Vacuum state in particle hole symmetric Hamiltonian Following this review, particle-hole is a unitary transformation that mixes creation and annihilation operators in fermionic systems. The operators transform in the following way:
$$ \hat{C}\hat{\psi}_I\hat{C}^{-1}=(U_C^*)_I^{\,\,J}\psi_J^\dagger $$
But how does the vacuum transforms, i.e., if I want to write an arbitrary state:
$$ |\alpha\rangle=\psi^\dagger_\alpha|0\rangle \\\qquad\qquad\,\,\qquad =\hat{C}^{-1}\hat{C}\psi_\alpha^\dagger\hat{C}^{-1}\hat{C}|0\rangle \\\hspace{29mm} =(U_C)_\alpha^{\,\,J}\hat{C}^{-1}\psi_J \hat{C}|0\rangle $$
Based on what I understand about particle-hole symmetry, I want to make an interpretation in the following way: $\hat{C}|0\rangle$ is the state with all electrons (particles) states filled, but is there a way of showing this formally? Is it the right interpretation?
 A: Here is a perhaps more direct argument: I suppose you define $|0\rangle$ to be the state with no fermion occupation, so it means that $n_I=\psi_I^\dagger\psi_I=0$ on this state. Define $N=\sum_I n_I$. Now you can check that $CN C^{-1}=N_I-N$, where $N_I$ is the total number of fermionic modes:
$$
C\sum_I\psi_I^\dagger \psi_I C^{-1}=\psi_J \sum_I(U_C)_{IJ}^*(U_C)_{IK}\psi_K^\dagger=\sum_J \psi_J\psi_J^\dagger = N_I-N.
$$
Therefore,
$C|0\rangle$ must have fermion number $N_I-0=N_I$. There is a unique state that has $N_I$ fermions, that is the one with each state filled.
A: On a more pedestrian level particle-hole transformation is
$$
c_k=\begin{cases}p_k, \text{ if }\epsilon_k>\epsilon_F,\\
h_k^\dagger, \text{ if }\epsilon_k<\epsilon_F\end{cases},
c_k^\dagger=\begin{cases}p_k^\dagger, \text{ if }\epsilon_k>\epsilon_F,\\
h_k, \text{ if }\epsilon_k<\epsilon_F\end{cases},
$$
where $c_k, c_k^\dagger$ are the electron annihilation and creation operators, whereas $p_k,p_k^\dagger$ and $h_k,h_k^\dagger$ are the annihilation and creation operators for particles and holes respectively.
Then if the vacuum is defined as filled by electrons up to energy $\epsilon_F$, that is
$$
c_k|0\rangle=0,\text{ if } \epsilon_k>\epsilon_F,
c_k^\dagger|0\rangle=0,\text{ if } \epsilon_k<\epsilon_F,
$$
then the same vacuum is elominated by the particle and hole annihilation operators:
$$p_k|0\rangle=0,h_k|0\rangle=0.$$
A: Another way to see that $C|0\rangle$ is the state with all states occupied is to start with the full state:
$$ |f\rangle = c_1^\dagger c_2^\dagger...c^\dagger_N |0\rangle$$
and then insert the unity many times:
$$ \hspace{35mm}|f\rangle = C^{-1}C c_1^\dagger C^{-1}C c_2^\dagger C^{-1}C ...C^{-1}Cc^\dagger_N C^{-1}C|0\rangle\\= C^{-1}c_1c_2...c_NC|0\rangle $$
The only way the second line doesn't vanish is if $C|0\rangle$ has all the states occupied.
