I am studying the time-reversal symmetry in the context of topological insulators.

As usual, the minimal non-trivial model to be considered is a two-level system with Hilbert space $\newcommand{\ket}[1]{|#1\rangle} \mathcal{H} = \text{span} \{\ket{0}, \ket{1}\}$.

Previously I always considered linear operators which can be represented by $2\times2$ matrices. For example \begin{equation} \hat \sigma^+ = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \end{equation}

I understand that this representation is basis dependent. For example, we can choose another basis in which \begin{equation} \hat \sigma^+ = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \end{equation}

However, I learnt that the time-reversal operator is antilinear and, as such, does not have such a $2\times2$ representation.

Thanks to a comment to this question, I now understand that the time-reversal operator, as any operator in the Hilbert space, can be represented by a matrix with sufficiently large dimension.

However, it seems that it does not have a representation with the minimal dimension of $2$.

Is this correct? Is there some physical interpretation of this fact?

  • 1
    $\begingroup$ If you represent every complex entry of a vector by a 2-vector (real and imaginary part), then charge conjugation acts by $\sigma_z$, and you have a consistent tensor structure $M\otimes\sigma_z$. What is the trouble? $\endgroup$ Commented May 10, 2020 at 11:44
  • $\begingroup$ @CosmasZachos I have no troubles, just questions :) So your answer is yes, but we need to enlarge the dimension of the representation, is this correct? $\endgroup$
    – fra_pero
    Commented May 10, 2020 at 12:24
  • $\begingroup$ Sure, that's it. Try a simple example. $\endgroup$ Commented May 10, 2020 at 13:58

1 Answer 1


Time reversal is antiunitary, hence involves complex conjugation -- not a linear operator, not representable by a matrix, and also basis dependent.

To consider why, start from the Schrodinger equation $ i \hbar \partial_t \Psi = H \Psi$. Now take its complex conjugate, and use $\partial_t = -\partial_{-t}$, to get $ i \hbar \partial_{-t} \Psi^\ast = H^\ast \Psi^\ast$. Since this is true for any $\Psi^\ast$, this tells you is that if $H$ time-evolves your states forward, then $H^\ast$ time-evolves them backward.

So if you time-evolve any state $\Psi$ by time t, then take its complex conjugate, then time-evolve it using $H^\ast$ instead of $H$, then take the complex conjugate, you will get back to $\Psi$. It is as if you have "reversed the flow of time" by complex conjugating the state as well as the Hamiltonian.

If you are working in a basis where your Hamiltonian has only real matrix elements, then you don't even need to change the Hamiltonian to reverse time, just take complex conjugate of the state. In such a situation, you could be tempted to say that the Hamiltonian is time-reversal invariant or time-reversal symmetric.

Note that any Hamiltonian is real in its eigenbasis. So is any Hamiltonian time-reversal invariant? Not really, we call Hamiltonians time-reversal invariant if a local basis can be found where they are real, and this basis can be found without diagonalizing the Hamiltonian. I guess the reason for this is that our usual notion of time-reversal should be "don't change positions but reverse momenta".

So the general form of time-reversal $T$ involves a transformation to the special basis via a unitary $\tau$ and complex conjugation $K$:

$T = \tau K$

The operator $\tau$ is basis dependent, and complex conjugation $K$ is to be understood in position basis (plus whatever convention you want for the orbital/spin degrees of freedom). This also includes the case where $\tau = \sigma_y$, which is the usual time-reversal operation for spin-1/2.


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