I Are there other types of operators that can produce real eigenvalues?

[TheCanadian]

I am learning that operators corresponding to observable quantities are Hermitian since the eigenvalues are real. This makes sense (at least intuitively) and I have seen corresponding proofs of why eigenvalues of Hermitian operators are always real. That is fine. But are there any other types of operators (i.e. non-Hermitian) that correspond to observable quantities? Are not Hermitian operators only one such type of matrix that can produce real eigenvalues for a given basis?

 

[S.G. Janssens]

But are there any other types of operators (i.e. non-Hermitian) that correspond to observable quantities? Are not Hermitian operators only one such type of matrix that can produce real eigenvalues for a given basis?

Nice question. Certainly there are matrices that are non-Hermitian and still have real spectrum: Consider a non-trivial upper-triangular or lower-triangular matrix with real entries on the diagonal. Whether there are quantum systems with, say, a Hamiltonian taking the form of such a matrix I do not know, but surely there are physicists here that do. I would also be curious.

 

[stevendaryl]

I am learning that operators corresponding to observable quantities are Hermitian since the eigenvalues are real. This makes sense (at least intuitively) and I have seen corresponding proofs of why eigenvalues of Hermitian operators are always real. That is fine. But are there any other types of operators (i.e. non-Hermitian) that correspond to observable quantities? Are not Hermitian operators only one such type of matrix that can produce real eigenvalues for a given basis?

It's generally assumed that observables have to correspond to Hermitian operators, but in some sense, this is just conventional. If you have any linear operator O, you can write it in the form A + i B, where A and B are Hermitian. (Let A = \frac{1}{2}(O + O^\dagger) and let B = \frac{1}{2i}(O - O^\dagger)). Then measuring O is equivalent to measuring the pair of observables A and B.

Now, let's suppose that O has real eigenvalues. That means there is a state |\psi\rangle such that (A+iB)|\psi\rangle = \lambda |\psi\rangle, where \lambda is real. Now, we can write:

\langle \psi|A+iB|\psi \rangle = \lambda \langle \psi | \psi \rangle = \lambda
(\langle \psi|A+iB|\psi \rangle)^* = \langle \psi|A^\dagger - iB^\dagger |\psi \rangle = \lambda^* = \lambda

Subtracting the two, we find: 2 i \langle \psi|B|\psi \rangle = 0. So |\psi\rangle can only be an eigenstate of O with a real eigenvalue if B |\psi\rangle = 0. If all of the eigenvalues of O are real, then that means that B |\psi\rangle is identically zero, which is only possible if B=0.

 

[DrDu]

In principle it is sufficient that the operator corresponding to an observable is diagonalizable, with the eigenvalues being eventually complex. These operators are called "normal operators". But each normal operator C can be written as $C=A+iB$ where A and B are commuting hermitian operators.

 

[vanhees71]

Also all unitary operators are normal (an operator is called normal is $\hat{A} \hat{A}^{\dagger}=\hat{A}^{\dagger} \hat{A}$, which is the case for a unitary operator since then $\hat{A} \hat{A}^{\dagger}=\hat{1}$ implies $\hat{A}^{\dagger} \hat{A}=\hat{1}$ too.

 

[S.G. Janssens]

It's generally assumed that observables have to correspond to Hermitian operators, but in some sense, this is just conventional. If you have any linear operator O, you can write it in the form A + i B, where A and B are Hermitian. (Let A = \frac{1}{2}(O + O^\dagger) and let B = \frac{1}{2i}(O - O^\dagger)). Then measuring O is equivalent to measuring the pair of observables A and B.

Now, let's suppose that O has real eigenvalues. That means there is a state |\psi\rangle such that (A+iB)|\psi\rangle = \lambda |\psi\rangle, where \lambda is real. Now, we can write:

\langle \psi|A+iB|\psi \rangle = \lambda \langle \psi | \psi \rangle = \lambda
(\langle \psi|A+iB|\psi \rangle)^* = \langle \psi|A^\dagger - iB^\dagger |\psi \rangle = \lambda^* = \lambda

Subtracting the two, we find: 2 i \langle \psi|B|\psi \rangle = 0.

