Subsections


A.22 Forces by particle exchange

As noted in chapter 7.5.2, the fundamental forces of nature arise from the exchange of particles. This addendum will illustrate the general idea. It will first derive the hypothetical Koulomb force due to the exchange of equally hypothetical particles called fotons.

The Koulomb potential provides a fairly simple model of a quantum field. It also provides a simple context to introduce some key concepts in quantum field theories, such as Green’s functions, variational calculus, Lagrangians, the limitation of the speed of light, description in terms of momentum modes, Fock space kets, annihilation and creation operators, antiparticles, special relativity, the imperfections of physicists, and Lorentz invariance. The Koulomb potential can also readily be modified to explain nuclear forces. However, that will have to wait until a later addendum, {A.41}.

In the current addendum, the Koulomb potential provides the starting point for a discussion of the electromagnetic field. The classical Maxwell equations for the electromagnetic field will be derived in a slightly unconventional way. Who needs to know classical electromagnetics when all it takes is quantum mechanics, relativity, and a few plausible guesses to derive electromagnetics from scratch?

To quantize the electromagnetic field is not that straightforward; it has unexpected features that do not occur for the Koulomb field. This book follows the derivation as formulated by Fermi in 1932. This derivation is the basis for more advanced modern quantum field approaches. These advanced theories will not be covered, however.

Essentially, the Fermi derivation splits off the Coulomb potential from the electromagnetic field. What is left is then readily described by a simple quantum field theory much like for the Koulomb potential. This is sufficient to handle important applications such as the emission or absorption of radiation by atoms and atomic nuclei. That, however, will again be done in subsequent addenda.

A word to the wise. While this addendum is on the calculus level like virtually everything else in this book, there is just quite a lot of mathematics. Some mathematical maturity may be needed not to get lost. Note that this addendum is not needed to understand the discussion of the emission and absorption of radiation in the subsequent addenda.


A.22.1 Classical selectostatics

The Koulomb force holds the sarged spotons and selectons together in satoms. The force is due to the exchange of massless particles called fotons between the sarged particles. (It will be assumed that the spoton is an elementary particle, though really it consists of three skarks.)

This subsection will derive the selectostatic Koulomb force by representing the fotons by a classical field, not a quantum field. The next subsection will explain classical selectodynamics, and how it obeys the speed of light. Subsection A.22.3 will eventually fully quantize the selectic field. It will show how quantum effects modify some of the physics expected from the classical analysis.

Physicists have some trouble measuring the precise properties of the selectic field. However, a few basic quantum ideas and some reasonable guesses readily substitute for the lack of empirical data. And guessing is good. If you can guess a self-consistent Koulomb field, you have a lot of insight into its nature.

Consider first the wave function for the exchanged foton in isolation. A foton is a boson without spin. That means that its wave function is a simple function, not some vector. But since the foton is massless, the Schrö­din­ger equation does not apply to it. The appropriate equation follows from the relativistic expression for the energy of a massless particle as given by Einstein, chapter 1.1.2 (1.2):

\begin{displaymath}
E^2= {\skew0\vec p}^{\,2} c^2
\end{displaymath}

Here $E$ is the foton energy, ${\skew0\vec p}$ its linear momentum, and $c$ the speed of light. The squares are used because momentum is really a vector, not a number like energy.

Quantum mechanics replaces the momentum vector by the operator

\begin{displaymath}
{\skew 4\widehat{\skew{-.5}\vec p}}= \frac{\hbar}{{\rm i}}...
...c{\partial}{\partial y} + {\hat k}\frac{\partial}{\partial z}
\end{displaymath}

Note the vector operator $\nabla$, called nabla or del. This operator is treated much like an ordinary vector in various computations. Its properties are covered in Calculus III in the U.S. system. (Brief summaries of properties of relevance here can be found in the notations section.)

The Hamiltonian eigenvalue problem for a foton wave function $\varphi_{\rm {f}}$ then takes the form

\begin{displaymath}
{\skew 4\widehat{\skew{-.5}\vec p}}\strut^{\,2} c^2 \varphi_{\rm {f}} = E^2 \varphi_{\rm {f}}
\end{displaymath}

A solution $\varphi_{\rm {f}}$ to this equation is an energy eigenstate. The corresponding value of $E$ is the energy of the state. (To be picky, the above is an eigenvalue problem for the square Hamiltonian. But eigenfunctions of an operator are also eigenfunctions of the square operator. The reverse is not always true, but that is not a concern here.)

Using the momentum operator as given above and some rearranging, the eigenvalue problem becomes

\begin{displaymath}
- \nabla^2 \varphi_{\rm {f}} = \frac{E^2}{\hbar^2c^2} \var...
...artial^2}{\partial y^2}
+ \frac{\partial^2}{\partial z^2} %
\end{displaymath} (A.101)

This is called the “time-independent Klein-Gordon equation” for a massless particle.

For foton wave functions that are not necessarily energy eigenstates, quantum mechanics replaces the energy $E$ by the operator ${\rm i}\hbar\partial$$\raisebox{.5pt}{$/$}$$\partial{t}$. That gives the time-dependent Klein-Gordon equation as:

\begin{displaymath}
- \nabla^2 \varphi_{\rm {f}} =
- \frac{1}{c^2}\frac{\partial^2\varphi_{\rm {f}}}{\partial t^2} %
\end{displaymath} (A.102)

Now consider solutions of this equation of the form

\begin{displaymath}
\varphi_{\rm {f}}({\skew0\vec r};t) = e^{-{\rm i}{\omega}t} \varphi_{\rm {fs}}({\skew0\vec r})
\end{displaymath}

Here $\omega$ is a positive constant called the angular frequency. Substitution in the time-dependent Klein-Gordon equation shows that this solution also satisfies the time-independent Klein-Gordon equation, with energy

\begin{displaymath}
E = \hbar\omega
\end{displaymath}

That is the famous Planck-Einstein relation. It is implicit in the association of $E$ with ${\rm i}\hbar\partial$$\raisebox{.5pt}{$/$}$$\partial{t}$.

