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 find a concept
that extends to electrons and positrons also. An electromagnetic field oscillating at frequency
nu has energy n h nu where 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’s 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 spring is stretched. If x is negative
the spring 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 always opposes 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 0 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 a spring stretched a length x, is one half k x 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 eye times h-bar times the slope in x . h-bar is Planck’s constant h over 2 pi.
This gives us the harmonic oscillator Schrewdinger equation. One half, minus h-bhar squared,
the slope of the slope in x, plus omega squared, x squared, applied to the wavefunction psi,
equals the energy times psi. To simplify the math, assume units in which
h-bar is one, and take omega equal one. Then the equation can be written as the slope
of the slope, which is effectively the curvature, of the wavefunction equals x squared minus
2 E times the wavefunction.
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 wavefunction blows
up for large absolute values of x, which cannot represent a valid quantum state.
But for certain values of E the wavefunction does flatten out for large x values. These
give us the valid quantum states of the harmonic oscillator.
The Schrewdinger equation can be solved analytically. The n-th wavefunction has the form of a constant
c-n times a polynomial H-n 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 percent. 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 wavefunctions, 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 wavefunction 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 0 is one half x 0 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 wavefunction times delta-x
. For the classical system, with x equals x-0
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/velocity is large, delta-t is relatively
small. Working out the details, we find that the
classical probability is delta-x over pi x-0, square root 1 minus x over x-0 squared.
Here’s a plot of these probability distributions for n equals 40. The corresponding classical
amplitude is 9 . We only plot positive x values. For 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’ve 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.
½ , 1 + ½ , 2 + ½ , and, in general, n + ½
Suppose the oscillator is in the n = 3 state. If we create a quantum of energy, this will
move us to the n = 4 state. If we destroy a quantum of energy, this will
move us to the n = 2 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 1 .
And an operator, call it “a” hat minus, which operates on state n to produce state
n minus 1 . More concretely, we have to find some mathematical
operation which when applied to wavefunction psi n produces psi n plus 1 .
And, some mathematical operation which when applied to psi n produces psi n minus 1 .
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 wavefunction 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 2 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 1 this gives us the creation operator one over square-root
of 2, x minus slope in x. Let’s see how this works with some numerical
examples. The blue curve is the n = 0 wavefunction. 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 2 delta.
Operating on the blue curve we obtain the red curve, which is psi-1.
Operating on the red curve we obtain the green curve, which is psi-2, although with a larger
amplitude. Operating on the green curve we obtain the
magenta curve, which is psi-3, although with an even larger amplitude.
Performing the operation analytically, we find that it transforms the wavefunction psi-n
into psi-n+1, times square-root of n plus 1 .
This is our creation operator. a-hat-plus operating on quantum state n, produces square-root
n plus 1 times quantum state n plus 1 .
Changing the minus sign to a plus in the a-hat-plus operator, we obtain the a-hat-minus operator.
This is 1 over square-root 2 times x plus slope in x.
Again, let’s look at this operation numerically. The blue curve is the wavefunction psi-2 .
Performing the numerical operation shown at top left, we obtain the red curve. This is
psi-1, although with a larger amplitude. Applying the operator to the red curve, we
obtain the green curve, which is psi-0, 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. 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 = 0 state. Analytically, the operation transforms the
psi-n wavefunction to the psi-n-minus-1 wavefunction, times square-root n .
This is our destruction operator. a-hat-minus operating on quantum state n, produces square-root
n times quantum state n minus 1 . 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
1 . And, the creation operator applied to state
n produces square-root n plus 1 times state n plus 1 .
Replacing n by n minus 1, we see that the creation operator applied to state n minus
1 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 ½ , the
operator n-hat plus ½ 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’ve 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.