# I Restriction on operators

1. Sep 28, 2016

### 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?

2. Sep 28, 2016

### Krylov

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.

3. Sep 28, 2016

### stevendaryl

Staff Emeritus
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 $B=0$.

4. Sep 28, 2016

### 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.

5. Sep 28, 2016

### 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.

6. Sep 28, 2016

### Krylov

Until here I understand you.

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

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

Last edited: Sep 28, 2016
7. Sep 28, 2016

### 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.

8. Sep 29, 2016

### TheCanadian

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.

9. Sep 30, 2016

### 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. :)

10. Sep 30, 2016

### TheCanadian

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.

11. Sep 30, 2016

### pac134

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
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. :)

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!

12. Oct 3, 2016

### stevendaryl

Staff Emeritus

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

13. Oct 4, 2016

### rubi

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.

14. Oct 4, 2016

### PeroK

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

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