In the tight binding model, $H=\sum_{r,r'}ta^{\dagger}_{r}a_{r'}+h.c.$. When conducting a time reversal transformation, what form will this Hamiltonian take? Or how can I express time reversal operator? When the Hamiltonian is transformed into momentum space, that is $H=\sum_{k}\epsilon_{k}a^{\dagger}_{k}a_{k}$, how does this Hamiltonian change under time reversal?

Moreover, when there are more than one valence electrons in a unit cell like graphene (honeycomb lattice), the Hamiltonian in real space is $H=\sum_{r,r'}ta^{\dagger}_{1,r}a_{2,r'}+h.c.$,where 1,2 denote sublattice in one unit cell. How does this one change under time reversal? And also in momentum space, where the Hamiltonian is now a $2\times 2 $ matrix, $H=\sum_{k}\sum_{i,j=1,2}H_{ij}a^{\dagger}_{i,k}a_{j,k}$, how does this one change? And what if I consider further hopping with terms like $a^{\dagger}_{1,r}a_{1,r'}$ involved?

In a word, how do we deal with time reversal in such systems? Does time reversal here have something to do with the one in field theory? Are they expressed the same? If so, there must be something like $\sigma_y$, what does this mean?


1 Answer 1


Starting with some background information from Wikipedia, we have that under time reversal the position is unchanged while the momentum changes sign.

In quantum mechanics we can express the action of time reversal on these operators as $\Theta\,\mathbf{x}\,\Theta^\dagger = \mathbf{x}$ and $\Theta\,\mathbf{p}\,\Theta^\dagger = -\mathbf{p}$. It is worth mentioning here that the time reversal operator, $\Theta$, is anti-unitary, which allows it to be expressed as $\Theta = UK$ where $U$ is unitary and $K$ is the complex conjugation operator.

As for the creation/annihilation operators used in second quantization the sign changes under $\Theta$ would suggest a transformation of $a_r \rightarrow a_r$ and $a_k \rightarrow a_{-k}$. If you are worried about the fact that $k$ represents a crystal momentum and not a true momentum you can just take the position transformation, which is perhaps more trustworthy, and use $a_k = \sum_r \, a_r \,\mathrm{exp}[ -\mathrm{i} k\cdot r]$ to verify $a_k \rightarrow a_{-k}$ directly.

Using these transformations you should be able to verify that the tight-binding Hamiltonian is invariant under time reversal in position and momentum space for a lattice with or without a basis. Keep in mind that you would generally take the complex conjugate of the coefficients in $H$, however in your case $t$ and $\epsilon_k$ are both real. Its important to remember though, mostly to make sure $H$ stays hermitian.

As far as your comment about $\sigma_y$, this is only necessary if you include spin. Spin changes sign under time reversal so $\Theta\,\mathbf{S}\,\Theta^\dagger = -\mathbf{S}$. In this case, we can formally write $\Theta = \mathrm{exp}[-i \pi J_y]\,K$, which is probably the relation you are alluding to.

According to J.J. Sakurai's Modern Quantum Mechanics one possible convention for the time-reversed angular momentum states is $\Theta | j,m\rangle = (-1)^m |j,-m\rangle$. This suggests that with spin indices the creation/annihilation operators transform like $a_{r,m} \rightarrow (-1)^m\,a_{r,-m}$ and $a_{k,m} \rightarrow (-1)^m \, a_{-k,-m}$ under time reversal. From what I understand, most spin Hamiltonians will be invariant under this transformation. An example when this is not the case would be in the presence of an external magnetic field which couples to the spins through a $\mathbf{S}\cdot \mathrm{B}-$like term.

It is interesting how even in the absence of an external field the groundstate of spin Hamiltonians can still sponanteously break the time reversal symmetry present in $H$, but rather than discuss this myself I will direct you to this very well written answer.

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    $\begingroup$ The time reversal operator apply on the second quantized operators in this answer seems not as in your answer. $\Psi_A$ and $\Psi_B$ in that answer are defined on a lattice, When applies time reversal operator, they are changed according to a linear combination (matrix $U_T$). Can you explain this point? $\endgroup$ May 4, 2021 at 11:44

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