Symmetry breaking and non-vanishing mean field parameters

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6 months ago by
https://imgur.com/eTXUFbW

that for a translational invariant lattice, the following operator
$\rho\left(\mathbf{Q}\right)=\sum_{\mathbf{k},\sigma}c^{\dagger}_{\mathbf{k}\sigma}c_{\mathbf{k}+\mathbf{Q}\sigma}$ρ(Q)=k,σckσck+Qσ    →  < ckσ ck+Qσ>=  $\frac{1}{Z}$1Z  ∑P  <P| ckσ ck+Q σ |P>  $\ne$ 0 ,                         (1)
should have a nonvanishing expectation value and this implies of the existence of density waves. In the next step, he shows that the thermal average of such operator is zero, due to the orthogonality of the thermal states in the sum, and therefore, no density wave can exist in a translational invariant lattice. Yet, since we know that the density waves do exist, he justifies the argument by saying that the summation over the thermal states is restricted. Such restriction stems from a symmetry breaking, like the Mexican hat phase transition. This phase transition falls the Hilbert space of the system into different sections, which interrupts that summation and does not allow it to run over all possible states. It can only run over low energy states, which are defined by the new symmetry of the system, and not all of them. As a consequence, one may obtain a nonvanishing and physically meaningful result. Do you think that this explanation, which is my perception from the text, is convincing?

I think there are several ambiguous points here that need to be resolved. I will go through them one by one.

1. The thermal average of the operator $c^\dagger_{k , \sigma} c_{k+Q , \sigma}$ has been calculated with respect to states $| P \rangle$, which are the common eigenbasis of the Hamiltonian and the translation operator $T(R) = \exp(R \cdot P)$, with P being the total momentum operator of the system and R the displacement operator (displacing all particles of the system by the amount of R ). This is a suitable eigenbasis for the translational invariant systems. Now the question is: the orthogonality problem that makes the thermal average sum to be vanishing, explained in the page 72 of the Bruus, cannot be solved truly by some sort of restricted sum. Because it is vanishing on all sections of the Hilbert space that the eigenbasis $| P \rangle$ covers.

#The Art of Mean-Field Theory