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:

  1. Clearly in above equation an extra imaginary $i\epsilon$ is added into $\hat{H_{0}}+\hat{V}$. Could anyone justify such an added term?

  2. 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 $\frac{i}{w-\hat{H_{0}}-\hat{V}+i\epsilon}$

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

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.

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  • $\begingroup$ So in numerical calculations, can I also take $\epsilon$ in the result to be zero? Besides, I see in some cases, $\epsilon$ is added to make sure the state decays into ground state. Is there any physical reason to add the imaginary $\epsilon$ apart from seeking a convergent integral? $\endgroup$ – xiang sun Jul 1 '19 at 14:11
  • 1
    $\begingroup$ Well theoretically speaking, you need the $\pm i\epsilon$ to make the integral converge. So numerically you would put a small non-zero value. The physical reason, as stated in the answer, is to make sure time causality is preserved (because of the $\theta(\pm t)$ functions). $\endgroup$ – fgoudra Jul 1 '19 at 14:24

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