I'm trying to evaluate the expectation value of some perturbation Hamiltonian $$H=\alpha \delta^3(\vec{r}),$$ where $\alpha$ is a positive constant, for the ground state wavefunction of the hydrogen atom $$\psi_{100}~\propto~ \exp[-r/a]$$ (I want to calculate the shift in the energy in first order perturbation theory). Why is it wrong to write:

$$\langle\psi_{100}|H|\psi_{100}\rangle ~\propto~ \int_0^\infty dr~ r^2 \exp[-2r/a] \alpha \delta(r) ~=~ 0$$

Does it have to do with either:

  1. I can't just write $\delta^3(\vec{r}) = \delta(r)$ in the integral over $r$, or

  2. I can't evaluate the delta function at zero, since it is at the endpoint of the integration limits [not inside the interval $(0,\infty)$]?


2 Answers 2


Hint: Rather than using spherical coordinates, which are singular where the 3D Dirac delta function has support, work instead in Cartesian coordinates $\vec{r}=(x,y,z)$ and use the defining property

$$ \iiint_{\mathbb{R}^3}\! d^3r ~f(\vec{r})~ \delta^3(\vec{r})~=~f(\vec{0}) $$

of the 3D Dirac delta function.


I want to post my attempt at the solution based on Qmechanic's hint (thanks!):

Rewriting in cartesian coordinates

$$ \psi_{100} \propto \exp[-\sqrt{x^2+y^2+z^2}/a] $$

$$ <\psi_{100}|H|\psi_{100}> \propto \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} dy \int_{-\infty}^{+\infty} dz \exp\left[-2\sqrt{x^2+y^2+z^2}/a\right] \alpha \delta(x)\delta(y)\delta(z) =\alpha$$

Hope thats correct, and sorry if this was trivial, certainly the expected answer.

I still don't fully understand why it's not possible to directly solve this in spherical coordinates, but I understand it is related to the difficulty in evaluating $\delta(r)$ at the endpoint $r=0$ (see CuriousKev's comment).

  • $\begingroup$ When changing to spherical coordinates you are really only integrating over the region covered by those coordinates. That region is all of space except a half-line through the origin. The integral of a continuous function is insensitive to removing such a small set, but a distribution like the Dirac delta can be sensitive to even single points. (This is one way of seeing that the Dirac delta can't be a function in the usual sense.) $\endgroup$ Sep 27, 2014 at 23:51
  • $\begingroup$ Thanks for the explanation. Could you use some representation of the Dirac delta to get around that? $\endgroup$
    – smörkex
    Sep 28, 2014 at 0:18
  • $\begingroup$ @Kurt If you wrote the Dirac delta as a limiting procedure (for example solve with a general gaussian and then after the integral take the limit as the width goes to zero while maintaining the area under the curve), that should work even in spherical coordinates. $\endgroup$
    – CuriousKev
    Sep 28, 2014 at 0:29

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