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?


2 Answers 2


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.

  • $\begingroup$ If $A$ is fermionic (odd) then $\langle A\rangle=0$. $\endgroup$ Commented Nov 20, 2016 at 17:02
  • $\begingroup$ 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. $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, $\langle\psi|a|\psi\rangle=\alpha^*\beta$. $\endgroup$ Commented Nov 20, 2016 at 17:15
  • $\begingroup$ 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. $\endgroup$ Commented Nov 20, 2016 at 18:02
  • $\begingroup$ 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. $\endgroup$ Commented Nov 20, 2016 at 20:09
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    $\begingroup$ @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. $\endgroup$ Commented Nov 23, 2016 at 0:05

Actually Wikipedia has an answer for your question,


which will tell you how to bulid a mean field approximation self-consistently based on the Bogoliubov inequality.

enter image description here

If you want to know more details about the fundamental inequality,you can go through the book,Statistical Mechanics: A Set Of Lectures,written by Feynman.

Hope it helps.

  • $\begingroup$ Thank you, So the mean field approx. is just equivalent to the variational method i.e choosing the wave function such that the expectation value of the Hamiltonian becomes as much small as possible ? And so this can describe why do both the Bogoliubov's mean field model and the BCS variational wave function give the same result in the theory of superconductivity? $\endgroup$
    – Hossein
    Commented Nov 20, 2016 at 20:25
  • $\begingroup$ In my opinion,the mean field approximation is just a method to let us to decouple the interacting many-body problem into an effective non-interacting many-body problem,which is actually a single-body problem.But in some strong correlation materials the method will drop the essential physics so it doesn't work well. $\endgroup$
    – Jack
    Commented Nov 21, 2016 at 4:18

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