltonian $$\hat{H}$$, then the matrix $$U$$ must satisfy $$UU^*\propto 1$$.

Any references would also be gratefully received.

1. Steven Weinberg, The Quantum Theory of Fields Volume 1: Foundations (Chapter 2, Appendix C)
2. C-K. Chiu, J. Teo, A. Schnyder, S. Ryu, Classification of topological quantum matter with symmetries, Rev. Mod. Phys. 88, 035005

No, the structure of the group containing time-reversal can be arbitrarily complicated. For example arXiv:1803.09336 and arXiv:1904.12884 find 3d QFTs with time-reversal satisfying $$T^2=C$$, where $$C$$ denotes charge-conjugation. The latter paper finds even more exotic symmetry structures, such as dihedral groups, etc.

In general, $$T$$ can satisfy arbitrarily complicated algebras, such as $$T^{2n}=1$$ for any integer $$n$$, and even non-abelian algebras (see e.g. arXiv:1712.08639). So any claim that time-reversal must satisfy a given structure is just false. It is just not true that in QFT time-reversal must satisfy $$T^2=\pm1$$, or any other specific condition. The condition satisfied by a given $$T$$ depends on the specific QFT you are working with. Some theories do indeed lead to $$T^2=1$$, but not all of them.

One statement you can make is the following. If time-reversal satisfies $$T^2=1$$ classically, then quantum-mechanically it can satisfy $$T^2=\pm1$$. The reason is that QM always gives a projective representation of the symmetry group, i.e., you could have in principle $$T^2=e^{i\theta}$$ for some phase $$e^{i\theta}\in U(1)$$. But $$T$$ is anti-unitary, so if you compute $$T^3=(T^2)T=T(T^2)$$ you get the condition $$e^{i\theta}=e^{-i\theta}$$, i.e., $$\theta=0,\pi$$. In other words, the only claim we can make is that if you begin with a QFT where you know that $$T^2=1$$ classically, then you can conclude that $$T^2=\pm1$$ quantum-mechanically.

(Another statement you can make is that the total symmetry group, including time-reversal, is always a $$\mathbb Z_2$$ extension of the unitary symmetry group. But this is almost a tautology and not very interesting; it just means that $$T^2$$, whatever this operator is, is a unitary symmetry, which is obvious. But it does not mean that it has to be the trivial symmetry.)

• @xzd209 Yes, you can have $T^2\neq1$ even in free theories. The first two references are actually free theories (they are abelian Chern-Simons, whose Lagrangian is quadratic). Nov 10, 2022 at 16:44
• @xzd209 Correct, the classification is not exhaustive, as far as I can see (but then again, it is often very hard for me to decipher what cond-mat people say, so the classification might be exhaustive under some extra assumptions that they make and that I missed). Nov 10, 2022 at 17:00
• @xzd209 I'm glad I could help. I would recommend unaccepting the answer for now, and waiting for the full week before awarding the bounty. That way other people can contribute too and you might get better answers! Nov 10, 2022 at 17:07
• and for an example where $T^2=\pm 1$ fails even more jarringly, in arxiv.org/abs/2208.04331 a non-invertible time reversal symmetry is constructed (and indeed, in a free theory, i.e. maxwell with a theta angle) Nov 10, 2022 at 17:36
• @AccidentalFourierTransform in the context of the paper linked in the question, it is sufficient to consider the case where $T^2 =\pm 1$ because if $T^2$ is something else (as you correctly point out it could be) then it must at least be unitary, and then you just diagonalize the bands in the eigenspaces of $T^2$ and thus reduce to the case where $T^2 = \pm 1$. Dec 15, 2022 at 21:47