Note however that there will also be a solution of the form

\begin{displaymath}
\varphi_{\rm {f}}({\skew0\vec r};t) = e^{{\rm i}{\omega}t} \varphi_{\rm {fs}}({\skew0\vec r})
\end{displaymath}

This solution too has energy $\hbar\omega$. The difference in sign in the exponential is taken to mean that the particle moves backwards in time. Note that changing the sign in the exponential is equivalent to changing the sign of the time $t$. At least it is if you require that $\omega$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E$$\raisebox{.5pt}{$/$}$$\hbar$ cannot be negative. If a particle moves backwards in time, it is called an antiparticle. So the wave function above describes an antifoton.

There is really no physical difference between a foton and an antifoton. That is not necessarily true for other types of particles. Quantities such as electric charge, lepton number, baryon number, strangeness, etcetera take opposite values for a particle and its antiparticle.

There is a very important difference between the Klein-Gordon equation and the Schrö­din­ger equation. The Schrö­din­ger equation describes nonrelativistic physics where particles can neither be destroyed nor created. Mass must be conserved. But the Klein-Gordon equation applies to relativistic physics. In relativistic physics particles can be created out of pure energy or destroyed following Einstein’s famous relationship $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$, chapter 1.

There is a mathematical consequence to this. It concerns the integral

\begin{displaymath}
\int\vert\varphi_{\rm {f}}\vert^2{\rm d}^3{\skew0\vec r}
\end{displaymath}

(In this addendum, integrals like this are over all space unless explicitly stated otherwise. It is also assumed that the fields vanish quickly enough at large distances that such integrals are finite. Alternatively, for particles confined in a large box it is assumed that the box is periodic, chapter 6.17.) Now for solutions of the Schrö­din­ger equation, the integral $\int\vert\varphi_{\rm {f}}\vert^2{\rm d}^3{\skew0\vec r}$ keeps the same value, 1, for all time. Physically, the integral represent the probability of finding the particle. The probability of finding the particle if you look in all space must be 1.

But fotons are routinely destroyed or created by sarged particles. So the probability of finding a foton is not a preserved quantity. (It is not even clear what finding a foton would mean in the first place.) The Klein-Gordon equation reflects that. It does not preserve the integral $\int\vert\varphi_{\rm {f}}\vert^2{\rm d}^3{\skew0\vec r}$. (There is one exception: if the wave function is purely described by particle states or purely described by antiparticle states, the integral is still preserved.)

But the Klein-Gordon equation does preserve an other integral, {D.32}. That is

\begin{displaymath}
\int \left\vert\frac{1}{c}\frac{\partial\varphi_{\rm {f}}}...
...t\nabla\varphi_{\rm {f}}\right\vert^2 {\rm d}^3{\skew0\vec r}
\end{displaymath}

Now if the number of fotons is not a preserved quantity, what can this preserved integral stand for? Not momentum or angular momentum, which are vectors. The integral must obviously stand for the energy. Energy is still preserved in relativity, even if the number of particles of a given type is not.

Of course, the energy of a foton wave function $\varphi_{\rm {f}}$ is also given by the Planck-Einstein relation. But wave functions are not observable. Still, fotons do affect spotons and selectons. That is observable. So there must be an observable foton field. This observable field will be called the foton potential. It will be indicated by simply $\varphi$, without a subscript f. Quantum uncertainty in the values of the field will be ignored in this subsection. So the field will be modeled as a classical (i.e. nonquantum) field.

And if there is an observable field, there must be an observable energy associated with that field. Now what could the expression for the energy in the field be? Obviously it will have to take the form of the integral above. What other options are there that are plausible? Of course, there will be some additional empirical constant. If the integral is constant, then any multiple of it will be constant too. And the above integral will not have units of energy as it is. The needed empirical constant is indicated by $\epsilon_1$ and is called, um no, the permissivity of space. It is a measure of how efficient the foton field is in generating energy. To be precise, for arcane historical reasons the constant in the energy is actually defined as half the permissivity. The bottom line is that the expression for the energy in the observable foton field is:

\begin{displaymath}
E_\varphi = \frac{\epsilon_1}{2}\int
\left\vert\frac{1}{...
...ft\vert\nabla\varphi\right\vert^2 {\,\rm d}^3{\skew0\vec r} %
\end{displaymath} (A.103)

That is really all that is needed to figure out the properties of classical selectostatics in this subsection. It will also be enough to figure out classical selectodynamics in the next subsection.

The first system that will be considered here is that of a foton field and a single spoton. It will be assumed that the spoton is pretty much located at the origin. Of course, in quantum mechanics a particle must have some uncertainty in position, or its kinetic energy would be infinite. But it will be assumed that the spoton wave function is only nonzero within a small distance $\varepsilon$ of the origin. Beyond that distance, the spoton wave function is zero.

However, since this is a classical derivation and not a quantum one, the term spoton wave function must not be used. So imagine instead that the spoton sarge $s_{\rm {p}}$ is smeared out over a small region of radius $\varepsilon$ around the origin.

