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.

  • $\begingroup$ 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. $\endgroup$
    – G. Smith
    Jan 10, 2021 at 7:13
  • $\begingroup$ @G.Smith That's why I put the relevant equations here. All I want to ask is about them. $\endgroup$
    – xiaohuamao
    Jan 10, 2021 at 9:17

1 Answer 1


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 $\bf r-r'$

There are inconsistence somewhere between my result and the authors', maybe from typos or different style of notation, but in general by this way you could get the proper result.

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.


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