Subsections


A.21 Photon type 2 wave function

In quantum mechanics, photons are the particles of the electromagnetic field. To actually use photons, something like a wave function for them is needed. But that is not quite trivial for a purely relativistic particle with zero rest mass like the photon. That is the primary topic of this addendum. It will be assumed throughout that the photon is in empty space.


A.21.1 The wave function

To see the problem with a photon wave function, a review of the wave function of the nonrelativistic electron is useful, chapters 3.1 and 5.5.1. The electron wave function can be written as a vector with two components:

\begin{displaymath}
\mbox{electron:} \quad
\skew{-1}\vec\Psi({\skew0\vec r};...
... r};t) \\
\Psi_-({\skew0\vec r};t)
\end{array}
\right)
\end{displaymath}

This wave function takes on two different meanings
1.
It gives the probability per unit volume of finding the electron at a given position with a given spin. For example, $\vert\Psi_+({\skew0\vec r};t)\vert^2{\,\rm d}^3{\skew0\vec r}$ gives the probability of finding the electron with spin-up in an vicinity of infinitesimal volume ${\,\rm d}^3{\skew0\vec r}$ around position ${\skew0\vec r}$. That is the Born statistical interpretation.
2.
It is the unobservable function that nature seems to use to do its quantum computations of how physics behaves.

Now a wave function of type 1 is not really meaningful for a photon. What would it mean, find a photon? Since the photon has no rest mass, you cannot bring them to a halt: there would be nothing left. And anything you do to try to localize the electromagnetic field is likely to just produce new photons. (To be sure, with some effort something can be done towards a meaningful wave function of type 1, e.g. [Sype, J.E. 1995 Phys. Rev. A 52, 1875]. It would have two components like the electron, since the photon has two independent spin states. But wave functions of that type are not widely accepted, nor useful for the purposes here.)

So what? A wave function of type 1 is not that great anyway. For one, it only defines the magnitudes of the components of the wave function. If you only define the magnitude of a complex function, you define only half of it. True, even as a type 2 wave function the classical electron wave function is not quite unique. You can still multiply either component by a factor $e^{{\rm i}\alpha}$, with $\alpha$ a real constant, without changing any of the physics. But that is not by far as bad as completely ignoring everything else besides the magnitude.

Furthermore, relativistic quantum mechanics has discovered that what we call an electron is something cloaked in a cloud of virtual particles. It is anybody’s guess what is inside that cloak, but it will not be anything resembling what we would call an electron. So what does it really mean, finding an electron within an infinitesimal volume around a point? What happens to that cloak? And to really locate an electron in an infinitesimal volume requires infinite energy. If you try to locate the electron in a region that is small enough, you are likely to just create additional electron-positron pairs much like for photons.

For most practical purposes, classical physics understands the particle behavior of electrons very well, but not their wave behavior. Conversely, it understands the wave behavior of photons very well, but not their particle behavior. But when you go to high enough energies, that distinction becomes much less obvious.

The photon most definitely has a wave function of type 2 above. In quantum electrodynamics, it may simply be called the photon wave function, [24, p. 240]. However, since the term already seems to be used for type 1 wave functions, this book will use the term “photon type 2 wave function.” It may not tell you where to find that elusive photon, but you will definitely need it to figure out how that photon interacts with, say, an electron.

What the type 2 wave function of the photon is can be guessed readily from classical electromagnetics. After all, the photon is supposed to be the particle of the electromagnetic field. So, consider first electrostatics. In classical electrostatics the forces on charged particles are described by an electric force per unit charge $\skew3\vec{\cal E}$. That is called the electric field.

But quantum mechanics uses potentials, not forces. For example, the solution of the hydrogen atom of chapter 4.3 used a potential energy of the electron $V$. In electrostatics, this potential energy is written as $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-e\varphi$ where $\vphantom0\raisebox{1.5pt}{$-$}$$e$ is the charge of the electron and $\varphi$ is called the electrostatic potential. This potential is not directly observable nor unique; you can add any constant to it without changing the observed physics.

