Subsections


A.23 Quantization of radiation

Long ago, the electromagnetic field was described in terms of classical physics by Maxwell, chapter 13. His equations have stood up well to special relativity. However, they need correction for quantum mechanics. According to the Planck-Einstein relation, the electromagnetic field comes in discrete particles of energy $\hbar\omega$ called photons. A classical electromagnetic field cannot explain that. This addendum will derive the quantum field in empty space. While the description tries to be reasonably self-contained, to really appreciate the details you may have to read some other addenda too. It may also be noted that the discussion here is quite different from what you will find in other sources, {N.12}.

First, representing the electromagnetic field using the photons of quantum mechanics is called “second quantization.” No, there is no earlier quantization of the electromagnetic field involved. The word second is there for historical reasons. Historically, physicists have found it hysterical to confuse students.

In the quantum description, the electromagnetic field is an observable property of photons. And the key assumption of quantum mechanics is that observable properties of particles are the eigenvalues of Hermitian operators, chapter 3. Furthermore, these operators act on wave functions that are associated with the particles.

Therefore, second quantization is basically straightforward. Find the nature of the wave function of photons. Then identify the Hermitian operators that give the observable electromagnetic field.

However, to achieve this in a reasonable manner requires a bit of preparation. To understand photon wave functions, an understanding of a few key concepts of classical electromagnetics is essential. And the Hermitian operators that act on these wave functions are quite different from the typical operators normally used in this book. In particular, they involve operators that create and annihilate photons. Creation and annihilation of particles is a purely relativistic effect, described by so-called quantum field theory.

Also, after the field has been quantized, then of course you want to see what the effects of the quantization really are, in terms of observable quantities.


A.23.1 Properties of classical electromagnetic fields

Classical electromagnetics is discussed in considerable detail in chapter 13.2 and 13.3. Here only a few selected results are needed.

The classical electromagnetic field is a combination of a so-called electric field $\skew3\vec{\cal E}$ and a magnetic field $\skew2\vec{\cal B}$. These are measures for the forces that the field exerts on any charged particles inside the field. The classical expression for the force on a charged particle is the so-called Lorentz force law

\begin{displaymath}
q \left(\skew3\vec{\cal E}+ \vec v\times \skew2\vec{\cal B}\right)
\end{displaymath}

where $q$ is the charge of the particle and $\vec{v}$ its velocity. However, this is not really important here since quantum mechanics uses neither forces nor velocities.

What is important is that electromagnetic fields carry energy. That is how the sun heats up the surface of the earth. The electromagnetic energy in a volume ${\cal V}$ is given by, chapter 13.2 (13.11):

\begin{displaymath}
E_{\cal V}={\textstyle\frac{1}{2}} \epsilon_0
\int_{\cal...
...}^2+c^2\skew2\vec{\cal B}^2\right){\,\rm d}^3{\skew0\vec r} %
\end{displaymath} (A.152)

where $\epsilon_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8.85 10$\POW9,{-12}$ C$\POW9,{2}$/J m is called the permittivity of space and $c$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 10$\POW9,{8}$ m/s the speed of light. As you might guess, the energy per unit volume is proportional to the square fields. After all, no fields, no energy; also the energy should always be positive. The presence of the permittivity of space is needed to get proper units of energy. The additional factor $\frac12$ is not so trivial; it is typically derived from examining the energy stored inside condensers and coils. That sort of detail is outside the scope of this book.

Quantum mechanics is in terms of potentials instead of forces. As already noted in chapter 1.3.2, in classical electromagnetics there is both a scalar potential $\varphi$ as well as a vector potential $\skew3\vec A$. In classical electromagnetics these potentials by themselves are not really important. What is important is that their derivatives give the fields. Specifically:

\begin{displaymath}
\skew3\vec{\cal E}= - \nabla \varphi - \frac{\partial \ske...
...}
\qquad
\skew2\vec{\cal B}= \nabla \times \skew3\vec A %
\end{displaymath} (A.153)

Here the operator

\begin{displaymath}
\nabla = {\hat\imath}\frac{\partial}{\partial x}
+ {\hat...
...c{\partial}{\partial y} + {\hat k}\frac{\partial}{\partial z}
\end{displaymath}

is called nabla or del. As an example, for the $z$ components of the fields:

\begin{displaymath}
{\cal E}_z = - \frac{\partial\varphi}{\partial z} - \frac{...
...c{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}
\end{displaymath}

