Proving the Existence of a Hydrogen Electron in Space via Wavefunction Integral

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Homework Statement



A wavefunction for a hydrogen electron is given by \Psi = - <br /> <br /> \sqrt{\frac{3}{8 \pi}} sin\theta e^{i \phi} (\frac{1}{2a^3})^{3/2} <br /> <br /> \frac{re^{-r/2a}}{a \sqrt{3}}

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

2. Homework Equations & attempt at solution

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

out.

Expressed in spherical polar coordinates, dV = r^2 sin \theta dr <br /> <br /> d\theta d\phi

The squared wavefunction,
<br /> {\Phi}^2 = \frac{1}{64\pi a^5} r^2 {sin}^2 \theta e^{2i\phi}<br />

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

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

by inspection,

<br /> <br /> \int^{\infty}{\0} r^4 e^{r/a} = 4a \int^{\infty}{\0} r^3 e^{r/a} = 4.3a^2 <br /> <br /> \int^{\infty}{\0} r^2 e^{r/a}... = 24a^5<br /> <br />


With respect to \theta,

\int^{\pi}{\0} {sin}^3 \theta d\theta = \frac{4}{3}

This gives us,

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

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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Shouldn't that be
\int |\Psi|^2 dV?

In other words, you want the absolute value squared, not the function. And
|e^{i\phi}|= 1[/itex]
 
Thanks, HallsofIvy, I think that explains that one. (Sorry, I mixed up psi/phi.) The rest of my solution is ok, right?

My text explicitly gave the condition required as

\int {\Psi}^2 dV = 1

Which leads me to another question. What is the physical significance of the modulus? (I'm not studying a text on quantum physics at the moment, I'm working on a mathematical physics text.)
 
|\Psi |^2 is interpreted to be a probability density function describing the probability of finding your particle in a specific state.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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