# What is the observable for the optical field?

Typically, observables in quantum mechanics are associated with Hermitian operators. However, Glauber argues in 1963 ([1]) that the electric field operator $$\hat{\mathbf{E}}(x,t)$$ is not the relevant observable for quantum optics. The electric field operator can be split into positive and negative frequency components (in the classical context, this is simply finding the analytic signal representation of the field) $$\hat{\mathbf{E}}(x,t) = \hat{\mathbf{E}}^+ (x, t) +\hat{\mathbf{E}}^- (x, t)$$ with $$\hat{\mathbf{E}}^+ (x, t) = \int_0 ^\infty \hat{\tilde{\mathbf{E}}}(x, \omega) e^{i \omega t} d\omega$$ and $$\hat{\mathbf{E}}^- (x, t) = {\hat{\mathbf{E}}^+}^\dagger (x, t)$$. Now, the positive frequency component $$\hat{\mathbf{E}}^+ (x, t)$$ is an annihilation operator, and thus is non-Hermitian. However, $$\hat{\mathbf{E}}^+ (x, t)$$ does have eigenstates (the coherent states), and this seems to have widespread implications across quantum optics. From now on I will use $$\hat{a}$$ as a shorthand to refer to $$\hat{\mathbf{E}}^+ (x, t)$$.

The Glauber-Sudarshan $$P$$ representation is used to classify fields as having a classical counterpart or not (i.e. is $$P(\alpha)$$ approximately a probability distribution), and indeed it seems intuitive that coherent states would constitute a pointer basis of the electromagnetic field, as coherent states often are pointer states in contexts with continuous degrees of freedom that are weakly constrained (i.e. not bound) and locally measured [2] (in this case that degree of freedom is essentially the phase). Furthermore, since measurements of the optical field relevant to Glauber were all performed using photodetectors (photoelectric effect), thinking that the annihilation operator is the relevant observable seems even more intuitive. Critically, Glauber's criteria for classicality of the field relies on the annihilation operator being the observable, as $$\hat{a}$$ is the operator used in the high-order coherence functions.

However, there seem to be many reasons to doubt that the annihilation operator is the observable, even in macroscopic experiments. For instance, the Casimir effect involves properties of the vacuum state of the field, no photon absorption necessary. In addition, there are also processes involving the Bogoliubov-transformed operators $$\hat{b} = \nu \hat{a} + \mu \hat{a}^\dagger$$ $$\hat{b}^\dagger = \nu^* \hat{a} + \mu^* \hat{a}^\dagger$$ with $$|\nu|^2 - |\mu|^2 = 1$$. The eigenstates of $$\hat{b}$$ are the squeezed states, which have non-classical $$P(\alpha)$$ representations. However, they would be "classical" if the coherent states of $$\hat{b}$$ are used to make a $$P(\beta)$$ representation over the eigenstates $$\hat{b} | \beta \rangle = \beta |\beta \rangle$$. If $$\hat{b}$$ merely represents another way to measure (interact with) the field, then is classicality dependent on the method of measurement?

My question is this: what is the latest thinking on this issue? Was Glauber correct or not in using $$\hat{a}$$ as the relevant observable? If other observables are more relevant in other contexts, does this mean that the "classicality" of the light field depends on what is measured?

Edit: While it is possible that they are not, I wonder whether How does a laser emit light in a coherent state? and Why is laser light described by a coherent state? are related questions.

There are many different observables in quantum optics. They produce different results when applied to different states. It depends on what you want to measure.

First, we need to point out that the individual ladder operators are not Hermitian and therefore do not represent observables, which are required to be Hermitian. Therefore, although the coherent states are eigenstates of the annihilation operator, it does not mean that the annihilation operator is an observable for the coherent state. The electric field operator is Hermitian because the electric field is real valued. It is a bit like the quadrature operators.

For the intensity, one can use the number operator $$\hat{n}=\hat{a}^{\dagger}\hat{a}$$. For a coherent state, it gives $$\langle n \rangle = \langle \alpha | \hat{n} | \alpha \rangle = |\alpha|^2 .$$ For a Fock state, it gives the number of photons in the state. The intensity is readily measured with a detector such as a PIN diode.

Another observable is the projection operator defined in terms of Fock states by $$\hat{P}=|n\rangle\langle n|$$. It projects out the coefficient for a specific Fock state in the expansion of a state in terms of the Fock basis. To perform such a measurement, one needs a photon-number-resolving detector.

Then there are the quadrature operators $$\hat{q}=(\hat{a}+\hat{a}^{\dagger})/\sqrt{2}$$ and $$\hat{p}=-i(\hat{a}-\hat{a}^{\dagger})/\sqrt{2}$$. They measure the average coordinate along two orthogonal directions on the phase space. Their second moments can be used to reveal information about squeezing in squeezed states. The measure quadrature, we need to use homodyne detection.

Obviously, this list is not exhaustive. Any Hermitian combination of ladder operators can in principle be used as an observable. However, not all such observable may be useful or easy to implement as physical measurement.

• Thank you very much for taking the time to think about my question! Unfortunately, I feel this answer does not quite satisfy me. It seems quite common in (especially older) quantum optics literature to privilege $\hat{a}$ as the relevant operator. It Commented Apr 17 at 18:50
• While clearly any Hermitian operator can be an "observable" in the sense that it's eigenvalues are real, this does not itself mean that a physical interaction exists to form a measurement in the eigenbasis defined by that observable. For example, discussing "superselection" in this context has been the source of some controversy in quantum optics, and these issues can be quite subtle. See link and related papers. Commented Apr 17 at 19:02
• It is not clear to me what it is that you are asking then. I am not aware of controversies concerning observables in quantum optics. If there is a specific issue concerning super selection that is controversial, then I suggest you edit your question to focus in this issue. Commented Apr 18 at 3:34