# What is the definition of particle-hole symmetry in condensed matter physics?

People often talk about particle-hole symmetry in solid state physics. What are the exact definition and physics picture of particle-hole symmetry? How to define the density of particles and holes?

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I asked this question myself and couldn't find a clear cut definition, but there are some articles and papers that circumscribe it. I think it is possible to get a sense of particle-hole symmetry out of them. These two describe the transformation: prb.aps.org/pdf/PRB/v84/i20/e205121 and sciencedirect.com/science/article/pii/S0375960197001631 . Describes symmetry in atomic limit (no jumping to or from impurity): raas.de/files/Raas_Carsten_PhD_Thesis.pdf Edit: re-entered the comment because there was a broken link and could'nt edit the comment anymore. – hauntergeist Jun 13 '14 at 11:43

If you have a superconductor with N electrons in them (some electrons form Cooper pairs and some electrons are unpaired), the BdG Hamiltonian is of size $2N\times2N$ and the state of the system is written as a vector that consists of $N$ annihilation operators for electrons an $N$ creation operators for electrons above each other with the annihilation operators first. You can view the bottom creation operators for the electrons as annihilation operators for holes. You then have a complete vector of annihilation operators of length $2N$. Holes do the exact opposite of electrons and therefore you expect that the energy levels are complete symmetric with respect to the Fermi level. While you have $N$ electrons in the system, there are $2N$ energy levels which form pairs.
For superconductors the particle hole symmetry operator has the form $P = \tau_x K$, where $\tau_x$ is a Pauli matrix and $K$ means complex conjugation. The Hamiltonian then satisfies the following equation: $$P^\dagger H P = H \quad\Rightarrow\quad \tau_x H^* \tau_x = H$$ and $P^2$ is the identity matrix. Other Hamiltonians can also have particle hole symmetry and the PHS operator may then different, but is always squares to $1$ or $-1$.