# Normalizing etc. Wavefunction of hydrogen atom

1. Jun 2, 2012

### Kentaxel

1. The problem statement, all variables and given/known data

Find the constant A such that the equation

$\psi(r,\theta,\varphi)=\sqrt{6\pi}A\sqrt{r}e^{-r/a}$

Wich describes one electron in a hydrogenatom, is normalized

3. The attempt at a solution
I figured this equation is seperable in the form

$\psi(r,\theta,\varphi)=R(r)Y_{l,m}(\theta,\varphi)$

Such that $Y_{l,m}$ is the first spherical harmonic

$Y_{0,0}=\frac{1}{\sqrt{4\pi}}$

Enabling me to write ψ in the form

$\sqrt{24}{\pi}A\sqrt{r}e^{-r/a}Y_{0,0}(\theta\varphi)$

and since the spherical harmonics are all normalized it is suficient to normalize this acording to

$\int\left|\sqrt{24}{\pi}A\sqrt{r}e^{-r/a}\right|^2 r^2 dr =1$

Which gives me the result

$A=\frac{1}{3\pi^2 a^4}$

the problem is just that there is no solution available and im not exactly 100% that this is correct so i would appreciate some input.

Last edited: Jun 2, 2012
2. Jun 2, 2012

### anjelin

Your solving way is right. But why you choose l=0, m=o. I think you have to write in general form i.e.,

I attached file.

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3. Jun 2, 2012

### Jorriss

Method seems fine. Viewing it as a spherical harmonic though is unnecessary. All you needed to do was integrate the given wave function over all space and the angular contributions would give 4pi (full solid angle).

4. Jun 2, 2012

### Kentaxel

The reason i choose to seperate it was because of the following questions which are

Show that ψ is not an energy eigenfunction.

and

Compute the probability to find the system in the groundstate of the hamilton operator when measuring the energy.

Where regarding the first one i thought that in writing the given wavefunction in that form it was obvious that it is not one of the solutions listed for an electron in a hydrogenic atom. This however feels allot like a guess on my part.

For the second one i used the fact that i knew l and m was 0 and tried to solve it by taking $\left|c_{1,0,0}\right| ^2 = \left|<\psi_{1,0,0}|\psi > \right| ^2$

Otherwise how would i know what groundstate psi to use?

Edit: actually since the ground state is non degenerate i suppose that would not matter?

Last edited: Jun 3, 2012