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Struggling with Baker-Campbell-Haussdorf-like formulas

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data

    I have been asked to prove [tex]e^{\alpha a}f(a,a^\dagger)e^{-\alpha a}=f(a,a^\dagger+\alpha)[/tex], where [tex]a[/tex] and [tex]a^\dagger[/tex] represent the annihilation and creation operators of the quantum harmonic oscillator, respectively.

    2. Relevant equations



    3. The attempt at a solution

    The only thing I have done is to use the Hadamard Lemma (http://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula). Any ideas will be greatly appreciated!
     
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  3. Oct 17, 2009 #2

    gabbagabbahey

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  4. Oct 17, 2009 #3
    I get

    [tex]e^{\alpha a}f(a,a^\dagger)e^{-\alpha a}=f(a,a^\dagger)+\frac{\alpha}{1!}[a,f(a,a^\dagger)]+\frac{\alpha^2}{2!}[a,[a,f(a,a^\dagger)]]+\ldots[/tex]

    but I don't know what to do next.
     
  5. Oct 17, 2009 #4
    It seems that I have to apply some kind of Taylor expansion but I'm not sure on how to do this.
     
  6. Oct 17, 2009 #5

    gabbagabbahey

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    Hint: Consider [itex]f(a,a^{\dagger})=a^{\dagger}[/itex] and [itex]f(a,a^{\dagger})=a[/itex]....what do these considerations tell you about the unitary transformation [itex]f(a,a^{\dagger})\to e^{\alpha a}f(a,a^{\dagger})e^{-\alpha a}[/itex]?
     
    Last edited: Oct 17, 2009
  7. Oct 17, 2009 #6
    [tex]
    e^{\alpha a}a^\dagger e^{-\alpha a}=a^\dagger+\frac{\alpha}{1!}[a,a^\dagger]+\frac{\alpha^2}{2!}[a,[a,a^\dagger]]+\ldots=a^\dagger+\frac{\alpha}{1!}=\alpha+a^\dagger
    [/tex]

    [tex]
    e^{\alpha a}a e^{-\alpha a}=a+\frac{\alpha}{1!}[a,a]+\frac{\alpha^2}{2!}[a,[a,a]]+\ldots=a
    [/tex]

    In this sense the formula seems to be working fine.

    Does it mean that the unitary transformation [itex] f(a,a^{\dagger})\to e^{\alpha a}f(a,a^{\dagger})e^{-\alpha a} [/itex] represents a displacement of [itex]\alpha[/itex] in the second argument of [itex] f [/itex]?
     
  8. Oct 17, 2009 #7

    gabbagabbahey

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    Yes, exactly. Under the transformation [itex]f(a,a^{\dagger})\to e^{\alpha a}f(a,a^{\dagger})e^{-\alpha a}[/itex], you have [itex]a\to a[/itex] and [itex]a^{\dagger}\to a^{\dagger}+\alpha[/itex] and hence [itex]f(a,a^{\dagger})\to f(a,a^{\dagger}+\alpha)[/itex]....nice and simple, no Taylor series needed:smile:
     
  9. Oct 17, 2009 #8
    Thanks a lot, I think I finally understand it!!
     
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