Does \(A e^{iA} = e^{iA} A\)?

russdot · 2009-10-26T23:55:34+00:00

If A is an operator, is it correct/allowed to say: Ae^{iA} = e^{iA}A Thanks

Thread starter russdot
Start date Oct 26, 2009
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The discussion centers on the commutation of the operator \(A\) with the exponential operator \(e^{iA}\). It is established that since \(e^{A}\) is defined via a power series, specifically \(e^{A} = 1 + A + \frac{A^{2}}{2!} + \frac{A^{3}}{3!} + \ldots\), and given that \(A\) commutes with itself, it follows that \(Ae^{iA} = e^{iA}A\) is indeed valid. This conclusion is based on the properties of operators in functional analysis.

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Oct 26, 2009

russdot

If A is an operator, is it correct/allowed to say:
[tex]Ae^{iA} = e^{iA}A[/tex]

Thanks

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Oct 26, 2009

Ben Niehoff

Science Advisor

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Think of how [itex]e^{A}[/itex] is defined.

Oct 26, 2009

russdot

Ah yes, so since [tex]e^{A}[/tex] is defined with a power series [tex]e^{A} = 1 + A + \frac{A^{2}}{2!} + \frac{A^{3}}{3!} + ...[/tex] and A commutes with itself then A would commute with [tex]e^{A}[/tex]
Thanks!