For a smeared out sarge, there will be a sarge density $\sigma_{\rm {p}}$, defined as the local sarge per unit volume. This sarge density can be expressed mathematically as

\begin{displaymath}
\sigma_{\rm {p}}({\skew0\vec r}) = s_{\rm {p}} \delta_\varepsilon^3({\skew0\vec r})
\end{displaymath}

Here $\delta_\varepsilon^3({\skew0\vec r})$ is some function that describes the detailed shape of the smeared-out sarge distribution. The integral of this function must be 1, because the sarge density $\sigma_{\rm {p}}$ must integrate to the total spoton sarge $s_{\rm {p}}$. So:

\begin{displaymath}
\int \delta_\varepsilon^3({\skew0\vec r}) {\,\rm d}^3{\skew0\vec r}= 1
\end{displaymath}

To ensure that the sarge density is zero for distances from the origin $r$ greater than the given small value $\varepsilon$, $\delta_\varepsilon^3({\skew0\vec r})$ must be zero at these distances. So:

\begin{displaymath}
\delta_\varepsilon^3({\skew0\vec r}) = 0 \quad\mbox{if}\qu...
...\vec r}\vert\mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon
\end{displaymath}

In the limit that $\varepsilon$ becomes zero, $\delta_\varepsilon^3({\skew0\vec r})$ becomes the so-called three-di­men­sion­al “Dirac delta function” $\delta^3({\skew0\vec r})$. This function is totally concentrated at a single point, the origin. But its integral over that single point is still 1. That is only possible if the function value at the point is infinite. Now infinity is not a proper number, and so the Dirac delta function is not a proper function. However, mathematicians have in fact succeeded in generalizing the idea of functions to allow delta functions. That need not concern the discussion here because “Physicists are sloppy about mathematical rigor,” as Zee [52, p. 22] very rightly states. Delta functions are named after the physicist Dirac. They are everywhere in quantum field theory. That is not really surprising as Dirac was one of the major founders of the theory. See section 7.9 for more on delta functions.

Here the big question is how the spoton manages to create a foton field around itself. That is not trivial. If there was a nonzero probability of finding an energetic foton well away from the spoton, surely it would violate energy conservation. However, it turns out that the time-independent Klein-Gordon equation (A.101) actually has a very simple solution where the foton energy $E$ appears to be zero away from the origin. In spherical coordinates, it is

\begin{displaymath}
\varphi_{\rm {f}} = \frac{C}{r} \quad\mbox{if}\quad r \ne 0
\end{displaymath}

Here $C$ is some constant which is still arbitrary about this stage. To check the above solution, plug it into the energy eigenvalue problem (A.101) with $E$ zero.

This then seems to be a plausible form for the observable potential $\varphi$ away from the spoton at the origin. However, while the energy of a $C$$\raisebox{.5pt}{$/$}$$r$ potential appears to be zero, it is not really. Such a potential is infinite at the origin, and you cannot just ignore that. The correct foton field energy is given by the earlier integral (A.103). For a steady potential, it can be written as

\begin{displaymath}
E_\varphi
= \frac{\epsilon_1}{2}\int \left(\nabla\varphi...
...phi\left(\nabla^2\varphi\right)
{\,\rm d}^3{\skew0\vec r} %
\end{displaymath} (A.104)

The final integral comes from an integration by parts. (See {A.2} and {D.32} for examples how to do such integrations by parts.) Note that it looks like the energy could be zero according to this final integral: $\nabla^2\varphi$ is zero if $\varphi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $C$$\raisebox{.5pt}{$/$}$$r$. That is true outside the small vicinity around the origin. But if you look at the equivalent first integral, it is obvious that the energy is not zero: its integrand is everywhere positive. So the energy must be positive. It follows that in the final integral, the region around the origin, while small, still produces an energy that is not small. The integrand must be not just nonzero, but large in this region.

All that then raises the question why there is a foton field in the first place. The interest in this subsection is in the selectostatic field. That is supposed to be the stable ground state of lowest energy. According to the above, the state of lowest energy would be when there is no foton field; $\varphi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

And so it is. The only reasonable way to explain that there is a nontrivial foton field in the ground state of the spoton-foton system is if the foton field energy is compensated for by something else. There must be an energy of interaction between the foton field and the spoton.

Consider the mathematical form that this energy could take in a given volume element ${\rm d}^3{\skew0\vec r}$. Surely the simplest possibility is that it is proportional to the potential $\varphi$ at the location times the sarge $\sigma_{\rm {p}}{\rm d}^3{\skew0\vec r}$. Therefore the total energy of spoton-foton interaction is presumably

\begin{displaymath}
E_{\varphi\rm {p}} = - \int \varphi({\skew0\vec r}) \sigma...
...elta^3_\varepsilon({\skew0\vec r}){\,\rm d}^3{\skew0\vec r} %
\end{displaymath} (A.105)

Note that this expression really defines the sarge $s_{\rm {p}}$. Sarge gives the strength of the coupling between spoton and foton field. Its units and sign follow from writing the energy as the expression above.

The question is now, what is the ground state foton field? In other words, for what potential $\varphi$ is the complete system energy minimal? To answer that requires “variational calculus.” Fortunately, variational calculus is just calculus. And you need to understand how it works if you want to make any sense at all out of books on quantum field theory.

Suppose that you wanted an equation for the minimum of some function $f$ depending on a single variable $x$. The equation would be that ${\rm d}{f}$$\raisebox{.5pt}{$/$}$${\rm d}{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at the position of the minimum $x_{\rm {min}}$. In terms of differentials, that would mean that the function does not change going from position $x_{\rm {min}}$ to a slightly different position $x_{\rm {min}}+{\rm d}{x}$:

\begin{displaymath}
{\rm d}f = \frac{{\rm d}f}{{\rm d}x} {\rm d}x = 0
\quad\mbox{at}\quad x = x_{\rm {min}}
\end{displaymath}

It is the same for the change in net energy $E_\varphi+E_{\varphi\rm {p}}$. Assume that $\varphi_{\rm {min}}$ is the desired potential at minimum net energy. Then at $\varphi_{\rm {min}}$ the net energy should not change when you change $\varphi$ by an infinitesimal amount ${\rm d}\varphi$. Or rather, by an infinitesimal amount $\delta\varphi$: the symbol $\delta$ is used in variational calculus instead of ${\rm d}$. That is to avoid confusion with any symbol ${\rm d}$ that may already be around.

