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I'm going through Sergio Dutra's Cavity Electrodynamics: The Strange Theory of Light in a Box. In equation (2.31) he computes:

$$\begin{aligned}\langle x|\hat{p}|x'\rangle&=i\hbar\int\frac{dk}{2\pi}\frac{e^{i(x-x')k}}{x-x'}\\\\ &=-\frac{i\hbar}{2\pi}\int dk\,e^{i(x-x')k}ik \\\\ &=\frac{\hbar}{i}\frac{\partial}{\partial x}\delta(x-x').\end{aligned}\tag{2.31}$$

The author claims that the second line arises via integration by parts, but I can't quite see how.

With integration by parts we have $$\int_a^b u dv = uv\rvert_a^b-\int_a^b vdu.$$ I assume we take $$u=\frac{e^{i(x-x')k}}{x-x'},\;dv=1.$$ This gives $$i\hbar\int\frac{dk}{2\pi}\frac{e^{i(x-x')k}}{x-x'}=\left.\frac{i\hbar}{2\pi}\frac{e^{i(x-x')k}}{x-x'}k\right\rvert_{k=-\infty}^{k=+\infty}-\frac{i\hbar}{2\pi}\int dk\,e^{i(x-x')k}ik.$$

If the boundary term vanishes, then I recover Dutra's derivation. However, I can't see why the term term should vanish. We have an increasing function $k$, multiplied by a complex oscillation, so the limit is not defined. Why should we take this to be zero, rather than some other constant? I understand that something weird must be happening, since the result is a distribution rather than a function.

EDIT: So having looked at some other answers on this site, my understanding that Dutra's derivation here isn't valid. As QMechanic says dividing a distribution by $x$ isn't defined. This stackexchange answer shows that the expression $\delta(x)/x$ can be manipulated to give multiple different answers.

Qmechanic
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Ruvi Lecamwasam
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2 Answers2

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Let's throw rigor aside and look at the pertinent term as a functional acting on a test function, in other words consider $$\int dx f(x)\left[ \frac{k~e^{ikx}}{x}\right]^{\infty}_{-\infty} = \left[k\int dx\frac{f(x)}{x}e^{ikx}\right]^{\infty}_{-\infty}$$ Assuming that $f(x)/x$ has a fourier transform and that it vanishes at infinity faster than $1/k$, the above integral vanishes.

AfterShave
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  • That was my first instinct. Unfortunately the integral is over k, not x, so it is k that's going to infinity. – Ruvi Lecamwasam Feb 24 '22 at 13:30
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    You misunderstand by answer, I am taking the surface term and integrating it over a test function in $x$. The idea is that these expressions are distributions and only their effect on test functions under the integral sign matters. – AfterShave Feb 24 '22 at 17:36
  • Oh I see, yes that makes sense, thanks! – Ruvi Lecamwasam Feb 25 '22 at 01:45
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  1. Ref. 1 heuristically justifies the first equality in eq. (2.31) by starting from the correct equation $$ (x-x^{\prime})\langle x|\hat{p}|x^{\prime}\rangle~=~i\hbar\delta(x-x^{\prime}),\tag{2.30}$$ and dividing both sides with $x-x^{\prime}$.

  2. However the expression $$ \frac{1}{x-x^{\prime}}\delta(x-x^{\prime})$$ is mathematical ill-defined in distribution theory, so the manipulations in eq. (2.31) does not make much mathematical sense.

  3. Nevertheless, the resulting relation LHS=RHS in eq. (2.31) is actually true, e.g. by using the Schrödinger representation for $\hat{p}$, as proven in many posts on this site.

References:

  1. Sergio Dutra, Cavity Electrodynamics: The Strange Theory of Light in a Box, 2005.
Qmechanic
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