Q:

Questions about perturbation theory

In time-dependent perturbation theory, when assuming $\hat{V}$ is time-independent, 
the time development operator is as:
$$\hat{U}(t,0)\theta(t)=e^{-i(\hat{H_{0}}+\hat{V})t}=\int \frac{dw}{2\pi}\frac{i}{w-\hat{H_{0}}-\hat{V}+i\epsilon}e^{-iwt}$$ 
where $\theta(t)$ is equal to 1 when $t\geq1$， and 0 otherwise.
My question is the following:

Clearly in above equation an extra imaginary $i\epsilon$ is added into $\hat{H_{0}}+\hat{V}$. Could anyone justify such an added term?
Even we admit that added term, I still did not figure out how the second equation holds:
$$e^{-i(\hat{H_{0}}+\hat{V}-i\epsilon)t}=e^{-i(\hat{H_{0}}+\hat{V}-i\epsilon)t}\int \frac{dw}{2\pi}e^{iwt}e^{-iwt} =\int\frac{dw}{2\pi}(e^{iwt}e^{-i(\hat{H_{0}}+\hat{V}-i\epsilon)t})e^{-iwt}.$$

I cannot see the terms in parentheses is equal to $\frac{i}{w-\hat{H_{0}}-\hat{V}+i\epsilon}$

When I did numerical calculations, how should I choose a proper value for $\epsilon$

A:

The $i\epsilon$ is introduced to make sure the Fourier Transform converges. If you take directly the $e^{-i(H_0+V)t}\theta(t)$ and you fourier transform it directly you get:
$$\int_{-\infty}^{\infty} dt\theta(t)e^{-i(H_0+V)t}e^{i\omega t} = \int_{0}^{\infty} dte^{-i(H_0+V-\omega)t}=\frac{e^{-i(H_0+V-\omega)t}}{-i(H_0+V-\omega)}\bigg|_{t=0}^{t=\infty}$$
The integral is purely imaginary exponential which does not converge if $t\rightarrow \infty$ ($e^{i\infty}=??$). The only way to make sure it converges is to arbitrarily add an imaginary term in the exponential to make the integrant go to $0$ as $t\rightarrow\infty$. Since $t>0$ we need to add a $
-i\epsilon$ so that:
$$\int_{-\infty}^{\infty} dt\theta(t)e^{-i(H_0+V-i\epsilon)t}e^{i\omega t} = \int_{0}^{\infty} dte^{-i(H_0+V-\omega)t}e^{-\epsilon t}=\frac{i}{\omega-H_0 -V+i\epsilon}$$
which now converges when $t\rightarrow\infty$. The result of this Fourier transform is the result you stated.
Thus you have your result by executing the inverse transform (and taking $\epsilon\rightarrow 0$):
$$e^{-i(H_0+V)t}\theta(t)=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\frac{i}{\omega-H_0 -V+i\epsilon}e^{-i\omega t}$$
Conversely, if you had $\theta(-t)$ in the initial integrant, you would have to add a $+i\epsilon$ instead. Thus, time causality is dictated by the sign of this infinitesimal term.

