1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Measuring angular momentum

  1. May 24, 2012 #1
    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. jcsd
  3. May 25, 2012 #2
    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.
     
  4. May 25, 2012 #3
    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 ([itex]\phi[/itex],[itex]\theta[/itex])-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 ([itex]\phi[/itex],[itex]\theta[/itex])-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] ?
     
  5. May 25, 2012 #4
    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.
     
  6. May 25, 2012 #5
    Very neat explanation, thank you so much it really helps allot!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook