# Are coherent states of light 'classical' or 'quantum'?

Coherent states of light, defined as

$$|\alpha\rangle=e^{-\frac{|\alpha|^2}{2}}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}|n\rangle$$

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'?

Coherent states are quantum states, but they have properties that mirror classical states in a sense that can be made precise.

To be concrete, let's consider coherent states in the context of the simple harmonic quantum oscillator which have precisely the expression you wrote in the question. One can demonstrate the following two facts (which I highly encourage you to prove to yourself);

• The expectation value of the position operator in a coherent state is \begin{align} \langle\alpha|\hat x|\alpha\rangle = \sqrt{\frac{\hbar}{2m\omega}}(\alpha + \alpha^*) \end{align}

• The time evolution of a coherent state is obtained by simply time evolving its eigenvalue by a phase; \begin{align} e^{-it \hat H/\hbar}|\alpha\rangle = |\alpha(t)\rangle, \qquad \alpha(t):=e^{-i\omega t}\alpha. \end{align} In other words, if the system is in a coherent state, then it remains in a coherent state!

If you put these two facts together, then you find that the expectation value of the position operator has the following time-evolution behavior in a coherent state: \begin{align} \langle\hat x\rangle_t:=\langle\alpha(t)|\hat x|\alpha(t)\rangle = \sqrt{\frac{\hbar}{2m\omega}}(e^{-i\omega t}\alpha + e^{i\omega t}\alpha^*) \end{align} but now simply write the complex number $\alpha$ in polar form $\alpha = \rho e^{i\phi}$ to obtain \begin{align} \langle \hat x\rangle = \sqrt{\frac{\hbar}{2m\omega}}2\rho\cos(\omega t-\phi) \end{align} In other words, we have shown the main fact indicating that coherent states behave "classically":

• The expectation value of the position of the system oscillates like the position of a classical simple harmonic oscillator.

This is one sense in which the coherent state is classical. Another fact is that

• Coherent states minimize qauntum uncertainty in the sense that they saturate the heisenberg uncertainty bound; \begin{align} \sigma_x\sigma_p = \frac{\hbar}{2} \end{align} To the extent that uncertainty is a purely quantum effect, minimization of this effect can be interpreted as maximizing "classicalness."
• It seems unlikely that the only thing truly quantum about coherent states is that they still obey the uncertainty principle, since this hardly seems to be the feature that is relevant in their application to quantum information processing. – Juan Miguel Arrazola Dec 6 '13 at 22:37
• It's not the only thing; did I seem to imply that somewhere in my answer? Coherent states are decidedly quantum in the sense that they are vectors (pure states) in Hilbert spaces modeling quantum systems. I think it's more accurate to say that they are quantum states with some properties that are strongly reminiscent of classical states. – joshphysics Dec 6 '13 at 23:06

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.

One last coherent state property that hasn't been talked about in the other answers is that it 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 whole set of quantum harmonic oscillator states which achieve saturation of the Heisenberg inequality.

If coherent state are indeed the most classical states (which means that the mean value of the EM fields obeys the classical Maxwell equations), the state used in the paper you mentioned are not coherent state (at least in the arXiv paper), but cat states !

The state $|\alpha\rangle+|-\alpha\rangle$ is not a coherent state ! It is the superposition of two classical state, which is really what we mean by quantumness.

Stated otherwise, coherent states form a basis which with you can write any quantum state, but that does not mean that all these states are as classical than a coherent state.

• My apologies, I actually didn't look past the title of the paper I linked to. However, I do know that coherent states play a role in many architectures for quantum computing. – Juan Miguel Arrazola Dec 4 '13 at 23:52

It is all about what meaning you put into the words "quantum" and "classical". Fock space and elements of this space are notions that belong to quantum theory of radiation and have no direct relation to states of radiation in classical electromagnetic theory, so the coherent state may be called "quantum" with good reason.

However, coherent states have properties very similar to those of harmonically-oscillating standing waves of electromagnetic field as used in classical theory of microwave cavities, so they are often called "classical-like" quantum states.

Coherent states are classical in a precise way which hasn't been stated explicitly yet, although Rod suggests at it.

