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 began by considering a classical electromagnetic wave, polarized in the x
direction and traveling in the z direction.
The vector potential A equals the unit vector in the x direction, times an arbitrary
amplitude b, times cosine k z minus omega t plus an arbitrary phase
theta.
In our units, with the speed of light equal to one, 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 a quantum state
that corresponds to the classical wave.
In addition to the field operator we need to specify the wavefunction, represented by
psi-ket.
Let's take this to be the state with n photons.
Weíll 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 wavefunction of a stationary
state with definite energy E satisfies i, h-bar time-derivative of the wavefunction
equals E times the wavefunction.
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 wavefunction.
To do that we need both the above ket and the following bra forms of the wavefunction.
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 will not 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 wavefunction, and then projecting the result onto the wavefunction.
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 wavefunction will produce two terms.
One each for the destruction and creation operators.
The destruction operator produces square-root n times the n-minus-1 state.
The creation operator produces square-root n-plus-1 times the n-plus-1 state.
The projections of the n-minus-1 and n-plus-1 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 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 photons 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 already treated a very similar problem in the quantum mechanics series, when
discussing solutions of the Schrodinger equation.
For a two-dimensional harmonic oscillator in a box.
We found stationary states with different numbers of energy quanta.
Here's the wavefunction for zero quanta.
And various excited states.
In all these cases the symmetry of the wavefunction 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 do 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 wonít explicitly show the time dependence.
Now, we need to determine the coefficients, b-n.
Applying the destruction operator, the n-state becomes square-root of n times the n-minus-one
state.
We can shift the index by replacing n with n-plus-one.
And then set this equal to c times the state psi.
The expressions are equal if b n-plus-one equals b n times c over square-root n-plus-one.
Letís set b-0 equal to some constant gamma.
Then b-1 equals gamma times c.
b-2 equals gamma times c-squared over square-root two.
b-3 equals gamma times c-cubed over square-root two times three.
And, in general, b-n equals gamma times c to-the-n over square-root n-factorial.
To solve for gamma, we enforce the requirement that the wavefunction is normalized to one.
This gives the sum of the squared-magnitudes of all coefficients equal to one.
Magnitude b-n squared is the probability that the field is in the state with n photons.
The sum of all these is simply the probability that the field has any possible number of
photons, which has to be one-hundred percent.
Substituting b-n 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
one.
A solution is gamma equals e-to-the minus one-half magnitude-c-squared.
Here's a plot of magnitude-b-n-squared versus n for magnitude-c-squared equals twenty.
We see that the most probable number of photons is near twenty, but there are significant
probabilities of the field having anywhere from about fifteen to twenty-five 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; twenty 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 b-0,
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.