Until here I understand you.

So |\psi\rangle can only be an eigenstate of O with a real eigenvalue if B |\psi\rangle = 0.

Sorry for this, but I am missing how $\langle \psi|B|\psi\rangle = 0$ implies that $B|\psi\rangle = 0$.

If all of the eigenvalues of O are real, then that means that B |\psi\rangle is identically zero, which is only possible if B=0.

I think you are assuming here that the corresponding eigenvectors of $O$ form a basis for the (underlying, unspecified) space?

 

[pac134]

Physicists assume that observables are represented by self-adjoint ('Hermitian') operators because they guarantee that the corresponding eigenvalues will be always non-negative. The theory was aparently structured in this way because, I think, this might have been somewhat obvious for them at the time. I mean, it is well-known in linear algebra that this is true. Now, there isn't in fact anything that says that there could not be an observable which is not self-adjoint. In fact, an interesing field began in the past decade or so called PT-symmetric quantum theory, where P stands for parity and T for time-reversal. I don't know much about it but it seems interesting. Apparently, if the Hamiltonian (which is an observable) is PT-symmetric (which in general will not be self-adjoint) then the eigenvalues are also guaranteed to be non-negative.
So here's the answer to your question.
Of course there are other operators which are not self-adjoint as was already pointed out, but there is a class of operators, namely those which are PT-symmetric, that seems to be also physically relevant. I suggest you look up in the repository arxiv.org, there are some introductory papers to the subject.

 

[TheCanadian]

Physicists assume that observables are represented by self-adjoint ('Hermitian') operators because they guarantee that the corresponding eigenvalues will be always non-negative. The theory was aparently structured in this way because, I think, this might have been somewhat obvious for them at the time. I mean, it is well-known in linear algebra that this is true. Now, there isn't in fact anything that says that there could not be an observable which is not self-adjoint. In fact, an interesing field began in the past decade or so called PT-symmetric quantum theory, where P stands for parity and T for time-reversal. I don't know much about it but it seems interesting. Apparently, if the Hamiltonian (which is an observable) is PT-symmetric (which in general will not be self-adjoint) then the eigenvalues are also guaranteed to be non-negative.
So here's the answer to your question.
Of course there are other operators which are not self-adjoint as was already pointed out, but there is a class of operators, namely those which are PT-symmetric, that seems to be also physically relevant. I suggest you look up in the repository arxiv.org, there are some introductory papers to the subject.

Thank you for the reference. I began looking into introductory papers and it's certainly very interesting. Although it appears it's an extension of quantum theory to the complex domain. Furthermore, the PT-symmetrixc Hamiltonian can be mapped to a Hermitian operator via a similarity transformation. I still have more reading to do, but I don't believe it quite answers my question as the operator corresponding to an observable is reducible to a Hermitian operator.

 

[pac134]

It certainly is an extension but I don't think it's a simple extention to the complex domain. Quantum mechanics already deals with complex numbers so I don't know what you mean by 'an extention to the complex domain'.

The PT-symmetric Hamiltonian cannot be mapped to a self-adjoint operator vie a similarity transformation since, it can be shown, that a self-adjoint operator is mapped, via a similarity transformation, to another self-adjoint operator. Hence any operator which is not self-adjoint will never be mapped to a self-adjoint operator via similarity transformation. Formally, the class of self-adjoint operators is closed under similarity transformations.

Ultimately I think your question amounts why mathematics describes the physical world. I mean, we use self-adjoint operators as observables because it works! Even if we find some motive behind this we would just change the question to another level. That's my view. :)

 

[TheCanadian]

It certainly is an extension but I don't think it's a simple extention to the complex domain. Quantum mechanics already deals with complex numbers so I don't know what you mean by 'an extention to the complex domain'.

The PT-symmetric Hamiltonian cannot be mapped to a self-adjoint operator vie a similarity transformation since, it can be shown, that a self-adjoint operator is mapped, via a similarity transformation, to another self-adjoint operator. Hence any operator which is not self-adjoint will never be mapped to a self-adjoint operator via similarity transformation. Formally, the class of self-adjoint operators is closed under similarity transformations.