Suppose you want to time-evolve the interaction between a coherent EM state and matter. This amounts to solving the Schroedinger equation for:

$$i\hbar\frac{d}{dt} |\psi \rangle= H |\psi \rangle$$ with $$\hat{H}= \hat{H}_{0A}+\sum_k\omega_k \hat{b}^\dagger_k \hat{b}_k+\sum_k \hat{\vec{\mu}}_A\cdot\hat{\vec{E_k}}$$ describing the interaction between some matter and a quantized EM field (in the dipole approximation), and the initial state $$|\psi_0\rangle=|0,\alpha\rangle$$ in which $|0\rangle$ is some initial state of the matter and $|\alpha \rangle$ labels the initial coherent state of the EM field.

If all this is true, there is a result due to Mollow (1) that says there is a canonical transformation to this system which maps it to the system:

$$\hat{H}'=\hat{H}+\sum_k \hat{\vec{\mu}}\cdot{\vec{E}_{k,\alpha}}(t)$$ and $$|\psi_0\rangle=|0,0\rangle$$

In other words, the Hamiltonian now has an additional time-dependent external potential- note there's no hat on the $\vec{E_{k,\alpha}}$ because it is not a field operator! This potential has the same amplitude and frequency as the initial coherent state. The quantized field is still present but starts in the vacuum state. What this means is that the system evolves just the same as an atom in the corresponding external potential, except also with the possibility of being influenced by photons that it emits itself (which will often be negligible in comparison to a large external field).

The conclusion is that coherent states are 'classical' in the sense that they can be replaced with an external potential. This justifies, among other things, the semiclassical model of light interacting with matter that is ubiquitous in atomic and condensed matter physics.

Out of the many quantum-mechanically possible states of an oscillator (be it a mechanical one or light waves), the ones we almost exclusively observe are the coherent states. In a way, they are the states where uncertainty is evenly distributed, such that every uncertain quantity scales as $\sqrt{N}$ for $N$ quanta (e.g. photons or energy quanta in an oscillator). Everything else tends to be somewhat hard to generate experimentally because one needs to couple states in a nontrivial manner, which is somewhat unusual for the typically non-interacting bosons that these quanta are. Where we succeed in engineering states significantly different from coherent states, we often emphasize that fact by calling them squeezed (with some uncertainties growing faster and some slower than $\sqrt{N}$ and hence looking like a circle squeezed into an ellipse when plotted together) or nonclassical states.

It is likely that the quantum (optics) protocols you found were rather intentionally based on coherent states simply because lasers (that emit coherent states) are comparatively easy to come by and easy to operate, compared to squeezed light sources. However, the earliest and easiest protocols for quantum cryptography assume (highly nonclassical and impractically difficult to build) single photon sources because their security against an eavesdropper potentially equipped with some unknown means to syphon off duplicate photons is easier to prove and their throughput higher when one does not need to make allowances for that.

When the mean number of photons is huge, the Heisenberg uncertainty becomes negligible and "disappears" (formally it looks like $\hbar\to 0$). Thus, such a coherent state becomes quite classical one.

• The mean number of photons in the state $|\alpha\rangle$ is $|\alpha|^2$, which can take any positive real value, huge or tiny. The arguments for classicalness of coherents states given in the other answers don't depend on $|\alpha|^2$ being big or small. – Bosoneando Jun 12 '15 at 22:26
• @Bosoneando I know. That is why there they use quote marks and other things in general case. – Vladimir Kalitvianski Jun 13 '15 at 19:22
• Why does this argument apply for a coherent state but not a harmonic oscillator eigenstate with a very large occupation number? – tparker Jul 11 '16 at 6:45
• @tparker: A pure state $\psi_n(x,t)$ is rather spread and does not "move" on average. The mean value of particle coordinate is, of course, unique, but observations will give an essential dispersion of $x$ around $\langle x \rangle$. To "narrow" the spread and make it "move", one may use a superposition of $\psi_n(x,t)$ with different $n$. A coherent state is such that it minimizes the spread. Then observations give something more "classical" - an optimally localized wave packet. – Vladimir Kalitvianski Jul 11 '16 at 9:24