# Eigenvalues of Hermitian opertors

• migwing007
In summary, he is looking for a proof that orthogonal eigenfunctions of a hermitian operator have distinct eigenvalues.
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!

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.

migwing007 said:
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).

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!

Last edited:

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?

strangerep said:
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

Last edited:

migwing007 said:
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

samalkhaiat said:
migwing007 said:
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.

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

## What is the definition of eigenvalues of Hermitian operators?

The eigenvalues of Hermitian operators are the special set of complex numbers that are associated with a given linear operator. They represent the values that an operator will return when acting on a corresponding eigenvector.

## Why are eigenvalues of Hermitian operators important in quantum mechanics?

In quantum mechanics, Hermitian operators represent physical observables, and their eigenvalues represent the possible outcomes of a measurement. Therefore, understanding the eigenvalues of Hermitian operators is essential for predicting and interpreting experimental results.

## What are the properties of eigenvalues of Hermitian operators?

Some important properties of eigenvalues of Hermitian operators include: they are real numbers, they are orthogonal, and they form a complete set of basis vectors. Additionally, all eigenvalues of a Hermitian operator are unique and non-degenerate.

## How do you find the eigenvalues of Hermitian operators?

The eigenvalues of Hermitian operators can be found by solving the characteristic equation, which is given by det(A-λI) = 0, where A is the Hermitian operator and λ is the eigenvalue. This equation can be solved using various methods such as diagonalization or the power method.

## What is the physical interpretation of the eigenvalues of Hermitian operators?

The physical interpretation of the eigenvalues of Hermitian operators is that they represent the possible outcomes of a measurement of a physical observable. The eigenvectors associated with these eigenvalues represent the different states that a system can be in, and the probability of obtaining a particular eigenvalue upon measurement is given by the squared magnitude of the corresponding eigenvector coefficient.

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