Normalizing Angular Wavefunctions: Troubleshooting and Solutions

[bon]

Homework Statement

Ok so I am told that the angular part of a system's wavefunction is:

Psi (theta, phi) = root2 cos(theta) -2i sin(theta) sin (phi)

Now I am trying to normalise it..

Homework Equations

The Attempt at a Solution

Psi * (theta, phi) = root2 cos(theta) + 2i sin(theta) sin (phi)

Therefore, when we integrate over all space, we need the integral of psi* psi sin(theta) d theta d phi, but I keep getting this to be 0! We get a cos^2theta sintheta which integrates over 2pi to give 0 and a sin^3 theta which also integrates over 2pi to give 0! what's going wrong!?
THanks

 

[LeonhardEuler]

You need to find the integral of
|\psi|^2=\psi\psi^{*}
to normalize, not the integral of \psi[/itex]. <br /> (I&#039;ve made the same mistake before, so I know your pain)

 

[bon]

Thanks, but that is what i found:

I found |psi|^2 = 2 cos^2 (theta) + 4 sin^2 (theta) sin^2 (phi)

but then when I integrate this I have the factor of sin theta that comes from the sin(theta) d theta d phi,

so both the theta integrals come out to 0?!

thanks!

 

[LeonhardEuler]

Can you show the integral because this is a non-negative function that is positive almost everywhere and the integral can not possibly be 0, so there is some other mistake in your integration.

 

[bon]

Well I am just integrating cos^2 theta sin (theta) d theta from 0 to 2pi = 0

and adding this to the integral of sin^3 (theta) d theta from 0 to 2pi = 0

thanks for your help

 

[LeonhardEuler]

Oh, your mistake is that you take the factor of integration to be sin(theta) over 0,2pi. In fact, depending on the convention, theta is the angle from the z-axis and the integral only goes to pi, or theta is the polar angle in the x-y plane and the integral factor is in fact sin(phi)

 

[bon]

aha thank you so much!

 

[bon]

One related question if you don't mind:

Im told a system's wavefunction is proportional to sin^2 theta and I'm asked what the possible results of measurements of L_z and L^2 and asked for probabilities of each...

How do I go about doing this?

Thanks again!

 

[LeonhardEuler]

I would start by representing L_z and L^2 in terms of phi and theta. For instance,
L_z = \i\hbar\frac{\partial}{\partial \phi}
and then putting this into the expectation formula

 

[bon]

Ok thanks!
Back to my first question - I've now worked out A = 1/root (8pi)
But looking at the wavefunction, i can see by inspection that Lz has to be 1, -1 or 0

now since A = A = 1/root (8pi), the amplitude to find +1 = amplitude to find -1 = A = 1/root (8pi)

and the amplitude to find 0 = root(2) /root (8pi)

but if i square and add these, it doesn't come to 1. what has gone wrong?

thanks again

 

[LeonhardEuler]

I'm not getting the same A as you, though it's entirely possible I made a mistake.

Edit: Did make a mistake. Read the coefficient of cos(theta) in psi as 1/root2 instead of root2 for some reason. I'll calculate angular momentum now.

 

[LeonhardEuler]

Ok, I'm sure I got the right answer now for L_z and it's not what you got. Make sure you're doing the \phi integral correctly, it comes out very easily.