So the requirement for the ground state potential is

\begin{displaymath}
\delta (E_\varphi+E_{\varphi\rm {p}}) = 0
\quad\mbox{whe...
...rm {min}}
\;\to\; \varphi=\varphi_{\rm {min}}+\delta\varphi
\end{displaymath}

Using the expressions (A.104) and (A.105) for the energies, that means that

\begin{displaymath}
\delta \left[
\frac{\epsilon_1}{2}\int \left(\nabla\varp...
...rm {min}}
\;\to\; \varphi=\varphi_{\rm {min}}+\delta\varphi
\end{displaymath}

The usual rules of calculus can be used, (see {A.2} for more details). The only difference from basic calculus is that the change $\delta\varphi$ may depend on the point that you look at. In other words, it is some arbitrary but small function of the position ${\skew0\vec r}$. For example,

\begin{displaymath}
\delta (\nabla\varphi)^2 = 2 (\nabla\varphi) \delta(\nabla...
...varphi) - \nabla(\varphi_{\rm {min}})
= \nabla\delta\varphi
\end{displaymath}

Also, $\varphi$ by itself is validly approximated as $\varphi_{\rm {min}}$, but $\delta\varphi$ is a completely separate quantity that can be anything. Working it out gives

\begin{displaymath}
\frac{\epsilon_1}{2} \int 2 (\nabla\varphi_{\rm {min}})
...
...lta^3_\varepsilon \delta \varphi {\,\rm d}^3{\skew0\vec r}= 0
\end{displaymath}

Performing an integration by parts moves the $\nabla$ from $\delta\varphi$ to $\nabla\varphi_{\rm {min}}$ and adds a minus sign. Then the two integrals combine as

\begin{displaymath}
- \int \left(\epsilon_1\nabla^2{\varphi_{\rm {min}}}
+ s...
...varepsilon \right) \delta\varphi {\,\rm d}^3{\skew0\vec r}= 0
\end{displaymath}

If this is supposed to be zero for whatever you take the small change $\delta\varphi$ in field to be, then the parenthetical expression in the integral will have to be zero. If the parenthetical expression is nonzero somewhere, you can easily make up a nonzero change $\delta\varphi$ in that region so that the integral is nonzero.

The parenthetical expression can now be rearranged to give the final result:

\begin{displaymath}
- \nabla^2\varphi
= \frac{s_{\rm {p}}}{\epsilon_1} \delta^3_\varepsilon %
\end{displaymath} (A.106)

Here the subscript min was left away again as the ground state is the only state of interest here anyway.

The above equation is the famous “Poisson equation” for the selectostatic potential $\varphi$. The same equation appears in electrostatics, chapter 13.3.4. So far, this is all quite encouraging. Note also that the left hand side is the steady Klein-Gordon equation. The right hand side is mathematically a forcing term; it forces a nonzero solution for $\varphi$.

Beyond the small vicinity of radius $\varepsilon$ around the origin, the spoton sarge density in the right hand side is zero. That means that away from the spoton, you get the time-independent Klein-Gordon equation (A.101) with $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That was a good guess, earlier. Assuming spherical symmetry, away from the spoton the solution to the Poisson equation is then indeed

\begin{displaymath}
\varphi = \frac{C}{r} \quad\mbox{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon
\end{displaymath}

But now that the complete Poisson equation (A.106) is known, the constant $C$ can be figured out, {D.2}. The precise field turns out to be

\begin{displaymath}
\varphi = \frac{s_{\rm {p}}}{4 \pi \epsilon_1 r}
\quad\mbox{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon
\end{displaymath}

For unit value of $s_{\rm {p}}$$\raisebox{.5pt}{$/$}$$\epsilon_1$ the above solution is called the “fundamental solution” or “Green’s function” of the Poisson equation. It is the solution due to a delta function.

If the spoton is not at the origin, but at some position ${\skew0\vec r}_{\rm {p}}$, you simply replace $r$ by the distance from that point:

\begin{displaymath}
\varphi^{\rm {p}} =
\frac{s_{\rm {p}}}{4 \pi \epsilon_1 \vert{\skew0\vec r}- {\skew0\vec r}_{\rm {p}}\vert} %
\end{displaymath} (A.107)

The superscript p indicates that this potential is created by a spoton at a position ${\skew0\vec r}_{\rm {p}}$. This solution of the Poisson equation will become very important in the Fermi derivation.

Now the net energy is of interest. It can be simplified by substituting the Poisson equation (A.106) in the expression (A.104) for the foton field energy and adding the interaction energy (A.105). That gives

\begin{displaymath}
E_\varphi + E_{\varphi\rm {p}} =
{\textstyle\frac{1}{2}}...
...delta^3_\varepsilon({\skew0\vec r}) {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

which simplifies to
\begin{displaymath}
E_\varphi + E_{\varphi\rm {p}} = - {\textstyle\frac{1}{2}}...
...lta^3_\varepsilon({\skew0\vec r}) {\,\rm d}^3{\skew0\vec r} %
\end{displaymath} (A.108)

Note that the spoton-foton interaction energy is twice the foton field energy, and negative instead of positive. That means that the total energy has been lowered by an amount equal to the foton field energy, despite the fact that the field energy itself is positive.

The fact that there is a foton field in the ground state has now been explained. The interaction with the spoton lowers the energy more than the field itself raises it.

Note further from the solution for $\varphi$ above that $\varphi$ is large in the vicinity of the spoton. As a result, the energy in the foton field becomes infinite when the spoton sarge contracts to a point. (That is best seen from the original integral for the foton field energy in (A.104).) This blow up is very similar to the fact that the energy in a classical electromagnetic field is infinite for a point charge. For the Koulomb field, the interaction energy blows up too, as it is twice the foton field energy. All these blow ups are a good reason to use a sarge density rather than a point sarge. Then all energies are normal finite numbers.

