Electromagnetic linear response theory in function integral language [closed]

Question

I'm being confused with the discussion in Altland&Simons' textbook, page P391-392. How is eq(7.46) derived? Specificly, Here we have the action: $$S[\bar{\psi},\psi,A]=\int dx \bar{\psi_\sigma}\left( \partial_\tau+\phi+\frac{1}{2m}(-i\nabla-A)^2-\mu+V_0 \right)\psi_\sigma+S_{int}[\bar{\psi},\psi]$$ using $j_\mu = \frac{\delta S_c[A]}{\delta A_\mu}$, I got a different identity: $$j_{0}^{A}=-i\hat{\rho}=-i\bar{\psi_\sigma}\psi_\sigma$$ $$j_i^A = -\hat{j_i} = \frac{1}{2m}\bar{\psi}_{\sigma}(+i\overset{\leftrightarrow}{\partial_{i}}+2A_i)\psi_\sigma$$ which seems different from 4-current vector in Euclidean geometry: $$j_\mu = (+i\rho,j_i)$$ Moreover,the response kernel $K_{\mu\nu}(x,x^{\prime})$ in functional derivative formalism should be: $$K_{\mu\nu}(x,x^{\prime})=-Z^{-1}\frac{\delta^2}{\delta A_\mu(x) \delta A_\nu(x^{\prime})}\vert_{A=0}\int D(\bar{\psi},\psi)\exp \left(-S[\bar{\psi},\psi,A] \right)$$ instead of the one given by Altland&Simons.

Answer 1

I think it is simply the usual minimal coupling prescription $$ \partial_{\tau} \rightarrow \partial_{\tau} + e \phi $$ $$ -i \nabla \rightarrow -i\nabla - e\vec{A} $$ Where the authors use the convention $e=1$ (or, better, they define $\phi \rightarrow e \phi$ and $\vec{A}\rightarrow e\vec{A}$). This is applied to the usual action for a system of particle.

Answer 2

If the following identities are asked for: $$j_{0}^{A}=i\bar\psi_{\sigma}^{}\psi_{\sigma}^{} \hspace{0.5cm} \text{and} \hspace{0.5cm}j_{i}^{A}=\frac{1}{2m}\bar\psi_{\sigma}^{}\left(-i\overset\leftrightarrow\partial_{i}^{}+A_{i}^{}\right)\psi_{\sigma}^{}$$ Then they are obtained from their definition: $$j_{\mu}^{A}=\frac{\delta}{\delta A_{\mu}^{}}S[\psi_{}^{},\psi_{}^{},\mathbf{A}]$$ with the action given by OP.