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Topic: Susskind's proof of orthogonality of eigenvectors
Replies: 2   Last Post: Jan 27, 2013 8:59 AM

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 Will Janoschka Posts: 28 Registered: 12/13/04
Re: Susskind's proof of orthogonality of eigenvectors
Posted: Jan 26, 2013 10:02 PM
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On Sat, 26 Jan 2013 21:01:30, Hetware <hattons@speakyeasy.net> wrote:

> http://www.youtube.com/watch?v=CaTF4QZ94Fk&list=ECA27CEA1B8B27EB67
>
> Lecture 3, beginning around 1:03:20.
>
> This is what I believe he intended:
>
> Begin with the assumption that we have two unique eigenvalues for a 2X2
> Hermitian matrix.
>
> M|a> = lambda_a|a>
>
> M|b> = lambda_b|b>
>
> Multiply the first by the conjugate of the second and the second by the
> conjugate of the first.
>
> <b|M|a> = lambda_a<b|a>
>
> <a|M|b> = lambda_b<a|b>
>
> Observe that:
>
> <a|M|b> = <b|M|a>*
>
> <a|b> = <b|a>*
>
> So, as I understand it:
>
> <a|M|b> = lambda_b<a|b> = <b|M|a>* = lambda_b<b|a>*
>
> Notice this is different from what Susskind presents. I have not
> conjugated lambda_b, whereas he did. I know he has already stated that
> the eigenvalues are real, so lambda_b*=lambda_b. Therefore, there is no
> difference in bedeutung (denotation). There is a difference in
> sinn(sense), however.
>
> I don't see the motivation for conjugating lambda_b where he did so. He
> isn't really conjugating both sides of the equation:
>
> <a|M|b> = lambda_b<a|b>
>
> That would result in:
>
> <a|M|b>* = (lambda_b<a|b>)* = <b|M|a> = lambda_b*<b|a>,
>
> if I'm not mistaken.
>
> One comment on the YouTube page says that he screwed up the presentation
> at that point. It certainly made me do a double-take, but if he had
> said something like "Now we rewrite <a|M|b> = lambda_b<a|b> in it's
> equivalent complex conjugate form by replacing all terms by equivalent
> complex conjugate terms." I believe his development would be
> procedurally valid.
>
> Does that make sense?

No! Nor do you! Complex conjugate in how many terms?
Four is not enough!

Date Subject Author
1/26/13 Will Janoschka
1/27/13 Sam Wormley

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