The final step to derive the classical Koulomb force is to add a selecton. The selecton is also sarged, so it too generates a field. To avoid confusion, from now on the field generated by the spoton will always be indicated by $\varphi^{\rm {p}}$, and the one generated by the selecton by $\varphi^{\rm {e}}$. The variational analysis can now be repeated including the selecton, {D.37.1}. That shows that there are three effects that produce the Koulomb force between the spoton and selecton:

1.
the selecton sarge interacts with the potential $\varphi^{\rm {p}}$ generated by the spoton;
2.
the spoton sarge interacts with the potential $\varphi^{\rm {e}}$ generated by the selecton;
3.
the energy in the combined foton field $\varphi^{\rm {p}}+\varphi^{\rm {e}}$ is different from the sum of the energies of the separate fields $\varphi^{\rm {p}}$ and $\varphi^{\rm {e}}$.

All three effects turn out to produce the same energy, but the first two energies are negative and the third positive. So the net energy change is the same as if there was just item 1, the interaction of the selecton sarge density $\sigma_{\rm {e}}$ with the potential $\varphi^{\rm {p}}$ produced by the spoton. That is of course given by a similar expression as before:

\begin{displaymath}
V_{\rm {ep}} = - \int \varphi^{\rm {p}}({\skew0\vec r}) \sigma_{\rm {e}}({\skew0\vec r}) {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

The expression for $\varphi^{\rm {p}}({\skew0\vec r})$ was given above in (A.107) for any arbitrary position of the spoton ${\skew0\vec r}_{\rm {p}}$. And it will be assumed that the selecton sarge density $\sigma_{\rm {e}}$ is spread out a bit just like the spoton one, but around a different location ${\skew0\vec r}_{\rm {e}}$. Then the interaction energy becomes

\begin{displaymath}
V_{\rm {ep}} = - \int \frac{s_{\rm {p}}}{4\pi\epsilon_1\ve...
...w0\vec r}-{\skew0\vec r}_{\rm {e}}) {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

Since $\varepsilon$ is assumed small, the selecton sarge density is only nonzero very close to the nominal position ${\skew0\vec r}_{\rm {e}}$. Therefore you can approximate ${\skew0\vec r}$ as ${\skew0\vec r}_{\rm {e}}$ in the fraction and take it out of the integral as a constant. Then the delta function integrates to 1, and you get

\begin{displaymath}
V_{\rm {ep}} = - \frac{s_{\rm {p}}s_{\rm {e}}}
{4 \pi \e...
...vert{\skew0\vec r}_{\rm {e}}-{\skew0\vec r}_{\rm {p}}\vert} %
\end{displaymath} (A.109)

That then is the final energy of the Koulomb interaction between the two sarged particles. Because the spoton and the selecton both interact with the foton field, in effect it produces a spoton-selecton interaction energy.

Of course, in classical physics you would probably want to know the actual force on say the selecton. To get it, move the origin of the coordinate system to the spoton and rotate it so that the selecton is on the positive $x$-​axis. Now give the selecton a small displacement $\partial{x}_{\rm {e}}$ in the $x$-​direction. Slowly of course; this is supposed to be selectostatics. Because of energy conservation, the work done by the force ${F_x}_{\rm {e}}$ during this displacement must cause a corresponding small decrease in energy. So:

\begin{displaymath}
F_x\strut_{\rm {e}} \partial x\strut_{\rm {e}} = - \partial V_{\rm {ep}}
\end{displaymath}

But on the positive $x$-​axis, $\vert{\skew0\vec r}_{\rm {e}}-{\skew0\vec r}_{\rm {p}}\vert$ is just the $x$-​position of the selecton $x_{\rm {e}}$, so

\begin{displaymath}
F_x\strut_{\rm {e}} = - \frac{\partial V_{\rm {ep}}}{\part...
...frac{s_{\rm {p}}s_{\rm {e}}} {4 \pi \epsilon_1 x_{\rm {e}}^2}
\end{displaymath}

It is seen that if the sarges have equal sign, the force is in the negative $x$-​direction, towards the spoton. So sarges of the same sign attract.

More generally, the force on the selecton points towards the spoton if the sarges are of the same sign. It points straight away from the spoton if the sarges are of opposite sign.

The Koulomb energy $V_{\rm {ep}}$ looks almost exactly the same as the Coulomb energy in electrostatics. Recall that the Coulomb energy was used in chapter 4.3 to describe the attraction between the proton and electron in a hydrogen atom. The difference is that the Coulomb energy has no minus sign. That means that while like sarges attract, like charges repel each other. For example, two spotons attract, but two protons repel.

Now a spoton must necessarily create a foton field that is attractive to spotons. Otherwise there should be no field at all in the ground state. And if spotons create fields that attract spotons, then spotons attract. So the Koulomb force is clearly right.

It is the Coulomb force that does not seem to make any sense. Much more will be said about that in later subsections.


A.22.2 Classical selectodynamics

According to the previous section the Koulomb energy between a spoton and a selecton is given by

\begin{displaymath}
V_{\rm {ep}} = -
\frac{s_{\rm {p}}s_{\rm {e}}}{4 \pi \ep...
... \vert{\skew0\vec r}_{\rm {e}}-{\skew0\vec r}_{\rm {p}}\vert}
\end{displaymath}

However, this result can only be correct in a stationary state like a ground state, or maybe some other energy state.

To see the problem, imagine that the spoton is suddenly given a kick. According to the Koulomb potential given above, the selecton notices that instantly. There is no time in the Koulomb potential, so there is no time delay. But Einstein showed that no observable effect can move faster than the speed of light. So there should be a time delay.

Obviously then, to discuss unsteady evolution will require the full governing equations for selectodynamics. The big question is how to find these equations.

The quantum mechanics in this book is normally based on some Hamiltonian $H$. But there is a more basic quantity for a system than the Hamiltonian. That quantity is called the “Lagrangian” ${\cal L}$. If you can guess the correct Lagrangian of a system, its equations of motion follow. That is very important for quantum field theories. In fact, a lot of what advanced quantum field theories really do is guess Lagrangians.

