Timeline for Correlation function of single annihilation/creation operator vanishes

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Mar 14, 2020 at 8:11	comment	added	user245141		No, I mean a standard Hamiltonian like the tight-binding one you wrote. Note that $a$ and $a^{\dagger}$ are not independent. If you take $a\to -a$ you must do the same for its hermitian conjugate, in order for it to remain the hermitian conjugate. In fact, for tight-binding Hamiltonians the symmetry is more general $a\to \exp(i\theta)a$ and this $U(1)$ symmetry reflects the conservation of number of particles. For a SC Hamiltonian this $U(1)$ symmetry breaks and only the $Z_2$ remains (multiplication by $-1$) as the number of particles if not conserved, but the parity of it is.
Mar 13, 2020 at 21:12	comment	added	yanscha		Oh wait, I think, I misunderstood you. With "quadratic in a" you mean quadratic in the annihilation operator, not like $a a^\dagger$, right? Then I see your argument! Thank you. But then still, I have no idea how to see, that $<a>$ is zero for example for a tight binding Hamiltonian $H=-t \sum_{<ij>}a_i^\dagger a_j$.
Mar 13, 2020 at 20:02	comment	added	yanscha		Ah that's how you can show it, thank you!! What about $<aa>$ though? It is also even under $a \rightarrow -a$, right?
Mar 13, 2020 at 19:44	history	answered	user245141	CC BY-SA 4.0