Quantum mechanics is all about the potentials. But the potentials are not unique. In particular, for any arbitrary function $\chi$ of position and time, you can find two different potentials $\varphi'$ and $\skew3\vec A'$ that produce the exact same electric and magnetic fields as $\varphi$ and $\skew3\vec A$. These potentials are given by

\begin{displaymath}
\varphi' = \varphi - \frac{\partial\chi}{\partial t}
\qquad
\skew3\vec A' = \skew3\vec A+ \nabla \chi %
\end{displaymath} (A.154)

(To check that the fields for these potentials are indeed the same, note that $\nabla$ $\times$ $\nabla\chi$ is zero for any function $\chi$.) This indeterminacy in potentials is the famous gauge property of the electromagnetic field. The arbitrary function $\chi$ is the gauge function.

Classical relativistic mechanics likes to combine the four scalar potentials in a four-di­men­sion­al vector, or four-vector, chapter 1.3.2:

\begin{displaymath}
{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt...
...\begin{array}{c}\varphi/c\\ A_x\\ A_y\\ A_z\end{array}\right)
\end{displaymath}


A.23.2 Photon wave functions

The wave function of photons was discussed in addendum {A.21}. A summary of the key results will be given here.

Superficially, the photon wave function ${\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}_{\kern-1pt\gamma}$ takes the same form as an electromagnetic potential four-vector like in the previous subsection. However, where classical potentials are real, the photon wave function is in general complex. And unlike physical potentials, the derivatives of the photon wave function are not physically observable.

Furthermore, to use the photon wave function in an reasonably efficient manner, it is essential to simplify it. The gauge property of the previous subsection implies that the wave function is not unique. So among all the possible alternatives, it is smart to select the simplest. And in empty space, as discussed here, the simplest photon wave function is of the form:

\begin{displaymath}
{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt...
...array}\right)
\qquad \nabla \cdot \skew3\vec A_\gamma = 0 %
\end{displaymath}

The gauge function corresponding to this wave function is called the Coulomb-Lorenz gauge. Seen from a moving coordinate system, this form of the wave function gets messed up, so do not do that.

The real interest is in quantum states of definite energy. Now for a nonrelativistic particle, wave functions with definite energy must be eigenfunctions of the so-called Hamiltonian eigenvalue problem. That eigenvalue problem is also known as the time-independent Schrö­din­ger equation. However, a relativistic particle of zero rest mass like the photon must satisfy a different eigenvalue problem, {A.21}:

\begin{displaymath}
- \nabla^2 \skew3\vec A_\gamma^{\rm {e}} = k^2 \skew3\vec ...
...\frac{p}{\hbar}
\qquad E = \hbar \omega \quad p = \hbar k %
\end{displaymath} (A.155)

Here $\skew3\vec A_\gamma^{\rm {e}}$ is the energy eigenfunction. Further $p$ is the magnitude of the linear momentum of the photon. The second-last equation is the so-called Planck-Einstein relation that gives the photon energy $E$ in terms of its frequency $\omega$, while the last equation is the de Broglie relation that gives the photon momentum $p$ in terms of its wave number $k$. The relation between frequency and wave number is $\omega$ $\vphantom0\raisebox{1.5pt}{$=$}$ $kc$ with $c$ the speed of light.

The simplest energy eigenfunctions are those that have definite linear momentum. A typical example of such an energy eigenfunction is

\begin{displaymath}
\skew3\vec A_\gamma^{\rm {e}} = {\hat k}e^{{\rm i}k y}
\end{displaymath} (A.156)

This wave function has definite linear momentum ${\hat\jmath}{\hbar}k$, as you can see from applying the components of the linear momentum operator $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$ on it. And it is an energy eigenfunction, as you can verify by substitution in (A.155). Further, since the wave function vector is in the $z$-​direction, it is called linearly polarized in the $z$-​direction.

Now the photon wave function ${\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}_{\kern-1pt\gamma}$ is not a classical electromagnetic potential. The “electric and magnetic fields” $\skew3\vec{\cal E}_\gamma$ and $\skew2\vec{\cal B}_\gamma$ that you would find by using the classical expressions (A.153) are not physically observable quantities. So they do not have to obey the classical expression (A.152) for the energy in an electromagnetic field.

