Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Raising and lowering operators on a ket?

  1. Dec 13, 2008 #1
    How do you use the S(+) and S(-) operators on integer kets, |1>, |-1>, |0>?

    I'm told the outcome of the ones which aren't zero will be something like h(bar)/sqrt(2) * |ket>

    Confused!? I thought operators are 2 x 2 matrices...

    Any help much appreciated,

    Philip
     
  2. jcsd
  3. Dec 13, 2008 #2

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    Operating with S+ on |+1> Will give you Zero if you are operating on S = 1 states.. you are not so clear in your question here, but I assume you have S = 1 states here..

    Operating with S- on |-1> = 0 as well.

    Operating with S- on |0> you use formulas:

    http://www.mt.luth.se/~nikle/Education/MTF067/Lectures/PL13.pdf

    (just replace L with S , these operators work the same no matter what kind of angular momentum you have)
     
  4. Dec 13, 2008 #3

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    And since you have S = 1 states here, you should have a 3x3 matrix
     
  5. Dec 14, 2008 #4
    Cheers for the answer, I have a question though:

    In the third slide you posted, what does the l = 3 correspond to? The question I am looking at says for spin 1. What is l and what is m. Should my l in the examples always be 1, as they are spin 1, and my m be either -1, 0 or 1?
     
  6. Dec 14, 2008 #5

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    yes :-)
     
  7. Dec 14, 2008 #6
    cool, so if it were for spin 1/2, would l = 1/2 and m = 1/2, 0 or 1/2?
     
  8. Dec 14, 2008 #7

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    No, think! For a spin 1/2 particle, which possible z-projections exists? What is the general relation for an Angularmomentum J and its z-projections m_J ?
     
  9. Dec 14, 2008 #8
  10. Dec 14, 2008 #9

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    Given an angular momentum [itex] j [/itex], which can have values: 0,1/2 ,1, 3/2 ,2 ,5/2, ...

    the z-component, [itex] m_j [/itex], can obtain values: [itex] m_j = -j, -j +1, \ldots , j-1, j [/itex]

    So if j = 1/2, what possible values can you have of the z-component? ;-)
     
  11. Dec 14, 2008 #10
    -5/2 to 5/2, why didn't i say that the first time!?! sorry for being dim
     
  12. Dec 14, 2008 #11

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    No you must have missunderstood. In your case, j = 1/2, what values of m_j can you have?
     
  13. Dec 14, 2008 #12
    no thats wrong
     
  14. Dec 14, 2008 #13
    -3/2 3/2

    miss print
     
  15. Dec 14, 2008 #14

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    now you are REALLY making a fool out of yourself ;-)

    If j = 1/2 and the general "formula" for possible m_j's for a given j are -j, -j+1, .. ,j -1, j

    What values of m_j can you have for j = 1/2 ?
     
  16. Dec 14, 2008 #15
    3/2 to 3/2???
     
  17. Dec 14, 2008 #16
    missed out the minus by mistake.
     
  18. Dec 14, 2008 #17

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    are you guessing or using the formula?

    if you are guessing, stop doing that and apply formula.

    if you use formula, tell me how you do it, you are doing something REALLY wrong.

    According to formula, for a given j, the maximum m_j is j, and the minumum is -j.
     
  19. Dec 14, 2008 #18
     
  20. Dec 14, 2008 #19

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    good :-)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Raising and lowering operators on a ket?
Loading...