QM - Part 1 - Eigenvalues of the Hermitian Operator A are real

**bugatti79** · March 12th 2011, 03:18 AM

Hi FOlks,

Studying this theorem 'Eigenvalues of the Hermitian Operator A are real' from sakurai's Modern QM's

$\hat A | a' \rangle = a'| a' \rangle$

If we multiply the above by $\langle a''|$ from the left and right respectively we get

$\langle a''|\hat A|a' \rangle= \langle a''|a'|a' \rangle$ (1)and

$ \hat A|a' \rangle \langle a''|=a'|a' \rangle \langle a''| $ (2)

If we flip the LHS of (1) we get

$ \langle a'|\hat A|a'' \rangle=\underbrace{\langle a'|(a')^*|a'' \rangle }_\textrm{?}= (a')^* \langle a'|a'' \rangle $

Should underbrace not be $\langle a'|(a')^*|a'' \rangle ^*$

Second question I have is how the following was arrived at?

$ \langle a'| \hat A|a'' \rangle =\langle a'|a''a'' \rangle = a'' \langle a'|a'' \rangle $

Thanks

**Ackbeet** · March 12th 2011, 04:04 AM

For the first question: what's under the underbrace is correct. Think of it this way:

You have the equation

$\hat{A}|a'\rangle=a'|a'\rangle.$ Taking the Hermitian adjoint of both sides yields

$\langle a'|\hat{A}=\langle a'|(a')^{*},$

since $\hat{A}^{\dagger}=\hat{A},$ by assumption. Now just right-multiply by $|a''\rangle,$ and you get the underbrace expression.

As for your second question, I'm guessing that $a''$ satisfies the eigenvalue equation

$\hat{A}|a''\rangle =a''|a''\rangle.$ And I think you have a typo there:

$\langle a'|\hat{A}|a''\rangle=\langle a'|a''|a''\rangle=a''\langle a'|a''\rangle,$

right?

**bugatti79** · March 12th 2011, 05:54 AM

Originally Posted by Ackbeet

For the first question: what's under the underbrace is correct. Think of it this way:

You have the equation

$\hat{A}|a'\rangle=a'|a'\rangle.$ Taking the Hermitian adjoint of both sides yields

$\langle a'|\hat{A}=\langle a'|(a')^{*},$

How did you arrive at the RHS using the Hermitian adjoint. I understand using it for the LHS

As for your second question, I'm guessing that $a''$ satisfies the eigenvalue equation

$\hat{A}|a''\rangle =a''|a''\rangle.$ And I think you have a typo there:

$\langle a'|\hat{A}|a''\rangle=\langle a'|a''|a''\rangle=a''\langle a'|a''\rangle,$

right?

I guess so, then we arrive at

$ 0= ((a')^*-a'') \langle a'| a''\rangle $

If we assume a' and a'' to be equal then we determine that a' must be real..that I understand.

I dont follow when a' not equal a'' ie the orthonormality...

$ | a' \rangle \neq | a'' \rangle \implies a' \neq a'' $ not sure how to continue...

thanks

**Ackbeet** · March 12th 2011, 10:54 AM

Originally Posted by bugatti79

How did you arrive at the RHS using the Hermitian adjoint? I understand using it for the LHS.

Well, the Hermitian adjoint is defined, at least in finite-dimensional Hilbert spaces, as the complex conjugate transpose. Scalars you can think of as 1-dimensional vectors. Transposing a scalar doesn't change it, but the complex conjugate part does (if it's complex, of course). Hence, the Hermitian adjoint of a scalar is just the complex conjugate. The fact that I'm right-multiplying by the scalar is unimportant, as scalars can pass through vectors no problem (that is, scalar multiplication of a vector is commutative).

I guess so, then we arrive at

$ 0= ((a')^*-a'') \langle a'| a''\rangle $

If we assume a' and a'' to be equal then we determine that a' must be real..that I understand.

I dont follow when a' not equal a'' ie the orthonormality...

$ | a' \rangle \neq | a'' \rangle \implies a' \neq a'' $ not sure how to continue...

thanks

I think you're confusing two different proofs.

1. If you're trying to show that the eigenvalues of an Hermitian operator are real, then you play around with

$\langle a'|A|a'\rangle,$ not $\langle a'|A|a''\rangle.$

The reason is that you have to know that $\langle a'|a'\rangle\not=0,$ which is true because eigenvectors, by definition, are nonzero.

