Is the annihilation operator an observable (it is non-Hermitian)?

Question

In most treatments of quantum mechanics that I have seen, observables of a quantum system are defined using Hermitian operators. The most intuitive reason for this is that Hermitian operators have real eigenvalues and orthogonal eigenvectors. However, there is a sense in which the annihilation operator may be meaningfully thought as an observable.

Intriguingly, in some of his seminal work, Roy Glauber practically says that the annihilation operator is an observable. He says [1]

It has become customary, in discussions of classical theory, to regard the electric field $\mathbf{E}(\mathbf{r},t)$ as the quantity one measures experimentally... Such an attitude can only be held in the classical domain, where quantum phenomena play no essential role... In measuring a classical field strength, $\mathbf{E}(\mathbf{r},t)$, we implicitly sum the effects of photon absorption and emission which are described individually by the fields $\mathbf{E}^+(\mathbf{r},t)$ and $\mathbf{E}^-(\mathbf{r},t)$.

Where quantum phenomena are important the situation is usually quite different. Experiments which detect photons ordinarily do so by absorbing them in one or another way. The use of any absorption process, such as photoionization, means in effect that the field we are measuring is the one associated with photon annihilation, the complex field $\mathbf{E}^+(\mathbf{r},t)$.

and later

It is important to bear in mind that such a detector for quanta measures the average value of the product $E_{\mu}^-(\mathbf{r},t)E^+_{\mu}(\mathbf{r},t)$ and not that of the square of the real field $E_{\mu}(\mathbf{r},t)$. Recording photon intensities with a single detector does not exhaust the measurments we can make upon the field, thought it does characterize, in principle, virtually all of the classic experiments of optics.

Was Glauber in fact saying that $E^+_{\mu}(\mathbf{r},t)$ (which is an annihilation operator) is an observable of the field? If so, was this correct?

In addition to Glauber's arguments, which are also found in textbooks on quantum optics, there are other reasons that it would make sense that the annihilation operator is an observable. For example, the eigenvectors of the annihilation operator (i.e. the coherent states) are pointer states for the optical field in many situations [2], especially the situation described by Glauber where photon absorption is used to measure the state of the field.

I want to mention a few other points that I would prefer answers to my question to speak to, if possible.

First, one might object that measurements always return a real number. Why can I not say that measuring in the basis of coherent state measures the real number $\theta$ (modulo $2\pi$), with $e^{i \theta}$ being the phase of the coherent state $|\alpha \rangle$? Perhaps this will be seen as begging the question, so I will also just ask: why must measurements return real numbers?

Second, I have seen it mentioned that $\hat{a}$ is not an observable because these measurements of phase I am discussing are "weak measurements", in contrast with strong projective measurements normally associated with Hermitian observables. Why would this disqualify $\hat{a}$ from being an observable? Is the term "weak observable" ever used?

Links: [1] https://journals.aps.org/pr/abstract/10.1103/PhysRev.130.2529 [2] https://journals.aps.org/prd/pdf/10.1103/PhysRevD.53.7327 [3] https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.85.3552

Note this this and this seem related, but my question seems different.

Edit: the part of my question that involved the position operator was nonsense. I removed it.

Answer 1

Is the annihilation operator an observable (it is non-Hermitian)?

No. It is, by definition, not an observable, since it is not Hermitian.

In most treatments of quantum mechanics that I have seen, observables of a quantum system are defined using Hermitian operators.

In all treatments of quantum mechanics that are not completely wrong, an "observable" is defined as a specific type of Hermitian operator (also known as "self-adjoint operator"). This means that if an operator is not Hermitian then it is not an Observable.

---

As an appeal to authority, consider what Stephen Weinberg says in his book "Lectures on Quantum Mechanics" in Section 3.3 "Observables":

"Now we come to the second postulate of quantum mechanics. This postulate requires that observable physical quantities... are represented as Hermitian operators on Hilbert space... An Hermitian operator is one that is linear and self-adjoint..."

---

One nice property of Hermitian operators is that their eigenvalues are real, just like measurement values.

In the case of a complex-valued electric field, do not forget that the physical field is the real part of the complex field. We use complex fields because it is often easier to deal with $e^{i\phi}$ than to deal with $\cos(\phi)$. But, at the end of the day, the measured quantity is still real. Of course, you can measure two real quantities $x$ and $y$ and force them into a complex number $z=x+iy$. This might be helpful computationally or theoretically, but the values you measure with your measuring apparatus are still real numbers.

---

...and later

It is important to bear in mind that such a detector for quanta measures the average value of the product $E_{\mu}^-(\mathbf{r},t)E^+_{\mu}(\mathbf{r},t)$ and not that of the square of the real field $E_{\mu}(\mathbf{r},t)$. Recording photon intensities with a single detector does not exhaust the measurments we can make upon the field, thought it does characterize, in principle, virtually all of the classic experiments of optics.

Was Glauber in fact saying that $E^+_{\mu}(\mathbf{r},t)$ (which is an annihilation operator) is an observable of the field?

No, that is not what he is saying, and that is not what he wrote. He said the average value of $E^+ E^-$ is what is measured. I am not entirely familiar with your notation, and you have not defined it in the body of your question (other than to say that "E^+ is an annihilation operator), but presumably $E^+ E^-$ is Hermitian. (Indeed, in typical notation, $a_i^\dagger a_i$ is Hermitian and is the mode $i$ number operator.)