# Fourier transform of propagator in Lehmann spectral representation

I need help showing that:

$$iG(k,t)=\int_{\mu/\hbar}^{+\infty} e^{-i\omega t}A(k,\omega)d\omega$$

knowing that $$G(k,\omega) = \int_{-\infty}^{+\infty} \frac{A(k,\omega')}{\omega -\omega'+i\eta\ sgn(\omega'-\frac{\mu}{\hbar})}d\omega'$$ where $$\eta$$ is a positive infinitesimal and $$A(k,\omega)$$ is the spectral function in Lehmann representation. (See here eq. 5 for a reference)

So I did the Fourier transform $$iG(k,t) =\frac{i}{2\pi}\int_{-\infty}^{+\infty} e^{-i\omega t}d\omega\int_{-\infty}^{+\infty} \frac{A(k,\omega')}{\omega -\omega'+i\eta\ sgn(\omega'-\frac{\mu}{\hbar})}d\omega'$$

and I think that I probably need to split the integral into two, one going from $$-\infty$$ to $$\mu/\hbar$$ and the other one from $$\mu/\hbar$$ to $$+\infty$$ , and then use residue method but can't manage to get to the final result.

Rewriting the labels in your last equation, we have $$iG(k,t)={i \over 2\pi}\int_{-\infty}^{+\infty}{d\omega A(k,\omega)\int_{-\infty}^{+\infty}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}}}$$ Let us recall the Cauchy's integral formula $$f(a)={1 \over 2\pi i}\int_{C}{{f(z) \over z-a}dz}$$ where $$C$$ is some closed non-intersecting curve, $$a$$ is enclosed by $$C$$, and $$f$$ is holomorphic. Therefore, to compute the integral over $$\omega'$$, we have to choose some well-designed curve. Let $$C=C_1+C_2$$ where $$C_1$$ is a line starting from $$-R$$ to $$+R$$ with $$R>0$$, and $$C_2=\{Re^{-i\theta} | \theta \in [0,\pi]\}$$. This curve is a semicircle on the complex plane for visualization. Given $$\omega > {\mu \over \hbar}$$, the point $$\omega-i\eta\text{sgn}(\omega-{\mu \over \hbar})$$ is enclosed by $$C$$. And the integral (there is an additional minus sign on the left hand side since the path along $$C$$ is clockwise) is $${-2 \pi i}e^{-i (\omega - i \eta \text{sgn}(\omega - {\mu \over \hbar}))t}=\int_{C_1}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}} + \int_{C_2}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}}$$ Notice if we make $$R \rightarrow +\infty$$, $$\lim_{R \rightarrow +\infty}{\int_{C_1}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}}} = \int_{-\infty}^{+\infty}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}}$$ In addition, for large enough $$R$$ and $$t>0$$, \begin{align} \bigg|\int_{C_2}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}}\bigg| & \leq \int_{0}^{\pi}{{|e^{-i Re^{-i\theta} \ \ t}| \over {1 \over 2}|Re^{-i\theta}t|} |Re^{-i\theta}(-it)|d\theta} \\ & = \int_{0}^{\pi}{2te^{-R\sin{\theta}t} d\theta} \end{align} When $$R \rightarrow +\infty$$, $$\int_{0}^{\pi}{2te^{-R\sin{\theta }t} d\theta} \rightarrow 0$$. Therefore, $$\int_{-\infty}^{+\infty}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}} = - 2\pi ie^{-i (\omega - i \eta \text{sgn}(\omega - {\mu \over \hbar}))t} = - 2\pi ie^{-i (\omega - i \eta)}$$ What if $$\omega < {\mu \over \hbar}$$? The process works similarly except the function $$f(\omega')={e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}$$ is holomorphic on the region enclosed by $$C$$ (the only pole at $$\omega+i\eta$$ is in the upper side of the complex plane). Therefore, we have $$0$$ on the left hand side instead of $$-2\pi ie^{-i (\omega - i \eta \text{sgn}(\omega - {\mu \over \hbar}))t}$$ at the start of computing the path integral along $$C$$. Therefore, if $$\omega < {\mu \over \hbar}$$, $$\int_{-\infty}^{+\infty}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}} = 0$$ This implies $$iG(k,\omega) = \int_{\mu \over \hbar}^{+\infty}{e^{-i\omega t}A(k,\omega)d\omega}$$
• Thanks this seem right, I was missing swapping the integrals at the beginning. But aren't you missing the initial i (see the difference between my last integral and your first one)? In exercise (eq.9) he asks to show that $iG$ is equal to the last integral in your answer, shouldn't I get a $-1$ using your method?
• If I use Jordan's lemma, since I'm closing in the lower side of the plane, I get $-2\pi i \cdot Residue$, which is your result with a minus, that gets to the right final result. So there's probably a sign error in your answer somewhere.
• Thank you for the correction! And yes, I made a sign error since my path integral along $C$ is clockwise, where I forgot to put an additional minus sign. It has been some time since I did these path integrals, and I got careless about them. The post has been updated. Jun 27, 2022 at 13:20
• I think the page is relatively friendly to non math major people. Also, swapping integral orders are quite common in physics, while most of the time they are done without proving their legitimacy. As you may see in this case, since $A(k,\omega)$ has an unknown form, we are actually unsure whether Fubini's theorem can be applied to it (perhaps still yes if we are willing to look into its form more carefully). However, I guess we do not care about it as much as mathematicians do. Jun 27, 2022 at 13:50