How do you normalize a wave function in three dimensions with spherical coordinates?
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$\begingroup$ Do you know what a volume integral is? $\endgroup$– Kyle KanosCommented Apr 20, 2015 at 2:30
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$\begingroup$ Yes. However my confusion is because of the r portion of the integral. I end up integrating 1 from 0 to infinity. Then when I move on to the theta integral I have infinity*sin(theta) from 0 to pi. $\endgroup$– disc otterCommented Apr 20, 2015 at 2:33
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2 Answers
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Since the wavefunction depends on r, which is the spherical coordinate representing the distance from the origin, we use spherical coordinates to perform the integration because it is most convenient. And yes, this is a triple integral, $\int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta\int_0^{\infty}r^2\Psi^*\Psi dr$. The wave function doesn't depend on the two angular coordinates, so it should be straight-forward to carry out if you've done triple integration before.
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$\begingroup$ One more quick thing. When I take the integral with respect to r, I have r^2/r^2 which goes to one. If psi star psi =A^2/r^2 and we multiply by r^2, how can I avoid getting 1 inside the integral? $\endgroup$ Commented Apr 20, 2015 at 2:05
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1$\begingroup$ That's exactly right, which means the wavefunction is not normalizable. $\endgroup$– mr blickCommented Apr 20, 2015 at 2:47
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$\begingroup$ So that means that the r integral = infinity? Which when I carry it all the way through mandates that the normalization constant A is also equal to infinity? Thus non-normalizable? $\endgroup$ Commented Apr 20, 2015 at 2:52
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$\begingroup$ Actually it would mean the normalization constant is 0, since we need the full integral to be 1. Since the r integral integrates to infinity, we need A^2*Infinity=1 which mean A=0. Thus, the full wavefunction is A/r = 0/r = 0 which isn't physically acceptable $\endgroup$– mr blickCommented Apr 20, 2015 at 2:59
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Integrate the complex square of the wavefunction $\Psi^*\Psi=\frac{A^*A}{r^2}$ over all space and set the result to 1. Since the wave function is given in spherical coordinates, it would be easiest to integrate in spherical coordinates using the volume element $dV=r^2\sin\theta drd\theta d\phi$ where $0\leq r\leq \infty$, $0\leq \theta\leq \pi$ and $0\leq \phi \leq 2\pi$.
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$\begingroup$ How do I go about doing the math then? Is dV representing a triple integral that depends on r, phi and theta? I don't quite understand why we convert it to spherical coordinates or how to do evaluate the integral once I do. $\endgroup$ Commented Apr 20, 2015 at 1:20