Clearly, an unobservable function $\varphi$ for the electromagnetic field sounds much like a wave function for the particle of that field, the photon. But actually, the electrostatic potential $\varphi$ is only part of it. In classical electromagnetics, there is not just an electric field $\skew3\vec{\cal E}$, there is also a magnetic field $\skew2\vec{\cal B}$. It is known that this magnetic field can be represented by a so-called vector potential $\skew3\vec A$.

The following relationships give the electric and magnetic fields in terms of these potentials:

\begin{displaymath}
\skew3\vec{\cal E}= - \nabla \varphi - \frac{\partial \ske...
...}
\qquad
\skew2\vec{\cal B}= \nabla \times \skew3\vec A %
\end{displaymath} (A.86)

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}

When both potentials are allowed for, the nonuniqueness becomes much larger. 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.87)

This indeterminacy in potentials is the famous “gauge property” of the electromagnetic field.

Finally, it turns out that 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}

That is the one. Quantum mechanics takes a four-vector potential of this form to be the type 2 wave function of the photon ${\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}_{\kern-1pt\gamma}$. It keeps the gauge property (A.87) for this wave function. However, note the following important caveat:

The photon wave function ${\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}_{\kern-1pt\gamma}$ should not be confused with the classical four-potential ${\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}$.
Wave functions are in general complex. The classical four-potential, and especially its physically observable derivatives, the electric and magnetic fields, must be real. Indeed, according to quantum mechanics, observable quantities correspond to eigenvalues of Hermitian operators, not to wave functions. What the operators of the observable electric and magnetic fields are will be discussed in addendum {A.23}.


A.21.2 Simplifying the wave function

To use the photon wave function in practical applications, it is essential to simplify it. That can be done by choosing a clever gauge function $\chi$ in the gauge property (A.87).

One very helpful simplification is to choose $\chi$ so that

\begin{displaymath}
\frac{1}{c} \frac{\partial \varphi_\gamma/c}{\partial t}
+ \nabla \cdot \skew3\vec A_\gamma = 0 %
\end{displaymath} (A.88)

where $c$ is the speed of light. This is called the “Lorenz condition.” A corresponding gauge function is a “Lorenz gauge.” The reason why the Lorenz condition is a good one is because all observers in inertial motion will agree it is true. (You can crunch that out using the Lorentz transform as given in chapter 1.2.1 (1.6). The four-vector ${\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}_{\kern-1pt\gamma}$ transforms the same way as the four-vector $\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over r}
\kern-1.3pt$. However, you will need to use the inverse transform for one of the two four-vectors. Alternatively, those familiar with index notation as briefly described in chapter 1.2.5 recognize the Lorenz condition as being simply $\partial_{\mu}A_\gamma^\mu$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That is unchanged going from one observer to the next, because the upper index transforms under the Lorentz transform and the lower index under the inverse Lorentz transform.)

To achieve the Lorenz condition, assume an initial wave function $(\varphi_\gamma',\skew3\vec A_\gamma')$ that does not satisfy it. Then plug the gauge property (A.87) into the Lorenz condition above. That shows that the needed gauge function $\chi$ must satisfy

\begin{displaymath}
- \frac{1}{c^2} \frac{\partial^2\chi}{\partial t^2} + \nab...
...hi_\gamma'/c}{\partial t}
+ \nabla\cdot\skew3\vec A_\gamma'
\end{displaymath}

This equation for $\chi$ is called an inhomogeneous Klein-Gordon equation. (More generically, it is called an inhomogeneous wave equation.)

