# Normalising a wavefunction

## Homework Statement

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

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

Now im trying to normalise it..

## 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
Gold Member
You need to find the integral of
$$|\psi|^2=\psi\psi^{*}$$
to normalize, not the integral of $$\psi[/itex]. (I've made the same mistake before, so I know your pain) Thanks, but that is what i found: I found |psi|^2 = 2 cos^2 (theta) + 4 sin^2 (theta) sin^2 (phi) but then when I integrate this I have the factor of sin theta that comes from the sin(theta) d theta d phi, so both the theta integrals come out to 0!?!! thanks!! LeonhardEuler Gold Member Can you show the integral because this is a non-negative function that is positive almost everywhere and the integral can not possibly be 0, so there is some other mistake in your integration. Well im just integrating cos^2 theta sin (theta) d theta from 0 to 2pi = 0 and adding this to the integral of sin^3 (theta) d theta from 0 to 2pi = 0 thanks for your help LeonhardEuler Gold Member Oh, your mistake is that you take the factor of integration to be sin(theta) over 0,2pi. In fact, depending on the convention, theta is the angle from the z-axis and the integral only goes to pi, or theta is the polar angle in the x-y plane and the integral factor is in fact sin(phi) aha thank you so much! One related question if you don't mind: Im told a system's wavefunction is proportional to sin^2 theta and I'm asked what the possible results of measurements of L_z and L^2 and asked for probabilities of each... How do I go about doing this? Thanks again! LeonhardEuler Gold Member I would start by representing L_z and L^2 in terms of phi and theta. For instance, [tex]L_z = \i\hbar\frac{\partial}{\partial \phi}$$
and then putting this into the expectation formula

Ok thanks!
Back to my first question - ive 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
Gold Member
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
Gold Member
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.