I'm sure the answer is yes, but how is this shown? Normally for a single spin-1/2 you have a time reversal operator: $-i \sigma_y \hat{K}$ where $\sigma_y$ is the second Pauli matrix and $\hat{K}$ is the conjugation operator. How is this generalized to two spins?

I am thinking of whether or not interactions like exchange ($J \hat{S}_1 \cdot \hat{S}_2$) or the hyperfine interaction (contact Fermi: $a \hat{S} \cdot \hat{I}$) break time reversal symmetry.

  • $\begingroup$ Interesting question. General time reversal is $T=UK$ where $K$ is complex conjugation and $U$ is some unitary matrix. Wiki gives a representation for a particle with definite spin (not necessarily 1/2). So I would recommend: go to the $|jm\rangle$ basis (rather than $|m_1 m_2\rangle$) and try a block diagonal $U$ based on wiki's. You'll have to check it satisfies all the properties of a good time reversal, but it should. $\endgroup$
    – Michael
    Sep 24, 2013 at 5:03
  • 4
    $\begingroup$ If the spin $S_i$ is odd under time reversal symmetry, $S_1.S_2$ should be even under time reversal symmetry. $\endgroup$
    – Trimok
    Sep 24, 2013 at 9:24
  • $\begingroup$ @Timrok, I see your point. In this paper, the opposite is assumed: arxiv.org/abs/1304.5096 $\endgroup$
    – BeauGeste
    Sep 24, 2013 at 18:46
  • $\begingroup$ From a quick look at the Lunde paper, I think what they're saying is that if you have a spin-spin interaction, then the mean field experienced by single particles violates time-reversal symmetry. The mean field always violates lots of symmetries that are present in the underlying two-body interaction. For example, in a nucleus, we have a mean field that is an attractive potential with a certain shape, centered on a certain point in space. This clearly violates translational invariance, even though the two-body interaction is symmetric under translational invariance. $\endgroup$
    – user4552
    Oct 6, 2013 at 14:43

1 Answer 1


I think the answer should be 'no'.

Because when we introduce the antiunitary time-reversal(TR) opeartor $T$ for spin-system, it should satisfy $T\mathbf{S}_iT^{-1}=-\mathbf{S}_i$ since angular-momentum should be sign reversed under TR(due to the classical correspondence). Thus, spin-spin interactions like $\mathbf{S}_i\cdot\mathbf{S}_j$ are invariant under TR.

The TR operator $T$ for the $N$-spin-$1/2$ system has a form $T=(-i)^N\sigma_1^y\sigma_2^y...\sigma_N^yK$, where $K$ is the conjugation operator. You can easily check that $T$ is antiunitary and satisfy $T\mathbf{S}_iT^{-1}=-\mathbf{S}_i$. Furthermore, $T^2=(-1)^N$, so for odd-number spin system(including single spin case), if the Hamiltonian has TR symmetry, we will arrive at the well-known Kramers theorem.


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