However, you make life a lot simpler for yourself if you normalize the photon wave functions so that they do satisfy it. That produces a normalized wave function and corresponding unobservable fields of the form, {A.21}:

\begin{displaymath}
\skew3\vec A_\gamma^{\rm n}= \frac{\varepsilon_k}{{\rm i}k...
...on_k}{{\rm i}k}
\nabla\times\skew3\vec A_\gamma^{\rm {e}} %
\end{displaymath} (A.157)

where the constant $\varepsilon_k$ is found by substitution into the normalization condition:
\begin{displaymath}
{\textstyle\frac{1}{2}} \epsilon_0 \int_{\rm all}
\left(...
...t\vert^2
\right){\,\rm d}^3{\skew0\vec r}
= \hbar\omega %
\end{displaymath} (A.158)

Consider how this works out for the example eigenfunction of definite linear momentum mentioned above. That eigenfunction cannot be normalized in infinite space since it does not go to zero at large distances. To normalize it, you have to assume that the electromagnetic field is confined to a big periodic box of volume ${\cal V}$. In that case the normalized eigenfunction and unobservable fields become:

\begin{displaymath}
\skew3\vec A_\gamma^{\rm n}= \frac{\varepsilon_k}{{\rm i}k...
...epsilon_k
= \sqrt{\frac{\hbar\omega}{\epsilon_0{\cal V}}} %
\end{displaymath} (A.159)

Another interesting example is given in {A.21.6}. It is a photon state with definite angular momentum $\hbar$ in the direction of motion. Such a photon state is called right-circularly polarized. In this example the linear momentum is taken to be in the $z$-​direction, rather than the $y$-​direction. The normalized wave function and unobservable fields are:

\begin{displaymath}
\skew3\vec A_\gamma^{\rm n}= \frac{\varepsilon_k}{{\rm i}k...
...epsilon_k
= \sqrt{\frac{\hbar\omega}{\epsilon_0{\cal V}}} %
\end{displaymath} (A.160)

It will be interesting to see what the observable fields for this photon state look like.


A.23.3 The electromagnetic operators

The previous subsection has identified the form of the wave function of a photon in an energy eigenstate. The next step is to identify the Hamiltonian operators of the observable electric and magnetic fields.

But first there is a problem. If you have exactly one photon, the wave functions as discussed in the previous subsection would do just fine. If you had exactly two photons, you could readily write a wave function for them too. But a state with an exact number $i$ of photons is an energy eigenstate. It has energy $i\hbar\omega$, taking the zero of energy as the state of no photons. Energy eigenstates are stationary, chapter 7.1.4. They never change. All the interesting mechanics in nature is due to uncertainty in energy. As far as photons are concerned, that requires uncertainty in the number of photons. And there is no way to write a wave function for an uncertain number of particles in classical quantum mechanics. The mathematical machinery is simply not up to it.

The mathematics of quantum field theory is needed, as discussed in addendum {A.15}. The key concepts will be briefly summarized here. The mathematics starts with Fock space kets. Consider a single energy eigenfunction for photons, like one of the examples given in the previous subsection. The Fock space ket

\begin{displaymath}
\big\vert i\big\rangle
\end{displaymath}

indicates the wave function if there are exactly $i$ photons in the considered energy eigenfunction. The number $i$ is called the occupation number of the state. (If more than one photon state is of interest, an occupation number is added to the ket for each state. However, the discussion here will stay restricted to a single state.)

The Fock space ket formalism allows wave functions to be written for any number of particles in the state. And by taking linear combinations of kets with different occupation numbers, uncertainty in the number of photons can be described. So uncertainty in energy can be described.

Kets are taken to be orthonormal. The inner product $\big\langle i_1\big\vert i_2\big\rangle $ of two kets with different occupation numbers $i_1$ and $i_2$ is zero. The inner product of a ket with itself is taken to be one. That is exactly the same as for energy eigenfunctions in classical quantum mechanics,

Next, it turns out that operators that act on photon wave functions are intrinsically linked to operators that annihilate and create photons. Mathematically, at least. These operators are defined by the relations

\begin{displaymath}
\widehat a\big\vert i\big\rangle = \sqrt{i} \big\vert i{-}...
...big\vert i{-}1\big\rangle = \sqrt{i} \big\vert i\big\rangle %
\end{displaymath} (A.161)

for any number of photons $i$. In words, the annihilation operator $\widehat a$ takes a state of $i$ photons and turns it into a state with one less photon. The creation operator $\widehat a^\dagger $ puts the photon back in. The scalar factors $\sqrt{i}$ are a matter of convenience. If you did not put them in here, you would have to do it elsewhere.

