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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.

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closed as unclear what you're asking by CR Drost, Jon Custer, GiorgioP, Rory Alsop, Phonon May 5 at 22:13

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    $\begingroup$ I flagged this as unclear what you're asking; please include any further details about what eq 7.46 is, what you are confused by, and any steps you have taken to try and derive the equation similarly, and where that has tripped you up. $\endgroup$ – CR Drost May 3 at 20:28
  • $\begingroup$ Sure. I 've said more about that $\endgroup$ – hehehehehehe May 4 at 3:29
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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.

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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.

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  • $\begingroup$ I think there are several typos as though, but the final conclusion eq(7.47) seems correct. Could you check my derivation here? $\endgroup$ – hehehehehehe May 4 at 3:33

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