Timeline for Why does the variational method simplify in this way when $H$ Hermitian?
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7 events
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| Aug 18, 2023 at 2:53 | comment | added | naturallyInconsistent | It makes less sense to me because I thought Ballentine was just going "btw, you know from your other studies that Hermitian matrices have such and so properties", i.e. you should not be trying to read too much into this particular argument. Which is why I was thinking I should just directly prove the missing bits; But now peek-a-boo did a far better presentation so just take that~ | |
| Aug 17, 2023 at 18:09 | comment | added | EE18 | I'm still afraid i don't understand. Please see my edit in the OP if you get the chance -- does that make my question make more sense? | |
| Aug 17, 2023 at 17:59 | comment | added | naturallyInconsistent | Well, in the one with two different vectors we dont actually know that the eigenvalues are real, i.e. we have to assume that they are complex, and that we do not know if the eigenvectors are orthogonal. Because of that, the set of (10.112a) and (10.112b) is actually much more difficult to work with. What I have supplied is a direct proof of the standard properties of Hermitian operators, i.e. their eigenvalues are real and the eigenvectors are the same on both sides. You will need a bit more work to show that when the eigenvalues are different, then orthogonality (and this extends in general). | |
| Aug 17, 2023 at 17:50 | comment | added | EE18 | But in this case I'm afraid I don't see how what you've written proves the statement in my comment, and perhaps that's because I don't follow what you're trying to prove. If we agree that Ballentine's claim is basically "if $H$ Hermitian, then extremization of $\Lambda(\phi,\psi)$ is equivalent to exeremiztion of $\Lambda(\psi,\psi)$", then how have we concluded that here? | |
| Aug 17, 2023 at 17:43 | comment | added | naturallyInconsistent | Yes, you can read it that way. But you might want to understand the proof, which I was supplying. You just need to understand my last 3 lines. Act on the bra line with the ket, and act on the ket line with the bra, and you get my result that both equations have to satisfy. This is the thing that forces the eigenvalues to be real and the eigenfunctions to be the same on both sides. | |
| Aug 17, 2023 at 16:38 | comment | added | EE18 | Yes it is a wonderful book! Callen, Ballentine, and Ashcroft and Mermin are what I am working through right now. At any rate, I am a bit confused about what you are proving here (i.e. your (result) line). I interpret Ballentine's line which begins "If $H = H^\dagger$..." as saying that if $H$ of (10.110) is Hermitian, then (as a theorem) the functional $\Lambda(\phi,\psi)$ (which is in general minimized with two different vectors $\phi,\psi$ is in fact minimized by just one vector, i.e. $\phi = \psi,\psi$, so we can just consider $\Lambda(\psi,\psi)$. Is this a correct reading? | |
| Aug 17, 2023 at 15:26 | history | answered | naturallyInconsistent | CC BY-SA 4.0 |