At first it may seem just weird that there are physical operators like that. But a bit more thought may make it more plausible. First of all, nature pretty much forces the Fock space kets on us. Classical quantum mechanics would like to number particles such as photons just like classical physics likes to do: photon 1, photon 2, ... But nature makes a farce out of that with the symmetrization requirement. It allows absolutely no difference in the way one photon occupies a state compared to another one. Indeed, nature goes to such a length of preventing us to, God forbid, make a distinction between one photon and another that she puts every single photon in the universe partly in every single microscopic photon state on earth. Now Fock space kets are the only way to express how many photons are in a given state without saying which photons. And if the symbols of nature are then apparently Fock space kets, the operators of nature are pretty unavoidably annihilation and creation operators. There is not much else you can do with Fock space kets than change the number of particles.

The annihilation and creation operators are not Hermitian. They cannot be taken unchanged to the other side of an inner product of kets. However, they are Hermitian conjugates: they change into each other when taken to the other side of an inner product:

\begin{displaymath}
\Big\langle \big\vert i_2\big\rangle \Big\vert\widehat a\b...
...gle i_1\big\vert \widehat a^\dagger \big\vert i_2\big\rangle
\end{displaymath}

To see this, note that the inner products are only nonzero if $i_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i_1-1$ because of the orthogonality of kets with different numbers of photons. And if $i_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i_1-1$, then (A.161) shows that all four inner products above equal $\sqrt{i_1}$.

That is important because it shows that Hermitian operators can be formed from combinations of the two operators. For example, $\widehat a+\widehat a^\dagger $ is a Hermitian operator. Each of the two operators changes into the other when taken to the other side of an inner product. So the sum stays unchanged. More generally, if $c$ is any real or complex number, $\widehat a{c}+\widehat a^\dagger {c}^*$ is Hermitian.

And that then is the basic recipe for finding the operators of the observable electric and magnetic fields. Take $\widehat a$ times the unobservable field of the normalized photon state, (A.157) with (A.158) and (A.155). Add the Hermitian conjugate of that. And put in the usual 1$\raisebox{.5pt}{$/$}$$\sqrt{2}$ factor of quantum mechanics for averaging states. In total

\begin{displaymath}
\fbox{$\displaystyle
\skew6\widehat{\skew3\vec{\cal E}}=...
...t a^\dagger \skew2\vec{\cal B}_\gamma^{\rm{n}*}\right)
$} %
\end{displaymath} (A.162)

You might wonder why there are two terms in the operators, one with a complex conjugate wave function. Mathematically that is definitely needed to get Hermitian operators. That in turn is needed to get real observed fields. But what does it mean physically? One way of thinking about it is that the observed field is real because it does not just involve an interaction with an $e^{{\rm i}({\vec k}\cdot{\skew0\vec r}-{\omega}t)}$ photon, but also with an $e^{-{\rm i}({\vec k}\cdot{\skew0\vec r}-{\omega}t)}$ antiphoton.

Of course, just because the above operators are Hermitian does not prove that they are the right ones for the observable electric and magnetic fields. Unfortunately, there is no straightforward way to deduce quantum mechanics operators from mere knowledge of the classical approximation. Vice-versa is not a problem: given the operators, it is fairly straightforward to deduce the corresponding classical equations for a macroscopic system. It is much like at the start of this book, where it was postulated that the momentum of a particle corresponds to the operator $\hbar\partial$$\raisebox{.5pt}{$/$}$${\rm i}\partial{x}$. That was a leap of faith. However, it was eventually seen that it did produce the correct classical momentum for macroscopic systems, chapter 7.2.1 and 7.10, as well as correct quantum results like the energy levels of the hydrogen atom, chapter 4.3. A similar leap of faith will be needed to quantize the electromagnetic field.


A.23.4 Properties of the observable electromagnetic field

The previous subsection postulated the operators (A.162) for the observable electric and magnetic fields. This subsection will examine the consequences of these operators, in order to gain confidence in them. And to learn something about the effects of quantization of the electromagnetic field.

Consider first a simple wave function where there are exactly $i$ photons in the considered photon state. In terms of Fock space kets, the wave function is then:

\begin{displaymath}
\Psi = c_i e^{-{\rm i}i\omega t} \big\vert i\big\rangle
\end{displaymath}

where $c_i$ is a constant with magnitude 1. This follows the Schrö­din­ger rule that the time dependence in a wave function of definite energy $E$ is given by $e^{-{{\rm i}}Et/\hbar}$, with in this case $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i\hbar\omega$.

