Commutator of exponential operators

  • Thread starter gentsagree
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  • #1
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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.
 

Answers and Replies

  • #3
96
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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]
 
  • #4
34,655
10,797
The first order is not the final result...
 
  • #5
96
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Second order does the job....thanks.
 
  • #6
34,655
10,797
Second order does the job....thanks.
I am surprised by that, as the final result has all orders, and just ignoring orders means changing the value of things.
 
  • #7
96
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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.
 
  • #8
34,655
10,797
Ah, okay.
 

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