# Why does photocurrent need $G^<$ among other Green functions?

In this highly-cited paper (or its pdf) on photoemission, Eq.(3) gives the current density in terms of $$G^<$$ $${\bf j}({\bf r},t) = 2\hbar\left( \frac{e\hbar}{2m} (\partial_{\bf r'} - \partial_{\bf r}) + \frac{ie^2}{m}{\bf A}({\bf r},t)\right) G^<({\bf r},t;{\bf r'},t)\big|_{{\bf r'}={\bf r}}.$$

It simply follows from Eq.(2) the electromagnetic coupling Hamiltonian $$H_1 = \int d{\bf r} \psi^\dagger({\bf r},t) \left( \frac{ie\hbar}{m}{\bf A}({\bf r},t)\cdot\partial_{\bf r} + \frac{e^2}{2m}{\bf A}^2({\bf r},t)\right)\psi({\bf r},t).$$ I think one can calculate $${\bf j}=\langle{\delta S/\delta {\bf A}\rangle}$$ where the action $$S$$ contains the coupling part $$H_1$$ and the expectation value leads to some Green's function.

But why is it the lesser one $$G^<$$? This is not obvious to me.

• What is $G^<$? The paper is behind the APS paywall, and, even if it were not, a question on PSE is supposed to be entirely self-contained and not require reading any linked resources. Jan 10, 2021 at 7:13
• @G.Smith That's why I put the relevant equations here. All I want to ask is about them. Jan 10, 2021 at 9:17

You could verify this by brute force. The current operator has this form: $${\bf j}(r,t)=-2\psi^\dagger({\bf r},t) \left( \frac{ie\hbar}{m}\partial_{\bf r} + \frac{e^2}{m}{\bf A}({\bf r},t)\right)\psi({\bf r},t).$$ (counting the spin degenarcy)
and r.h.s. is $$2\left( \frac{e\hbar}{2m} (\partial_{\bf r'} - \partial_{\bf r}) + \frac{ie^2}{m}{\bf A}({\bf r},t)\right) i\langle\psi^\dagger(r',t')\psi(r,t)\rangle\big|_{{\bf r'}={\bf r}}\\ =2\left( -\frac{e\hbar}{m}\partial_{\bf r} + \frac{ie^2}{m}{\bf A}({\bf r},t)\right) i\langle\psi^\dagger(r',t')\psi(r,t)\rangle\big|_{{\bf r'}={\bf r}}\\ =2\langle\psi^\dagger(r',t')\left( -i\frac{e\hbar}{m}\partial_{\bf r} - \frac{e^2}{m}{\bf A}({\bf r},t)\right) \psi(r,t)\rangle\big|_{{\bf r'}={\bf r}}\\ =-2\langle\psi^\dagger(r,t)\left( i\frac{e\hbar}{m}\partial_{\bf r} + \frac{e^2}{m}{\bf A}({\bf r},t)\right) \psi(r,t)\rangle$$ You could replace $$\partial_{\bf r'}$$ by $$-\partial_{\bf r}$$ since $$G^<({\bf r},t;{\bf r'},t')$$ only depends on $$\bf r-r'$$
PS: since I have no enough reputation to add a comment, so I do that here: maybe it would be more convenient if you directly show $$G^<({\bf r},t;{\bf r'},t')=i\langle\psi^\dagger(r',t')\psi(r,t)\rangle$$ in your question.