To get at the Lagrangian for selectodynamics, consider first the motion of the spoton for a given foton field $\varphi$. The Hamiltonian of the spoton by itself is just the energy of the spoton. As discussed in the previous subsection, a spoton has a potential energy of interaction with the given foton field

\begin{displaymath}
E_{\varphi\rm {p}} = - \int \varphi({\skew0\vec r};t) \sigma_{\rm {p}}({\skew0\vec r};t)
{\,\rm d}^3{\skew0\vec r} %
\end{displaymath} (A.110)

Here $\sigma_{\rm {p}}$ was the sarge density of the spoton. The foton potential and sarge density can now of course also depend on time.

However, to discuss the dynamics of the spoton, it is easier to consider it a point particle located at a single moving point ${\skew0\vec r}_{\rm {p}}$. Therefore it will be assumed that the sarge density is completely concentrated at that one point. That means that the only value of the foton field of interest is the value at ${\skew0\vec r}_{\rm {p}}$. And the sarge distribution integrates to the net spoton sarge $s_{\rm {p}}$. So the above energy of interaction becomes approximately

\begin{displaymath}
E_{\varphi\rm {p}} \approx - \varphi_{\rm {p}} s_{\rm {p}}...
...varphi_{\rm {p}} \equiv \varphi({\skew0\vec r}_{\rm {p}};t) %
\end{displaymath} (A.111)

In terms of the components of position, this can be written out fully as

\begin{displaymath}
E_{\varphi\rm {p}} \approx - \varphi(r_{\rm {p}}\strut_1,r_{\rm {p}}\strut_2,
r_{\rm {p}}\strut_3;t) s_{\rm {p}}
\end{displaymath}

Note that in this addendum the position components are indicated as $r_{\rm {p}}\strut_1$, $r_{\rm {p}}\strut_2$, and $r_{\rm {p}}\strut_3$ instead of the more familiar $r_{\rm {p}}\strut_x$, $r_{\rm {p}}\strut_y$, and $r_{\rm {p}}\strut_z$ or $x_{\rm {p}}$, $y_{\rm {p}}$, and $z_{\rm {p}}$. That is in order that a generic position component can be indicated by $r_{\rm {p}}\strut_i$ where $i$ can be 1, 2, or 3.

In addition to the interaction energy above there is the kinetic energy of the spoton,

\begin{displaymath}
E_{\rm {p,kin}} = {\textstyle\frac{1}{2}} m_{\rm {p}} \vec v_{\rm {p}}^{\,2}
\end{displaymath}

Here $m_{\rm {p}}$ is the mass of the spoton and $\vec{v}_{\rm {p}}$ its velocity,

\begin{displaymath}
\vec v_{\rm {p}} \equiv \frac{{\rm d}{\skew0\vec r}_{\rm {p}}}{{\rm d}t}
\end{displaymath}

The kinetic energy can be written out in terms of the velocity components as

\begin{displaymath}
E_{\rm {p,kin}} = {\textstyle\frac{1}{2}} m_{\rm {p}}
\l...
...ac{{\rm d}r_{\rm {p}}\strut_i}{{\rm d}t} \mbox{ for } i=1,2,3
\end{displaymath}

Now the Hamiltonian of the spoton is the sum of the kinetic and potential energies. But the Lagrangian is the difference between the kinetic and potential energies:

\begin{displaymath}
\Lag_{\rm {p}}(\vec v_{\rm {p}},{\skew0\vec r}_{\rm {p}})
...
...m {p}}^{\,2} + \varphi({\skew0\vec r}_{\rm {p}};t)s_{\rm {p}}
\end{displaymath}

This Lagrangian can now be used to find the equation of motion of the spoton. This comes about in a somewhat weird way. Suppose that there is some range of times, from a time $t_1$ to a time $t_2$, during which you want to know the motion of the spoton. (Maybe the spoton is at rest at time $t_1$ and becomes again at rest at time $t_2$.) Suppose further that you now compute the so-called action integral

\begin{displaymath}
{\cal S}\equiv \int_{t_1}^{t_2} \Lag_{\rm {p}}(\vec v_{\rm {p}},{\skew0\vec r}_{\rm {p}}) {\,\rm d}t
\end{displaymath}

If you use the correct velocity and position of the spoton, you will get some number. But now suppose that you use a slightly different (wrong) spoton path. Suppose it is different by a small amount $\delta{\skew0\vec r}_{\rm {p}}$, which of course depends on time. You would think that the value of the action integral would change by a corresponding small amount. But that is not true. Assuming that the path used in the original integral was indeed the right one, and that the change in path is infinitesimally small, the action integral does not change. Mathematically

\begin{displaymath}
\delta {\cal S}= 0 \quad\mbox{at the correct path}
\end{displaymath}

Yes, this is again variational calculus. The action may not be minimal at the correct spoton path, but it is definitely stationary at it.

Probably this sounds like a stupid mathematical trick. But in the so-called path integral approach to quantum field theory, the action is central to the formulation.

For classical physics the action by itself is pretty useless. However, with some manipulations, you can get the evolution equations for your system out of it, {A.1}. They are found as

\begin{displaymath}
\frac{{\rm d}}{{\rm d}t}
\left(\frac{\partial{\cal L}}{\...
...c{\partial{\cal L}}{\partial r_{\rm {p}}\strut_i}
\right) %
\end{displaymath} (A.112)

Here $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, or 3 gives the equation in the $x$, $y$, or $z$ direction, respectively.

Note that for the governing equations it does not matter at all what you take the times $t_1$ and $t_2$ in the action to be. They are pretty vaguely defined anyway. You might want to let them go to minus and plus infinity to get rid of them.

The next step is to write out the governing equation (A.112) in terms of physical quantities. To do that correctly, the trick is that the Lagrangian must be treated as a function of velocity and position, as independent variables. In reality velocity and position are not independent; velocity is the derivative of position. But when differentiating the Lagrangian you are supposed to forget about that. Consider how this works out for the $x$-​component, $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1,

\begin{displaymath}
\frac{\partial{\cal L}}{\partial v_{\rm {p}}\strut_1} = m_...
...{p}}\strut_3;t)}
{\partial r_{\rm {p}}\strut_1} s_{\rm {p}}
\end{displaymath}

That are simple differentiations taking the given Lagrangian at face value.

