Fermion parity operator is defined as $$ \hat{\mathcal{Q}}=\exp(i\pi\sum_j \hat{n}_j) = (-1)^{\sum_j \hat{n}_j} $$ And also if $\sum_j \hat{n}_j = \sum_j c^{\dagger}_{j}c_j=N $ is constant then it commutes with Fermionic Hamiltonion, i.e $$ [\hat{\mathcal{Q}},\hat{H}]=0 $$ What is the physical meaning of this Fermion parity operator?


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The Fermionic parity operator $(-1)^N$ literally tells you is there an odd number of particles or even number of particles in the system.

Sure $[H, Q]=0$ if the particle number is conserved. But actually even if particle number is not conserved, we still have Fermionic parity conserved, i.e. $[H,Q]=0$ still holds. That means, if you start with a system of even(odd) number of Fermions, you will always end up with even(odd) number of Fermions.

Fermionic parity is always conserved as long as Fermionic operators come up in pairs in our Hamiltonian. That is, the Hamiltonian contains only terms like $c_j^\dagger c_j,c_jc_j,c^\dagger_jc^\dagger_j,c_i^\dagger c^\dagger_jc_jc_i$, etc. If this condition is satisfied, the expansion of $U=\exp(-itH)$ will only contains those terms as well, and thus Fermions are always created or annihilated in pair. You will see $c_j^\dagger c_j$ as number operator, $c_i^\dagger c_j$ as hopping term, $c_i^\dagger c^\dagger_jc_jc_i$ as interacting term and terms like $c_jc_j$, $c_j^\dagger c_j^\dagger$ in superconductivity, which lead to non-conservation of particle number(but still Fermionic parity conserves).

It seems that a Fermionic system always preserved this symmetry, which actually made some problem hard to solve. So a more interesting question is "Why does a fermionic Hamiltonian always obey fermionic parity symmetry?", i.e. why is there no term like $c_j+c_j^\dagger$? See the detailed discussions there.

Additionally, another operator similar to $\hat{Q}$ in your question is $Q_j=\exp(i\pi\sum_{i<j}n_i)$ in 1d systems, which can make Bosonic operators into Fermionic ones(this is only possible in 1d). It is very useful in JW transformation.


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