In Zeidler's book on QFT, page 94 there is a definition for a Laplace transform that reads

\begin{equation} (\mathcal{L} f)(\mathcal{E}) = \int_0^\infty e^{i\mathcal{E}t/\hbar}f(t) dt, \end{equation}

where $\mathcal{E}_0=E_0-i\Gamma_0=E_0-i\hbar \gamma_0$ and $\mathcal{E}=E-i\Gamma$. Consider a model for a "truncated" damped wave

\begin{equation} y(t)=A\theta(t)e^{-\gamma_0t}e^{-iw_0t}, \end{equation}

where $\theta(x)$ is the Heaviside function and $\gamma,w\in\mathbb{R}$. According to the definition of the Laplace transform, the $y(t)$ becomes

\begin{equation} A\int_0^\infty e^{i\mathcal{E}t/\hbar}e^{-i\mathcal{E}_0t}dt = A\int_0^\infty e^{-i(E_0-E)t/\hbar}e^{\gamma t-\gamma_0 t}dt. \end{equation}

I'm confused with this last expression. If I understand correctly, we are considering "truncated" waves such that $\gamma_0 \leq 0$ (this is written in page 93). Then, in order for the integral to converge we need $\gamma>-\gamma_0$. If this is true, the exponent $\gamma t-\gamma_0 t$ does not go to the zero as $t\rightarrow \infty$. Did I screwed up in a sign, or the definition of the Lapace transform has a wrong sign?

  • $\begingroup$ Can you fix your initial definition? $\mathcal{E}_0$ appears nowhere in it. $\endgroup$ – probably_someone Sep 3 at 21:00
  • $\begingroup$ Done, the missing definition was for $\mathcal{E}$. $\endgroup$ – user2820579 Sep 3 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.