Orthonormal basis and operators

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Discussion Overview

The discussion revolves around the relationship between orthonormal bases and Hermitian operators in the context of linear algebra and quantum mechanics. Participants explore whether every orthonormal basis corresponds to the eigenvectors of some Hermitian operator, and the conditions under which this holds true. The conversation includes theoretical considerations, implications for operator construction, and the nature of eigenvalues.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose that given an orthonormal basis in a Hilbert space, one can construct a Hermitian operator with those basis vectors as eigenvectors by defining a diagonal matrix with real entries.
  • Others argue that all vectors are eigenvectors of the identity operator, which is Hermitian, suggesting that orthonormal bases can indeed correspond to eigenvectors of Hermitian operators.
  • A participant questions whether the operator generating the orthonormal basis is unique or if there are multiple operators with the same eigenvectors differing by eigenvalues.
  • It is noted that there exists an infinite set of Hermitian operators with the same basis of eigenvectors, differing by their eigenvalues.
  • Some participants express concern about the possibility of non-real eigenvalues corresponding to the orthonormal basis, emphasizing that non-real eigenvalues would imply the operator is not Hermitian.
  • A later reply clarifies that the original question may not have been about the uniqueness of the Hermitian operator but rather if at least one such operator exists for a given orthonormal basis.

Areas of Agreement / Disagreement

Participants generally agree that there is a connection between orthonormal bases and Hermitian operators, but there is no consensus on the uniqueness of the operator or the implications of eigenvalues being real. The discussion remains unresolved regarding the conditions under which an orthonormal basis can be guaranteed to correspond to eigenvectors of a Hermitian operator.

Contextual Notes

Limitations include the dependence on the definitions of inner products and Hilbert spaces, as well as the implications of eigenvalue properties on the nature of the operators discussed.

friend
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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.
 
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friend said:
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.

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

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.
 
friend said:
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.

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

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

Eugene.

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:
friend said:
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?

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.
 
friend said:
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.

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.
 
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
Thank you all for your replies.
 
  • #11
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
sridhar said:
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.

If the eigenvalues are not real then the designed operator will not be hermitean.
 
  • #13
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
sridhar said:
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.

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
OOO said:
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.

should have quoted! am still not used to posting properly.
Apologies
 
  • #16
meopemuk said:
There is an infinite set of Hermitian operators with the same basis of eigenvectors. They are different by their eigenvalues.

Eugene.
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
 

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