Normalising phi for the Hydrogen atom.

1. Aug 19, 2009

mrausum

This is most likely very simple, but I can't figure it out.

http://www.sussex.ac.uk/physics/teaching/btv/Lect02_2006.pdf

Step 5 they've got an equation for $$\Phi$$. They then normalise it to get A = $$\frac{1}{\sqrt{2\pi}}$$. Every time I do the integral I get:

$$A^2.^{2\pi}_{0}[ \frac{exp(2i\sqrt{\Lambda}\Phi)}{2i\sqrt{\Lambda}}] = 1$$

Which makes the integral go to zero when you rewrite the exp using Euler's and and take into account $$\sqrt{\Lambda}$$ must be an integer?

Last edited: Aug 19, 2009
2. Aug 19, 2009

Edgardo

What is the absolute square of phi?

3. Aug 19, 2009

mrausum

The probability density function. So?

4. Aug 19, 2009

Edgardo

No, I mean write down phi and then compute |phi|^2. What do you get for |phi|^2?

5. Aug 19, 2009

mrausum

$$A^2.exp(2i\sqrt{\Lambda}\Phi)$$

6. Aug 19, 2009

Edgardo

No, look again. It is the absolute square.

7. Aug 19, 2009

mrausum

So I can't take the two inside the exp? I don't really see why not? So i'm guessing you're saying that I have to do the integral of phi multiplied by its conjugate?

Thanks anyway, that's solved it :)

Last edited: Aug 19, 2009