# Can the time-reversal operator of a two-level system be represented by a $2\times2$ matrix?

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 $$$$\hat \sigma^+ = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$$$

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

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?

• 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? Commented May 10, 2020 at 11:44
• @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? Commented May 10, 2020 at 12:24
• Sure, that's it. Try a simple example. Commented May 10, 2020 at 13:58

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.