Suppose we had a lattice Hamiltonian $H$ which was symmetric under the particle-hole transformation
$$ c_n \mapsto U^\dagger c_nU=(-1)^nc^\dagger _n$$
such that $[H,U] = 0$, where $c_n$ are Fermionic operators obeying $\{ c_n , c_m^\dagger \} = \delta_{nm}$ and $\{ c_n,c_m \} = \{ c^\dagger_n, c^\dagger_m \} = 0$ where the indices label the lattice sites of the system.
If the ground state $|\Omega \rangle$ of $H$ is unique, then it would be an eigenstate of the particle-hole transformation as $U |\Omega\rangle = e^{i \theta} |\Omega\rangle$, where the eigenvalue is some phase as $U$ is unitary. From this, if I calculated the nearest-neighbour correlation, I would find
\begin{align*}
\langle \Omega | c_n^\dagger c_{n+1} |\Omega \rangle & = \langle \Omega | U^\dagger c_n^\dagger c_{n+1} U |\Omega \rangle \\
& = \langle \Omega |( U^\dagger c_n^\dagger U)(U^\dagger c_{n+1} U) |\Omega \rangle \\
&= (-1)^{2n+1}\langle \Omega| c_n c_{n+1}^\dagger |\Omega \rangle \\
&=- \langle \Omega |(-c_{n+1}^\dagger c_n) |\Omega \rangle \\
&= \langle \Omega| c^\dagger_{n+1} c_n |\Omega \rangle \\
&= \langle \Omega | c_n^\dagger c_{n+1} |\Omega \rangle^*
\end{align*}
therefore I find it is real. A similar calculation would also show that this implies half-filling $\langle \Omega| c^\dagger_n c_n |\Omega \rangle = \frac{1}{2}$ as shown in this answer.
This calculation relied upon the fact that $|\Omega\rangle $ is unique, however many systems will have ground state degeneracy in which case $U |\Omega \rangle \neq e^{i\theta} |\Omega\rangle$ in general. One can always find a state in the ground state subspace that is an eigenstate however, and the above results will apply.
(Edit) Example Hamiltonian
An example of an interacting Hamiltonian that has particle-hole symmetry is given by
$$ H = -t \sum_n c_n^\dagger c_{n+1} + \lambda \sum_n c^\dagger_n c_{n+1} c_{n+2}^\dagger c_{n+3} + \mathrm{h.c.} = H_0 + H_\mathrm{int} $$
where $t,\lambda \in \mathbb{R}$. We have
\begin{align*}
U^\dagger H_0 U & = -t \sum_n (-1)^{2n+1} c_n c^\dagger_{n+1} + \mathrm{h.c.} \\
& = t \sum_n c_n c^\dagger_{n+1} + \mathrm{h.c.} \\
& = -t\sum_n c_{n+1}^\dagger c_n + \mathrm{h.c} \\
& = H_0
\end{align*}
And for the interaction part
\begin{align*}
U^\dagger H_\mathrm{int} U & = \lambda \sum_n (-1)^{4n + 6} c_n c^\dagger_{n+1} c_{n+2} c^\dagger_{n+3} + \mathrm{h.c.}\\
& = \lambda \sum_n c_{n+3} c^\dagger_{n+2} c_{n+1} c_n^\dagger + \mathrm{h.c.} \\
& = H_\mathrm{int}
\end{align*}
where I have interchanged parts with the hermitian conjugate and anti-commuted fermions past each other.
As this Hamiltonian is interacting, I am not sure how I would derive the correlation matrix of this, but all I would like to know is if this system has half-filling and real nearest-neighbour correlations in the ground state.
