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Re: Susskind's proof of orthogonality of eigenvectors
Posted:
Jan 26, 2013 10:02 PM


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. > > Ma> = lambda_aa> > > Mb> = lambda_bb> > > Multiply the first by the conjugate of the second and the second by the > conjugate of the first. > > <bMa> = lambda_a<ba> > > <aMb> = lambda_b<ab> > > Observe that: > > <aMb> = <bMa>* > > <ab> = <ba>* > > So, as I understand it: > > <aMb> = lambda_b<ab> = <bMa>* = lambda_b<ba>* > > 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: > > <aMb> = lambda_b<ab> > > That would result in: > > <aMb>* = (lambda_b<ab>)* = <bMa> = lambda_b*<ba>, > > 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 doubletake, but if he had > said something like "Now we rewrite <aMb> = lambda_b<ab> 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!



