I'm not sure if this is the appropriate section, perhaps my question is better suited for Linear Algebra. At any rate, here goes.(adsbygoogle = window.adsbygoogle || []).push({});

Regarding a proof of the orthogonality of eigenvectors corresponding to distinct eigenvalues of some Hermitian operator [itex]A[/itex]:

Given [itex]A|\phi_1\rangle = a_1|\phi_1\rangle[/itex] and [itex]A|\phi_2\rangle = a_2|\phi_2\rangle[/itex] we have

[tex]\begin{array}{lll}

0 & = & \langle\phi_1|A|\phi_2\rangle -

\langle\phi_2|A|\phi_1\rangle^* \\

& = & a_1\langle\phi_2|\phi_1\rangle - a_2\langle\phi_1|\phi_2\rangle^* \\

& = & (a_1 - a_2)\langle\phi_2|\phi_1\rangle \\

\end{array}

[/tex]

Now, I understand that this shows the inner product of the eigenvectors to be 0 (given distinct eigenvalues) and therefore that the eigenvectors are orthogonal. But, if someone could please help, how do we arrive at line 2 of the equivalence from line 1. That is, replacing the operator with its corresponding eigenvalues. There appears to me to be some gymnastics with the positions of multiplicands. If someone could possibly detail this for me (and please assume I am an idiot, not such a stretch really) I would again be eternally grateful and would show just how grateful I am by possibly responding with a thank you. No, make that most definitely responding with a thank you.

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# Problem with proof of orthogonality of eigenvectors for Hermitian

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