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Harmonic Oscillator, Ladder Operators, and Dirac notation

  1. May 15, 2007 #1
    Defining the state [itex]| \alpha > [/itex] such that:
    [tex]| \alpha > = Ce^{\alpha {\hat{a}}^{\dagger}} | 0 >\ ,\ C \in \mathbf{R};\ \alpha \in \mathbf{C};[/tex]

    Now, [itex]| \alpha >[/itex] is an eigenstate of the lowering operator [itex]\hat{a}[/itex], isn't it?

    In other words, the statement that [itex] \hat{a} | \alpha >\ =\ \alpha | \alpha > [/itex] is true, right?
    (for the eigenvalue [itex]\alpha[/itex]?)

    Is that the correct use of the Dirac notation?

    How might one go about proving the above?
     
    Last edited: May 16, 2007
  2. jcsd
  3. May 17, 2007 #2
    Anybody? I'd really appreciate some advice here.

    Cheers.
     
  4. May 17, 2007 #3
    I got the same result. Was there step where you had

    [tex]
    a (a^\dagger)^k = k (a^\dagger)^{k-1} + (a^\dagger)^k a
    [/tex]

    It looks like it's an eigenstate of lowering operation then. Pathological state...

    I'm not sure about what is precisly correct notation, looked fine to me, but at least the latex symbols \langle and \rangle look better than < and > :wink:

    Adding with edit:

    Hups, I didn't notice your last question. I thougth you had calculated that, and were wondering how such strange state could exist. Steps to prove it are these. Use series expansion of the exponential. Prove the commutation rule of [tex]a[/tex] and [tex](a^\dagger)^k[/tex] with induction (or with some other technique). Calculate.
     
    Last edited: May 17, 2007
  5. May 17, 2007 #4
    I wouldn't call coherent states "pathological states".

    The way you would go about proving it would probably be by construction. Start by declaring a state [tex]|\alpha \rangle[/tex] to be an eigenstate of the lowering operator, then expand that state in the basis of the harmonic oscillator and see what comes out. Then you can equate it to the exponential of the raising operator by looking at what comes out and saying "AH HA!".

    That's probably the least algebraically intensive method of solving the problem, unless you really like wrestling with commutators.
     
  6. May 17, 2007 #5
    I didn't know anything about coherent states, but the calculation wasn't too difficult so I replied. I'll take the pathological comment back.
     
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