# Hermitian Operator Questions

1. Dec 30, 2012

### Sekonda

Hey,

I have the following question on Hermitian operators

Initially I thought this expectation value would have to be zero as the eigenvectors are mutually orthogonal due to Hermitian Operator and so provided the eigenvectors are distinct then the expectation would be zero... Though I'm not sure if this is correct.

Anyway, any help is appreciated.
Cheers,
Sk

2. Dec 30, 2012

### kevinferreira

Yes, it is indeed zero.

3. Dec 30, 2012

### cattlecattle

$\langle \omega_2 | \Lambda \Omega | \omega_1\rangle =\langle \omega_2|\Lambda \omega_1|\omega_1\rangle = \omega_1\langle \omega_2|\Lambda|\omega_1\rangle\\ \langle \omega_2 | \Omega\Lambda|\omega_1\rangle = \langle\omega_2|\omega_2\Lambda|\omega_1\rangle= \omega_2\langle\omega_2|\Lambda|\omega_1\rangle$
From $[\Omega,\Lambda]=0$
$\Longrightarrow \omega_1\langle\omega_2|\Lambda|\omega_1\rangle = \omega_2\langle\omega_2|\Lambda|\omega_1\rangle \Longrightarrow \langle\omega_2|\Lambda|\omega_1\rangle=0$

4. Dec 31, 2012

### Sekonda

Ahh thanks cattlecattle I had written down pretty much this but wasn't sure if it was right, but now I do!

Cheers, thanks man
SK

5. Dec 31, 2012

### DocZaius

So if A|V1> = a|v1> where A is a Hermitian operator, then <v1|A = a<v1| ?

I would have thought <v1|A = a*<v1| ...

Is my first line above true only if A is Hermitian? What if it was merely square?

6. Dec 31, 2012

### dextercioby

Yes, if A is hermitean (I call it self-adjoint), then the normal a* from your second line would be equal to a, since the eigenvalues of hermitean operators are real.

What do you mean 'square' ?

7. Dec 31, 2012

### DocZaius

Not self adjoint but of the right dimension to satisfy A|v1> and <v1|A being valid operations. A would still have eigenvalue a for |v1>, but just not be self adjoint.

edit: I shouldn't have said square. I should have just said what if A|v1> = a|v1> and A is not hermitian? Then <v1|A = a*<v1| ? And it seems from dextercioby that that is the case.

Last edited: Dec 31, 2012
8. Dec 31, 2012

### cattlecattle

No, the general result is if
$\Omega|\omega\rangle=\omega|\omega\rangle$
then
$\langle\omega|\Omega^\dagger=\langle\omega|\omega^*$
There is usually no relation between $\Omega|\omega\rangle$ and $\langle\omega|\Omega$

9. Dec 31, 2012

### DocZaius

Thanks!

10. Jan 2, 2013

### aim1732

why was it called expectation value?I mean if it were not identically zero then how would you look at the expectation value thing? Or is it that <ω2|Λ|ω1> is an expectation value?