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

1 Answer 1


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.)

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    $\begingroup$ @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). $\endgroup$ Nov 10, 2022 at 16:44
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    $\begingroup$ @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). $\endgroup$ Nov 10, 2022 at 17:00
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    $\begingroup$ @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! $\endgroup$ Nov 10, 2022 at 17:07
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    $\begingroup$ 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) $\endgroup$ Nov 10, 2022 at 17:36
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    $\begingroup$ @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 $T^2 = \pm 1$. $\endgroup$ Dec 15, 2022 at 21:47

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