# General method of deriving the mean field theory of a microscopic theory

What's the most general way of obtaining the mean field theory of a microscopic Hamiltonian/action ? Is the Hubbard-Stratonovich transformation the only systematic method? If the answer is yes then what does necessitate our mean field parameter to be a Bosonic quantity ? Is the reason that all of directly physical observable quantities should commute?

In second quantization, the mean field approximation consists in approximating some combination of operators $A$ by a $c$-number $\langle A\rangle$. For example, for Bose-Einstein condenstate $A=b_0$, for Cooper pairing $A= a_{\mathbf{p}\uparrow}a_{-\mathbf{p}\downarrow}$, in the Hartree-Fock approximation $A=a_{\mathbf{p}}^+a_{\mathbf{p}}$, in the charge density wave state $A=a_{\mathbf{p}+\mathbf{q}}^+a_{\mathbf{p}}$. Here $a_i$ and $b_i$ are fermionic and bosonic annihilation operators.

The same in the Hubbard-Stratonovich transformation: we can couple any desired combination $A$ of field operators to auxiliary field.

In all mentioned cases, $A$ is indeed bosonic. Since bosonic operators can have large occupation numbers $\langle A\rangle\gg1$, deviations of $A$ from $\langle A\rangle$ can be relatively small, and also noncommutativity of $A$ and $A^+$ can be neglected (if we have the right choice of $A$): $$\frac{\sqrt{\langle (A-\langle A\rangle)^2\rangle}}{\langle A\rangle}\rightarrow0,\qquad\frac{\langle[A,A^+]\rangle}{\langle A\rangle}\rightarrow0.$$ This makes the approximation accurate.

For fermionic quantities $A$, the mean field approximation does not have much sense because the occupation number is limited, $\langle A\rangle\sim 1$. Therefore, $A$ will be strongly fluctuating with respect to $\langle A\rangle$, and the approximation will be inaccurate.

Update number 2 following the discussion in comments

Average value of the fermionic opearator can be nonzero, $\langle A\rangle\neq0$ only at coherent mixtires of different numbers of fermions in the system, e.g. $|\psi\rangle=\alpha|N\rangle+\beta|N+1\rangle$. Generally, such superpositions of integer- and half-integer angular momentum states are forbidden by superselection rules.

In any case, the average $\langle A\rangle$ of a fermionic quantity $A$ is either zero or remains small in the macroscopic limit, so it is a bad candidate for a mean field order parameter.

• If $A$ is fermionic (odd) then $\langle A\rangle=0$. Commented Nov 20, 2016 at 17:02
• I don't understand why. It can be nonzero if the system state is a coherent superposition of states with different particle numbers, e.g. $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, $\langle\psi|a|\psi\rangle=\alpha^*\beta$. Commented Nov 20, 2016 at 17:15
• yes, if you allow mixing fermionic and bosonic states. But such superpositons are forbidden by a corresponding superselection rule. journals.aps.org/pr/abstract/10.1103/PhysRev.88.101 - Ignoring this would make quantum mechanics inconsistent because even and odd states transform differently under a rotation by 360 degrees. Commented Nov 20, 2016 at 18:02
• Yes, I understood the point on boson-fermion superselection, but I think the states with $\langle a\rangle\neq0$ are still possible in open systems, for example in a semiconductor quantum dot coupled to leads or other dots (as in charge qubits), where the tunneling Hamiltonian $~|N\rangle\langle N+1|+\mbox{h.c.}$ mixes the states with different electron numbers. Commented Nov 20, 2016 at 20:09
• @AlexeySokolik The tunnelling Hamiltonian doesn't create coherent superpositions of states with a different total number of fermions. We are talking about electrons here, after all, the total number is always strictly conserved (in a low-energy setting). At best you can create (locally, e.g. when looking just at one quantum dot) an incoherent mixture of fermion number states, but a fermionic operator will still have vanishing expectation value in such a state. Commented Nov 23, 2016 at 0:05