So I have a part of the energy eigenvalue equation that look like this: $$ \delta(\hat{x})|n\rangle $$

Where n is the energy basis of the Hamiltonian I'm considering. To deal with this, I tried projecting onto a complete set of position states and I obtained:

$$ \int^{\infty}_{-\infty}|x\rangle\langle x|\delta(\hat{x})|n\rangle dx $$

Which becomes:

$$ \int^{\infty}_{-\infty}\delta(x) \phi_n(x) |x\rangle dx $$

Where $\phi_n(x)$ is the position representation of the energy eigenstate.

And this is where I am getting confused. Obviously with the delta function we should be able to collapse the integral, evaluating $\phi_n(x)$ at zero, but I am confused about what to do with the other ket, or if there is anything I can do with it. I would love to be able to move it outside of the integral, but I don't know if there is any justification in that.

  • $\begingroup$ It's late and I'm tired but isn't the final integral just a scaled position eigenstate for $x=0$? $\phi_n(0)|x=0\rangle $ $\endgroup$ – Alfred Centauri Sep 16 '14 at 1:36

As Alfred Centauri said, I don't know that there's much more to do here than to say

$$ \int^{\infty}_{-\infty} dx \ \delta(x) \phi_n(x) |x\rangle = \phi_n(0) |x=0\rangle $$

You could instead write this as $$ \sum_m \phi_n(0) |m\rangle \langle m |x=0\rangle = \phi_n(0) \sum_m \phi^*_m(0) |m\rangle $$


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