There is another reason why you want to satisfy the Lorenz condition. The photon is a purely relativistic particle with zero rest mass. Then following the usual ideas of quantum mechanics, in empty space its wave function should satisfy the homogeneous Klein-Gordon equation, {A.14} (A.43):

\begin{displaymath}
\fbox{$\displaystyle
- \frac{1}{c^2} \frac{\partial^2 {\...
...ightarrow$\hspace{0pt}}}\over A}_{\kern-1pt\gamma} = 0
$} %
\end{displaymath} (A.89)

Unfortunately, that is not automatic. In general, gauge transforms mess up this equation. However, as long as gauge transforms respect the Lorenz condition, they also respect the Klein-Gordon equation. So reasonably speaking, normal photon wave functions, the ones that do satisfy the Klein-Gordon equation, should be exactly the ones that also satisfy the Lorenz condition.

Maxwell’s classical electromagnetics provides additional support for that idea. There the Klein-Gordon equation for the potentials also requires that the Lorenz condition is satisfied, {A.36}.

Since the inhomogeneous Klein-Gordon equation for the gauge function $\chi$ is second order in time, it still leaves two initial conditions to be chosen. These can be chosen such as to make the initial values for $\varphi_\gamma$ and its time-derivative zero. That then makes $\varphi_\gamma$ completely zero, because it satisfies the homogeneous Klein-Gordon equation.

And so the fully simplified photon wave function becomes:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{Coulomb-Lorenz gauge:} \qquad...
...}\right)
\qquad \nabla \cdot \skew3\vec A_\gamma = 0
$} %
\end{displaymath} (A.90)

The final condition applies because of the Lorenz condition (A.88). Using an expensive word, the final condition says that $\skew3\vec A_\gamma$ must be solenoidal. A gauge function that makes $\skew3\vec A_\gamma$ solenoidal is called a “Coulomb gauge.”

It should be noted that the Coulomb gauge is not Lorentz invariant. A moving observer will not agree that the potential $\varphi_\gamma$ is zero and that $\skew3\vec A_\gamma$ is solenoidal. In real life that means that if you want to study a process in say a center-of-mass system, first switch to that system and then assume the Coulomb gauge. Not the other way around. The Coulomb-Lorenz gauge is too helpful not to use, although that is possible, [24, p. 241].


A.21.3 Photon spin

Now that the photon wave function has been simplified the photon spin can be determined. Recall that for the electron, the two components of the wave function correspond to its two possible values of the spin angular momentum ${\widehat S}_z$ in the chosen $z$-​direction. In particular, $\Psi_+$ corresponds to ${\widehat S}_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\hbar$, and $\Psi_-$ to ${\widehat S}_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\frac12\hbar$. Since the wave function of the photon is a four-di­men­sion­al vector, at first it might therefore look like the photon should have spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. That would make ${\widehat S}_z$ one of $\frac32\hbar$, $\frac12\hbar$, $-\frac12\hbar$, or $-\frac32\hbar$. But that is not true.

The simplified wave function (A.90) has only three nontrivial components. And the gauge property requires that this simplified wave function still describes all the physics. Since the only nontrivial part left is the three-di­men­sion­al vector $\skew3\vec A_\gamma$, the spin of the photon must be 1. The possible values of the spin in the $z$-​direction ${\widehat S}_z$ are $\hbar$, 0, and $\vphantom0\raisebox{1.5pt}{$-$}$$\hbar$. The photon is a vector boson like discussed in addendum {A.20}.

However, that is not quite the end of the story. There is still that additional condition $\nabla\cdot\skew3\vec A_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 to satisfy. In principle this constraint allows another component of the wave function to be eliminated. However, all three remaining components are spatial ones. So it does not make much sense to eliminate one and not the other. More importantly, it is known from relativity that $\skew3\vec A$ behaves like a normal three-di­men­sion­al vector under rotations of the coordinate system, not like a two-di­men­sion­al spinor like the electron wave function. That is implicit in the fact that the complete four-vector transforms according to the Lorentz transform, chapter 1.3.2. The spin is really 1.

Still, the additional constraint does limit the angular momentum of the photon. In particular, a photon does not have independent spin and orbital angular momentum. The two are intrinsically linked. What that means for the net angular momentum of photons is worked out in subsection A.21.7.

