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I was asked to evaluate the amplitude for finding a relativistic particle at (x,t) when it was completely localized at the origin (0,0) earlier within the framework of 1st quantization and single particle wavefunction (i.e not involving any quantum field). Naively, the amplitude is represented by the following integral

\begin{equation} \int \frac{d^{3}k}{(2\pi )^3}e^{ikx-E_{k}t} \end{equation} where $E_{k}=\sqrt(k^2+m^2)$ After performing all the angular integrals, I end up with the follwing \begin{equation} \frac{1}{4\pi^2ix}\int_{0}^{\infty}ke^{-i\sqrt{k^2+m^2}t}(e^{ikx}-e^{-ikx}) \end{equation}

Naively, it seems that this integral does not converge because the integrand is a growing function multiplied by an oscillating phase, especially for the case m=0. Can I evaluate this integral with some standard special functions?

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  • $\begingroup$ Could you elaborate on why you are using this integral? The QFT propagator has a different form per se. Did you forget the factor of $(2 E_p)^{-1/2}$, or was it cancelled by something? $\endgroup$
    – Prof. Legolasov
    Commented Feb 9, 2018 at 2:12
  • $\begingroup$ Thanks for your comment! As I mentioned in the post, I am discussing the propagation amplitude within the framework of first quantization. Basically, it is just the relativistic counterpart of <x|U(t)|0>, with Ek being the relativistic dispersion instead of the quadratic one in nonrelativistic QM (which can be easily evaluated by gaussian integral). U(t) is the familiar exp(-iHt). I know that the propagation amplitude resulting from field quantization has a 1/sqrt(Ep) $\endgroup$
    – Heng Fai Chang
    Commented Feb 10, 2018 at 2:03
  • $\begingroup$ the propagator takes the same form both in QFT and in first-wuantized approach. My question remains open. $\endgroup$
    – Prof. Legolasov
    Commented Feb 10, 2018 at 2:05
  • $\begingroup$ if we expand the time evolution operator U(t) = $\int dp exp(-iEp*t) |p><p| $and sandwich it with <x| and |0> as usual in calculating the position transition amplitude <x|U(t)|y> in ordinary QM, we won't have the 1/sqrt(Ep) which comes from the mode expansion of the KG field $\endgroup$
    – Heng Fai Chang
    Commented Feb 10, 2018 at 2:09

1 Answer 1

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This is the familiar problem in the path integral approach to QFT. What one can do is the analytic continuation. There are two ways to do that.

  1. Make the parameters complex like $m^2\rightarrow m^2 - i \epsilon$. Here $\epsilon$ is a very small number. In this particular case , we need $\epsilon $ to be negative no. What it will do is that it makes oscillatory term multiply by exponential damping term. Any other function in the integrand other that exponential growth function will have good convergence property. So in this particular case, the integral converges and just do the integral usual ways without worry about convergence. In the $m^2 \rightarrow 0$ limit, it follows exactly same just replace $0-i \epsilon$. After doing integral take $\epsilon \rightarrow 0$ limit appropriately.

  2. Make $t$ to be complex $t\rightarrow t- i \epsilon$. Here also same thing happens. But the first way is more useful and easy to do.

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