Ultimately I think your question amounts why mathematics describes the physical world. I mean, we use self-adjoint operators as observables because it works! Even if we find some motive behind this we would just change the question to another level. That's my view. :)

To clarify:
"These PT-symmetric theories may be viewed as analytic continuations of conventional theories from real to complex phase space."

With regards to mapping a PT-symmetric Hamiltonian:
"The reality of the spectrum of $H$ is also equivalent to the presence of an exact antilinear symmetry of the system. Therefore, a diagonalizable Hamiltonian having an exact antilinear symmetry, e.g., PT-symmetry, is quasi-Hermitian. In particular, it may be mapped to a Hermitian Hamiltonian $H′$ via a similarity transformation $H → H′ = UHU^{−1}$ where the operator $U$ satisfies $⟨⟨·,·⟩⟩η+ = ⟨U·,U·⟩$, i.e., $U$ is a unitary operator mapping $H_i$ onto $H_i′$. This shows that indeed the quantum system defined by the Hamiltonian $H$and the Hilbert space $H_i$ may as well be described by the Hermitian Hamiltonian $H′$ and the original Hilbert space $H_i′$, [14]."

Correct me if I am mistaken, but the PT-symmetric Hamiltonian is quasi-Hermitian and can undergo a similarity transformation to become Hermitian.

Yes, my question is along those lines. I'm just curious as to what other types of operators also work.

 

[pac134]

To clarify:
"These PT-symmetric theories may be viewed as analytic continuations of conventional theories from real to complex phase space."

Alright! So you were talking about a complex phase space. That's interesting!

Regarding the similarity transformation I also buy the argument since the inner product is defined with respect to a pseudo-metric, which is not the usual case in quantum mechanics.

This similarity transformation is not what people will interpret in general, because in quantum mechanics the metric is trivial (identity matrix). So the statement

PT-symmetric Hamiltonian is quasi-Hermitian and can undergo a similarity transformation to become Hermitian.

is right. My only point being that we should also say that the space has a non-trivial metric, otherwise people might interpret erroneously as I did. :)

Yes, my question is along those lines. I'm just curious as to what other types of operators also work.

What I conclude from the discussion is: yes! There are other operators. An example being this extention of quantum mechanics.
I would be glad if these studies would provide some new physics. Otherwise this extension is 'just' mathematics.

Thanks for the information!

 

[stevendaryl]

Sorry for this, but I am missing how $\langle \psi|B|\psi\rangle = 0$ implies that $B|\psi\rangle = 0$.

You're right. I'll have to think about whether my argument can be fixed. Sorry for spreading an incorrect argument.

 

[rubi]

You're right. I'll have to think about whether my argument can be fixed. Sorry for spreading an incorrect argument.

A counterexample would be $O = \begin{pmatrix} 0&1 \\ 0&0\end{pmatrix}$, which is a ladder operator for spin $\frac{1}{2}$. You get $A=\frac{1}{2} \sigma_1$ and $B=\frac{1}{2} \sigma_2$, where $\sigma_i$ are the Pauli matrices. The only eigenvalue of $O$ is $0$ (which is real), but $\sigma_i$ don't have $0$ as an eigenvalue and hence don't annihilate any vectors apart from the null vector.

Your argument can be fixed if you additionally require that $A$ and $B$ commute, which is equivalent to the condition of normality. In that case, there is a version of the spectral theorem that works just like the well-known spectral theorem for self-adjoint operators. In my example, the Pauli matrices of course don't commute, so the normality condition is violated.

 

[PeroK]

I am learning that operators corresponding to observable quantities are Hermitian since the eigenvalues are real. This makes sense (at least intuitively) and I have seen corresponding proofs of why eigenvalues of Hermitian operators are always real. That is fine. But are there any other types of operators (i.e. non-Hermitian) that correspond to observable quantities? Are not Hermitian operators only one such type of matrix that can produce real eigenvalues for a given basis?

Just a minor point. The eigenvalues of an operator are independent of the basis.