(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A wavefunction for a hydrogen electron is given by [tex]\Psi = -

\sqrt{\frac{3}{8 \pi}} sin\theta e^{i \phi} (\frac{1}{2a^3})^{3/2}

\frac{re^{-r/2a}}{a \sqrt{3}}[/tex]

Prove that the electron exists in space, ie, [tex]\int {\Psi}^2= 1 [/tex]

2. Relevant equations & attempt at solution

Apologise in advance for the shortcuts, these equations are terrible to type

out.

Expressed in spherical polar coordinates, [tex] dV = r^2 sin \theta dr

d\theta d\phi [/tex]

The squared wavefunction,

[tex]

{\Phi}^2 = \frac{1}{64\pi a^5} r^2 {sin}^2 \theta e^{2i\phi}

[/tex]

With respect to r, [tex] \int^{\infty}{\0} r^4 e^{r/a} = 24 a^5 [/tex]

This is a pain to do due to iterated application of integration by parts, but

by inspection,

[tex]

\int^{\infty}{\0} r^4 e^{r/a} = 4a \int^{\infty}{\0} r^3 e^{r/a} = 4.3a^2

\int^{\infty}{\0} r^2 e^{r/a}... = 24a^5

[/tex]

With respect to [tex]\theta[/tex],

[tex] \int^{\pi}{\0} {sin}^3 \theta d\theta = \frac{4}{3} [/tex]

This gives us,

[tex]

\int {\Phi}^2 dV = \frac{1}{2\pi} \int^{2\pi}{\0} e^{2i\phi} d\phi

[/tex]

I'm stuck at this point. How do I proceed? Was my earlier working correct?

If the earlier integration was right, then the last integral must be equal to 2pi.

Exploration

From using traditional methods the answer I actually get is 0. How does the pi term come about.

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# Homework Help: Wavefunction integral

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