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Raising and lowering operators on a simple harmonic oscillator

  1. Jun 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi, I'm currently studying for a quantum mechanics exam but I am stuck on a line in my notes:

    [tex]Ha\left|\Psi\right\rangle =\hbar\omega\left(a^{t}a a + \frac{a}{2}\right)\left|\Psi\right\rangle[/tex]

    [tex]Ha\left|\Psi\right\rangle =\hbar\omega\left(\left(a a^{t} - 1\right)a + \frac{a}{2}\right)\left|\Psi\right\rangle[/tex]

    Where H is the harmonic oscillator hamiltonian

    [tex]H = \frac{P^{2}}{2m} + \frac{1}{2}m\omega^{2} x^{2}[/tex]

    and [tex] a, a^{t}[/tex] are the lowering and raising operators respectively.

    [tex] a^{t} = \frac{1}{\sqrt{2\hbar m \omega}}\left(m \omega x - i p\right)[/tex]

    [tex] a= \frac{1}{\sqrt{2\hbar m \omega}}\left(m \omega x + i p\right)[/tex]

    The part I dont understand is how they got from [tex] a^{t}a a[/tex] to [tex]\left(a a^{t} - 1\right)a [/tex]

    I'm not afraid of doing a bit of calculating but I really need a hint as to what method I should use.
    Last edited: Jun 9, 2010
  2. jcsd
  3. Jun 9, 2010 #2


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    Are you sure your first equation is correct?

    The Hamiltonian has one of two forms



    Whichever form you use, if you tack on an extra at on the right, you don't end up with what you have. Check your notes farther up.
  4. Jun 9, 2010 #3
    Apologies I mistyped when I wrote


    I meant to say


    does this make more sense now?

    It now looks just like your second hamiltonian multiplied by "a" to me.
  5. Jun 9, 2010 #4


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    Now that you got the first line right, you can get from the first to the second line using the commutator

    [tex][a, a^t]=1[/tex]
  6. Jun 9, 2010 #5
    thanks so much! this has been bothering me all day!

    I used [tex]\left[ a,a^{t}\right]= a a^{t}-a^{t} a =1[/tex]

    and re-arranged to get

    [tex] a a^{t}-1=a^{t} a [/tex]

    which I then sub straight into the first line :-)
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