For now it may already be noted that the photon has no state of zero net angular momentum. A state of zero angular momentum needs to look the same from all directions. That is a consequence of the relationship between angular momentum and symmetry, chapter 7.3. Now the only vector wave functions that look the same from all directions are of the form ${\hat\imath}_rf(r)$. Here $r$ is the distance from the origin around which the angular momentum is measured and ${\hat\imath}_r$ the unit vector pointing away from the origin. Such a wave function cannot satisfy the condition $\nabla\cdot{\hat\imath}_rf(r)$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That follows from applying the divergence theorem for a sphere around the origin.


A.21.4 Energy eigenstates

Following the rules of quantum mechanics, {A.14}, photon states of definite energy $E$ take the form

\begin{displaymath}
\skew3\vec A_\gamma = c_0 \skew3\vec A_\gamma^{\rm {e}} e^{-{\rm i}E t/\hbar}
\end{displaymath}

Here $c_0$ is an arbitrary constant. More importantly $\skew3\vec A_\gamma^{\rm {e}}$ is the energy eigenfunction, which is independent of time.

Substitution in the Klein-Gordon equation and cleaning up shows that this eigenfunction needs to satisfy the eigenvalue problem, {A.14},

\begin{displaymath}
\fbox{$\displaystyle
- \nabla^2 \skew3\vec A_\gamma^{\rm...
...{p}{\hbar}
\qquad E = \hbar \omega \quad p = \hbar k
$} %
\end{displaymath} (A.91)

Here $p$ is the magnitude of the linear momentum of the photon. The so-called Planck-Einstein relation gives the energy $E$ in terms of the photon frequency $\omega$, while the de Broglie relation gives the momentum $p$ in terms of the photon wave number $k$.


A.21.5 Normalization of the wave function

A classical wave function for a particle is normalized by demanding that the square integral of the wave function is 1. That does not work for a relativistic particle like the photon, since the Klein-Gordon equation does not preserve the square integral of the wave function, {A.14}.

However, the Klein-Gordon equation does preserve the following integral, {D.36.1},

\begin{displaymath}
\int_{\rm all}
\left(
\left\vert \frac{\partial\skew3\...
...ht)
{\,\rm d}^3{\skew0\vec r}= \mbox{the same for all time}
\end{displaymath}

Reasonably speaking, you would expect this integral to be related to the energy in the electromagnetic field. After all, what other scalar physical quantity is there to be preserved?

Consider for a second the case that $\skew3\vec A_\gamma$ was a classical potential $\skew3\vec A$ instead of a photon wave function. Then the above integral can be rewritten in terms of the electric and magnetic fields $\skew3\vec{\cal E}$ and $\skew2\vec{\cal B}$ as, {D.36.1},

\begin{displaymath}
\int_{\rm all} \left(\left\vert\skew3\vec{\cal E}\right\ve...
...ht)
{\,\rm d}^3{\skew0\vec r}= \mbox{the same for all time}
\end{displaymath}

Now classical physics does not have photons of energy $\hbar\omega$. All it has are electric and magnetic fields. Then surely the integral above must be a measure for the energy in the electromagnetic field? What is more logical than that the energy in the electromagnetic field per unit volume would be given by the square magnitudes of the electric and magnetic fields? No fields, no energy.

Of course, there needs to be an additional constant; the integral above does not have units of energy. If you check, you find that the permittivity of space $\epsilon_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8.85 10$\POW9,{-12}$ C$\POW9,{2}$/J m has the right units to be the constant. Actually, it turns out that the correct constant is $\frac12\epsilon_0$. But that is not a fundamental issue; classical physics could just as well have defined $\epsilon_0$ as half of what it did.

Now the photon wave function is not physically observable and does not have to conform to the rules of classical physics. But if you have to choose a normalization constant anway? Why not choose it so that what classical physics would take to be the energy is in fact the correct energy $\hbar\omega$? It is likely to simplify your life a lot.

