How are charge conjugation and sublattice symmetries related?

Question

Let $\psi$ be a $2n$-component fermion wavefunction, and $H$ be the single-particle Hamiltonian.

What I call sublattice symmetry is some unitary transformation $U_s$ such that $U_s^{\dagger} H \, U_s = - H$. All energy levels are then mirrored around $E = 0$ (except for possible zero-energy Majorana?) To me this seems to already imply a charge conjugation symmetry: for any positive energy excitation, one can find a hole excitation in the negative energy band that has exactly the same energy.

But I am having trouble linking this to the standard formulation of the charge conjugation symmetry. My understanding is:

1. Start with $i \partial_t \psi = H \psi$
2. Take complex conjugate $-i \partial_t \psi^* = H^* \psi^*$
3. If there exists a unitary $U_c$ so that

$$ -i\partial_t U_c \psi^{*} = \left(U_c H^{*} U_{c}^{\dagger}\right) U_{c} \psi^* = - H \, U_c \psi^{*} $$

then $H$ has a charge conjugation symmetry.

Does the existence of $U_s$ imply the existence of $U_c$? Or the other way round? How?

For a concrete example, the above is realized in (3+1)-dimensional Dirac equation. The Dirac Hamiltonian reads $$ \begin{split} H &= \vec{\alpha} \cdot (-i\nabla) + \beta m; \\ \alpha_i &= \gamma^0 \gamma^i; \; \beta = \gamma^{0}. \end{split} $$ Left-multiply the Schroedinger equation by $\gamma^{0}$ and one recovers the usual Dirac equation in covariant form. The two unitary matrices are $$ \begin{split} U_c &= i\gamma^{2}; \\ U_s &= \beta \alpha_1 \alpha_2 \alpha_3. \end{split} $$

Answer 1

Whatever, I figure it out myself. Charge conjugation and sublattice symmetries are equivalent.

First, if there is a superconducting order, the BdG Hamiltonian naturally comes with both symmetries. In fact, complex conjugation of the Nambu spinor $\Psi = (\psi, \psi^*)^{t}$ is just a unitary transformation: $$ \Psi^* = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \Psi. $$ So charge conjugation and "sublattice" (whatever it actually means for BdG Hamiltonian) are two facets of the same thing.

If the system is normal, you don't usually use the Nambu spinor and then there is no way to realize complex conjugation with a unitary transformation. So charge conjugation (which involve c.c.) doesn't seem immediately related to the sublattice symmetry, but they are.

We can just go to a basis that diagonalizes the Hamiltonian. Actually, I only need to make $H$ real, but the diagonal basis conveniently fits the bill.

Go back to Schroedinger equation and it's c.c.: $$ -i \partial_t \psi^{*} = H^{*} \psi^{*} = H \psi^{*} $$

Now the charge conjugation condition is $\lbrace U_c, H^{*} \rbrace = \lbrace U_c, H \rbrace = 0$, exactly the same as the sublattice condition $\lbrace U_s, H \rbrace = 0$.

--edit--

To be more precise, let's say $K$ diagonalize $H$: $$ K H K^{\dagger} = \tilde{H} = \text{diag}(E_1, E_2, \dots) $$ then the unitary operator $J = K^{t} K$ is complex conjugation of $H$: $$ J H J^{\dagger} = H^{*}. $$ $J$ always exists for any Hermitian $H$. Now we can write $$ U_c = J U_s. $$ So the two symmetries mutually imply each other.