Are Coherent States of Light Classical or Quantum?

A coherent state is a well-defined quantum state of a single mode of light. It is often described as the most classical-like quantum state of that mode, but it is not a classical electromagnetic field: it has photon-number uncertainty and field-measurement noise. Which label is appropriate depends on what property you mean by “classical.”

What a coherent state is

For a mode with annihilation operator $a$, a coherent state $|\alpha\rangle$ is defined by

$$a|\alpha\rangle=\alpha|\alpha\rangle,$$

where $\alpha$ is a complex number. Its squared magnitude gives the mean photon number, and, when $\alpha\ne 0$, its phase gives the phase of the oscillating field. In the photon-number basis, it is a superposition of number states:

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

The state is not a state with a fixed number of photons. Measuring photon number gives a Poisson distribution,

$$P(n)=e^{-\bar n}\frac{\bar n^n}{n!},\qquad \bar n=|\alpha|^2,$$

so $\operatorname{Var}(n)=\bar n$. For $\bar n>0$, the relative size of the photon-number fluctuations is $\sqrt{\operatorname{Var}(n)}/\bar n=1/\sqrt{\bar n}$.

In what sense it is classical-like

The expectation value of the electric field in a coherent state oscillates like a classical wave. For an ideal single mode, its time dependence has the familiar sinusoidal form, with amplitude and phase set by $\alpha$. This makes coherent states useful for describing laser light in many situations, especially when the mean photon number is large.

Coherent states also have the minimum uncertainty allowed for the two field quadratures. The quadratures are the field-mode analogues of position and momentum. Their uncertainties are nonzero, but balanced: neither quadrature is squeezed below the vacuum-noise level. In a phase-space picture, a coherent state is a localized, circular Gaussian centered at the classical complex amplitude.

A further technical sense of “classical” comes from the Glauber–Sudarshan $P$ representation. A coherent state has a nonnegative $P$ distribution concentrated at one amplitude. More generally, a mixture of coherent states with an ordinary nonnegative probability distribution can often be treated as classical light in quantum optics. Coherent states sit at the boundary of this description: they have no extra statistical spread in the distribution of coherent amplitudes, but they still have quantum vacuum fluctuations.

In what sense it remains quantum

A coherent state is a vector in a quantum state space and is subject to quantum measurement rules. Its photon number is uncertain, and its quadratures cannot both be perfectly sharp. These facts are not removed by calling it classical-like.

The word “classical” also does not mean “made of no photons.” A coherent state contains a superposition of different photon numbers, and photon-counting measurements produce discrete outcomes. For a strong coherent state, the relative size of the count fluctuations becomes small, so many measurements are accurately described by a classical wave plus small noise. At low intensity, the discreteness and fluctuations are more consequential.

Nor is every state of light that looks wave-like a coherent state. A number state $|n\rangle$, for example, has exactly $n$ photons but no definite optical phase in the same sense. Squeezed states can reduce the noise in one quadrature below the coherent-state level, at the cost of increased noise in the other; their phase-space behavior is therefore nonclassical by common optical criteria.

Worked example: a coherent state with four mean photons

Take $\alpha=2$, chosen real for simplicity. Then

$$\bar n=|\alpha|^2=4,\qquad \operatorname{Var}(n)=4,$$

and the standard deviation in a photon count is $\sqrt{4}=2$. The probability of measuring exactly four photons is

$$P(4)=e^{-4}\frac{4^4}{4!}\approx 0.195.$$

So the state does not contain exactly four photons; four is the mean of repeated measurements. The relative standard deviation is $2/4=0.5$, which is substantial. If instead $\bar n=10{,}000$, the standard deviation is $100$ and the relative fluctuation is $100/10{,}000=0.01$. In that high-intensity regime, the average field is a good classical approximation for many purposes, even though the underlying state remains quantum.

Common mistakes

• Calling a coherent state a classical field: It is a quantum state whose average field follows classical equations under suitable conditions. The two are not identical.
• Assuming it has a definite photon number: Its photon number is Poisson-distributed. A number state, not a coherent state, has an exact photon count.
• Assuming all laser light is exactly coherent: An idealized single-mode laser is often modeled as coherent, but real lasers can have finite linewidth, multiple modes, and other imperfections. The coherent-state approximation is useful, not universal.
• Treating “classical” as one test: Classical-like field evolution, a positive $P$ representation, and absence of entanglement are different properties. State the criterion being used when the distinction matters.

Practical answer

If “classical” means that the mean field behaves like a classical electromagnetic wave and the state has no sub-vacuum quadrature noise, a coherent state is classical-like. If “quantum” means a state with quantum measurement statistics and irreducible fluctuations, it is quantum. In quantum optics, the most precise short description is: a coherent state is a quantum state with classical-like behavior.