So, the photon wave function normalization that will be used in this book is:

\begin{displaymath}
\fbox{$\displaystyle
{\textstyle\frac{1}{2}} \epsilon_0 ...
...t^2
\right){\,\rm d}^3{\skew0\vec r}
= \hbar\omega
$} %
\end{displaymath} (A.92)

Here $\skew3\vec{\cal E}_\gamma^{\rm n}$ and $\skew2\vec{\cal B}_\gamma^{\rm n}$ are what classical physics would take to be the electric and magnetic fields for the normalized photon energy eigenfunction $\skew3\vec A_\gamma^{\rm n}$. Specifically,

\begin{displaymath}
\skew3\vec{\cal E}_\gamma^{\rm n}= {\rm i}k c \skew3\vec A...
... B}_\gamma^{\rm n}= \nabla \times \skew3\vec A_\gamma^{\rm n}
\end{displaymath}

(To be sure, classical physics would take $\skew3\vec{\cal E}$ to be minus the time derivative of the potential $\skew3\vec A$. But for an energy eigenstate, the time derivative gives a simple factor $\vphantom0\raisebox{1.5pt}{$-$}$${\rm i}\omega$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$${{\rm i}}kc$.) The functions $\skew3\vec{\cal E}_\gamma^{\rm n}$ and $\skew2\vec{\cal B}_\gamma^{\rm n}$ will be referred to as “unobservable fields” to avoid confusion with the observable electric and magnetic fields.

Assume that you start with an unnormalized energy eigenfunction $\skew3\vec A_\gamma^{\rm {e}}$. Then the normalized functions are usually most conveniently written as

\begin{displaymath}
\fbox{$\displaystyle
\skew3\vec A_\gamma^{\rm n}= \frac{...
...}{{\rm i}k}
\nabla\times\skew3\vec A_\gamma^{\rm{e}}
$} %
\end{displaymath} (A.93)

Here the constant $\varepsilon_k$ is to be found by substitution into the normalization condition (A.92).


A.21.6 States of definite linear momentum

The simplest quantum states for photons are states of definite linear momentum ${\skew0\vec p}$. And to make it even simpler, it will be assumed that the $z$-​axis is chosen in the direction of the linear momentum.

In that case, the photon wave function takes the form

\begin{displaymath}
\skew3\vec A_\gamma^{\rm {e}} = \skew3\vec A^0 e^{{\rm i}k z} \qquad {\skew0\vec p}= {\hat k}\,\hbar k
\end{displaymath}

Here $\skew3\vec A^0$ is a constant vector. That this wave function has definite linear momentum ${\skew0\vec p}$ may be verified by applying the linear momentum operator ${\skew 4\widehat{\skew{-.5}\vec p}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$ on it. And substitution into the eigenvalue problem (A.91) verifies that it is an energy eigenfunction.

The vector $\skew3\vec A^0$ is not completely arbitrary; its $z$-​component must be zero. That is in order that $\nabla\cdot\skew3\vec A_\gamma^{\rm {e}}$ is zero as the Coulomb-Lorenz gauge requires. So the wave function can be written as

\begin{displaymath}
\skew3\vec A_\gamma^{\rm {e}} = A^0_x\,{\hat\imath}e^{{\rm i}k z} + A^0_y\,{\hat\jmath}e^{{\rm i}k z}
\end{displaymath}

The bottom line is that there are only two independent states, even though the wave function is a three-di­men­sion­al vector. The wave function cannot have a component in the direction of motion. It may be noted that the first term in the right hand side above is called a wave that is “linearly polarized” in the $x$-​direction. Similarly, the second term is a wave that is linearly polarized in the $y$-​direction. There is no longitudinal polarization of photons possible.

There is another useful way to write the wave function:

\begin{displaymath}
\skew3\vec A_\gamma^{\rm {e}} = c_1 ({\hat\imath}+{\rm i}{...
...z}
+ c_2 (-{\hat\imath}+{\rm i}{\hat\jmath}) e^{{\rm i}k z}
\end{displaymath}

where $c_1$ and $c_2$ are constants. The first term in this expression is called “right-circularly polarized.” It has angular momentum $\hbar$ in the $z$-​direction. (To see why is a matter of rotating the coordinate system around the $z$-​axis, {A.20}. The exponential does not change in such a rotation.) Similarly, the second state has angular momentum $\vphantom0\raisebox{1.5pt}{$-$}$$\hbar$ in the $z$-​direction and is called left-circularly polarized. There is no state with angular momentum zero in the $z$-​direction. In fact, it is exactly the missing $z$-​component of $\skew3\vec A^0$ that would provide such a state, {A.20}.

