Normally, for a (linear) operator $L$ and a DE

$$ Lu(x) = f(x) $$

the Green function is defined as

$$ LG(x,s) = \delta(x-s) $$

and it is found that

$$ u(x) = \int G(x,s) f(s) ds $$

is the general solution of the DE.

Now, I've read some texts about Green functions in many-body theory (example), but the form is unfamiliar to me.

Can you explain how those Green functions are introduced? I.e., why are objects of the form $\frac{1}{E - H_0 \pm i\eta}$ called Green functions (examples here and here)?


1 Answer 1


Because these are actually Fourier transform of the usual Green functions. Consider the Schrödinger equation : $$ \hat{\mathcal{H}}|\Psi(t)\rangle=\mathrm{i}\partial_t|\Psi(t)\rangle $$ The general solution $|\Psi(t)\rangle$ of such equation for a time-independant hamiltonian $\hat{\mathcal{H}}$ can be expressed in terms of Green function $G(x',x,t)$ : $$ \Psi(x,t)=\langle x|\Psi(t)\rangle=\langle x|e^{-\mathrm{i}t\hat{\mathcal{H}}}|\Psi(t=0)\rangle=\int\mathrm{d}x'\,G(x',x,t)\,\Psi(x',t=0) $$ where $G(x',x,t)=\langle x'|e^{-\mathrm{i}t\hat{\mathcal{H}}}|x\rangle$. The last equality is obtained by introducing the closure identity : $$ \int\mathrm{d}x'\,|x'\rangle\langle x'|=\hat{1} $$

One can then define a Green operator : $$ \hat{G}(t)=-\mathrm{i}\,\Theta(t)\,e^{-\mathrm{i}t\hat{\mathcal{H}}} $$ where $\Theta$ stands for the Heaviside step function which is here to ensure the causality of the solution $\Psi(x,t)$.

Then, one can compute the Fourier transform of such operator, which sometimes is called resolvent operator : $$ \hat{G}(\epsilon)=\int\mathrm{d}t\,\hat{G}(t)\,e^{\mathrm{i}\epsilon t}=-\mathrm{i}\int\mathrm{d}t\,\Theta(t)\,e^{\mathrm{i}t(\epsilon-\hat{\mathcal{H}})} $$

Then one can express the $\Theta$ function in terms of its Fourier transform : $$ \Theta(t)=\int\frac{\mathrm{d}\omega}{2\pi\mathrm{i}}\,\frac{e^{-i\omega t}}{\omega-\mathrm{i}\eta} $$ where $\eta$ is a positive infinitesimal parameter.

Taking all of this together, you will find that : $$ \hat{G}(\epsilon)=-\frac{1}{2\pi}\int\mathrm{d}\omega\,\frac{1}{\omega-\mathrm{i}\eta}\int\mathrm{d}t\,e^{\mathrm{i}t(\epsilon-\omega-\hat{\mathcal{H}})} $$ It is possible to recognize with the Fourier transform of the Dirac distribution : $$ \delta(\omega)=\frac{1}{2\pi}\int\mathrm{d}t\,e^{\mathrm{i}\omega t} $$ that : $$ \hat{G}(\epsilon)=-\int\mathrm{d}\omega\,\frac{1}{\omega-\mathrm{i}\eta}\;\delta(\omega+\epsilon-\hat{\mathcal{H}})=\frac{1}{\epsilon+\mathrm{i}\eta-\hat{\mathcal{H}}} $$ which is what you are looking for.

  • $\begingroup$ Thank you for the thorough answer. I understood everything except the second formula. Can you explain why this $\Psi$ is truly a solution of the Schrödinger equation? $\endgroup$
    – Minethlos
    Jun 9, 2015 at 14:40
  • $\begingroup$ See edit $\uparrow$ for more precision. $\endgroup$
    – dolun
    Jun 9, 2015 at 16:02
  • $\begingroup$ It seems to me that there is a minus sign mistake, as you have $it(\epsilon - \omega -H)$ in the exponential, and then this becomes $\omega + \epsilon - H$ in the delta function. $\endgroup$
    – jarm
    Nov 20, 2015 at 11:57

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