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QM: Angular momentum

  1. Apr 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    A particle is in the state [itex]\left| {ml} \right\rangle [/itex], and I need to find the eigenvalues for the operator L2 and Lz.

    If we start by looking at L2, then I know that the eigenfunctions of this operator are the spherical harmonics, and I know how to write L2 in spherical coordinates. But finding the eigenenergies this way seems to be a long task.

    What is the easiest way of doing this?

    Regards,
    Niles.
     
  2. jcsd
  3. Apr 19, 2009 #2

    Ben Niehoff

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    It's much, much simpler than you're making it. The state ket [itex]|l,m \rangle[/itex] is already written in terms of the eigenvalues you need.
     
  4. Apr 20, 2009 #3
    For L2 I know from the WWW that the eigenvalues are [itex]\hbar^2 l(l+1)[/itex], but how can I see this from the state ket?
     
    Last edited: Apr 20, 2009
  5. Apr 20, 2009 #4

    malawi_glenn

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    write L^2 in terms of ladder operators L+ L- and Lz, and do a sandwhich with the bra <lm|
     
  6. Apr 20, 2009 #5
    Hello, Niles! I wish I could do some help.

    You are infact using Dirac notation (DN). All the possible states of a certain quantum system form a Hilbert space (HS). Vectors in HS (generally complex rather than real) are denoted via the bra $|>$, and if one wants to represent a special state, one will enrich the formulism of the bra. For example, we use $|\phi>$ to represent the state of by the wavefunction $\phi$. As to eigen states, the corresponding eigenvalue or quantum number are generally put into the bra $|>$. Hence, $|x'>$ means the eigen state of the coordinate $x$(with eigenvalue x'), $|p'>$ means the eigen state of momemtum(with eigenvalue p'), $|En>$ or $|n>$ means the eigen state of energy(with eigenvalue En), and
    $ |m,L> $ , as you put above, means the co-eigenstate of $(L_z, L^2)$ , with eigen values $ m\hbar $ and $ L(L+1)\hbar^2$ respectively.

    And please also notice that, in Dirac notations above, we simply employ an abstract state vector, and refer to no concrete quantum mechanical representation.

    Hence, indeed you know the answer if you know the meaning of Dirac notation.
     
  7. Apr 20, 2009 #6
    But my mission is to prove that the ket |m,l> has the eigenvalues (m \hbar, l(l+1)\hbar), respectively.

    I'll try writing L2 with ladder-operators and Lz.
     
  8. Apr 20, 2009 #7
    This is what I have so far:

    L2 = L+L-+Lz2-\hbar Lz.

    This gives us:

    [tex]
    M = \left\langle {lm} \right|L_ + L_ - \left| {lm} \right\rangle + \left\langle {lm} \right|L_z^2 \left| {lm} \right\rangle - \hbar \left\langle {lm} \right|L_z \left| {lm} \right\rangle = 0 + \hbar ^2 m^2 - \hbar ^2 m = \hbar ^2 (m^2 - m)
    [/tex]

    where M is the eigenvalue of L2. Am I on the right track?
     
  9. Apr 20, 2009 #8
    I see, I will have a try to solve your problem.
     
  10. Apr 20, 2009 #9

    malawi_glenn

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    recall that the forum rules forbids you to post full solutions to problems here.
     
  11. Apr 20, 2009 #10
    How is it that the integer l even comes into play when finding the eigenvalues of L2? In my approach (which I am fairly certain is correct), only m shows up.
     
  12. Apr 20, 2009 #11

    malawi_glenn

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    hint: write L+L- = L^2 - Lz^2 + \hbar Lz

    and act on the ket with m_max.
     
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