The expectation value of the electric field at a given position and time is then by definition

\begin{displaymath}
\langle \skew3\vec{\cal E}\rangle =
\Big\langle\Psi\Big\...
...*}\right)
c_i e^{-{\rm i}i\omega t} \big\vert i\big\rangle
\end{displaymath}

That is zero because $\widehat a$ and $\widehat a^\dagger $ turn the right hand ket into $\big\vert i{-}1\big\rangle $ respectively $\big\vert i{+}1\big\rangle $. These are orthogonal to the left hand $\big\langle i\big\vert$. The same way, the expectation magnetic field will be zero too.

Oops.

Zero electric and magnetic fields were not exactly expected if there is a nonzero number of photons present.

No panic please. This is an energy eigenstate. Often these do not resemble classical physics at all. Think of a hydrogen atom in its ground state. The expectation value of the linear momentum of the electron is zero in that state. That is just like the electric and magnetic fields are zero here. But the expectation value of the square momentum of the hydrogen electron is not zero. In fact, that gives the nonzero expectation value of the kinetic energy of the electron, 13.6 eV. So maybe the square fields need to be examined here.

Come to think of it, the first thing to check should obviously have been the energy. It better be $i\hbar\omega$ for $i$ photons. Following the Newtonian analogy for the classical energy integral (A.152), the Hamiltonian should be

\begin{displaymath}
H = {\textstyle\frac{1}{2}} \epsilon_0 \int_{\rm all} \lef...
...widehat{\skew2\vec{\cal B}}^2\right){\,\rm d}^3{\skew0\vec r}
\end{displaymath}

This can be greatly simplified by plugging in the given expressions for the operators and identifying the integrals, {D.40}:
\begin{displaymath}
\fbox{$\displaystyle
H = {\textstyle\frac{1}{2}}\hbar\om...
...t a^\dagger \widehat a+ \widehat a\widehat a^\dagger )
$} %
\end{displaymath} (A.163)

Now apply this Hamiltonian on a state with $i$ photons. First, using the definitions (A.161) of the annihilation and creation operators

\begin{displaymath}
\widehat a^\dagger \widehat a\big\vert i\big\rangle = \wid...
...1}\big\vert i{+}1\big\rangle ) = (i+1)\big\vert i\big\rangle
\end{displaymath}

This shows that the operators $\widehat a$ and $\widehat a^\dagger $ do not commute; their order makes a difference. In particular, according to the above their commutator equals
\begin{displaymath}[\widehat a,\widehat a^\dagger ]\equiv \widehat a\widehat a^\dagger - \widehat a^\dagger \widehat a= 1 %
\end{displaymath} (A.164)

Anyway, using the above relations the expression for the Hamiltonian applied on a Fock space ket becomes

\begin{displaymath}
H \big\vert i\big\rangle = \hbar\omega(i+{\textstyle\frac{1}{2}}) \big\vert i\big\rangle
\end{displaymath}

The factor in front of the final ket is the energy eigenvalue. It is more or less like expected, which was $i\hbar\omega$ for $i$ photons. But there is another one-half photon worth of energy.

That, however, may be correct. It closely resembles what happened for the harmonic oscillator, chapter 4.1. Apparently the energy in the electromagnetic field is never zero, just like a harmonic oscillator is never at rest. The energy increases by $\hbar\omega$ for each additional photon as it should.

Actually, the half photon vacuum energy is somewhat of a problem. If you start summing these half photons over all infinitely many frequencies, you end up, of course, with infinity. Now the ground state energy does not affect the dynamics. But if you do measurements of the electric or magnetic fields in vacuum, you will get nonzero values. So apparently there is real energy there. Presumably that should affect gravity. Maybe the effect would not be infinite, if you cut off the sum at frequencies at which quantum mechanics might fail, but it should certainly be extremely dramatic. So why is it not observed? The answer is unknown. See chapter 8.7 for one suggestion.

The vacuum energy also has consequences if you place two conducting plates extremely closely together. The conducting plates restrict the vacuum field between the plates. (Or at least the relatively low energy part of it. Beyond say the X-ray range photons will not notice the plates.) Because of the restriction of the plates, you would expect the vacuum energy to be less than expected. Because of energy conservation, that must mean that there is an attractive force between the plates. That is the so-called “Casimir force.” This weird force has actually been measured experimentally. Once again it is seen that the half photon of vacuum energy in each state is not just a mathematical artifact.

