1
$\begingroup$

Suppose I have a 3D system of spinless fermions described by the following two-band model Hamiltonian:

$$ H(\vec{k})=\vec{d}(\vec{k}) \cdot \vec{\sigma} $$

where $\vec{d}=\left(-\sin k_{x},-\sin k_{y},m-\cos k_{z}\right)$, where $m$ is some constant and the hopping is assumed to be unity.

I am trying to figure out what is the appropriate expression for the time reversal operator in this case, and whether this system is invariant under time reversal. Does the fact that this system describes spinless fermions mean that the time reversal operator is $T=K$ where $K$ is complex conjugation? I've seen that is some cases $T$ also includes an exponent of a Pauli matrix (typically $\sigma_y$), but from what I gather it mostly applies to particles with spin. Assuming $T$ is indeed just $K$ in this case, what would be the simplest way of showing that this Hamiltonian is invariant under time reversal? I understand that it should be something like this

$$ TH\left(\vec{k}\right)T^{-1}=\vec{d}\left(\vec{k}\right)\cdot K\vec{\sigma}K^{-1} =...=H\left(-\vec{k}\right) $$

but because $K$ is nonstandard operator (it's anti-linear and also basis dependent), I'm not sure how exactly it affects the Pauli matrices.

$\endgroup$
0

1 Answer 1

0
$\begingroup$

To have time-reversal symmetry, there needs to be a unitary operator $U$ such that

$ U\sigma^x U^{-1}=-\sigma^x, U\sigma^y U^{-1}= \sigma^y, U\sigma^z U^{-1}=\sigma^z$.

These follow from $T=UK$, and I choose a basis for Pauli matrices so that the complex conjugation $K$ only changes $\sigma^y\rightarrow -\sigma^y$.

However, you can easily see that such a $U$ does not exist. For example, under the $U$,

$ U\sigma^x \sigma^y\sigma^z U^{-1}= - \sigma^x \sigma^y \sigma^z. $

However, $\sigma^x\sigma^y\sigma^z = i$, which commutes with any unitary.

Another way to see this is that when $k_z=0$, this Hamiltonian describes a massive Dirac fermion in 2+1, which breaks time reversal (unless $m=1$, but then there is still parity anomaly).

$\endgroup$
1
  • $\begingroup$ @dnrk Yes, that is right. $\endgroup$
    – Meng Cheng
    Aug 18, 2023 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.