Commutator of exponential operators

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Discussion Overview

The discussion revolves around the computation of the commutator of exponential operators, specifically [a, e^{-iHt}], given the relation [H, a] = -Ea. Participants explore the implications of Taylor expanding the exponential and the significance of different orders in the expansion.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about computing the commutator [a, e^{-iHt}] and expresses confusion over obtaining -iEta to first order.
  • Another participant questions why the initial result seems wrong, prompting further exploration of the calculation.
  • A participant references an expected result of e^{-iEt} based on a related calculation involving e^{iHt}ae^{-iHt} and the commutation relation [H, a] = -Ea.
  • There is a discussion about the sufficiency of first-order terms in the Taylor expansion, with one participant noting that second-order terms are necessary to achieve the desired result.
  • Another participant expresses surprise that second-order terms are needed, highlighting concerns about ignoring orders affecting the final value.
  • One participant clarifies that using two orders allows for the combination of terms to form an exponential, indicating that the vanishing of the first commutator complicates the computation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the sufficiency of first-order versus second-order terms in the Taylor expansion for accurately computing the commutator. There are differing views on the implications of these orders on the final result.

Contextual Notes

Limitations include the dependence on the Taylor expansion and the unresolved nature of higher-order terms in the context of the commutation relation.

gentsagree
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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.
 
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Why does it seem wrong?
 
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]
 
The first order is not the final result...
 
Second order does the job...thanks.
 
gentsagree said:
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.
 
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.
 

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