Need help proving that an infinite double sum is 1

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Homework Help Overview

The problem involves proving that the expression e^{iB} is unitary, given that B is a self-adjoint matrix. The discussion centers around the properties of self-adjoint matrices and their eigenvalues, as well as the manipulation of infinite series.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods to prove unitarity, including direct approaches using eigenvalues and eigenvectors, as well as attempts to simplify an infinite double sum. Questions arise regarding the manipulation of series and the implications of self-adjoint properties.

Discussion Status

Some participants have suggested alternative methods for proving unitarity, while others express confusion about these suggestions. There is an ongoing exploration of how to simplify the series and whether it converges to one, with references to external tools for confirmation.

Contextual Notes

Participants are navigating the complexities of infinite series and the properties of self-adjoint matrices. There is a mention of needing to change variables in the series to facilitate simplification, indicating a focus on mathematical manipulation rather than direct solutions.

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Homework Statement



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

The Attempt at a Solution



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

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

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.
 
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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.
 
I'm sorry, I'm a little lost by what you mean by showing the e^(iB) is unitary directly.
 
mjordan2nd said:
I'm sorry, I'm a little lost by what you mean by showing the e^(iB) is unitary directly.

If U is unitary that means <Ux,Uy>=<x,y>, right? It preserves the inner product.
 
Last edited:
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.
 
mjordan2nd said:
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.

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

\sum_k \sum_m

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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