1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

A problem with Ladder operators

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data

    The problem is to show that,




    It's not hard to manipulate [tex]\hat{a_{+}}\hat{a_{-}}|\alpha>=\alpha|\alpha>[/tex] into the form,


    But I am unable to make the connection from this to,


    I know it's just using the properties of the eigenfunctions/values of a Hermatian operator at this point, but I seem to be missing exactly what that is.

    What am I missing?
  2. jcsd
  3. Oct 29, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

    Well, this is an eigenvalue equation for the operator [itex]\hat{a}_+\hat{a}_-[/itex], with eigenvalue [itex]\alpha+1[/itex] and eigenstate [itex]\hat{a}_+|\alpha\rangle[/itex]...but compare this to your original eigenvalue equation for this operator....surely if [itex]\alpha+1[/itex] is the eigenvalue, the eigenstate must be [itex]|\alpha+1\rangle[/itex] (or at least a scalar multiple of it).....doesn't that tell you everything you need to know about [itex]\hat{a}_+|\alpha\rangle[/itex]?:wink:
  4. Oct 29, 2009 #3
    From what you've done, using both ladder operators on the ket, what does that say about [itex]N=\hat{a}_+\hat{a}_-[/itex] and [itex]\hat{a}_\pm[/itex]??
  5. Oct 29, 2009 #4
    Thanks very much for the replys.

    gabba, That did occur to me, but I wasn't willing to make the concession that,

    [tex]\hat{a_{+}}\hat{a_{-}}|\alpha>=\alpha|\alpha> [/tex]

    Was a general property and not [tex]\alpha[/tex] specific.Is this a property of NORMALIZED eigenvectors(for which I should have specified |alpha> is defined as)? If so I suppose that would explain the [tex]A_{\alpha}[/tex] as a normalization constant.

    jd, I know [tex]\hat{a_{+}}\hat{a_{-}}[/tex] is Hermatian although neither are individually..... I'm not sure if that's what you mean.
  6. Oct 29, 2009 #5
    I was trying to guide you with less words than what gabba said: If


    and [itex]N|n\rangle=n|n\rangle[/itex], then [itex]\hat{a}_\pm|n\rangle[/itex] are multiplicative eigenstates of [itex]|n\pm1\rangle[/itex].
  7. Oct 29, 2009 #6
    Thank you that helps, I'll have to stare at that for awhile to let it sink in.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook