# Why is the auxiliary field from Hubbard-Stratonovich transformation for Hubbard model the same as the mean-field ferromagnetic order parameter?

This question comes from the Fradkin's Field Theory of Condensed Matter Physics, P50 (3.122), $$M = \sqrt{\dfrac{4U}{3}} \phi$$ where $M$ is the ferromagnetic order parameter and $\phi$ is the auxiliary field from the Hubbard-Stratonovich transformation. The book argues that because the above equation is correct, the mean field theory which is derived from the Hartree-Fock approach is equivalent to the the saddle point approximation formalism for H-S transformation auxiliary field Lagrangian. But I can not understand the equation.

• – leongz Mar 2 '17 at 8:38

You can always derive connection between auxiliary fields in saddle point and averages of fermion operators they are coupled to. For example, take a full decoupled action: $$S=\frac{1}{2m}\sum\phi(q)\phi(-q)+g\sum\phi(q)M(-q)+S_f,$$ where $\phi$ is an auxiliary field, $S_f$ denotes fermonic part of action and $M(-q)=\frac{1}{2}\sum_{k\alpha\beta}\bar{\psi}_\alpha(k-q)\psi_\beta(k)\tau^z_{\alpha\beta}$ is just a FT of spin in z direction. Now, consider a variational derivative with respect to $\phi_q$ in a saddle point (where $\frac{\delta}{\delta \phi_q} S=0$ holds):
$$\frac{1}{Z}\frac{\delta }{\delta \phi_q}Z = \frac{1}{m}\phi_{MF}(-q)+g\left<M(-q)\right>_{MF} = -\frac{1}{Z}\int D(\bar{\psi}, \psi)e^{-S}\frac{\delta }{\delta \phi_q}S = 0$$ Which means that: $$-\frac{1}{gm}\phi(q) = \left<M(q)\right>.$$