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I am studying Relativistic Quantum Mechanics from my professor's notes. When calculating the second order perturbative correction to the transition coefficient $T_{fi}$* in a scattering process by a potential of finite duration, the following term appears

$$\sum_{n\neq i}\int_{-\infty}^\infty dt e^{i(E_f-E_n)t}\int_{-\infty}^tdt'e^{i(E_f-E_n)t'}.$$

My professor notes that the $dt'$ integral does not make sense unless we interpret it as an expression of the form

$$\lim_{ε\to 0^+}\int_{-\infty}^tdt'e^{i(E_f-E_n-iε)t'}.$$

I understand why the integral expression does not make sense as well as why the interpretation does, but I would really appreciate any physical reasoning for acting so, any indication as to what might gone wrong in the formulation of the derivation** and more than anything, any mathematical arguments for why this is alright (for example something indication that this is the only way of giving meaning to such troublesome expressions).

*With this I mean the long time value the coefficient of the the f state of a wave function which purely on the i state before the potential is turned on.

**Which is not presented here, but I hope that it is a standard one and that it is more or less given away (to someone familiar with the subject) by the notation and the results. If not, please ask me to clarify.

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I struggle with this concept as well, although I have found some comfort in casual functions.

You're given

$$\int_{-\infty}^{t}dt' e^{i w t'} =\int_{-\infty}^{\infty}dt' \theta(t-t') e^{i w t'} = e^{i w t}\int_{-\infty}^{\infty}dt' \theta(-t') e^{i w t'} = e^{i w t} \tilde{f}_{-}(w)$$

which calls for the fourier transform of a theta function, which I'm preemptively calling $\tilde{f}_{-}(w)$, to be defined below. In classical analysis such an integral does not exist, so I will appeal to distribution theory, defining both $f(t') = 1$ and $f_{-}(t')=\theta(-t') f(t')$, where $f_{-}(t')$ is called the advanced part of $f(t')$. Causality comes in to play when splitting functions into advanced and retarded parts, which have poles exclusively above and below the real axis, respectively.

Then take a laplace transform

$$F[z] = \int_{-\infty}^{\infty}dt' e^{i z t'} f_{-}(t')$$

which has nicer convergence properties due to the imaginary frequency $z$. Since $f_{-}$ is causal, one can show the relation

$$\lim_{\epsilon \to 0^+} F[w - i\epsilon] = \tilde{f}_{-}(w)$$

where the limit is a distribution limit and I have not bothered tracking overall signs. From here

$$e^{i w t} \tilde{f}_{-}(w) = e^{i w t}\int_{-\infty}^{\infty}dt' e^{i (w-i\epsilon) t'} f_{-}(t') = \int_{-\infty}^{\infty}dt' e^{i (w-i\epsilon) t'} f_{-}(t'-t) = \int_{-\infty}^{\infty}dt' \theta(t-t') e^{i (w-i\epsilon) t'}$$

which is the desired result given $w = E_f - E_n$. Hopefully this offers some insight into how what your professor has does is more than just inserting a convergence factor and is in fact necessary in order to preserve the causal structure of your theory.

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  • $\begingroup$ This actually looks very helpful, but I need some clarification. At first glance I got a bit confused by thinking $f_-(ω)$ and $f_-(t)$ to be the same function. Now I am only confused about the definition of $f_-(ω)$. Do you define it to be $lim_{ε\to 0^+} F[ω-iε]$? If so, then why is this it not defined using any other path approaching $ω$ from the negative imaginary part half-plane? Or is it the whole point that this limit is the same for all such ways of approaching ω? $\endgroup$ – T.T. Jan 5 at 9:09
  • $\begingroup$ *...why is it not defined... $\endgroup$ – T.T. Jan 5 at 9:45
  • $\begingroup$ I've edited the text to include $\tilde{f}$ when I want to speak of the fourier transform of $f$, I was abusing notation when distinguishing the two by their arguments and I hope this alleviates some confusion. $\endgroup$ – Fraguh Jan 5 at 19:29
  • $\begingroup$ Since we're approaching the real line from below, which is the region of analyticity, I suspect there is some freedom of path, however I'm only familiar with this particular equality. $\endgroup$ – Fraguh Jan 5 at 19:34
  • $\begingroup$ Ok, thanks for the reply. Referencing causal functions and distribution limits is comforting enough. $\endgroup$ – T.T. Jan 6 at 3:26

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