2. If you're trying to prove that the eigenvectors of differing eigenvalues are orthogonal, then you play around with

$\langle a'|A|a''\rangle,$ where the eigenvalues of the two eigenvectors there are different.

I would definitely prove that the eigenvalues are real before proving that the eigenvectors of differing eigenvalues are orthogonal.

**bugatti79** · March 12th 2011, 01:40 PM

Originally Posted by Ackbeet

Well, the Hermitian adjoint is defined, at least in finite-dimensional Hilbert spaces, as the complex conjugate transpose. Scalars you can think of as 1-dimensional vectors. Transposing a scalar doesn't change it, but the complex conjugate part does (if it's complex, of course). Hence, the Hermitian adjoint of a scalar is just the complex conjugate. The fact that I'm right-multiplying by the scalar is unimportant, as scalars can pass through vectors no problem (that is, scalar multiplication of a vector is commutative).

I think you're confusing two different proofs.

1. If you're trying to show that the eigenvalues of an Hermitian operator are real, then you play around with

$\langle a'|A|a'\rangle,$ not $\langle a'|A|a''\rangle.$

The reason is that you have to know that $\langle a'|a'\rangle\not=0,$ which is true because eigenvectors, by definition, are nonzero.

2. If you're trying to prove that the eigenvectors of differing eigenvalues are orthogonal, then you play around with

$\langle a'|A|a''\rangle,$ where the eigenvalues of the two eigenvectors there are different.

I would definitely prove that the eigenvalues are real before proving that the eigenvectors of differing eigenvalues are orthogonal.

Thanks, things are becoming clearer but slowly :-) Here is my working from start to finish

$\hat A|a' \rangle =a'|a' \rangle$ the hermitian adjoint is of this is $\langle a' | \hat A^{\dagger} = \langle a'|(a')^*$ but $\hat A = \hat A^{\dagger} \therefore$

$\langle a'| \hat A = \langle a' | (a')^*$ Now right x by $| a'' \rangle$ giving $\langle a'| \hat A| a'' \rangle = \langle a' |(a')^*| a'' \rangle$ (1)

Similarly assuming that a'' satisfies the eigen value equation then $\hat A|a'' \rangle = a''| a'' \rangle$ , left x this by $\langle a'|$ giving

$ \langle a'|\hat A| a'' \rangle = \langle a'| a''|a'' \rangle = a'' \langle a'| a'' \rangle $ (2)

subtracting 2 from 1 we get

$ 0=\langle a'|(a')^*|a'' \rangle - a''\langle a'|a'' \rangle = ((a')^*-a'') \langle a'|a'' \rangle $

According to Sakurai a' and a'' can be the same or different. If we assume the same we get

$ |a' \rangle =|a'' \rangle \implies \langle a'|a'' \rangle=\langle a'|a' \rangle > 0$ if

$ |a' \rangle \neq | 0 \rangle \implies $

$(a')^*-a'=0 \implies a'=(a')^* \implies$ a' is real

THat is my working. The bit I am unsure about is how do we know

$ \langle a'|a' \rangle > 0 $ ?

My final task will be to show the a' and a'' are orthogonal.

Thanks!

**topsquark** · March 12th 2011, 02:13 PM

Originally Posted by bugatti79

$ \langle a'|a' \rangle > 0 $ ?

It is not true in general that $\langle a' | a' \rangle > 0$ . For the purposes of the proof all we need is $| \langle a' | a' \rangle | \neq 0$

-Dan

Edit: Note carefully that if $| a'' \rangle \neq | a' \rangle$ we can't yet say anything about it being zero or not. This may be positive, negative, or zero. Your next step in the proof is to show that $\langle a'' | a' \rangle = 0$ if they have different eigenvalues. (At least, not for the non-degenerate case.)

**Ackbeet** · March 12th 2011, 03:59 PM

Originally Posted by topsquark

It is not true in general that $\langle a' | a' \rangle > 0$ .

It is true in general for eigenvectors, which are all the vectors being used at the moment.

**topsquark** · March 12th 2011, 08:31 PM

Originally Posted by Ackbeet

It is true in general for eigenvectors, which are all the vectors being used at the moment.