Because of the infinite energy, some authors describe the vacuum as a “seething cauldron” of electromagnetic waves. These authors may not be aware that the vacuum state, being a ground state, is stationary. Or they may not have access to a dictionary of the English language.

The next test of the field operators is to reconsider the expectation electric field when there is uncertainty in energy. Also remember to add another half photon of energy now. Then the general wave function takes the form:

\begin{displaymath}
\Psi = \sum_i c_i e^{-{\rm i}(i+\frac12)\omega t} \big\vert i\big\rangle
\end{displaymath}

The expectation value of the electric field follows as

\begin{displaymath}
\langle \skew3\vec{\cal E}\rangle = \sum_i \sum_j \frac{1}...
...
c_j e^{-{\rm i}(j+\frac12)\omega t} \big\vert j\big\rangle
\end{displaymath}

Using the definitions of the annihilation and creation operators and the orthonormality of the kets, this can be worked out further to
\begin{displaymath}
\langle \skew3\vec{\cal E}\rangle
= Ce^{-{\rm i}\omega t...
...
C \equiv \frac{1}{\sqrt{2}} \sum_i c_{i-1}^* c_i \sqrt{i} %
\end{displaymath} (A.165)

Well, the field is no longer zero. Note that the first term in the electric field is more or less what you would expect from the unobservable field of a single photon. But the observable field adds the complex conjugate. That makes the observable field real.

The properties of the observable fields can now be determined. For example, consider the photon wave function (A.159) given earlier. This wave function had its linear momentum in the $y$-​direction. It was “linearly polarized” in the $z$-​direction. According to the above expression, the observable electric field is:

\begin{displaymath}
\langle \skew3\vec{\cal E}\rangle = {\hat k}\varepsilon_k
...
...{{\rm i}(ky-\omega t)} + C^* e^{-{\rm i}(ky-\omega t)}\right)
\end{displaymath}

The first term is roughly what you would want to write down for the unobservable electric field of a single photon. The second term, however, is the complex conjugate of that. It makes the observable field real. Writing $C$ in the form $\vert C\vert e^{{\rm i}\alpha}$ and using the Euler formula 2.5 to clean up gives:

\begin{displaymath}
\langle \skew3\vec{\cal E}\rangle
= {\hat k}2 \varepsilon_k \vert C\vert \cos(ky-\omega t + \alpha)
\end{displaymath}

That is a real electromagnetic wave. It is still polarized in the $z$-​direction, and it travels in the $y$-​direction.

The corresponding magnetic field goes exactly the same way. The only difference in (A.159) is that ${\hat k}$ gets replaced by ${\hat\imath}$. Therefore

\begin{displaymath}
\langle c \skew2\vec{\cal B}\rangle
= {\hat\imath}2 \varepsilon_k \vert C\vert \cos(ky-\omega t + \alpha)
\end{displaymath}

Note that like for the photon wave function, the observable fields are normal to the direction of wave propagation, and to each other.

As another example, consider the “circularly polarized” photon wave function (A.160). This wave function had its linear momentum in the $z$-​direction, and it had definite angular momentum $\hbar$ around the $z$-​axis. Here the observable fields are found to be

\begin{eqnarray*}
& \langle \skew3\vec{\cal E}\rangle = \sqrt{2} \varepsilon_k...
...-\omega t +\alpha)+{\hat\jmath}\cos(kz-\omega t +\alpha)\right]
\end{eqnarray*}

Like for linearly polarized light, the electric and magnetic fields are normal to the direction of wave propagation and to each other. But here the electric and magnetic field vectors rotate around in a circle when seen at a fixed position $z$. Seen at a fixed time, the end points of the electric vectors that start from the $z$-​axis form a helix. And so do the magnetic ones.

The final question is under what conditions you would get a classical electromagnetic field with relatively little quantum uncertainty. To answer that, first note that the square quantum uncertainty is given by

\begin{displaymath}
\sigma_\skew3\vec{\cal E}^2 = \langle\skew3\vec{\cal E}^2\rangle - \langle\skew3\vec{\cal E}\rangle^2
\end{displaymath}

(This is the square of chapter 4.4.3 (4.44) multiplied out and identified.)

