 Our field quantization technique led us to a quantum description in which the field is an operator and the wave function is a list of the number of photons present in each electromagnetic mode. The field operator is a sum over all possible modes of an expression containing one destruction and one creation operator for the corresponding photons. This representation seems rather abstract. How do we use these expressions to do physics and make useful predictions? And what does it mean to say the electromagnetic field is an operator? To get a feeling for how our formalism works, let's consider the classical limit of a quantum field. Recall that we begin by considering a classical electromagnetic wave polarized in the x direction and traveling in the z direction. The vector potential A equals a unit vector in the x direction times an arbitrary amplitude B times cosine kz minus omega t plus an arbitrary phase theta. In our units, with the speed of light equal to 1, omega equals k. After quantization, the electromagnetic field becomes an operator, a linear combination of the destruction and creation operators A hat minus and A hat plus. Let's try to work backward and see if we can use our formalism to find the quantum state that corresponds to the classical wave. In addition to the field operator, we need to specify the wave function represented by psi ket. Let's take this to be the state with n photons. We'll explicitly show the time dependence, e to the minus i, n plus one half omega t. This is consistent with the quantum mechanical requirement that the wave function of a stationary state with definite energy e satisfies i h bar time derivative of the wave function equals e times the wave function. Each of the n photons has energy h bar omega and the zero point field energy is one half h bar omega. It seems reasonable that when we have many photons, that is when n gets big, this quantum state might behave somewhat like a classical electromagnetic wave. We want to calculate the expected value of the electromagnetic field for this wave function. To do that, we need both the above ket and the following bra forms of the wave function. The bra form is just the complex conjugate of the ket form with time dependence e to the plus i, n plus one half omega t. Now, we already have a problem. When we multiply these conjugate exponential factors together, we'll get their squared magnitude, which is just one. So all time dependence will vanish. Therefore, we cannot end up with anything like the classical expression. Let's follow through the calculation. In quantum mechanics, the expected value of an operator is obtained by applying it to the wave function and then projecting the result onto the wave function. With the vanishing of the time dependence, we end up with the field operator between bra and ket versions of the n quantum state. The field operator acting on the wave function will produce two terms, one each for the destruction and creation operators. The destruction operator produces square root n times the n minus one state. The creation operator produces square root n plus one times the n plus one state. The projections of the n minus one and n plus one states onto the n state vanish. So the expected value of the field operator also vanishes, no matter how many photons are present. Now, this doesn't mean that the field vanishes, only that at a given time and place, the field is equally likely to be positive or negative with an average value of zero. This will be the case if the n photons are, quote, incoherent. That is, they do not oscillate in phase with each other. In fact, natural light typically behaves like this. Let's look at a cartoon representation of incoherent versus coherent light. Suppose we have a number of excited atoms that can emit photons of a given frequency. If all atoms decay randomly through spontaneous emission, the phases of the resulting photons will be randomly distributed. So at a given time and place, the total field, being the sum of a large number of random contributions, will have an expected value of zero. That's the nature of incoherent light. We're interested in coherent light, such as produced in a laser. If the atoms decay through stimulated emission, the photon phases will be aligned. So at a given time and place, the total field, ideally, will be the sum of n identical contributions and have a definite expected value. Actually, we've already treated a very similar problem in the quantum mechanic series when discussing solutions of the Schrödinger equation. For a two-dimensional harmonic oscillator in a box, we found stationary states with different numbers of energy quanta. Here's the wave function for zero quanta, and various excited states. In all these cases, the symmetry of the wave function results in the mean position of the particle being fixed in the center, with no variation through time. But we also found solutions corresponding more closely to a classical particle on a spring, oscillating back and forth in the box. These quasi-classical states did not contain a definite number of energy quanta, but are instead a particular superposition of the stationary states. We'll call these coherent states. A 1963 paper by Glauber showed how to construct these for the electromagnetic field. Let's look for a quantum state psi, such that when we apply the destruction operator, we obtain the same state multiplied by some constant c. We write this as a superposition of the n photon states. To keep the math compact, we won't explicitly show the time dependence. Now we need to determine the coefficients bn. Applying the destruction operator, the n state becomes square root of n times the n minus 1 state. We can shift the index by replacing n with n plus 1, and then set this equal to c times the state psi. The expressions are equal if bn plus 1 equals bn times c over square root n plus 1. Let's set b0 equal to some constant gamma. Then b1 equals gamma times c. b2 equals gamma times c squared over square root 2. b3 equals gamma times c cubed over square root 2 times 3. And in general, bn equals gamma times c to the n over square root n factorial. To solve for gamma, we enforce the requirement that the wave function is normalized to 1. This gives the sum of the squared magnitudes of all coefficients equal to 1. Magnitude bn squared is the probability that the field is in the state with n photons. The sum of all of these is simply the probability that the field has any possible number of photons, which has to be 100%. Substituting bn equals gamma times c to the n over square root n factorial. We get magnitude gamma squared times the sum of magnitude c squared to the nth power over n factorial. This summation is the so-called power series for e to the magnitude c squared. So our normalization condition is magnitude squared gamma times e to the magnitude squared c equals 1. A solution is gamma equals e to the minus one-half magnitude c squared. Here's a plot of magnitude bn squared versus n for magnitude c squared equals 20. We see that the most probable number of photons is near 20. But there are significant probabilities of the field having anywhere from about 15 to 25 photons. From this distribution we can calculate the expected or mean number of photons denoted by n inside angled brackets. We find this equals magnitude c squared 20 for the case plotted. The width of the distribution can be quantified by the root mean squared deviation denoted by sigma n. This equals magnitude c, which equals square root of the expected value of n, about 4.5 for the case shown. Treating the mean value of n as a signal, sigma n can be thought of as quantum noise. In many applications, magnitude c squared is large enough that quantum noise is negligible by comparison. Quantum noise does play a significant role in fiber optic digital communication for which it sets a limit on the bit error rate proportional to the magnitude squared b0, which equals e to the minus expected number of photons. Thus lower bit error rates require a higher average number of photons per data bit.