Can anyone rigorously prove this? $$\int dx \, \delta(x-\alpha)\delta^{\prime} (x-\beta) = \delta^{\prime} (\alpha-\beta).$$

  • $\begingroup$ Is the answer always the same for all of the various definitions of the the Dirac delta in terms limits of functions? $\endgroup$ – Victor V Albert Dec 3 '15 at 1:58
  • $\begingroup$ How is this a physics question? $\endgroup$ – ACuriousMind Dec 3 '15 at 1:58
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    $\begingroup$ $\langle\alpha|\hat{p}|\beta\rangle = ...$ :) $\endgroup$ – Victor V Albert Dec 3 '15 at 2:00
  • $\begingroup$ Actually, now that I wrote it that way, this has to be true. Thanks! $\endgroup$ – Victor V Albert Dec 3 '15 at 2:03
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    $\begingroup$ This is generally considered ill-defined, since you cannot multiply distributions. But maybe it can be viewed as a convolution? $\endgroup$ – Ian Dec 3 '15 at 3:34

In THIS ANSWER, I provided a primer on the Dirac Delta and the Unit Doublet distributions.

Here, we have for any test function $f$, the distribution $\int_{-\infty}^\infty\delta(x-\alpha)\delta'(x-\beta)\,dx$ satisfies the following:

$$\begin{align} \int_{-\infty}^\infty f(\alpha)\left(\int_{-\infty}^\infty\delta(x-\alpha)\delta'(x-\beta)\,dx\right)\,d\alpha&=\int_{-\infty}^\infty \delta'(x-\beta)\left(\int_{-\infty}^\infty f(\alpha)\,\delta(x-\alpha)\,d\alpha\right)\,dx\\\\ &=\int_{-\infty}^\infty \delta'(x-\beta)\,f(x)\,dx\\\\ &=-f'(\beta)\\\\ &=\int_{-\infty}^\infty \delta'(\alpha-\beta)\,f(\alpha)\,d\alpha \end{align}$$

We also have

$$\begin{align} \int_{-\infty}^\infty f(\beta)\left(\int_{-\infty}^\infty\delta(x-\alpha)\delta'(x-\beta)\,dx\right)\,d\beta&=\int_{-\infty}^\infty \delta(x-\alpha)\left(\int_{-\infty}^\infty f(\beta)\,\delta'(x-\beta)\,d\beta\right)\,dx\\\\ &=\int_{-\infty}^\infty \delta(x-\alpha)\,f'(x)\,dx\\\\ &=f'(\alpha)\\\\ &=\int_{-\infty}^{\infty}\delta'(\alpha -\beta)f(\beta)\,d\beta \end{align}$$

Therefore, in terms of Generalized Functions, we have the equivalence


Using a more compact notation, we have

$$\begin{align} \langle f,\langle \delta'_{\beta},\delta_{\alpha}\rangle \rangle_{\alpha}&=\langle \delta'_{\beta},\langle f,\delta_{\alpha}\rangle_{\alpha} \rangle\\\\ &=\langle \delta'_{\beta},f\rangle\\\\ &=-f'(\beta) \end{align}$$

A similar notation is applicable where we take the distribution over $\beta$.

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    $\begingroup$ This doesn't compute. What if you evaluate in a different order, i.e. $\langle \delta_{\alpha},\langle f,\delta'_{\beta}\rangle \rangle$? $\endgroup$ – A.S. Dec 3 '15 at 5:05
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    $\begingroup$ The two results are not equal unless $\alpha=\beta$, so I don't see how your definition of $\langle \delta'_{\beta},\delta_{\alpha}\rangle$ works. In some sense $\langle \delta'_{\beta},\delta_{\alpha}\rangle=0$ for $\alpha\ne \beta$. $\endgroup$ – A.S. Dec 3 '15 at 5:19
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    $\begingroup$ Why not consider both as a fixed number then? That's a much more natural assumption than assuming one is variable and the other is a parameter - especially since $\alpha$ and $\beta$ are not symmetric. Either both are parameters or both are variables. $\endgroup$ – A.S. Dec 3 '15 at 5:28
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    $\begingroup$ Mark, it clarifies the mechanics of what follows but not the original choice to treat one of $\alpha,\beta$ as a fixed parameter and the other as a variable of a distribution. The OP treats both as a parameter (as I think was intended), so the real question is if we can define $\int_R\delta(x-\alpha)\delta'(x-\alpha)$ - or more generally $\delta\delta'$ - which is a product of two distribution with non-disjoint singular supports. $\endgroup$ – A.S. Dec 3 '15 at 6:02
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    $\begingroup$ @VictorVAlbert Sure. Choose the one that best facilitates your specific problem. And remember, I provided only a small sample of regularizations - just the tip of the iceberg. - Mark $\endgroup$ – Mark Viola Dec 3 '15 at 17:08

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