Does any Hamiltonian with spectrum mirrored about 0 have a sublattice/chiral symmetry?

Question

This is a follow-up of my previous question on time reversal.

Again, I don't think this should be the case, because, for one thing, supposedly we can have charge conjugation symmetry alone.

Consider the second quantize fermion operator $\psi_I$, where subscript $I$ labels position/momentum, spin, orbit, what have you. The action of time reversal, charge conjugation and sublattice (T, C, S) are

$$ \begin{split} T \psi_I T^{\dagger} &= [U_t]_{IJ} \psi_J \\ C \psi_I C^{\dagger} &= \psi_J^{*} [U_c]_{JI} \\ S \psi_I S^{\dagger} &= \psi_J^{*} [U_s]_{JI}\\ \end{split} $$ Here $T$ and $S$ are antiunitary. $S$ is $TC$ with possibly an extra phase shift.

If we do a unitary change of basis $\tilde{\psi}_I = V_{IJ} \psi_J$, then these $U_t$, $U_c$ and $U_s$ matrices transform as $$ \begin{split} \tilde{U}_t &= V^{*} U_t V^{\dagger} \\ \tilde{U}_c &= V U_c V^{T} \\ \tilde{U}_s &= V U_s V^{\dagger} \end{split} $$

We see that $U_c$ and $U_t$ have weird transformation laws. Therefore one has to first specify how $T$ and $C$ acts on one basis and stick to the definition, and then check whether a given Hamiltonian is invariant against the specific choice of $T$ and $C$. This was pretty much the conclusion of my previous question.

However, given that $U_s$ just transforms nicely, I would expect the S-symmetry to be "basis-neutral", so to speak. In other words, I should be able to take a Hamiltonian, in any basis, and just look for an eligible $U_s$.

(I reckon the error is probably in the above paragraph, but I am not sure what is wrong.)

The condition for sublattice symmetry is that, given the Hamiltonian $H = \psi^{*}_I \mathcal{H}_{IJ} \psi_J$, the unitary $U_s$ satisfies $$ U_s \mathcal{H} U_s^{\dagger} = -\mathcal{H}; \, U_s^2 = 1 $$

Assuming $\mathcal{H}$'s eigenvalues are either $\pm E$ pair or $0$. I can always diagonalize it with some unitary $J$, so that $J \mathcal{H} J^{\dagger}$ is real and diagonal.

And then I can construct a permutation matrix $K$ that will swap every pair of eigenvalues, while leaving all zeros in place. Obviously $K K^{\dagger} = K^2 = 1$. Now $$ K J \mathcal{H} J^{\dagger} K^{\dagger} = K \tilde{\mathcal{H}} K^{\dagger} = - \tilde{\mathcal{H}} = - J \mathcal{H} J^{\dagger}. $$

So I can construct $U_s = J^{\dagger} K J$ that meets all the requirements.

To summarize:

1. Unlike the time reversal $U_t$, $U_s$ transform unitarily.

2. Therefore, if I find a $U_s$ in any basis, the Hamiltonian has an S-symmetry.

3. I can always cook up a $U_s$.

Again, please shoot me down.

Answer 1

I would have rather posted this as a comment, but I am new and don't have enough reputation.

I may be reading my own past confusion into this question; please forgive me if that is the case.

First of all, you are writing your Hamiltonian as a bilinear form, with a vector of operators on the right, a matrix in the middle, and the Herm. transfo. of the vector of operators on the left. When any antiunitary operator acts on the fermion operators, daggered operators are swapped for daggerless operators.

This leads to my assumptions:

• You are working in a fermion-doubled basis, and your $\psi_I$ vectors contain both particle creation and particle annihilation operators.
• Your original Hamiltonian may contain pairs of creation operators or annihilation operators rather than the usual terms with one of each.

Again, forgive me if these assumptions are incorrect, but I'm hoping that this answer will still clear up confusion even if they aren't. The reason for my assumptions is that one can't guarantee that the Hamiltonian, after performing an anitunitary transfo., remains a bilinear form between the same two vectors (Herm. conj. on the left) unless the vector contains the fermion operators and their Herm. transfo.s, as that antiunitary transfo. will swap daggered and daggerless operators.

In this case, your Hamiltonian will always have an antisymmetric spectrum, with one energy correspondingto the creation of a quasiparticle and one corresponding to annihilation. In the usual (pre-fermion-doubled) basis, we do not see these negative eigenvalues, because (in a certain intuitive sense) we are only finding energies corresponding to the creation operators.

For example, consider the spinful Hamiltonian $H = c^{\dagger}(a\sigma_0+b\sigma_z)c$, with eigenvalues $a+b$ and $a-b$. This can be written as a bilinear form (what's inside the parentheses). If we try to perform conjugation with $K$, we end up with a bilinear form, but not between $c^{\dagger}$ and $c$, but between $c^T$ and $c^*$.

If we extend our vectors to include both daggered and daggerless operators via, for example $\psi^T = (c^T_{\downarrow}\;\;c^T_{\uparrow}\;\; c^*_{\downarrow}\;\;c^*_{\uparrow})$, $K$ can be viewed as a matrix whose role is to swap the upper two and lower two components of $\psi$, and the new Hamiltonian (ignoring a constant term) will have paired eigenvalues $\pm(a+b)$ and $\pm(a-b)$. This also means there has to exist a particle-antiparticle-exchange operation that anticommutes (up to a constant) with the Hamiltonian.

This also occurs more naturally if you rewrite your fermion operators in terms of Majorana operators corresponding to the Hermitian and anti-Hermitian parts of your original operators. You are no longer looking at eigenvalues of an $N\times N$ matrix acting on a complex vector, but a $2N\times 2N$ matrix acting on a real vector. Because the original matrix has to be Hermitian and because the fermion operators are fermionic (anticommuting), the new $2N\times 2N$ matrix must have paired eigenvalues, antisymmetric about some constant energy. Clearly this Majorana picture is unitarily equivalent to the setup described in the previous paragraph, although perhaps my notation obscures this.