Normalising phi for the Hydrogen atom.

[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 [tex]\Phi[/tex]. They then normalise it to get A = [tex]\frac{1}{\sqrt{2\pi}}[/tex]. Every time I do the integral I get:

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

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

 

[Edgardo]

What is the absolute square of phi?

 

[mrausum]

What is the absolute square of phi?

The probability density function. So?

 

[Edgardo]

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

 

[mrausum]

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

[tex]A^2.exp(2i\sqrt{\Lambda}\Phi)[/tex]

 

[Edgardo]

No, look again. It is the absolute square.

 

[mrausum]

No, look again. It is the absolute square.

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 :)