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Commutator of exponential operators

  1. Jan 5, 2014 #1
    How do I compute the commutator [itex][a,e^{-iHt}][/itex], knowing that [itex][H,a]=-Ea[/itex]?

    I tried by Taylor expanding the exponential, but I get [itex]-iEta[/itex] to first order, which seems wrong.
     
  2. jcsd
  3. Jan 5, 2014 #2

    mfb

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    Why does it seem wrong?
     
  4. Jan 5, 2014 #3
    The answer I was expecting to get is something like [itex]e^{-iEt}[/itex]. This follows from the fact that I was trying to replicate the following calculation

    [tex]e^{iHt}ae^{-iHt}=e^{-iEt}a[/tex]

    which is said to be justified by [itex][H,a]=-Ea[/itex].

    If I take my result to be true, I get

    [tex]e^{iHt}ae^{-iHt}=e^{iHt}(e^{-iHt}a-iEta)=a-ie^{iHt}Eta[/tex]
     
  5. Jan 5, 2014 #4

    mfb

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    The first order is not the final result...
     
  6. Jan 5, 2014 #5
    Second order does the job....thanks.
     
  7. Jan 5, 2014 #6

    mfb

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    I am surprised by that, as the final result has all orders, and just ignoring orders means changing the value of things.
     
  8. Jan 5, 2014 #7
    It does the job in the sense that with two orders I can get back two terms which I can combine back together to form an exponential by recognising they are the first two terms in its Taylor expansion. I thought, naively, that one order would have been enough, but the fact that the first commutator vanishes doesn't make it work.
     
  9. Jan 5, 2014 #8

    mfb

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    Ah, okay.
     
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