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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?

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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.

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  • $\begingroup$ Thanks a lot, franz. I think you're missing a minus sign in the exponent in the second displayed equation. Also, it seems like in the second equality of your third displayed equation you are assuming that the wavefunction is real. I guess I'm thinking of the wavefunction as a vector of complex numbers, i.e., in a given basis. Can you comment on this? It's not so clear to me. $\endgroup$
    – Daniel
    Commented Mar 1, 2021 at 17:17
  • $\begingroup$ @Daniel You are right, I corrected the sign in the equation you were refering. This expression is of course only valid, if we assume the Hamiltonian to be time-independent. I do not assume that the wavefunction has to be real, it is of complex nature. $\endgroup$
    – franz
    Commented Mar 3, 2021 at 10:54
  • $\begingroup$ There is no equation $(1)$. Consider to correct this. $\endgroup$ Commented Jun 19, 2022 at 20:07
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    $\begingroup$ Dear @JasonFunderberker, equation (1) can be found within the text in the middle of my answer. $\endgroup$
    – franz
    Commented Jun 20, 2022 at 8:19
  • $\begingroup$ It's not quite clear to me how you arrived at equation 2, could you expand on that? $\endgroup$
    – Green05
    Commented May 28, 2023 at 10:54

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