# Measuring angular momentum

1. May 24, 2012

### Kentaxel

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

I have a wavefunction ψ(r), I want to know what the probability of obtaining a specific value, say 0, Is upon measuring the z-component of the angular momentum Lz

What do i do?

3. The attempt at a solution

My first thought is to apply the Lz-operator to my function and see if this yields the value im looking for, and if it does not the probability is simply zero. However when i actually do the calculations i don't even get my original function back, and i'm not sure how to interpret this result. i.e. I expected that i would get somethibg like Lzψ(r)=lψ(r) where l is the eigenvalue of Lz. But maybe this is a faulty reasoning?

Edit: Maybe what i'm wondering is really what happens if you try to measure a variable of wich my statevector is not an eigenfunction (as seems to be the case) or would such an action actually colapse the function to an eigenfunction of my L operator? In any case What sort of result is to be expected of the corresponding eigenvalue? nothing? 0? Or what?

Last edited: May 25, 2012
2. May 25, 2012

### Morgoth

you have several ways to work. Do you have the form of your ψ? if you do try writting it in the form of [SOMETHING that Lz cannot see]*[Eigenfunctions of Lz], which are the spherical harmonics Y[l,m].

try thinking in the same way you would, if you had ψ= Σ αη φη(χ) and were asked for energy Ek....

φη(χ) eigenfunctions of Hamiltonian.

3. May 25, 2012

### Kentaxel

Yes i see what you meen, i was thinking of it the wrong way. The function is in fact already given in a manner where it's ($\phi$,$\theta$)-dependence can easily be separated from its r-dependance. so all i have to do is calculate clm =<lm|Y> for whatever Y[l,m] corresponds to the desired eigenvalue of Lz and then square that c wich would give the probability? Specificaly <l0|Y> for 0 and so on.

But there are allot of eigenfunctions that would correspond to m = 0, would these all give me the same coefficient c?

Furthermore as i understand it, the only way to obtain a value for Lz is if it's possible to separate the ($\phi$,$\theta$)-dependence so that i can be sure that this part is a linear combination of the spherical harmonics? Or can i deduce this in any other way? Actually how can i ever be sure that the function is at all a linear combination of the Y[l,m] ?

4. May 25, 2012

### Morgoth

if you have for example- I will denote Y[l,m]-
Y[1,0], Y[0,0]
both will participate in giving you m=0.
if the one is 1/3 and the other is 1/3, then the total probability will be 2/3 :)

Generally you can find it by:
|<Your wanted state | ψ>|2
using dirac's notation since I see you used it too...

eg
|<l 0|ψ>|2 will give you the possibility of your state ψ to fall (to be) on the state
|l 0> it's like seeing |ψ>,|l,0> as vectors, and taking the inner product of them gives you the projection of one over the other.
writing the above in integral form (going to the {r} representation):
|∫Y*[l,0] ψ(r) d3r|2

how did I do this?
I used that I= ∫|r><r|d3r
<l 0|I|ψ>= ∫<l 0|r><r|ψ>d3r= ∫<r|l 0>*<r|ψ>d3r = ∫Y*[l,0] ψ(r) d3r

Now since you can always write ψ(r)= R(r) Y(θ,φ) you will have that R(r) is not seen by the Y* but Y* sees the Y[l,m](θ,φ)s you have in your state, leaving only those that have the same quantum numbers as it alive (orthogonal and normalised)

how can you be sure that the function is a linear combination of the Y[lm]?
Because they form a complete set of basis of your space (which is the space of θφ), so you can form out of them all possible states.

5. May 25, 2012

### Kentaxel

Very neat explanation, thank you so much it really helps allot!