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?

  • 1
    $\begingroup$ If we talk about condensed matter, then to claim that a Hamiltonian satisfies particle-hole symmetry one needs to define holes, which implies choosing a ground state. In other words, the ground state is already implicitly present in the initial Hamiltonian. $\endgroup$
    – Roger V.
    Jun 30, 2022 at 9:38
  • $\begingroup$ @RogerVadim Thank you for your comment. Could you expand upon this a bit further please? The transformation $U$ I have defined above surely has not chosen a ground state? $\endgroup$ Jun 30, 2022 at 9:42
  • $\begingroup$ You don't specify a Hamiltonian, so it is hard to discuss it... However, holes in condensed matter are usually introduced via annihilation of electrons under the Fermi sea (e.g., see this answer). This requires defining the ground state as a filled Fermi shell (or filled valence band in semiconductors/insulators). In other words, "vacuum" in condensed matter is not a physical vacuum of QFT, but a ground state. $\endgroup$
    – Roger V.
    Jun 30, 2022 at 9:49
  • $\begingroup$ @RogerVadim I have prodived an example Hamiltonian, please see my edits. $\endgroup$ Jun 30, 2022 at 10:02
  • $\begingroup$ Indeed, you can formally define particle-hole symmetry for a Hamiltonian itself, just as exchanging creation and annihilation operators. However, you have to transform not only the Hamiltonian, but also the states, and I think that only at half-filling the ground state will transform into itself. $\endgroup$
    – Roger V.
    Jun 30, 2022 at 10:08

2 Answers 2


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.


While I realise your question is in the second quantized picture and this may not be the answer you're looking for, I just wanted to offer a simple argument in the first quantized picture that would demonstrate the symmetry of PHS Hamiltonians about $E = 0$ (half filling).

Define $\hat{\mathcal{P}}$ as the particle-hole operator in the second quantized picture. The action of $\hat{\mathcal{P}}$ is defined as

\begin{equation} \hat{\mathcal{P}}\hat{c}^{}_{A}\hat{\mathcal{P}}^{-1} = \sum_{B} \hat{c}_{B}^{\dagger}(U_{P})_{B, A} \end{equation}

for some unitary matrix $U_{P}$ (i.e. we turn an electron into a hole and vice-versa). $\mathcal{P}$ is also linear in the sense $\mathcal{P}i\mathcal{P}^{-1} = i$ The action of $\mathcal{P}$ on the Hamiltonian, $H$, is therefore

\begin{equation} \begin{split} \hat{\mathcal{P}}\hat{H}\hat{\mathcal{P}}^{-1} &= \sum_{A, B} \hat{\mathcal{P}} \hat{c}_{A}^{\dagger} \hat{\mathcal{P}}^{-1} \hat{\mathcal{P}} H^{}_{A, B} \hat{\mathcal{P}}^{-1} \hat{\mathcal{P}} \hat{c}^{}_{B} \hat{\mathcal{P}}^{-1}\\ & =\sum_{A, B} \sum_{C, D} (U_{P}^{})^{\dagger}_{A, C} \hat{c}_{C} H_{A, B} \hat{c}_{D}^{\dagger} (U_{P}^{})^{}_{D, B} \\ &= \sum_{A, B}\sum_{C, D} \delta^{}_{C, D}(U_{P})^{\dagger}_{A, C}H^{}_{A, B}(U_{P})^{}_{D, B} - \hat{c}^{\dagger}_{D}(U_{P})^{}_{D, B}H^{}_{A, B}(U_{P})^{\dagger}_{A, C}\hat{c}^{}_{C} \\ & = \sum_{A, B}\sum_{C, D} (U_{P})^{\dagger}_{A, C}H^{}_{A, B}(U_{P})^{}_{C, B} - \hat{c}^{\dagger}_{D}(U_{P})^{}_{D, B}H^{T}_{B, A}(U_{P})^{\dagger}_{A, C}\hat{c}^{}_{C} \\ & = \sum_{A, B}\left( \delta^{}_{A, B}H_{A, B} - \sum_{C, D}\hat{c}^{\dagger}_{D}(U_{P})^{}_{D, B}H^{T}_{B, A}(U_{P})^{\dagger}_{A, C}\hat{c}^{}_{C} \right) \\ & = \text{Tr}(H) - \sum_{C, D} \hat{c}^{\dagger}_{D}H^{}_{D, C}\hat{c}^{}_{C}, \end{split} \end{equation}

where $H$ is the first quantized Hamiltonian. For the Hamiltonian to respect PHS we require $\hat{\mathcal{P}}\hat{H}\hat{\mathcal{P}}^{-1} = \hat{H}$ and therefore

\begin{equation} \begin{split} &\text{Tr}(H) = 0 \\ U_{P}H^{*}&U_{P}^{\dagger} = -H. \end{split} \end{equation}

These conditions are complementary in the sense that the second condition implies the spectrum is symmetric about $E=0$ (try it!) which implies the first. Indeed, diagonalizing the Hamiltonian, $H = \Lambda D \Lambda^{-1}$, we find $\text{Tr}(H) = \text{Tr}(\Lambda D \Lambda^{-1}) = \text{Tr}(D) = 0$. Therefore, our system has $n$ filled and $n$ empty bands and is at half-filling.

  • $\begingroup$ Thank you for your answer. You have worked with a Hamiltonian that is quadratic in the field operators, however would the same conclusions apply if the second-quantised Hamiltonian was interacting, i.e. quartic? $\endgroup$ Jun 30, 2022 at 10:18
  • $\begingroup$ Interesting question - off the top of my head I'm not sure, but the treatment above could be easily extended to an interaction term, e.g. $\hat{\mathcal{P}}\hat{c}_{i}^{\dagger}\hat{c}_{j}^{\dagger}\hat{c}_{k}\hat{c}_{l}\hat{\mathcal{P}}^{-1} = \hat{\mathcal{P}}\hat{c}_{i}^{\dagger}\hat{\mathcal{P}}^{-1} \hat{\mathcal{P}}\hat{c}_{j}^{\dagger}\hat{\mathcal{P}}^{-1} \hat{\mathcal{P}}\hat{c}_{k}\hat{\mathcal{P}}^{-1} \hat{\mathcal{P}}\hat{c}_{l}\hat{\mathcal{P}}^{-1}$ $\endgroup$
    – Niall
    Jun 30, 2022 at 10:41

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