Problem with proof of orthogonality of eigenvectors for Hermitian

[inquire4more]

I'm not sure if this is the appropriate section, perhaps my question is better suited for Linear Algebra. At any rate, here goes.

Regarding a proof of the orthogonality of eigenvectors corresponding to distinct eigenvalues of some Hermitian operator $A$:
Given $A|\phi_1\rangle = a_1|\phi_1\rangle$ and $A|\phi_2\rangle = a_2|\phi_2\rangle$ we have
$$\begin{array}{lll} 0 & = & \langle\phi_1|A|\phi_2\rangle - \langle\phi_2|A|\phi_1\rangle^* \\ & = & a_1\langle\phi_2|\phi_1\rangle - a_2\langle\phi_1|\phi_2\rangle^* \\ & = & (a_1 - a_2)\langle\phi_2|\phi_1\rangle \\ \end{array}$$
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.

 

[rachmaninoff]

In line two, the order inside the inner products is backwards. The left terms should read

$$\langle\phi_1|A|\phi_2\rangle \\ & = a_1\langle\phi_1|\phi_2\rangle$$

Applying A to the bra on the left, with eigenvalue a1.

Likewise

$$\langle\phi_2|A|\phi_1\rangle^* \\ & = a_2\langle\phi_2|\phi_1\rangle^*$$

Or you could obtain your result for line two, by taking complex conjugates and multiplying everything by (-1); I don't know why you did that?

 

[inquire4more]

In line two, the order inside the inner products is backwards. The left terms should read

$$\langle\phi_1|A|\phi_2\rangle \\ & = a_1\langle\phi_1|\phi_2\rangle$$

Applying A to the bra on the left, with eigenvalue a1.

Likewise

$$\langle\phi_2|A|\phi_1\rangle^* \\ & = a_2\langle\phi_2|\phi_1\rangle^*$$

Or you could obtain your result for line two, by taking complex conjugates and multiplying everything by (-1); I don't know why you did that?

Thank you. I arrived at precisely the same result, only arriving at the given line 2 by the conjugation and multiplication by -1. This is not my proof. It is one given in a textbook entitled Quantum Mechanics: A Modern Development authored by a Leslie Ballentine. I cannot determine why line 2 was arrived at, but I nevertheless assume I am missing something, hence my question.

 

[dextercioby]

Nevertheless, Ballentine's book is a good written one.

But i still like Gallindo & Pascual's volumes more than any other book on the subject.

Daniel.

 

[inquire4more]

Nevertheless, Ballentine's book is a good written one.

Right, no dispute there. I quite like the book. My issue is with the development of the proof. I am simply not grasping the development of the proof as concerns line 2 and was wondering if perhaps someone might shed some light on this for me.

 

[Doc Al]

$$\begin{array}{lll} 0 & = & \langle\phi_1|A|\phi_2\rangle - \langle\phi_2|A|\phi_1\rangle^* \\ & = & a_1\langle\phi_2|\phi_1\rangle - a_2\langle\phi_1|\phi_2\rangle^* \\ & = & (a_1 - a_2)\langle\phi_2|\phi_1\rangle \\ \end{array}$$

How strange! While each line is true, it is a bizarre sequence. I can't find my copy of Ballentine handy, but I suspect it's just something not caught in editing. A more natural sequence would be:
From the first line:
$$0 = \langle\phi_1|A|\phi_2\rangle -\langle\phi_2|A|\phi_1\rangle^*$$

Then:
$$0 = a_2 \langle\phi_1|\phi_2\rangle - a_1\langle\phi_2|\phi_1\rangle^*$$

$$0 = a_2 \langle\phi_1|\phi_2\rangle - a_1\langle\phi_1|\phi_2\rangle$$

$$0 = (a_2 - a_1) \langle\phi_1|\phi_2\rangle$$

(I don't think you're missing anything.)

 

[inquire4more]

How strange! While each line is true, it is a bizarre sequence. I can't find my copy of Ballentine handy, but I suspect it's just something not caught in editing...
(I don't think you're missing anything.)

Haha...thanks. I find it so interesting that line 2, though undoubtedly less natural, is in fact correct and can't help but wonder if Ballentine chose to state it in this form for some reason. Honestly, I can understand some minor typo passing the editor with greater ease than some quite true but very peculiar form of a statement. Oh well, my hang up. Afraid my mind isn't going to let this one go for awhile. Thanks for letting me know I wasn't quite so dunce here guys.

 

[qbert]

Part of the magic of Hermitian operators
is you can consider them opperating to the
left or right

so if a1 is an eigenvalue of A for vector |1> you have

A |1>= a1 |1> and <1|A = a1 <1|

to see it just conjugate the first equation and
realize A-conjugate = A and eigenvalues of
Hermitian operators are real.

 

[inquire4more]

Right, qbert, I follow you. What I cannot understand is the point to conjugating the first equivalence so as to arrive at the second one. Doc Al typed what I arrived at when working over the proof, which very naturally seems more, well, natural. Is there some reason to go by way of the second line which I am not seeing?

 

[qbert]

I agree the sequence seems unesc. complicated.
A professor my friend had however only used operators
acting to the left. (don't know why -- but he defined
hermition operators on bra's...)

You don't have to conjugate the first to arrive at the second
$$\begin{array}{lll} 0 & = & (\langle\phi_1|A)|\phi_2\rangle - (\langle\phi_2|A)|\phi_1\rangle^* \\ & = & a_1 \langle\phi_1|| \phi_2\rangle - a_2 \langle\phi_2||\phi_1\rangle^* \\ & = & a_1\langle\phi_2|\phi_1\rangle - a_2\langle\phi_1|\phi_2\rangle^* \\ & = & (a_1 - a_2)\langle\phi_2|\phi_1\rangle \\ \end{array}$$

But to me the proof seems needlesssly complex.