Bloch Hamiltonian and chiral symmetry

Question

Let $h_k$ be a "first-quantized" (single-particle) Hamiltonian for free fermions with two internal states e.g. spin states, $c_{k,\uparrow}$ and $c_{k,\downarrow}$ respectively. $h_k$ is hence a $2x2$ matrix, and can be written as $$h_k=d_0 \mathbb{1}+\vec{d}\cdot\vec{\sigma},$$ where $\vec{d}=(a,\;b,\;c)$ is a vector of scalar parameters and $\vec\sigma$ is the vector containing the three Pauli matrices. I will assume that $d_0=0$, so $h_k=\vec{d}\cdot\vec\sigma$. Trivially, $E^{\pm}_{k}=\pm\sqrt{a^2+b^2+c^2}$, so the bands are symmetric around $E=0$. This Hamiltonian is symmetric under chiral (or sublattice) symmetry if there is a unitary matrix $U$ that anticommutes with the Hamiltonian, $$U h_k U^\dagger=-h_{k}.$$ If this condition is satisfied, it can be shown that $E(k)=-E(k)$, i.e. bands are symmetric with respect to $E=0$.

However, one can only find such a unitary matrix if some component of $\vec{d}$ is $0$. If all components of $\vec{d}$ are non-vanishing, then such matrix $U$ does not exist. However, even in the general case where there is not chiral symmetry, the condition $E^{+}_k=E^-_k$ is satisfied by construction.

My question is therefore, why does $E^+_k=E^-_k$ (i.e. bands symmetric around E=0) hold even in the absence of chiral symmetry?

Answer 1

There is "no chiral symmetry" in the sense that one cannot find a matrix $U$ that satisfies the condition $$ U h_k U^{\dagger} = - h_k \quad \forall \, k. $$ That for all $k$ part is the important bit.

For each individual $k$, however, $\vec{d}\cdot\vec{\sigma}$ is always unitarily equivalent to $\sigma_z$. So the eigenvalues always come in pair. In fact, let the unitary matrix $J(k)$ diagonalize $h_k$, so that $$ J(k) h_k J(k)^{\dagger} = E \, \sigma_z. $$ You can verify that $U(k) = J(k)^{\dagger} \sigma_x J(k)$ is unitary, Hermitian, self-inverse and satisfies $$ U(k) h_k U(k)^{\dagger} = - h_k. $$

Such $U(k)$ exists for every $k$, but there isn't a one-size-fit-all $U$.