# Propagator in Wave Mechanics Laplace-Fourier transform

In my Modern quantum mechanics, J. J. Sakurai p.119-120, when considering the integral of the propagator $$K$$ in whole space, he gets:

$$G(t)= \int d^3 x' K(\textbf{x'},t;\textbf{x'},0) = \sum_a \exp \left(\frac{-iE_{a^{'}}t }{\hbar} \right).$$ Then he considers the Laplace-Fourier transform of G(t): $$\begin{equation} \label{eq1} \begin{split} \tilde{G}(E) = & -i\int^{\infty}_0 dt G(t)\exp(iEt/\hbar) \\ & = -i\int^{\infty}_0dt\sum_{a^{'}} \exp(-iE_{a^{'}}t/\hbar) \exp(iEt/\hbar)/ \hbar. \end{split} \end{equation}$$

But why does he say that this integrand oscillates indefinitely, and the result can be changed into a definite one by making the change $$E \rightarrow E + i \varepsilon$$ to obtain: $$\tilde{G}(E)=\sum_{a^{'}}\frac{1}{E-E_{a'}}~?$$

• $e^{i x}=\cos{x}+i\sin{x}$ hence imaginary exponentials oscillate Sep 29, 2018 at 21:03

This is known as the $$\epsilon$$ prescription. We have:

\begin{equation} \begin{aligned} \tilde{G}(E) =& - \frac{i}{\hbar} \sum_{a^{'}} \int^{\infty}_0 dt \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right]\\ =& - \frac{i}{\hbar} \sum_{a^{'}} \frac{\hbar}{i} \left( \frac{1}{E - E_{a'}} \right) \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right] \Bigg|_0^{\infty} \end{aligned} \end{equation}

Now, the problem is that we cannot evaluate the upper limit of the integral because

$$\lim_{t \to \infty}\exp (itx)$$

does not exist for real $$x$$. The limit does not converge to a definite value; it keeps oscillating instead.

The trick to force convergence upon the integral is to add to $$E$$ an infinitesimal constant $$\epsilon > 0$$:

$$E \rightarrow E + i \epsilon$$

Now, the upper limit of the integral goes to zero because of exponential suppression: $$\begin{equation} \lim_{t \to \infty}- \frac{i}{\hbar} \sum_{a^{'}} \frac{\hbar}{i} \left( \frac{1}{(E + i \epsilon) - E_{a'}} \right) \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right] \exp \left[ \frac{-\epsilon t}{\hbar} \right] = 0 \end{equation}$$

$$\begin{equation} \lim_{t \to 0} \frac{i}{\hbar} \sum_{a^{'}} \frac{\hbar}{i} \left( \frac{1}{(E + i \epsilon) - E_{a'}} \right) \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right] \exp \left[ \frac{-\epsilon t}{\hbar} \right] = \sum_{a^{'}} \left( \frac{1}{(E + i \epsilon) - E_{a'}} \right) \end{equation}$$
The $$\epsilon$$ prescription was used for convergence of the integral. The integral has now been done so we can safely take $$\epsilon \to 0$$ in the end to get:
$$\tilde{G}(E) = \sum_{a^{'}} \frac{1}{E - E_{a'}}$$
Analytic continuations such as this $$\epsilon$$ prescription are used in many places in physics. One of the key motivations is convergence.