# Orthonormal basis and operators

1. Oct 11, 2007

### friend

I hope this is the forum to ask this question.

We all know that the eigenvectors of a Hermitian operator form an orthonormal basis. But is the opposite true as well. Are the vectors of an orthonormal basis always the eigenvectors of some Hermitian operator? Or do we need added restrictions to make it so, such as an inner product and dual spaces being the complex conjugate of the normal space? Thanks.

2. Oct 11, 2007

### meopemuk

Hi friend,
welcome to the Forum!

If you are talking about an orthonormal basis, then you assume that the inner product is already defined (orthogonality and normalization of vectors requires presence of the inner product) and you are in a Hilbert space. Given such a basis you can simply define in this basis a diagonal matrix with arbitrary real numbers on the diagonal and you get a Hermitian operator whose eigenvectors are basis vectors and eigenvalues are those diagonal entries.

Eugene.

3. Oct 12, 2007

### kanato

All vectors are eigenvectors of the identity operator, and the identity is Hermitian. So, yes.

4. Oct 12, 2007

### friend

Thanks. So are you saying that the operator that generates this basis may not be unique (you say "arbitrary real numbers on the diagonal), or is it unique up to a scalar multiple? Thanks.

5. Oct 12, 2007

### meopemuk

There is an infinite set of Hermitian operators with the same basis of eigenvectors. They are different by their eigenvalues.

Eugene.

6. Oct 12, 2007

### friend

Is this because any orthonormal basis from an operator can be transformed to any other basis in the same space that are the eigenvectors of some other operator?

Last edited: Oct 12, 2007
7. Oct 12, 2007

### meopemuk

Yes. You should understand that there is equivalence between Hermitian operators and orthonormal sets of eigenvectors and eigenvalues. Once you get a Hermitian operator you can find a unique (up to degeneracy) basis of eigenvectors and real eigenvalues attached to them. Inversely, once you specified an arbitrary orthonormal basis and assigned an arbitrary real eigenvalue to each basis vector, then you have a unique Hermitian operator.

Eugene.

8. Oct 12, 2007

### dextercioby

Given an ortonormal basis in a Hilbert space, there exist an infinity of compact selfadjoint operators whose eigenvectors are precisely the vectors in the given orthonomal basis. The simplest of these operators is the identity operator.

9. Oct 12, 2007

### Manchot

Yes. You can always construct a Hermitian operator with a weighted sum of the outer product of each basis vector with itself. To do so, just make the weighting coefficients real. (Incidentally, you could make a unitary matrix as well, by making the weighting coefficients have unit modulus.)

10. Oct 12, 2007

### friend

Thank you all for your replies.

11. Oct 15, 2007

### sridhar

clarification

Is it not possible that some eigen values corresponding to the vectors in the basis that we have picked are not real values. Even though all the vectors in the basis are orthogonal. Just a thought.

12. Oct 15, 2007

### OOO

If the eigenvalues are not real then the designed operator will not be hermitean.

13. Oct 15, 2007

### sridhar

exactly! which is why this question does not arise.

"Are the vectors of an orthonormal basis always the eigenvectors of some Hermitian operator? " (mentioned in the first post)

that is all i was saying.

14. Oct 15, 2007

### OOO

Okay, I missed the connection to the first post. But I think thread starter did not want to ask if the construction from an orthonormal basis always yields a hermitean operator, but if there exists always at least one hermitean operator with these eigenvectors.

Nevertheless an important point.

15. Oct 15, 2007

### sridhar

should have quoted!!! am still not used to posting properly.
Apologies

16. Oct 16, 2007

### reilly

Let me expand this correct idea a bit. Given a complete set of orthonormal of basis vectors, |k>, where k labels the eigenstates -- can be a finite or infinite set. Then for example the spectral resolution, O = Sum over k {|k> k <k|} has the property that O |k> = k |k>. And O is Hermitian. But, the same will be true for F(O), where F is an arbitrary real function. So
F(O) |k> = F(k)|k>. That's about it.
Regards,
Reilly Atkinson