Time-reversal symmetry for spin Hamiltonian

Question

In the topology online course by TU Delft, the time-reversal operator acting on a system of spin-1/2 particles is introduced as $$ \mathcal T = i\sigma_y\mathcal K. $$ I understand this acts on the matrix $H$ in $$ \mathcal H = \vec c^\dagger H\vec c, $$ and $\sigma_y$ acts on the spin label.

My question is, does it have any significance that $i\sigma_y$ is chosen? Why not $\sigma_x$ or $\sigma_y$ only? Does it matter?

Answer 1

In general the representation of the time-reversal operator depends on the system that you consider. Let us start with some basic remarks on time-reversal symmetry prior to the spinor question that you are asking.

Time-reversal means to go backwards in time, $t \rightarrow -t$.

When we want to know the time-evolution of a quantum-mechanical state, we need to look at the time-dependent Schrödinger equation.

\begin{equation} - \frac{\hbar}{i} \frac{\partial \psi}{\partial t} = \mathcal{H} \psi \end{equation}

The solutions of this equation are known and can be represented by

\begin{equation} \psi(\vec{r}, t) = e^{-\frac{i}{\hbar} \mathcal{H}t} \psi(\vec{r}, 0) \quad. \end{equation}

Using the time reversal operator $\mathcal{T}$ on the wave function above, results in

\begin{equation} \mathcal{T} \psi(\vec{r}, t) = \psi(\vec{r}, -t) = \psi^*(\vec{r}, t) \quad . \end{equation}

From the action of the time reversal operator on the wave function we see, that it leads to a complex conjugation of the wave function. Thus, we can simply construct it in the case of spin-less wave functions as $\mathcal{T} = K$, where $K$ is the operator for complex conjugation. More generally we can write $$\mathcal{T} = UK \tag{1} \quad ,$$ where $U$ is a unitary operator. The determination of $U$ in the case of spin-$\frac{1}{2}$ particles, leaves us with the explicit representation for the time-reversal operator.

There are some basic properties, that every time-reversal operator needs to possess.

• energy conservation: $[\mathcal{T}, \mathcal{H}] = 0$
• anti-linearity: $\mathcal{T} i = -i$, where $i$ is the imaginary unit (in fact: one can show that $\mathcal{T}$ is antiunitary $\mathcal{T} \mathcal{T}^{\dagger} = - 1$)

So let us now consider the case of wave functions incorporation the spin degree of freedom.

For a spin-$\frac{1}{2}$ particle we need to fulfill the relation $$\mathcal{T} \, \vec{\sigma} \, \mathcal{T}^{-1} = -\vec{\sigma} \tag{2} \quad ,$$ where $\vec{\sigma} = (\sigma_x, \sigma_y, \sigma_z)^{T}$ is the Pauli-vector. This relation basically coincides with flipping the spin due to time-reversal ($\mathcal{T} \vec{S} = -\vec{S} \sim \mathcal{T} \vec{\sigma}$).

If we choose $\sigma_z$ to be diagonal, the matrices take the form:

$$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\, \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix},\, \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\quad .$$

If we now use the form of the time-reversal that we found for the spin-less case we find

$$K \sigma_x K^{-1} = \sigma_x,\quad K \sigma_y K^{-1} = -\sigma_y,\quad K \sigma_z K^{-1} = \sigma_z \tag{3} \quad .$$

From $(1)$, $(2)$ and $(3)$ we can simply show that $U = c \sigma_y$. As we require unitarity for $U \Rightarrow UU^{\dagger} = 1$ $\Rightarrow$ $|c|^2 = 1$ and we can choose $c = 1$.

This leaves us with the desired result. \begin{equation} \boxed{\mathcal{T} = \sigma_y K} \end{equation}

PS: that your time-reversal operator contains an additional $i$ could be because of the chosen representation of the Pauli matrices but either way your operator fulfills the criteria to serve as time-reversal operator, as $|i|^2 = 1$.

To develop a deeper understanding, I encourage you to read the relevant chapters in Dresselhaus' Group Theory - Application to the Physics of Condensed Matter.