Are Coherent States of Light Classical or Quantum?

I gave this answer on Physics Stack Exchange to the following question:

Coherent states of light, defined as:$$|\alpha\rangle=e^{-\frac{|\alpha|^2}{2}}\sum_{n=0}^\infty \frac{\alpha^n}{n!}|n\rangle$$

for a given complex number $\alpha$ and where $|n\rangle$ is a Fock state with $n$ photons, are usually referred to as the most classical states of light. On the other hand, many quantum protocols with no classical analog such as quantum key distribution and quantum computing can be implemented with coherent states.

In what sense or in what regime should we think of coherent states as being ‘classical’ or ‘quantum’?

and the answer, slightly edited, is as follows:

Coherent states, although strictly quantum, are “isomorphic” to classical states. They are also isomorphic in the same way to one-photon states.

There are bijective maps between any pair of the following three sets: (i) the set of all quantum coherent states (ii) the set of all one-photon states and (iii) and the set of all solutions of Maxwell’s equations. I speak more about this statement in my answer here and also this one here. So you can think of any solution of Maxwell’s equations as defining either a classical state or a quantum coherent state. When we do the latter, we exploit following the special property of the coherent state: it is uniquely and wholly defined by the means of the $\vec{E}$ and $\vec{H}$ observables as functions of space and time. So, although these means superficially aren’t the same as the quantum state, in the same way that many classical probability density functions, e.g. Gaussian are defined by more parameters than only their means, for the special case of coherent states they can be interpreted as such (just as the classical exponential and Poisson probability distributions are uniquely defined by their means).

So, if you like, the coherent states are how we consistently embed the classical states into the much bigger, quantum theory of the light fields. This is the “window” from the classical to the quantum World. This standpoint also underlies the radical difference between the complexities of classical and quantum states: for a quantisation volume, there are countably infinite $\aleph_0$ electromagnetic modes $\left\{(\vec{E}_j,\,\vec{H}_j)\right\}_{j=0}^\infty$. $\aleph_0$ then is a measure of the “complexity” of thus basis, which is both the basis of one-photon states and also the basis for a classical superposition of modes solving Maxwell’s equations. On the other hand, members of the basis for all the Fock states are countably infinite sequences of natural numbers like $\left.\left|n_1, n_2, n_3,\cdots\right.\right>$ so the basis itself has the same cardinality $\aleph_1$ as the continuum. The classical state space is the direct sum of the one photon subspaces, the general quantum state space the tensor product a countable product of countably infinite subspaces. But also see my discussion of Hilbert state space separability here.

A coherent state of a lone quantum harmonic oscillator can be defined as an eigenvector of the annihilation operator $a = \sqrt{2}^{-1}\left(\sqrt{\frac{m\,\omega}{\hbar}}\hat{x}+i\,\sqrt{\frac{\hbar}{m\,\omega}}\,\hat{p}\right)$ and, as such, both (i) saturates the Heisenberg inequality (i.e. $\Delta x\,\Delta p = \hbar$) and (ii) shares out the uncertainty equally between the two dimensionless position and momentum observables $\sqrt{\frac{m\,\omega}{\hbar}}\hat{x}$ and $\sqrt{\frac{\hbar}{m\,\omega}}\,\hat{p}$: so it achieves minimum uncertainty product and has no preference for where the measurement error arises. In normalized $x,\,p$ quantum phase space (Wigner distribution space), its uncertainty regions are thus minimum area disks, the reason why it is often spoken of as the “most classical state” that can be.

It can be represented as the image of the harmonic oscillator’s quantum ground state $ \left.\left|0\right.\right>$ under the action of the *displacement operator* $D(\alpha) = \exp\left(\alpha\, a + \alpha^*\,a^\dagger\right)$. This operator “displaces” the ground state in Wigner phase space along the vector $({\rm Re}(\alpha),\,{\rm Im}(\alpha))$ but otherwise leaves it unchanged. One can generalize the coherent state to the bigger set of *squeezed states* with the following property. A further operation by the squeeze operator $S(\beta) = \exp\left(\beta\, a – \beta^*\,a^\dagger\right)$ leaves the distribution centred at the same point and still achieving the minimum uncertainty product (i.e. the Heisenberg inequality saturates to an equality), but imparts a “preference” to the accuracy of measurements from one of the observables $\hat{x},\,\hat{p}$ at the expense of accuracy in the other in a so-called squeezed state. States of the form $S(\beta)\, D(\alpha)\,\left.\left|0\right.\right>$ are the exactly the whole set of quantum harmonic oscillator states which achieve saturation of the Heisenberg inequality.