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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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  • $\begingroup$ 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. $\endgroup$ – hauntergeist Jun 13 '14 at 11:43
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We find particle hole symmetry (PHS) for example in superconductors, where you can take for e.g. the Bogoliubov-de Gennes (BdG) Hamiltonian as a mean-field approximation. This is the only experimental example I know. There maybe other systems with particle hole symmetry but I don't know about them. Here, I will use superconductors as an example for the explanation.

If you have a superconductor with $N$ electrons (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, and $N$ creation operators for electrons, above one being 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 behave(in the sense of creation and annihilation) 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.

More technical now:

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 be then different, but it always squares to $1$ or $-1$.

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