Propagators in Frequency Space and Dyson's Expansion

Question

I am reading the book Topics in Advanced Quantum Mechanics by Holstein. In the book (chapter 1.2) he shows that the time evolution operator $U(t)$ (for a Hamiltonian $H = H_0 + V$, with $V$ time independent) can be written as a Fourier transform $$\int_{-\infty}^{\infty}\frac {d\omega}{2\pi}e^{-i\omega t}\frac{i}{\omega - H_0 -V +i\epsilon}$$ where we define $K(\omega)$ as $$K(\omega) = \frac{i}{\omega - H_0 -V+i\epsilon}.$$ One can then expand $K(\omega)$ $$-iK(\omega) = -i(K^0(\omega) +K^0(\omega)(-iV)K^0(\omega)+\cdots,$$ where $K^0(\omega) = \frac i {\omega - H_0+i\epsilon}$, in analogy with the time representation of Dyson's series or the Born series. Afterwords, one could Fourier transform to obtain the original Dyson series. Upon working through the calculation, I don't understand how one gets back the integral over time in the Dyson series. For example, the first order term in the Dyson series is $$-i\int_0^t dt_1e^{-iH_0(t-t_1)}V(t_1)e^{-iH_0t_1}$$ but the Fourier transform of the frequency space representation gives $$\int_{-\infty}^{\infty}\frac {d\omega}{2\pi}K^0(\omega)VK^0(\omega),$$ which doesn't seem to be equal to the first integral. Holstein's justification for this is that (dropping the bounds on the $\omega$ integrals) $$\int \frac {d\omega}{2\pi} e^{-i\omega t}K^0(\omega)VK^0(\omega) = \int \frac {d\omega}{2\pi} \int \frac {d\omega'}{2\pi}\int _0^t dt_1 e^{-i\omega (t-t_1)}e^{-i\omega't_1}K^0(\omega)VK^0(\omega'),$$ which if true gives us the first order term in Dyson's series. My Question is, why is this true? If it isn't true, how is it that we recover the time representation of the Dyson series from it's frequency space representation?

Answer 1

From RHS of your last equation, firstly, you can take out the factor involving $t_1$, i.e. $\int_{-\infty}^{\infty}dt_1e^{-i(\omega'-\omega)t_1}=2\pi \delta(\omega'-\omega)$. Then the integral over $\int d\omega'$ will replace $\omega'$ by $\omega$ because of the delta function $\delta(\omega'-\omega)$ and the result is just the LHS of your last equation.

Answer 2

Just factorize all t1 values, and integrate over the integrand. You then replace the substituent value from the delta function and you will get your formal equation

Answer 3

It seems to me that there is a typo in Holstein's book. The operator $V$ is not supposed to be time-dependent (except, of course, in the interaction picture). So e.g. in Eq. (4.18), it is supposed to be $V$ instead of $V(t_1)$.

Then, everything is easy: since $V$ on the right hand side of your final equation does not depend on $t_1$, you can carry out the $t_1$-integration and obtain the left hand side immediately.