Validity of mean-field approximation

In mean-field approximation we replace the interaction term of the Hamiltonian by a term, which is quadratic in creation and annihilation operators. For example, in the case of the BCS theory, where

$$\sum_{kk^{\prime}}V_{kk^{\prime}}c_{k\uparrow}^{\dagger}c_{-k\downarrow}^{\dagger}c_{-k^{\prime}\downarrow}c_{k^{\prime}\uparrow}\to\sum_{k}\Delta_{k}c_{k\uparrow}^{\dagger}c_{-k\downarrow}^{\dagger} + \Delta_{k}^{\star}c_{-k\downarrow}c_{k\uparrow}\text{,}$$

with $\Delta_{k}=\sum_{k^{\prime}}V_{kk^{\prime}}\langle c_{-k^{\prime}\downarrow}c_{k^{\prime}\uparrow}\rangle\in\mathbb{C}$. Then, in books, like this by Bruss & Flensberg, there is always a sentence like "the fluactuations around $\Delta_{k}$ are very small", such that the mean-field approximation is a good approximation. But we known for example in the case of the 1D Ising model the mean-field approximation is very bad.

My question: Is there a inequality or some mathematical conditions which says something about the validity of the mean-field approach? Further, is there a mathematical rigoros derivation of the mean-field approximation and the validity of it?

You can introduce the would-be'' bosonic mean field exactly, using the Hubbard-Stratonich (a.k.a partial bosonization) method, see wikipedia and Interacting fermions on a lattice and Hubbard-Stratonovich transformation and mean-field approximation .

The mean field approximation correspond to performing the integral over the bosonic field using the stationary phase approximation. The fermion action is bilinear and can be performed exactly. Corrections to the stationary phase approximation correspond to fluctuations around the mean field. These are small if the four-fermion coupling in the BCS theory is small. If the coupling is strong it may be possible to justify the mean field approximation by a large N (N component vector field) or large d (number of dimensions) approximation.

Schmatically, the effective action of the bosonized theory is $$S = {\rm Tr}\{\log[G_0^{-1}G(\phi)]\} -\frac{\phi^2}{g}$$ where $g$ is the coupling, $\phi$ is the bosonic field, and $$G(\phi) = \left( \begin{matrix} p_0-\epsilon_p & \phi \\ \phi & p_0+\epsilon_p \end{matrix} \right)$$ is the propagator. I also define $G_0=G(0)$. Now $$\left. \frac{\delta S}{\delta\phi}\right|_{\phi_0} = 0$$ is the MFA gap equation $$\phi_0 = g\int \frac{d^3p}{(2\pi)^3} \frac{\phi_0}{\sqrt{\epsilon_p^2+\phi_0^2}} .$$ Corrections can be found by expanding $S$ around $\phi=\phi_0+\delta \phi$. This will give higher loops containing $G(\phi_0)$. The expansion parameter is g. In physical units $1/g$ is the logarithm of the Fermi energy over the gap, $g\sim [\log(E_F/\phi_0)]^{-1}$.

Near $T_c$ corrections to mean field (=Landau-Ginsburg) are controlled by the Ginsburg criterion, as explained in Valrio92's answer. In weak coupling BCS the Ginsburg window is small, and mean field is accurate, except very close to $T_c$.

• With "Corrections to the stationary phase approximation" you mean high order terms in the Taylor expansion of the functional? – Lars Milz May 24 '16 at 8:51
• Added some details. – Thomas May 24 '16 at 13:10

Mean field theory is only good when fluctuations are small, which means that the free energy of a fluctuation must be much smaller than the total free energy.

The free energy of the typical fluctuation is of order $kT$ and its size is determined by the correlation length $\xi$, and is of order $\xi^d$, with $d=$ dimension:

$$F_{fluct}\sim \frac{kT}{\xi^d}\sim \mid t\mid^{\nu d}$$

Having used $\xi \sim \mid t \mid^{-\nu}$ where $t=(T-T_c)/T_c$ and $T_c$ is the critical temperature. To obtain the total free energy we must integrate twice the specific heat, $c\sim \mid t \mid^{-\alpha}$. Imposing that $F_{fluct}/F$ goes to $0$ for $t\to 0$ we obtain

$$d \nu > 2-\alpha$$

For example in the Ising model we have $\alpha=0$ and $\nu=1/2$, so that the condition is $d>4$. This is why the mean field approximation is bad for the Ising Model in less than four dimensions.

• Thanks, this is nice answer. However, I am more interested in the case of BCS theory and Hartree-Fock theory. Here it should work in a different way or? – Lars Milz May 23 '16 at 12:10
• These are quite general statistical-mechanical arguments and should hold irrespective of the model. What I stated is sometimes known as the Ginzburg Criterion. – valerio May 23 '16 at 12:29
• You are right, but what is the condition for the Hartree-Fock Approximation? Here we have no criticial exponents, just a fermionic System. – Lars Milz May 23 '16 at 21:59
• BCS theory (which is an Hartree-Fock approximation) is a microscopic theory of superconductivity, which is a critical phenomenon. So you will have a critical temperature and critical exponents. As a matter of fact, you will have the same critical exponents of the (macroscopic) Landau-Ginzburg theory, since they are both mean field theories. See for example pag. 766 here: lassp.cornell.edu/clh/Book-sample/7.3.pdf – valerio May 23 '16 at 22:24
• I think you explain the criterion for validity of the MFA near $T_c$, but the question is about $T=0$. – Thomas May 24 '16 at 1:30

Mean-field theory is exact (in the thermodynamic limit) in the case of long-range interaction (which is not the case for the nearest-neighbor Ising model). Therefore, mean-field theory is exact for BCS, where you have an effective long-range interaction. As for rigorous results, Bogoliubov rigorously proved that in the ground state (zero temperature) the energies per particle coincide for BCS and the mean-field theory in the thermodynamic limit. Later, Bogoliubov Jr. rigorously proved the same for arbitrary temperatures (and free energies per particle). See the bibliography, e.g., at http://arxiv.org/abs/1507.00563 (the discussion is around page 43).

• Not sure what you mean. 1) BCS is zero range. 2) MFA is certainly not exact for long range forces like gauge bosons, spin fluctuation etc. 3) You may mean truly infinite range (zero range in momentum space), but this is not realized in nature, – Thomas May 24 '16 at 1:37
• @Thomas: "The BCS interaction takes place exclusively at zero momentum, and as such involves an infinite range interaction between pairs. This long-range aspect of the model permits the exact solution of the BCS Hamiltonian using mean field theory."(books.google.com/… ). And it's a model, not nature:-) – akhmeteli May 24 '16 at 4:20
• s-wave BCS scatters (k,-k) into (k',-k') , so the momentum transfer q=k-k' is of order k_F, which is a short range interaction. – Thomas May 24 '16 at 4:29
• I guess you refer to the fact that the pair momentum is zero, but that does not make the MFA exact – Thomas May 24 '16 at 4:36
• @Thomas: Within the model, and in the thermodynamic limit, it does. – akhmeteli May 24 '16 at 4:48