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Need help proving that an infinite double sum is 1

  1. Dec 3, 2012 #1
    1. The problem statement, all variables and given/known data

    I am asked to prove that [itex]e^{iB}[/itex] is unitary if B is a self-adjoint matrix.

    3. The attempt at a solution

    In order to prove this I am attempting to show [itex]e^{iB} \widetilde{e^{iB}} = 1[/itex]. Using the assumption that B is self-adjoint I have been able to show that

    [tex]

    e^{iB} \widetilde{e^{iB}} = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m i^{m+n} B^{m+n}}{m!n!}.

    [/tex]

    I have looked at the first few powers of B and shown that the coefficients go to 0. I expect that this is the case in general for m+n>0, however I am having difficulty proving this (for m,n=0 the term is 1). Can anyone point me in the right direction? Thanks.
     
  2. jcsd
  3. Dec 3, 2012 #2

    Dick

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    I don't think that's the easiest way to prove this. If B is self-adjoint then B has a complete set of orthonomal eigenvectors with real eigenvalues. If v is an eigenvector then Bv=λv then e^(iB)v=e^(iλ)v. Pick a basis of eigenvectors and show e^(iB) is unitary directly.
     
  4. Dec 3, 2012 #3
    I'm sorry, I'm a little lost by what you mean by showing the e^(iB) is unitary directly.
     
  5. Dec 3, 2012 #4

    Dick

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    If U is unitary that means <Ux,Uy>=<x,y>, right? It preserves the inner product.
     
    Last edited: Dec 3, 2012
  6. Dec 3, 2012 #5
    I see what you're saying. I will attempt that.

    I'm still interested in seeing how one would simplify the sum. Wolfram alpha confirms that the sum goes to one, so if anyone could enlighten me about this I would appreciate it.
     
  7. Dec 3, 2012 #6

    micromass

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    If you want to specifically work with the series, then you should try to "change variables" in the series. Let k=m+n, then you should get series of the form

    [tex]\sum_k \sum_m[/tex]

    The inner sum (summed of m) is a finite sum. Try to get that in the form of a binomial sum.
     
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