1
$\begingroup$

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 $(0,\infty)$]?

|cite|improve this question
$\endgroup$
5
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
1
$\begingroup$

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).

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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.) $\endgroup$ – Robin Ekman Sep 27 '14 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 '14 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 '14 at 0:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.