## Exponential of Operators

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### Exponential of Operators

I didn't know whether to post this in 'School' or here — I settled for here since it's a science question and I'm just studying for the exam, and this has the best chance of someone spotting it.
Essentially, I'm trying to prove that:

Exp(A+B)=Exp(A)Exp(B)Exp(-(1/2)[A,B])

Where A and B are operators, and [A,B] is the commutator AB-BA.
I obviously need to expand is as the infinite sum Exp(x)=Sum[x^n/n!,{n,0,inf}], but I haven't been able to do anything with that, so ... halp please!
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

yurell

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### Re: Exponential of Operators

Does the commutator commute with A and B?

Anyway, I'd suggest working with the log of the equation. Also, in case you are unaware, you are essentially looking at a Campbell-Baker-Hausdorff formula.
SU3SU2U1

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### Re: Exponential of Operators

Oh, right, forgot to add A and B don't commute, but [A,[A,B]] = [B,[A,B]] = 0
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

yurell

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### Re: Exponential of Operators

If A and B could commute, sum of (A+B)^n / n! would look exactly like eA eB. If A and B can't commute, where do the commutators come in? Can you turn that into the exponential?
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Charlie!

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### Re: Exponential of Operators

I'd be tempted by Charlie!'s approach. In fact, I'd be tempted to find a proof that e^(x+y) = e^x e^y proven via the power series that assumes x and y commute, then see if I could modify the proof to extract out (xy-yx) terms.

I'm relatively sure if x and y commute, then e^x and e^y commute, but not absolutely so.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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