2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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: \begin{equation} S=\frac{1}{2m}\sum\phi(q)\phi(-q)+g\sum\phi(q)M(-q)+S_f, \end{equation} 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):

\begin{equation} \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 \end{equation} Which means that: \begin{equation} -\frac{1}{gm}\phi(q) = \left<M(q)\right>. \end{equation}

I hope it answers your question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.