 Now, we turn to the question of how to rigorously represent the creation and destruction of particles. At first, we'll focus on photons, but we'll hold out hope that we'll find a concept that extends to electrons and positrons also. An electromagnetic field oscillating at frequency new has energy NH new, or H is Planck's constant, and N is the number of energy quanta, or photons. When a photon is created, N increments by one. When a photon is destroyed, N decrements by one. This is a physical system characterized by a single frequency of oscillation and a variable amount of energy. A particle on a spring is a classical system sharing these properties. The oscillation frequency is fixed by the particle's mass and the spring stiffness. The more you initially stretch the spring, the more energy you store in the system, and the larger is the resulting oscillation amplitude. But the oscillation frequency remains the same. Let's analyze this so-called harmonic oscillator in some detail. X denotes the mass position, and also how much the string is stretched. If X is negative, the string is compressed, and X equals zero is the equilibrium position of the mass. Newton's second law gives the equation of motion. Mass times acceleration equals force, with the double dots denoting acceleration. The spring force is minus a constant K times the stretch X. The minus sign indicates that the force is always opposed to the mass displacement. To simplify things, suppose the mass is one and call the spring constant K omega squared. Then the equation of motion contains a single parameter omega, and has solutions of the form X equals an initial displacement X zero times cosine omega t, which are oscillations of the type shown in the animation. Omega is proportional to the oscillation frequency nu. Stiffer springs have larger values of K and omega, and oscillate at higher frequencies. The kinetic energy is one half mass times velocity squared. The mass is one, and the velocity is denoted by X dot. This equals one half p squared, where p is the momentum, mass times velocity. The potential energy stored in the spring stretch to length X is one half KX squared, which is also one half omega squared X squared. So one half times the quantity p squared, plus omega squared X squared, equals the total energy of the mass spring system. Now, we analyze the system quantum mechanically. We replace X, P, and E by operators. The X and E operators are just multiplication by X and E. The momentum operator is the imaginary unit I times H bar times the slope in X. H bar is Planck's constant H over 2 pi. This gives us the harmonic oscillator Schrodinger equation. One half minus H bar squared, the slope of the slope in X, plus omega squared X squared, applied to the wave function psi, equals the energy times psi. To simplify the math, assume units in which H bar is one, and take omega equal to one. Then the equation could be written as the slope of the slope, which is effectively the curvature, of the wave function equals X squared, minus 2E times the wave function. As is typical in quantum mechanics, this equation has physically acceptable solutions only for a discrete set of energy values. This is apparent if we solve the equation numerically for varying values of E, the so-called shooting method we introduced in the quantum mechanics chemistry video. For most values of E, the wave function blows up for large absolute values of X, which cannot represent a valid quantum state. But for certain values of E, the wave function does flatten out for large X values. These give us the valid quantum states of the harmonic oscillator. The Schrodinger equation can be solved analytically. The n-th wave function has the form of a constant Cn times a polynomial hn times a bell curve exponential, where the constants are fixed by the condition that the sum of the particle probability over all X values is 100%. The lowest energy is one-half. The next energy is one plus one-half. And so on. The n-th energy level is n plus one-half. For simplicity, our analysis assumed units in which h-bar and omega are both one. In standard units, the n-th energy level is n plus one-half times h-bar omega. Here are the first four wave functions with energies one-half, one and one-half, two and one-half, and three and one-half. Note that the lowest energy is not zero. The uncertainty principle does not allow the particle to be at a known location, X equals zero, with a known momentum P equals zero. We know that the wave function gives the probability amplitude for finding the particle at a particular X value. It's interesting to compare the quantum and classical probabilities. Assuming our special case h-bar and omega both equal to one, the energy of a classical harmonic oscillator with oscillation amplitude X zero is one-half X zero squared. The possible energies of the quantum version are n plus one-half. If for a given value of n we equate these, we have classical and quantum versions of the same system with the same energies. For the quantum system, the probability of finding the particle within an interval delta X centered on coordinate X is the magnitude squared of the wave function times delta X. For the classical system with X equals X zero cosine of t, the probability of finding the particle within some coordinate interval delta X is proportional to the corresponding time duration delta t. Where the slope of the cosine is small, that is, where the particle is moving slowly, delta t is relatively large. Conversely, where the slope slash velocity is large, delta t is relatively small. Working out the details, we find that the classical probability is delta X over pi X zero square root one minus X over X zero squared. Here's a plot of these probability distributions for n equals 40. The corresponding classical amplitude is nine. We only plot positive X values or negative X values, the plot is simply the mirror image