To evaluate this uncertainty requires the expectation value of the square electric field. That can be found much like the expectation value (A.165) of the electric field itself. The answer is

\begin{displaymath}
\langle \skew3\vec{\cal E}^2 \rangle
= 2 D_0 \vert\skew3...
... e^{2{\rm i}\omega t}(\skew3\vec{\cal E}_\gamma^{\rm {n}*})^2
\end{displaymath}

where

\begin{displaymath}
D_0 \equiv \frac{1}{2} \sum_i \vert c_i\vert^2 (i+{\textst...
...quiv \frac{1}{2} \sum_i c_{i-1}^* c_{i+1} \sqrt{i} \sqrt{i+1}
\end{displaymath}

Note that when this is substituted into the integral (A.152) for the energy, the $D_0$ term gives half the expectation value of the energy. In particular, the coefficient $D_0$ itself is half the expectation value of the $i+\frac12$ number of photons of energy. The other half comes from the corresponding term in the magnetic field. The $D_1$ terms above integrate away against the corresponding terms in the magnetic field, {D.40}.

To determine the uncertainty in the electric field, it is convenient to write the expectation square electric field above in real form. To do so, the coefficient $D_1$ is written in the form $\vert D_1\vert e^{2{\rm i}\beta}$. Also, the square unobservable electric field $(\skew3\vec{\cal E}_\gamma^{\rm {n}})^2$ is written in the form $\vert\skew3\vec{\cal E}_\gamma^{\rm {n}}\vert^2e^{2{\rm i}\gamma}$. Here $\gamma$ will normally depend on position; for example $\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ $ky$ for the given example of linearly polarized light.

Then the expectation square electric field becomes, using the Euler formula (2.5) and some trig,

\begin{displaymath}
\langle \skew3\vec{\cal E}^2 \rangle
= 2 (D_0-\vert D_1\...
...al E}_\gamma^{\rm {n}}\vert^2 \cos^2(\gamma-\omega t + \beta)
\end{displaymath}

with $D_0$ and $D_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert D_1\vert e^{2{\rm i}\beta}$ as given above. Similarly the square of the expectation electric field, as given earlier in (A.165), can be written as

\begin{displaymath}
\langle \skew3\vec{\cal E}\rangle^2
= 4 \vert C\vert^2 \...
... i}\alpha} = \frac{1}{\sqrt{2}} \sum_i c_{i-1}^* c_i \sqrt{i}
\end{displaymath}

For a field without quantum uncertainty, $\langle\skew3\vec{\cal E}^2\rangle$ and $\langle\skew3\vec{\cal E}\rangle^2$ as given above must be equal. Note that first of all this requires that $\vert D_1\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $D_0$, because otherwise $\langle\skew3\vec{\cal E}^2\rangle$ does not become zero periodically like $\langle\skew3\vec{\cal E}\rangle^2$ does. Also $\beta$ will have to be $\alpha$, up to a whole multiple of $\pi$, otherwise the zeros are not at the same times. Finally, $\vert C\vert$ will have to be equal to $D_0$ too, or the amplitudes will not be the same.

However, regardless of uncertainty, the coefficients must always satisfy

\begin{displaymath}
\vert D_1\vert \mathrel{\raisebox{-.7pt}{$\leqslant$}}D_0 ...
...t}{$\leqslant$}}{\textstyle\frac{1}{2}} (D_0+\vert D_1\vert)
\end{displaymath}

The first inequality applies because otherwise $\langle\skew3\vec{\cal E}^2\rangle$ would become negative whenever the cosine is zero. The second applies because $\langle\skew3\vec{\cal E}\rangle^2$ cannot be larger than $\langle\skew3\vec{\cal E}^2\rangle$; the square uncertainty cannot be negative. For quantum certainty then, the above relations must become equalities. However, a careful analysis shows that they cannot become equalities, {D.40}.

So there is always some quantum uncertainty left. Maximum uncertainty occurs when the number of photons has a definite value. Then $D_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $C$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

If there is always at least some uncertainty, the real question is under what conditions it is relatively small. Analysis shows that the uncertainty in the fields is small under the following conditions, {D.40}:

In that case classical electric and magnetic field result with little quantum uncertainty. Note that the above conditions apply for photons restricted to a single quantum state. In a real electromagnetic field, many quantum states would be occupied and things would be much messier still.

It may also be noted that the above conditions bear a striking resemblance to the conditions that produce a particle with a fairly coherent position and momentum in classical quantum mechanics, chapter 7.10.