There are still only two independent states. But another way of thinking about that is that the spin angular momentum in the direction of motion cannot be zero. The relative spin in the direction of motion, $m_s$$\raisebox{.5pt}{$/$}$$s$ is called the “helicity.” It turns out that for a particle with zero rest mass like the photon, the helicity can only be 1 (right handed) or -1 (left handed), [24, p. 65].

Note further that the angular momenta in the $x$ and $y$ directions are uncertain. It so happens that the angular momentum in the direction of motion commutes with all three components of linear momentum, chapter 4.5.4. So it can have definite values. But the $x$ and $y$ angular momenta do not commute.

For later use, it is necessary to normalize the wave function using the procedure described in the previous subsection. To do so, it must be assumed that the photon is in a periodic box of volume ${\cal V}$, like in chapter 6.17. In infinite space the wave function cannot be normalized, as it does not become zero at infinity. For the right-circularly polarized wave function as given above,

\begin{displaymath}
\fbox{$\displaystyle
\skew3\vec A_\gamma^{\rm n}= \frac{...
...ilon_k = \sqrt{\frac{\hbar\omega}{\epsilon_0{\cal V}}}
$} %
\end{displaymath} (A.94)

In order to compare to the classical electromagnetic wave in chapter 7.7.1, another example is needed. This photon wave function has its linear momentum in the $y$-​direction, and it is linearly polarized in the $z$-​direction. Then an unnormalized energy eigenfunction is

\begin{displaymath}
\skew3\vec A_\gamma^{\rm {e}} = {\hat k}e^{{\rm i}k y}
\end{displaymath}

The normalized eigenfunction and unobservable fields are in that case
\begin{displaymath}
\fbox{$\displaystyle
\skew3\vec A_\gamma^{\rm n}= \frac{...
...ilon_k = \sqrt{\frac{\hbar\omega}{\epsilon_0{\cal V}}}
$} %
\end{displaymath} (A.95)

Note that $\skew3\vec{\cal E}_\gamma^{\rm n}$, $\skew2\vec{\cal B}_\gamma^{\rm n}$, and the linear momentum are all orthogonal. That will reflect in the observable fields associated with the photon state. For the circularly polarized state, the electric and magnetic fields are not orthogonal. However, the observable fields will be.

For a general direction of the wave motion and its linear polarization, the above expession becomes

\begin{displaymath}
\fbox{$\displaystyle
\skew3\vec A_\gamma^{\rm n}= \frac{...
...ilon_k = \sqrt{\frac{\hbar\omega}{\epsilon_0{\cal V}}}
$} %
\end{displaymath} (A.96)

Here ${\vec k}$ and the unit vectors ${\hat\imath}_{\cal E}$ and ${\hat\imath}_{\cal B}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\vec k}$ $\times$ ${\hat\imath}_{\cal E}$$\raisebox{.5pt}{$/$}$$k$ are all orthogonal

For convenience, the density of states as needed for Fermi’s golden rule will be listed here. It was given earlier in chapter 6.3 (6.7) and 6.19:

\begin{displaymath}
\frac{{\rm d}N}{{\rm d}E} = \frac{\omega^2}{\hbar\pi^2c^3} {\cal V}
\end{displaymath}


A.21.7 States of definite angular momentum

It is often convenient to describe photons in terms of states of definite net angular momentum. That makes it much easier to apply angular momentum conservation in the emission of radiation by atoms or atomic nuclei. Unfortunately, angular momentum states are a bit of a mess compared to the linear momentum states of the previous subsection. Fortunately, engineers are brave.

