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Creation and annihilation operators

  1. Nov 9, 2012 #1
    Hi all

    1. The problem statement, all variables and given/known data
    Show:

    [tex] (a^\dagger a)^2=a^\dagger a^\dagger a a +a^\dagger a[/tex]

    wheres:
    [tex] a= \lambda x +i \gamma p [/tex]
    [tex] a^\dagger= \lambda x -i \gamma p [/tex]




    2. Relevant equations
    -

    3. The attempt at a solution

    Well, I haven't got much.

    I just tried to use the stuff given, put it into my equation and solve it, but I don't get to the right side.

    I calculated a+a first

    [tex] a^\dagger a ={\lambda}^2x^2 + \frac {1}{2} I + \gamma^2 p^2[/tex]

    But when I now try to calculate the square of that term I get lost. If I square it I get to:

    [tex] (a^\dagger a)^2= \lambda^4x^4+\gamma^4p^4 +\lambda^2 \gamma^2 (x^2p^2+p^2x^2)-\lambda^2 x^2 -\gamma^2 p^2 +\frac 1 4 I[/tex]


    Can anyone help me with this? I don't know what to do now/ If I'm on the right way.

    Thanks for your help

    edit: I is the identity matrix
     
    Last edited: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2
    (a[itex]^{+}[/itex]a)[itex]^{2}[/itex]
    = a[itex]^{=}[/itex]aa[itex]^{+}[/itex]a
    =a[itex]^{+}[/itex](aa[itex]^{+}[/itex])a.

    Now try to get aa[itex]^{+}[/itex] in terms of a[itex]^{+}[/itex]a.
     
  4. Nov 9, 2012 #3
    what does your a^= mean? I don't know this sign.

    Thanks for your help

    edit:

    [tex] aa^\dagger=\lambda^2 x^2 + \frac 1 2 I +\gamma^2 p^2[/tex]
     
  5. Nov 9, 2012 #4
    I used the plus sign instead of the dagger sign.

    You are supposed to see a with + as a superscript wherever you see a with ^ as a superscript.

    Sorry for the trouble in the notation used.
     
  6. Nov 9, 2012 #5
    I see a = not a + but I think it's just a typing mistake.

    so what you mean

    [tex]...= a^\dagger a a^\dagger a=a^\dagger (aa^\dagger) a[/tex]

    is that what you meant?

    I calculated my aa+ (see my post) and a+a is the same, though there's a minus infront of the I
     
  7. Nov 9, 2012 #6
    "The same"? That difference in sign makes all the difference. Can you now relate [itex]aa^{\dagger}[/itex] to [itex]a^{\dagger}a[/itex]?
     
  8. Nov 9, 2012 #7
    With "the same" I meant the rest of the terms. Unlucky word choice from my side I guess.
    What do you mean with "can you now relate aa+ to a+a"
    do you mean that I should calculate [tex] (aa^\dagger) (a^\dagger a)[/tex]

    thanks for the help
     
  9. Nov 9, 2012 #8
    No, I meant the relationship between [itex]aa^{\dagger}[/itex] and [itex]a^{\dagger}a[/itex] ie express one of them in terms of the other.
     
  10. Nov 9, 2012 #9
    Yes that is what I meant.

    Allow me to use a+ to mean a with + as a superscript.

    So aa+ = a+a + 1.

    Now one more step.
     
  11. Nov 9, 2012 #10
    Do you mean something like

    [tex] a^\dagger a = 2\lambda^2 x^2 +2 \gamma^2 x^2 -a a^\dagger[/tex]

    edit: nevermind, fail lol.

    I guess you meant what grzz just said. I have to eat dinner now. I'll try later or tomorrow and come back when I have further questions (I'm pretty sure I will have some :( )

    Thanks so far for the help guys.
     
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