# Is number of photons undefined for a classical EM wave?

Physics Asked by Ruslan on December 26, 2020

I have read somewhere (several times in different places, and can’t find this now) that for a classical EM wave (or an approximation thereof) number of photons is not defined. I.e. if we have a state with definite number of photons $$n$$, then this state is not even close to a classical EM wave, however large $$n$$ could be.

Is it true? If yes, how can this be proved?

Typically speaking, when we speak of a classical EM wave, we mean one with a well defined electric and magnetic fields as a function of time $$E(t)$$ and $$B(t)$$. As operators, these fields take the form of

$$E(t) propto a^dagger e^{iomega t} + a e^{-iomega t}$$ $$B(t) propto a^dagger e^{iomega t} + a e^{-iomega t}$$

If we evaluate the expectation values of the electric field operator on a state with definite photon number $$vert n rangle$$, we obtain

$$langle n vert E(t) vert n rangle propto langle n vert a^dagger e^{iomega t} + a e^{-iomega t} vert n rangle = 0$$

Thus, light with definite photon number has an expected Electromagnetic field of zero and behaves completely different than the classical light we're used to which has well-defined electric and magnetic fields that oscillate in time.

So the question is: what quantum state more accurately represents the behavior of (coherent) classical light?

The answer was given by Glauber and is the coherent state which has the defining property of being an eigenstate of the lowering operator $$a$$,

$$a vert alpha rangle = alpha vert alpha rangle$$

For this state, the electric field $$langle E(t) rangle$$ now goes as

$$langle alpha vert E(t) vert alpha rangle propto langle alpha vert a^dagger e^{iomega t} + a e^{-iomega t} vert alpha rangle = alpha^* e^{iomega t} + alpha e^{-i omega t} \=vert alpha vert cos(omega t - phi)$$

where $$alpha = vertalphavert e^{iphi}$$. Thus, a coherent state has a well-defined and oscillating electric field as you would expect from a coherent classical EM wave.

The expectation value for the number of photons of a coherent state is also possible to evaluate as

$$langle n rangle = langle alpha vert a^dagger a vert alpha rangle = vert alpha vert^2$$

However, unlike a definite number state, the variance $$langle n^2 rangle - langle n rangle^2$$ is nonzero and is given by $$vert alpha vert^2$$ as well. Therefore the standard deviation of the photon number is $$sqrt{n}$$ and one does not have a fixed photon number in the EM wave, although the relative standard deviation becomes very small for large $$n$$. This result clearly differs from a photon number eigenstate which has identically zero variance in the photon number. In fact, the coherent state has a photon number distribution that is Poissonian.

Correct answer by KF Gauss on December 26, 2020

To make this easier to understand to someone not too fluent in QFT, let's use a simple analogy with a system with a finite number of degrees of freedom. Namely, consider $$N$$ particles of mass $$m$$ connected by springs with spring constant $$k$$ to each other (and the first one to an immovable wall), arranged on a line, with equilibrium state of each particle being at a distance $$a$$ from its nearest neighbor. The potential energy will look like

$$U=frac{momega^2}2(x_1^2+(x_2-x_1-a)^2+...+(x_N-x_{N-1}-a)^2),$$

where

$$omega=sqrt{frac km}.$$

This function of positions $$x_i$$ is a quadratic form, thus a simple rotation in the $$N$$-dimensional space (or an equivalent change of variables) can bring it to the form

$$U=frac m2(omega_1^2y_1^2+omega_1^2y_2^2+...+omega_N^2y_N^2),$$

where $$omega_i$$ are now eigenfrequencies of the system (AKA normal modes), and $$y_i$$ are the corresponding displacements of these modes (i.e. the amplitudes of the terms in the Fourier sum corresponding to their associated frequencies). Each mode is a standing wave with a different wavelength (standing due to the boundary condition at $$x_1$$—being connected to the wall).

Since the kinetic energy is invariant under such a rotation, this lets us solve the equations of motion (in classical mechanics) and the Schrödinger's equation (in quantum mechanics) by simple separation of variables: the system is a set of independent linear oscillators—normal modes.

The energy in mode $$i$$, being continuous in the classical system, is discrete in the quantum case, with the quantum being famously equal to $$hbaromega_i$$. This quantum of collective motion of the particles is known in semiconductor physics as phonon, and it is in many ways analogous to photon in QED, when we consider an infinite 3D lattice of our masses $$m$$, in the limit of $$ato0$$.

The closest state of a quantum harmonic oscillator to a classical motion is the coherent state. It is such a kind of state where the probability density retains its shape at all times, and only oscillates so that its peak traces the classical trajectory. The very fact that probability density changes in time already implies that this state is not an eigenstate of the Hamiltonian of the oscillator. And since all eigenstates of the Hamiltonian have a definite number of quanta of energy (and conversely, a state with a definite number of these quanta is an eigenstate), this means that in a coherent state (except the ground state) the number of quanta doesn't have a definite value.

By the same logic, if we consider an eigenstate of our lattice of masses connected by springs, it'll have constant probability density at any energy, which of course is far from the expected classical wave-like oscillation in time. Actual classical wave-like motion must have at least one mode contain a superposition of non-degenerate eigenstates of the Hamiltonian.

Returning to the photons and EM fields, we can see by analogy that similarly, if we want to have e.g. $$vec E$$-field oscillate in a classical-like fashion, it must have a superposition of number states, making number of photons undefined.

Answered by Ruslan on December 26, 2020

I have read somewhere (several times in different places, and can't find this now) that for a classical EM wave (or an approximation thereof) number of photons is not defined

It will depend on the context. If you have an electromagnetic wave of fixed frequency , its energy is defined by the Poynting vector. Photons composing this beam have $$energy = h*ν$$ so it is easy to get a number of photons crossing a crossectional area. To convince you see this experiment one photon at a time, where the photons are detected on a frame.

1. Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

As you see the photons are counted , and on the rightmost frame one sees the interference pattern of the classical wave emergent from the addition of the quantum mechanical wave functions of those photons. So your statement is not correct:

if we have a state with definite number of photons n, then this state is not even close to a classical EM wave, however large n could be.

is not correct for the following reason:

The above experiment is with a coherent laser beam, and it clearly shows in the last frame the classical wave leaving the level of the two slits has countable photons. That it is a quantum mechanical state dependent on the probabilities of the total wavefunction is a common story at the quantum level, but it is another story.

Now for many frequency light the statement can hold because one has to know the distribution of frequencies to be able to estimate the energy of the component photons and count them.

Answered by anna v on December 26, 2020

Yes, a quantum state with a specific fixed number for photons is a Fock state (or number state). It is indeed not a classical state. Classical light is often thermal or it can be a coherent state, such as produced by a laser.

There are in principle many ways to see this. One way is to use a beam splitter. When you send classical light through a beam splitter, you just end up with two states that can be expressed as a tensor product. On the other hand if you send a Fock state through a beam splitter you'll end up with an entangled state.

Perhaps the best way to observe the difference experimentally is to use quantum state tomography. This can be done with the aid of quantum homodyne measurements. The result will give you a description of the phase space distribution of the state. Such a distribution will then tell you what the properties of the state are.

Answered by flippiefanus on December 26, 2020