Yes. I got thinking about that later. Thanks for the catch.

-Dan

**bugatti79** · March 13th 2011, 12:38 AM

Originally Posted by Ackbeet

It is true in general for eigenvectors, which are all the vectors being used at the moment.

To prove orthonormality

$|a' \rangle \neq |a'' \rangle \implies a' \neq a''$

2 things can happen

1) we established $(a')^*=a'$ or

2) $a'-a'' \neq0 \implies$
$\langle a'|a'' \rangle =0 \implies |a' \rangle and |a'' \rangle$ are orthonormal

This look right? :-)

**Ackbeet** · March 14th 2011, 04:53 AM

Originally Posted by bugatti79

To prove orthonormality

$|a' \rangle \neq |a'' \rangle \implies a' \neq a''$

What happens in the degenerate case? You can have Hermitian operators with degenerate eigenvalues (the identity, for example). In that case, you'd have more than one eigenvector per eigenvalue, thus rendering your implication false.

2 things can happen

1) we established $(a')^*=a'$ or

2) $a'-a'' \neq0 \implies$
$\langle a'|a'' \rangle =0 \implies |a' \rangle and |a'' \rangle$ are orthonormal

This look right? :-)

I think you're ok with 2), but like I said, I think your 1) needs a bit more work.

**bugatti79** · March 14th 2011, 03:31 PM

Originally Posted by Ackbeet

What happens in the degenerate case? You can have Hermitian operators with degenerate eigenvalues (the identity, for example). In that case, you'd have more than one eigenvector per eigenvalue, thus rendering your implication false.

I think you're ok with 2), but like I said, I think your 1) needs a bit more work.

Ok, perhaps what you are suggesting is beyond the scope of sakurai's material as attached. Thanks for the great help once again! :-)

**Ackbeet** · March 14th 2011, 03:37 PM

Aha. Your implication is going the wrong way. You should have

Originally Posted by bugatti79

To prove orthonormality

$a' \neq a''\implies |a' \rangle \neq |a'' \rangle$

2 things can happen

1) we established $(a')^*=a'$ or

2) $a'-a'' \neq0 \implies$
$\langle a'|a'' \rangle =0 \implies |a' \rangle and |a'' \rangle$ are orthonormal

This look right? :-)

Sakurai's argument is correct, in any case.

[EDIT]: See below for a correction.

**bugatti79** · March 15th 2011, 12:14 AM

Originally Posted by Ackbeet

Aha. Your implication is going the wrong way. You should have

Sakurai's argument is correct, in any case.

Not sure I follow regarding my implication going the wrong way…Is the above correct?

Thanks

**Ackbeet** · March 15th 2011, 01:58 AM

Ok, in looking at Sakurai's argument, I would phrase it a little differently. In proving the reality condition of the eigenvalues, I would say that you assume $a'=a'',$ AND you assume $|a'\rangle=|a''\rangle.$ Then the reality condition follows. The double assumption sidesteps the degenerate case nicely, and is probably what Sakurai had in mind anyway. Trust me, the degenerate case is not beyond Sakurai! By "degenerate case", I just mean that you have eigenvalues of algebraic multiplicity greater than one. They are multiple roots of the characteristic equation $\det(A-\lambda I)=0.$

Then, when you prove the orthogonality condition, ALL you have to assume is that the eigenvalues are different. Because you just proved they are all real, the result follows.

So please ignore what I said earlier about the "implication going the wrong way". What you really need is to replace the implication with the double assumption that the eigenvalues are the same, AND the eigenvectors are the same. Then you get the desired result.

Does that make sense?

**bugatti79** · March 16th 2011, 04:07 AM

I think that is making good sense now, thanks! So multiple roots would be like repeated roots.

Originally Posted by Ackbeet

So please ignore what I said earlier about the "implication going the wrong way". What you really need is to replace the implication with the double assumption that the eigenvalues are the same, AND the eigenvectors are the same. Then you get the desired result.

So for orthonormality we also take the double assumption but in the sense the eigen values AND eigenkets are NOT equal ie

$a' \neq a'' AND |a' \rangle \neq |a'' \rangle \implies a'-a'' \neq 0 \implies \langle a'|a'' \rangle =0 \implies |a' \rangle and |a'' \rangle$ are orthonormal

Cheers :-)

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