My question
In the case of no degneracy we can say that particle-hole symmetry implies half filling $\langle c^\dagger_n c_n \rangle = \frac{1}{2}$ and the correlation $\langle c_n^\dagger c_{n+1} \rangle \in \mathbb{R}$. However, in the case of degeneracy the above derivation would not apply unless we chose a state in the ground state subspace that is an eigenstate of $U$. My question is, am I allowed to choose the particle-hole symmetric ground state over another as "special" and can I always assume that particle-hole symmetry implies half-filling and real nearest-neighbour correlations?
Best Answer
I think this is quite a standard question in condensed matter. Particle-hole symmetry always implies half-filling, but whether it implies $\langle c^{\dagger}_{n+1}c_n \rangle$ to be real depends on what your particle-hole symmetry really is. If the transformation is what you defined in the question, the answer is yes.
Consider the partition function $e^{-\beta \hat{H}}$ where $\beta>0$. The correlation function you care is $$\langle c^{\dagger}_{n+1}c_n \rangle = {\text{Tr}\big(c^{\dagger}_{n+1}c_n e^{-\beta \hat{H}}\big) \over \text{Tr}\big(e^{-\beta \hat{H}}\big)}$$ Note that when $\beta \rightarrow +\infty$, we have the ground state limit given whether or not the ground state is non-degenerate. Since the Hamiltonian $\hat{H}$ is invariant under particle-hole transformation, we have \begin{align} \text{Tr}\big(c^{\dagger}_{n+1}c_n e^{-\beta \hat{H}}\big)^{\dagger} & = \text{Tr}\big(e^{-\beta \hat{H}}c^{\dagger}_nc_{n+1}\big) \\ & = \text{Tr}\big(c^{\dagger}_nc_{n+1} e^{-\beta \hat{H}}\big) \\ & = \text{Tr}\big(\hat{U}^{\dagger}c^{\dagger}_nc_{n+1} e^{-\beta \hat{H}}\hat{U}\big) \\ & = \text{Tr}\big(\big(\hat{U}^{\dagger}c^{\dagger}_n\hat{U}\big)\big(\hat{U}^{\dagger}c_{n+1} \hat{U}\big)\big(e^{-\beta \hat{U}^{\dagger}\hat{H}\hat{U}}\big)\big) \\ & = -\text{Tr}\big(c_nc^{\dagger}_{n+1} e^{-\beta \hat{H}}\big) \\ & = \text{Tr}\big(c^{\dagger}_{n+1}c_n e^{-\beta \hat{H}}\big) \end{align} This implies $\langle c^{\dagger}_{n+1}c_n \rangle$ is real. You can also use the same approach to show $\langle c^{\dagger}_nc_n \rangle={1 \over 2}$ and $\text{Re}\big(\langle c^{\dagger}_{n+2}c_n \rangle\big)=0$.
So it is a good time to answer your question. Whether we have $\text{Tr}\big(c^{\dagger}_{n+1}c_n e^{-\beta \hat{H}}\big)^{\dagger}=\text{Tr}\big(c^{\dagger}_{n+1}c_n e^{-\beta \hat{H}}\big)$ depends on the form of your particle-hole transformation. If you have $c_n$ transformed by $\hat{U}^{\dagger}c_n\hat{U}=c^{\dagger}_n$, we can no longer show $\langle c^{\dagger}_{n+1}c_n \rangle$ is real ($\text{Re}\big(\langle c^{\dagger}_{n+1}c_n \rangle\big)$ is in fact $0$). However, we still can prove $\langle c^{\dagger}_nc_n \rangle={1 \over 2}$ and $\text{Re}\big(\langle c^{\dagger}_{n+2}c_n \rangle\big)=0$. The very particular case is $\langle c^{\dagger}_nc_n \rangle$. Since the phase given by the particle-hole transformation in $c^{\dagger}_nc_n$ will be cancelled out, no matter which transformation is given, we keep the same result $\langle c^{\dagger}_nc_n \rangle={1 \over 2}$. However, the particle-hole transformation is defined in your question is useful when the lattice of the Hamiltonian is bipartite, including the hexagonal lattice and the square lattice.