However, when you do the remaining time derivative in (A.112) you have to do it properly, treating the velocity as the function of time that it is. That gives the final equation of motion as

\begin{displaymath}
m_{\rm {p}} \frac{{\rm d}v_{\rm {p}}\strut_1}{{\rm d}t} =
...
...}}\strut_3;t)}
{\partial r_{\rm {p}}\strut_1} s_{\rm {p}} %
\end{displaymath} (A.113)

Note that the left hand side is mass times acceleration in the $x$-​direction. So the right hand side must be the selectic force on the spoton. This force is called the Sorentz force. It is seen that the Sorentz force is proportional to the derivative of the foton potential, evaluated at the position of the spoton. If you compare the Sorentz force with the force in electrostatics, you see that the force in electrostatics has an additional minus sign. That reflects again that equal sarges attract, while equal charges repel.

So far, it was assumed that the foton field was given. But in reality the foton field is not given, it depends on the motion of the spoton. To describe the field, its energies must be added to the Lagrangian too. The total energy in the foton field was given in the previous subsection as (A.103). Using some shorthand notation, this becomes

\begin{displaymath}
E_\varphi = \frac{\epsilon_1}{2}\int \frac{1}{c^2} \varphi_t^2
+ \sum_{i=1}^3 \varphi_i^2 {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

The shorthand is to indicate derivatives by subscripts, as in

\begin{displaymath}
\varphi_t \equiv \frac{\partial\varphi}{\partial t}
\qquad
\varphi_i \equiv \frac{\partial\varphi}{\partial r_i}
\end{displaymath}

with $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, or 3 for the $x$, $y$, and $z$ derivatives respectively. For example, $\varphi_1$ would be the partial $x$-​derivative of $\varphi$.

Actually, even more concise shorthand will be used. If an index like $i$ occurs twice in a term, summation over that index is to be understood. The summation symbol will then not be shown. That is called the Einstein summation convention. So the energy in the foton field will be indicated briefly as

\begin{displaymath}
E_\varphi = \frac{\epsilon_1}{2}
\int \frac{1}{c^2} \varphi_t^2 + \varphi_i^2 {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

(Note that $\varphi_i^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varphi_i\varphi_i$, so $i$ occurs twice in the second term of the integrand.) All this is done as a service to you, the reader. You are no doubt getting tired of having to look at all these mathematical symbols.

Now the first term in the energy above is a time derivative, just like $\vec{v}_{\rm {p}}$ was the time derivative of the spoton position. So this term has presumably the same sign in the Lagrangian, while the sign of the other term flips over. That makes the total selectodynamic Lagrangian equal to

\begin{displaymath}
\Lag_{\varphi\rm {p}} = \frac{\epsilon_1}{2}
\int \frac{...
...}m_{\rm {p}}\vec v_{\rm {p}}^2 + \varphi_{\rm {p}}s_{\rm {p}}
\end{displaymath}

The last two terms are as before for a given field.

However, for the final term it is now desirable to go back to the representation of the spoton in terms of a sarge density $\sigma_{\rm {p}}$, as in (A.110). The final term as written would lead to a nasty delta function in the analysis of the field. In the sarge density form the term can be brought inside the integral to give the complete Lagrangian as

\begin{displaymath}
\Lag_{\varphi\rm {p}} = \int \frac{\epsilon_1}{2}\left(
...
... r}+ {\textstyle\frac{1}{2}} m_{\rm {p}} \vec v_{\rm {p}}^2 %
\end{displaymath} (A.114)

Note that there is no longer a subscript $p$ on $\varphi$; it is the integration against the sarge density that picks out the value of $\varphi$ at the spoton.

An integrand of a spatial integral in a Lagrangian is called a “Lagrangian density” and indicated by the symbol $\pounds $. In this case:

\begin{displaymath}
\pounds = \frac{\epsilon_1}{2}\left(\frac{1}{c^2}\varphi_t^2-\varphi_i^2\right)
+ \varphi\sigma_{\rm {p}} %
\end{displaymath} (A.115)

When differentiating this Lagrangian density, $\varphi$ and its derivatives $\varphi_t$ and $\varphi_i$, (with $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, and 3), are to be considered 5 separate independent variables.

The action principle can readily be extended to allow for Lagrangian densities, {D.37}. The equations of motion for the field are then found to be

\begin{displaymath}
\frac{\partial}{\partial t}
\left(\frac{\partial\pounds ...
...varphi_i}\right) =
\frac{\partial\pounds }{\partial\varphi}
\end{displaymath}

Working this out much like for the equation of motion of the spoton gives, taking $\varepsilon_1$ to the other side,

\begin{displaymath}
\frac{1}{c^2}\frac{\partial^2\varphi}{\partial t^2}
- \f...
...hi}{\partial r_i^2}
= \frac{\sigma_{\rm {p}}}{\epsilon_1} %
\end{displaymath} (A.116)

This is the so-called Saxwell wave equation of selectodynamics. If there is also a selecton, say, its sarge density can simply be added to the spoton one in the right hand side.

To check the Saxwell equation, first consider the case that the system is steady, i.e. independent of time. In that case the Saxwell wave equation becomes the Poisson equation of the previous subsection as it should. (The second term is summed over the three Cartesian directions $i$. That gives $\nabla^2\varphi$.) So the spoton produces the same steady Koulomb field (A.107) as before. So far, so good.

