Space Energy Propagator

• I
Homework Helper
Gold Member

Summary:

What is the space energy propagator?

Main Question or Discussion Point

This is section 16.3 of QFT for the Gifted Amateur. I understand the concept of the spacetime propagator ##G^+(x, t, x', t')##, but the following propagator is introduced without any explanation I can see:
$$G^+(x, y, E) = \sum_n \frac{i\phi_n(x)\phi_n^*(y)}{E - E_n}$$
It would be good to have an explanation of what this is and what role it plays. Thanks.

JD_PM

Related Quantum Physics News on Phys.org
vanhees71
Gold Member
2019 Award
This is the representation of the "spacetime propagtor" in terms of energy eigenfunctions. It's quite easy to prove. I'd like to write ##\tilde{G}^+(x,y,E)##, because it's a different function than ##G^+(x,t,x',t')##. You have of course
$$G^+(x,t,x',t')=G^+(x,x',t-t'),$$
because obviously the textbook author considers a Hamiltonian that's not explicitly time dependent and thus the system is time-translation invariant. Then you have
$$G^+(x,t,x',t')=\int_{\mathbb{R}} \frac{\mathrm{d} E}{2 \pi} G^+(x,x',E) \exp[-\mathrm{i} E(t-t')]. \qquad (**)$$
Further I also guess that ##G^+## is the retarded propagator. It's necessary to know that, because the definition of ##\tilde{G}^+## is incomplete without telling, how to deal with the poles at ##E=E_n##. For the retarded propgator you get
$$\tilde{G}^{+}(x,y,E)=\sum_n \frac{\mathrm{i} \phi_n(x) \phi_n^*(y)}{E-E_n+\mathrm{i} 0^+}. \qquad (*)$$
Then you have
$$(\mathrm{i} \partial_t -\hat{H}) G^+(x,t,x',t') = \int_{\mathrm{R}} \frac{\mathrm{d} E}{2 \pi} \exp[-\mathrm{i} E (t-t')] [E-\hat{H}] \sum_n \frac{\mathrm{i} \phi_n(x) \phi_n(x')}{E-E_n + \mathrm{i} 0^+} =\int_{\mathrm{R}} \frac{\mathrm{d} E}{2 \pi} \exp[-\mathrm{i} E (t-t')] \sum_n \mathrm{i} \psi_n(x) \psi_n^*(x') = \mathrm{i} \delta(t-t') \delta^{(3)}(\vec{x}-\vec{x}').$$
In the last step I've used the completeness of the energy eigenbasis.

With the choice of the regularization of the poles given (*), ##G^+## is choosen to be the retarded Green's function, i.e.,
$$G^+(x,t,x',t') \propto \Theta(t-t'),$$
as one can see, when taking the Fourier integral in (**) by closing the integration path in the complex energy plane with a large circle. For ##t-t'>0## you have to close the circle in the lower plane, including all the poles. So there you get some non-zero value, but for ##t-t'<0## you have to close the contour in the upper plane, and this gives 0, because there are no poles in the upper plane (thanks to the ##+\mathrm{i} 0^+## in the denominator). For ##t-t'<0## you can solve the integral very easily using the Theorem of Residues, leading finally to
$$G^+(x,t,x',t')=\Theta(t-t') \sum_{n} \phi_n(x) \phi_n^*(x') \exp[-\mathrm{i} E_n(t-t')].$$
Using ##\partial_t \Theta(t-t')=\delta(t-t')## it's easy to show again that this is indeed the retarded Green's function using the completeness of the energy eigenbasis.

JD_PM and PeroK
Homework Helper
Gold Member
Thanks. So, to summarise: ##\tilde G(x, y, E)## is the Fourier transform of ##G(x, y, t)##.

Where would you use ##\tilde G(x, y, E)##? Is it an intermediate stepping stone to get to ##G(p, E)##?

vanhees71