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Quantum Mechanics Ladder Operator and Dirac Notation

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm given the eigenvalue equations
    L[tex]^{2}[/tex]|[tex]\ell[/tex],m> = h[tex]^2\ell(\ell + 1)|\ell[/tex],m>
    [tex]L_z[/tex]|[tex]\ell[/tex],m> = m|[tex]\ell[/tex]
    [tex]L_{\stackrel{+}{-}}[/tex]|[tex]\ell[/tex],m> = h[tex]\sqrt{(\ell \stackrel{-}{+} m)(\ell \stackrel{+}{-} m + 1)}[/tex]|[tex]\ell[/tex], m [tex]\stackrel{+}{-}[/tex] 1>

    Compute <[tex]L_{x}[/tex]>.

    2. Relevant equations

    Know that [tex]L_x = (1/2)(L_+ + L_-)[/tex].

    3. The attempt at a solution

    Need to compute <[tex]\ell[/tex],m|[tex]L_x[/tex]|[tex]\ell[/tex],m>.

    = (1/2)(<[tex]\ell[/tex],m)([tex]L_+[/tex] + [tex]L_-[/tex])([tex]\ell[/tex],m>)

    = (1/2)(<[tex]\ell[/tex],m|[tex]L_+[/tex]|[tex]\ell[/tex],m> + <[tex]\ell[/tex],m|[tex]L_-[/tex]|[tex]\ell[/tex],m>)

    I don't think I have a good grasp on how to work with dirac notation, so this as far as I can get before I get stuck. Thanks in advance for any help!
     
  2. jcsd
  3. Nov 17, 2008 #2
    You are on the right lines. Now you need to evaluate what the L+ operator acting on the state |l,m> is, and the same for L-. then you will have something which has the form < | > + < | >. From this point, the answer is straightforward.

    Thats about as precise as i can be without doing it for you, i think, sorry if its a little abstract.
     
  4. Nov 17, 2008 #3
    When I'm evaluating the ladder operators for the state |l,m>, how do I deal with the fact that the corresponding eigenvalue is for |l,m+-1>?
     
  5. Nov 17, 2008 #4

    malawi_glenn

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    what is <l=1,m=1|l=1,m=0> for instance?

    or what was your latest question reffering to?
     
  6. Nov 17, 2008 #5
    That's what my question was about. I don't understand what it means to have <l_a, m_a | l_b, m_b>. If they were the same l's and m's, it should be 1, yes? But otherwise, I'm lost...
     
  7. Nov 17, 2008 #6

    malawi_glenn

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    If they are not same state, then it is ?.... guess ;-)
     
  8. Nov 17, 2008 #7
    Just 0?
     
  9. Nov 17, 2008 #8

    malawi_glenn

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    YES! :-)

    |L_a , M_b > is same state as |L_c, M_d> if and only if a=c and b=d
     
  10. Nov 17, 2008 #9
    What is the physical interpretation of <quantum numbers "a" | quantum numbers "b">? Is it the probability that a system in the "b" state will simultaneously be in the "a" state?
     
  11. Nov 17, 2008 #10

    malawi_glenn

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    [tex]<a|b> = \int _V\psi _a(\vec{r})^*\psi_b(\vec{r})d^3r[/tex]

    Where the [itex]\psi _a(\vec{r})[/itex] is the Wavefunction for the state a in position space.

    Now you see why dirac formalism is superior, you don't need to know or write all the eigenfunction to the operators and so on, just label the state by the things that are important for you at the moment.

    If you wanted to write your angular momentum states bra-ket in "old style":
    [tex]<L_c , M_d |L_a , M_b > = \int Y_{L_c}^{M_d}^*(\theta,\phi)Y_{L_a}^{M_b}}(\theta,\phi)\sin \theta d\theta d\phi[/tex]

    where [itex]Y_{L_c}^{M_d}(\theta,\phi)[/itex] is the so called "Spherical harmonic" function.
     
    Last edited: Nov 17, 2008
  12. Nov 17, 2008 #11
    Ohhh, okay. Thanks so much for all your help, I think things are becoming much clearer!
     
  13. Nov 17, 2008 #12

    malawi_glenn

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    good luck and please ask questions here again.
     
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