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: $$\tilde{G}(E)=\sum_{a^{'}}\frac{1}{E-E_{a'}}~?$$

  • $\begingroup$ $e^{i x}=\cos{x}+i\sin{x}$ hence imaginary exponentials oscillate $\endgroup$
    – Triatticus
    Sep 29, 2018 at 21:03
  • 1
    $\begingroup$ I recommend this question and it's accepted answer, because it actually gives some insight into why the $E \rightarrow E + i \epsilon$ thing makes sense. $\endgroup$
    – DanielSank
    Dec 27, 2022 at 17:14
  • $\begingroup$ The only real reason why "it works" is because it agrees with experiments. At least so far it does. $\endgroup$ May 6, 2023 at 16:14

1 Answer 1


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}

while the lower limit (with minus sign included) gives:

\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.

  • $\begingroup$ Just an additional note: Doing the integrals in Euclidean time via Wick rotations also gives you the same results (and is more mathematically satisfactory, according to me). $\endgroup$ Oct 12, 2018 at 11:00
  • $\begingroup$ For a physical picture of why adding $i \epsilon$ to the energy actually makes sense see this question and it's accepted answer. $\endgroup$
    – DanielSank
    Dec 27, 2022 at 17:15

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