Eigenvalues of Hermitian opertors

[migwing007]

I'm looking for a proof of the fact that orthogonal eigenfunctions of a Hermitian operator have distinct eigenvalues. I know the proof the converse: that eigenfunctions belonging to distinct eigenvalues are orthogonal.
thanks alot!

 

[Matterwave]

I don't believe that it is true that orthogonal eigenfunctions have distinct eigenvalues. Certainly in terms of degeneracy, one can employ the Gram-Schmidt orthogonalization procedure to get orthogonal eigenfunctions which have the same eigenvalue...

I'm pretty bad at math though, so I wouldn't take my word on it lol.

 

[George Jones]

I'm looking for a proof of the fact that orthogonal eigenfunctions of a Hermitian operator have distinct eigenvalues.

As Matterwave has said, this it not true. Consider the matrix
$$\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$$

$\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$ and $\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$ are orthogonal eigenvectors that both are associated with the eigenvalue 1 (as is any linear combination of these vectors).

 

[migwing007]

This is really bothering me

I'm remembering from a while ago having to prove the converse of a theorem from Griffiths, but I can't remember which one of the following it is:

Prove converse of:

1.) The eigenfunctions of a Hermitian operator are complete,

2.) Eigenfunctions of Hermitian operators belonging to distinct eigenvalues are orthogonal, or

3.) Eigenvalues of Hermitian operators are real

thanks alot!

 

[strangerep]

migwing007,

It's not clear (to me anyway) what you're asking. Maybe try stating the
theorem or its converse in exactly the wording that you want to prove?

 

[migwing007]

migwing007,

It's not clear (to me anyway) what you're asking. Maybe try stating the
theorem or its converse in exactly the wording that you want to prove?

Well, the converse of the first one would be: if the eigenfunctions of an operator are complete then the operator is hermitian.

the second would be: if the eigenfunctions of an operator belonging to distinct eigenvalues are orthogonal then the operator is hermitian.

and the third is: If the eigenvalues of an operator are real then the operator is hermitian
Actually I don't think 2 and 3 are correct so that leaves #1.

I've looked online but can't seem to find it

Hope that clarifies it

 

[samalkhaiat]

Well, the converse of the first one would be: if the eigenfunctions of an operator are complete then the operator is hermitian.

the second would be: if the eigenfunctions of a hermitian operator belong to distinct eigenvalues then the eigenfunctions are orthogonal.

and the third is: If the eigenvalues of an operator are real then the operator is hermitian



Actually I don't think 2 and 3 are correct so that leaves #1.

I've looked online but can't seem to find it

Hope that clarifies it

If the eigenvectors of an operator A form a complete orthonormal set {$|i \rangle$}, then you can write the operator A (in terms of its eigenvalues) as
$$A = \sum a_{i}|i \rangle \langle i |$$
Thus
$$\{A = A^{\dagger}\} \Leftrightarrow \{ a_{i} = a_{i}^{\ast}\}$$
So, your first statement is wrong while 2 and 3 are correct.

sam

 

[A. Neumaier]

If the eigenvectors of an operator A form a complete orthonormal set {$|i \rangle$}, then you can write the operator A (in terms of its eigenvalues) as
$$A = \sum a_{i}|i \rangle \langle i |$$
Thus
$$\{A = A^{\dagger}\} \Leftrightarrow \{ a_{i} = a_{i}^{\ast}\}$$
So, your first statement is wrong while 2 and 3 are correct.

No, only 2 is correct. The operator that maps a|up>+b|down> to b|up> has real eigenvalues but is not Hermitian.

 

[bcrowell]

Migwing007 started four separate threads on the same topic. I've deleted two and merged two.