Before diving in, it is a good idea to look first at a spinless particle. Assume that this hypothetical particle is in an energy eigenstate. Also assume that this state has square orbital angular momentum $l(l+1)\hbar^2$ where $l$ is called the azimuthal quantum number. And that the state has orbital angular momentum in the $z$-​direction $m_l\hbar$ where $m_l$ is called the magnetic quantum number. Then according to quantum mechanics, chapter 4.2.3, $l$ must be a nonnegative integer and $m_l$ must be an integer no larger in magnitude than $l$. Also, in spherical coordinates $(r,\theta,\phi)$, figure 4.7, the angular dependence of the energy eigenfunction must be given by the so-called spherical harmonic $Y_l^{m_l}(\theta,\phi)$. If in addition the particle is in empty space, the energy eigenfunction takes the general form, {A.6},

\begin{displaymath}
\psi = j_l(kr) Y_l^{m_l}(\theta,\phi)
\qquad\mbox{with}\qquad - \nabla^2 \psi = k^2 \psi
\end{displaymath}

Here $k$ is a constant related to the energy of the particle and whether it is relativistic or not, {A.14} (A.44). Further $j_l$ is the so-called “spherical Bessel function of the first kind of order $l$,” {A.6}. The parity of the eigenfunction is positive if $l$ is even and negative if $l$ is odd, {D.14}. The eigenfunction is of order $r^l$ near the origin. That is only nonzero at the origin $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 if $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That is important if there is, say, a vanishingly small atom is located at the origin. All states except $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 are virtually zero at such an atom. So the atom only has a decent chance to interact with the particle if the particle is in a state $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. End discussion of the hypothetical spinless particle.

Now the photon is a particle with spin 1. Its wave function is essentially a vector $\skew3\vec A_\gamma$. The angular momentum states and parity for such particles were discussed in {A.20}. But the photon is a special case because it must be solenoidal, it must satisfy $\nabla\cdot\skew3\vec A_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Normally, for three-di­men­sion­al vectors you expect three types of angular momentum states, like in {A.20}. But for the photon there are only two types.

The two types of photon energy eigenfunctions with definite net angular momentum are, {D.36.2} and with drums please,

\begin{displaymath}
\fbox{$\displaystyle
\skew3\vec A_\gamma^{\rm{E}}
= \n...
...}\times \nabla j_\ell(kr) Y_\ell^{m_\ell}(\theta,\phi)
$} %
\end{displaymath} (A.97)

Here $\ell$ is the azimuthal quantum number of the net photon angular momentum, orbital plus spin. And $m_\ell$ is the corresponding net magnetic quantum number.

The azimuthal quantum number $\ell$ is at least 1; the expressions above produce zero for $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. ($Y_0^0$ is just a constant and the gradient of a radial function is in the direction of ${\skew0\vec r}$.) The photon energy is related to the wave number $k$ as ${\hbar}kc$ with $c$ the speed of light, (A.91). That is really the Planck-Einstein relation, because $kc$ is the photon frequency $\omega$.

The parity of the electric multipole wave functions is negative if $\ell$ is odd and positive if $\ell$ is even, {D.36.2.7}. The parity of the magnetic multipole wave functions is exactly the other way around. From that it can be seen, {D.36.2.8}, that magnetic multipole wave functions have orbital angular momentum $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell$. The electric ones have uncertainty in orbital angular momentum, with nonzero probabilities for both $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell-1$ and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell+1$.

Atomic or nuclear transitions in which a photon in a state $\skew3\vec A_\gamma^{\rm {E}}$ is emitted or absorbed are called “electric multipole” transitions. They are indicated as E$\ell$ transitions.

In particular, for net angular momentum $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, they are called ${\rm {E1}}$ or electric dipole transitions. That is the normal kind. However, as discussed in chapter 7.4, such transitions may not be able to satisfy conservation of angular momentum and parity. Since the photon in the state has $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, transitions in which the atomic angular momentum changes by more than one unit cannot be accomodated. Neither can transitions in which the atomic or nuclear momentum parity does not change, because the ${\rm {E1}}$ photon has odd parity.

