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Ladder Operators acting upon N Ket

  1. Jan 9, 2010 #1
    I can't seem to find information regarding this anywhere.

    I understand why when the ladder operators act upon an energy eigenstate of energy E it produces another eigenstate of energy E [tex]\mp\hbar \omega[/tex]. What I don't understand is why the following is true:

    [tex]\ a \left| \psi _n \right\rangle &= \sqrt{n} \left| \psi _{n-1} \right\rangle [/tex]
    [tex]\ a^{\dagger} \left| \psi _n \right\rangle &= \sqrt{n+1} \left| \psi _{n+1} \right\rangle [/tex]

    I don't really even know what [tex] \left| \psi _n \right\rangle[/tex] represents, though I think it is something to do with the state of a system. How can you derive the above property?

    I am a bit of a beginner to SHO in quantum theory.
    Last edited: Jan 9, 2010
  2. jcsd
  3. Jan 9, 2010 #2
    Any help would be greatly appreciated, I am sure I am missing something simple. Thanks
  4. Jan 9, 2010 #3
    [tex]\psi_n[/tex] is a stationary state of the harmonic oscillator with energy [tex]E = (n+1/2)\hbar \omega[/tex]

    Ladder operators [tex]a_{+}, a_{-}[/tex] raise the energy of the state from n to n+1 or lower n to n-1.

    Since the energy changes the state must change as well since the state is characterized by its energy. If you raise the energy by n+1 you raise the state to [tex]\psi_{n+1}[/tex]

    So you can sort of see that if a ladder operator acts on a stationary wave function in state n it will raise the state to n+1 by some proportionality constant:

    [tex]\ a_+ \left| \psi _n \right\rangle \alpha \left| \psi _{n+1} \right\rangle[/tex]

    The two are proportional. Mathematically you can show the proportionality constants are [tex]\sqrt{n+1} and \sqrt{n}[/tex] respectively.
  5. Jan 9, 2010 #4
    So would I be right in saying that this works because any constant multiple of an eigenstate must also be an eigenstate.
    i.e. if the constant of proportionality is beta then this works because

    [tex]H (a^\dagger \left| \psi _n \right\rangle) = (E + \hbar\omega) (a^\dagger \left| \psi _n\right\rangle)[/tex]
    [tex]\beta \ H \left| \psi _{n+1} \right\rangle = \beta (E + \hbar\omega) \left| \psi _{n+1} \right\rangle[/tex]

    To find beta would I then have to take the modulus squared of [tex]
    \ a_+ \left| \psi _n \right\rangle
    [/tex] and set it to be equal to 1(normalising it). If that is correct what do I use for [tex]
    \left| \psi _n \right\rangle
    [/tex] ?

    Sorry if I have misunderstood you.
  6. Jan 9, 2010 #5


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    Have you shown that the energies of the states are given by [tex]E_n = (n+1/2)\hbar\omega[/tex] and that the Hamitonian can be written as [tex]H=\hbar\omega(a^\dagger a+1/2)[/tex]?

    The eigenstates [tex]|\psi_n>[/tex] are assumed to be normalized so that [tex]<\psi_n|\psi_n>=1[/tex]. With a little fiddling, you can calculate what [tex]a|\psi_n>[/tex] and [tex]a^\dagger|\psi_n>[/tex] are.
  7. Jan 10, 2010 #6
    yes I have. I think I understand

    If I write [tex]
    H=\hbar\omega(a^\dagger a\ +1/2) \
    or \ H= \hbar\omega(aa^\dagger\ -1/2)

    And then act upon an eigenvector [tex]
    [/tex], which returns [tex]
    E_n = (n+1/2)\hbar\omega

    I can show that
    a^\dagger a|\psi_n>=(n+1)|\psi_n>[/tex]

    I know that the identities solve this but I cannot prove it from here, since the constants of proportionality are functions of n. What can I do?
  8. Jan 10, 2010 #7


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    Try calculate the norms of [itex]a\left|\psi_n\right>[/itex] and [itex]a^\dagger\left|\psi_n\right>[/itex].
  9. Jan 10, 2010 #8
    ahhhhhh you legend. Makes so much sense now, sorry for being a bit dense. Thanks so much!!
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