of this. The quantum probability oscillates about the classical probability except near the classical displacement limit. Now, let's recall what we're after. We want to find a way to describe the creation and destruction of photons. These are the quanta of the electromagnetic field. We found that the energy quantum of the harmonic oscillator is, in standard units, H bar omega. And the allowable energy levels in terms of this quantum are one half, one plus one half, two plus one half, and in general, n plus one half. Suppose the oscillator is in the n equals three state. If we create a quantum of energy, this will move us to the n equals four state. If we destroy a quantum of energy, this will move us to the n equals two state. Using the Dirac ket notation to represent the quantum state with n energy quanta, we want to find an operator, call it a hat plus, which operates on state n to produce state n plus one. And an operator call it a hat minus, which operates on state n to produce state n minus one. More concretely, we have to find some mathematical operation which when applied to wave function psi n produces psi n plus one. And some mathematical operation, which when applied to psi n, produces psi n minus one. We'll call these the creation and destruction operators. One approach to finding them is to apply known operators, such as position and momentum, to the psi n wave function and see if there is some way to combine the results to get our desired operation. We find that a useful combination is one over square root two times position operator minus imaginary unit times momentum operator. The momentum operator is minus i h bar slope in x. And with h bar equals one, this gives us the creation operator one over square root of two x minus slope in x. Let's see how this works with some numerical examples. The blue curve is the n equals zero wave function. A numerical version of the creation operation is shown at upper left. For every x value, we compute x times psi minus the slope in x of psi. We approximate the slope by the change in psi over the change in x. The change in psi is psi at x plus some small number delta minus psi at x minus delta. The change in x is two delta. Operating on the blue curve, we obtain the red curve, which is psi one. Operating on the red curve, we obtain the green curve, which is psi two, although with a larger amplitude. Operating on the green curve, we obtain the magenta curve, which is psi three, although with an even larger amplitude. Performing the operation analytically, we find that it transforms the wave function psi n into psi n plus one times square root of n plus one. This is our creation operator. A hat plus operating on quantum state n produces square root n plus one times quantum state n plus one. Changing the minus sign to a plus sign in the A hat plus operator, we obtain the A hat minus operator. This is one over square root of two times x plus slope in x. Again, let's look at this operation numerically. The blue curve is the wave function psi two. Performing the numerical operation shown at top left, we obtain the red curve, and this is psi one, although with a larger amplitude. Applying the operator to the red curve, we obtain the green curve, which is psi zero with a larger amplitude. Now we face an interesting situation. The green curve is the lowest energy state. The destruction operator moves us to the next lowest energy state. So what happens if we apply it to the green curve? The result is the magenta curve, identically zero. This doesn't mean a quantum state with zero energy. It means no quantum state at all, nothing. The operator is telling us that there is no quantum state with energy lower than the n equals zero state. Analytically, the operation transforms the psi n wave function to the psi n minus one wave function times square root of n. This is our destruction operator. A hat minus operating on quantum state n produces square root n times quantum state n minus one. As a footnote, most textbooks omit the minus sign, denoting the operator just by a hat and calling it the annihilation operator. But we'll use the creation and destruction terminology and keep the minus sign to emphasize the respective raising and lowering operations. We might think that the square root factors produced by the creation and destruction operators are a bug and try to find some way to get rid of them, but they are actually a feature. The destruction operator applied to state n produces square root n times state n minus one. And the creation operator applied to state n produces square root n plus one times state n plus one. Replacing n by n minus one, we see that the creation operator applied to state n minus one produces square root n times state n. Therefore, applying the destruction operator to state n and then applying the creation operator to that result produces n times state n. This leads us to define the number operator n hat as a hat plus a hat minus. n hat applied to state n produces n times state n. We can think of the number operator as representing an observation of the number of energy quanta in the system. Since the energy is simply n plus a half, the operator n hat plus a half applied to state n produces the energy times state n. This is the so-called Hamiltonian operator, h hat, which represents the system energy. So, we found operators that at least for the harmonic oscillator represent creation and destruction of energy quanta and we've identified an operator corresponding to observation of the number of energy quanta. We seem to be heading in the right direction, but our results apply to a single particle moving in a single dimension at a single point with a single oscillation frequency. We want to describe a three-dimensional field with oscillations in three dimensions at every point in space with an infinite number of oscillation frequencies. Clearly, we still have plenty of work to do.