Such transitions may be accomodated by transitions in which photons in different states are emitted or absorbed, using the photon angular momenta and parities as noted above. Electric multipole transitions with $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 are called ${\rm {E2}}$ or electric quadrupole transitions. Those with $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 are ${\rm {E3}}$ or electric octupole ones, with $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4 ${\rm {E4}}$ or electric hexadecapole ones, with $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 5 ${\rm {E5}}$ or electric triakontadipole ones, for $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 6 ${\rm {E6}}$ or electric hexacontatetrapole ones and so on until your knowledge of latin and greek powers of 2 runs out.

Similarly, transitions in which photons in a state $\skew3\vec A_\gamma^{\rm {M}}$ are emitted or absorbed are called “magnetic multipole transitions.” The same latin applies.

Like the states of definite linear momentum in the previous subsection, the states of definite angular momentum cannot be normalized in infinite space. To deal with that, it will be assumed that the photon is confined inside a sphere of a very large radius $r_{\rm {max}}$. As a boundary condition on the sphere, it will be assumed that the Bessel function is zero. In terms of the wave functions, that works out to mean that the magnetic ones are zero on the sphere, but only the radial component of the electric ones is.

The normalized wave function and unobservable fields for electric multipole photons are then, subsection A.21.5 and {D.36},

\begin{displaymath}
\fbox{$\displaystyle
\skew3\vec A_\gamma^{\rm En}
= \f...
...rac{2\hbar\omega}{\ell(\ell+1)\epsilon_0r_{\rm{max}}}}
$} %
\end{displaymath} (A.98)

(The expression for the magnetic field arises because for a solenoidal vector $\nabla$ $\times$ $\nabla\times$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$\nabla^2$, and that produces a factor $k^2$ according to the energy eigenvalue problem.)

The normalized wave function and unobservable fields for magnetic multipole photons are

\begin{displaymath}
\fbox{$\displaystyle
\skew3\vec A_\gamma^{\rm Mn}
= \f...
...rac{2\hbar\omega}{\ell(\ell+1)\epsilon_0r_{\rm{max}}}}
$} %
\end{displaymath} (A.99)

Assume now that there is an atom or atomic nucleus at the origin that interacts with the photon. An atom or nucleus is typically very small compared to the wave length of the photon that it interacts with. Phrased more appropriately, if $R$ is the typical size of the atom or nucleus, then $kR$ is typically small. The atom or nucleus is just a tiny speck at the origin.

Now the wave functions $\skew3\vec A_\gamma^{\rm {E}}$ are larger at small radii than the $\skew3\vec A_\gamma^{\rm {M}}$. In particular, the $\skew3\vec A_\gamma^{\rm {E}}$ are of order $r^{\ell-1}$ while the $\skew3\vec A_\gamma^{\rm {M}}$ are of order $r^\ell$, one power of $r$ smaller. These powers of $r$ reflect the lowest measurable orbital angular momentum of the states.

A glance at the unobservable fields of electric multipole photons above then shows that for these photons, the field is primarily electric at the atom or nucleus. And even the electric field will be small unless $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, in other words, unless it is an electric dipole photon.

For the magnetic multipole photons, it is the magnetic field that dominates at the atom or nucleus. And even that will be small unless $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, which means a magnetic dipole photon. Note that the magnetic field acts as if it had one unit or orbital angular momentum less than the photon; the magnetic field is essentially the wave function of an electric multipole photon.

For later reference, the density of states as needed for Fermi’s golden rule will be listed here, {D.36.2.6}:

\begin{displaymath}
\fbox{$\displaystyle
\frac{{\rm d}N}{{\rm d}E} \approx \frac{1}{\hbar\pi c} r_{\rm{max}}
$} %
\end{displaymath} (A.100)

This approximation applies for large cut-off radius $r_{\rm {max}}$, which should always be valid.