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Homework Help: Commutator question. [A,B] =.lambda proof

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm having troubles with this proof.
    given two operators A &B , such that [A,B] = λ where λ is complex,and μ is also complex, show that

    exp{μ(A+B)} = exp{μA}exp{μB}exp{(-μ^2λ)/2}

    2. Relevant equations

    [A,B] = λ.
    [A,B] = AB-BA = λ

    3. The attempt at a solution
    I have tried expanding the right side of the equation into three different exponentials by subbing in λ=AB-BA.
    I get exp{μ(2A+2B-μAB+μBA}
    but i seem to have a factor of two extra...
    I'm not even sure if i am on the right track, so if anyone could give me a starting point it would be greatly appreciated!
  2. jcsd
  3. Nov 14, 2013 #2
    It seems like maybe you tried adding the exponents on the right hand side. That's not allowed. If [itex]e^Ae^B[/itex] generally is not equal to [itex]e^{A+B}[/itex]. That would contradict what you are trying to show. Anyway, I'm not sure the best way to solve this problem, but what you are trying to show is.

    [tex]\sum_{n=0}^\infty \frac{\mu^nA^n}{n!} \sum_{m=0}^\infty \frac{\mu^mB^m}{m!} \sum_{a=0}^\infty \frac{\left(-1\right)^a\mu^{2a}\lambda^a}{2^a a!}[/tex]
    [tex]=\sum_{b=0}^\infty \frac{\mu^b (A+B)^b}{b!}[/tex]

    You could try seeing if you can figure out where at the terms come from in the first expression.
    [itex](A+B)^0= 1[/itex]
    [itex](A+B)^1= A + B[/itex]
    [itex](A+B)^2= A^2 + 2AB - [A,B]+ B^2 = A^2 + 2AB + B^2 - \lambda[/itex]
    [itex](A+B)^3= (A+B)\left( A^2 + 2AB + B^2 - \lambda\right) [/itex]

    However I'm really not sure exactly how to get this result.
  4. Nov 14, 2013 #3


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    Science Advisor
    Gold Member

    That's the Baker-Campbell-Hausdorff formula. Search with Google and you will probably find help in the proof.
  5. Nov 15, 2013 #4
    Thank you guys.
    Yes , my TA said that you can use the Baker -Campbell-Hausdorff formula.

    I will be able to solve it now, thanks!!
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