How about the force on a selecton in this field? Of course, the force on a selecton is a Sorentz force of the same form as (A.113),

\begin{displaymath}
F_x\strut_{\rm {e}} =
\frac
{\partial\varphi(r_{\rm {e...
...}}\strut_3;t)}
{\partial r_{\rm {e}}\strut_1} s_{\rm {e}} %
\end{displaymath} (A.117)

In the steady case, the relevant potential at the selecton is the electrostatic one (A.107) produced by the spoton as given in the previous subsection (Strictly speaking you should also include the field produced by the selecton itself. But this self-interaction produces no net force. That is fortunate because if the selecton was really a point sarge, the self-interaction is mathematically singular.) Now minus the potential (A.107) times the selecton sarge $s_{\rm {e}}$ gave the energy $V_{\rm {ep}}$ of the spoton-selecton interaction in the previous subsection. And minus the derivative of that gave the force on the selecton. A look at the force above then shows it is the same.

So in the steady case the Saxwell equation combined with the Sorentz force does reproduce selectostatics correctly. That means that the given Lagrangian (A.114) contains all of selectostatics in a single concise mathematical expression. At the minimum. Neat, isn’t it?

Consider next the case that the time dependence cannot be ignored. Then the time derivative in the Saxwell equation (A.116) cannot be ignored. In that case the left hand side in the equation is the complete unsteady Klein-Gordon equation. Since there is a nonzero right-hand side, mathematically the Saxwell equation is an inhomogeneous Klein-Gordon equation. Now it is known from the theory of partial differential equations that the Klein-Gordon equation respects the speed of light. As an example, imagine that at time $t$ = 0 you briefly shake the spoton at the origin and then put it back where it was. The right hand side of the Saxwell equation is then again back to what it was. But near the origin, the foton field $\varphi$ will now contain additional disturbances. These disturbances evolve according to the homogeneous Saxwell equation, i.e. the equation with zero right hand side. And it is easy to check by substitution that the homogeneous equation has solutions of the form

\begin{displaymath}
\varphi = f(x - c t)
\end{displaymath}

That are waves traveling in the x-direction with the speed of light $c$. The wave shape is the arbitrary function $f$ and is preserved in time. And note that the $x$-​direction is arbitrary. So waves like this can travel in any direction. The perturbations near the origin caused by shaking the spoton will consist of such waves. Since they travel with the speed of light, they need some time to reach the selecton. The selecton will not notice anything until this happens. However, when the perturbations in the foton field do reach the selecton, they will change the foton field $\varphi$ at the selecton. That then will change the force (A.117) on the selecton.

It follows that selectodynamics, as described by the Lagrangian (A.114), also respects the speed of light limitation.


A.22.3 Quantum selectostatics

The previous subsections derived the Koulomb force between sarged particles. This force was due to foton exchange. While the derivations used some ideas from quantum mechanics, they were classical. The effect of the fotons took the form of a potential $\varphi$ that the sarged particles interacted with. This potential was a classical field; it had a definite numerical value at each point. To be picky, there really was an undetermined constant in the potential $\varphi$. But its gradient $\nabla\varphi$ produced the fully determined Sorentz force per unit sarge (A.117). This force can be observed by a sarged spoton or selecton.

However, that very fact violates the fundamental postulates of quantum mechanics as formulated at the beginning of this book, chapter 3.4. Observable values should be the eigenvalues of Hermitian operators that act on wave functions. While the foton potential was loosely associated with a foton wave function, wave functions should not be observable.

Now if classically every position has its own observable local potential $\varphi$, then in a proper quantum description every position must be associated with its own Hermitian operator $\widehat\varphi$. In the terminology of addendum {A.15.9}, the foton field $\widehat\varphi$ must be a quantum field; an infinite amount of operators, one for each position.

The objective in this subsection is to deduce the form of this quantum field. And the type of wave function that it operates on. The results will then be used to verify the Koulomb force between stationary sarges as found the first subsection. It is imperative to figure out whether like sarges still attract in a proper quantum description.

Doing this directly would not be easy. It helps a lot if the field is written in terms of linear momentum eigenstates.

In fact, typical quantum field theories depend very heavily on this trick. However, often such theories use relativistic combined energy-momentum states in four-di­men­sion­al space-time. This subsection will use simpler purely spatial momentum states. The basic idea is the same. And it is essential for understanding the later Fermi derivation of the Coulomb potential.

Linear momentum states are complex exponentials of the form $e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}$. Here ${\vec k}$ is a constant vector called the wave number vector. The momentum of such a state is given in terms of the wave number vector by the de Broglie relation as ${\skew0\vec p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar{\vec k}$. (It may be noted that the $e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}$ states need an additional constant to properly normalize them, chapter 6.18. But for conciseness, in this addendum that normalization constant will be absorbed in the constants multiplying the exponentials.)

If a field $\varphi$ is written in terms of linear momentum states, its value at any point ${\skew0\vec r}$ is given by:

\begin{displaymath}
\varphi({\skew0\vec r}) = \sum_{{\rm all\ }{\vec k}} c_{{\vec k}}\, e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}
\end{displaymath}

Note that if you know the coefficients $c_{{\vec k}}$ of the momentum states, it is equivalent to knowing the field $\varphi$. Then the field at any point can in principle be found by doing the sum.

The expression above assumes that the entire system is confined to a very large periodic box, as in chapter 6.17. In infinite space the sum becomes an integral, section 7.9. That would be much more messy. (But that is the way you will usually find it in a typical quantum field analysis.) The precise values of the wave number vectors to sum over for a given periodic box were given in chapter 6.18 (6.28); they are all points in figure 6.17.

The first subsection found the selectostatic potential $\varphi^{\rm {p}}$ that was produced by a spoton, (A.107). This potential was a classical field; it had a definite numerical value for each position. The first step will be to see how this potential looks in terms of momentum states. While the final objective is to rederive the classical potential using proper quantum mechanics, the correct answer will need to be recognized when written in terms of momentum states. Not to mention that the answer will reappear in the discussion of the Coulomb potential. For simplicity it will be assumed that the spoton is at the origin.

According to the first subsection, the classical potential was the solution to a Poisson equation; a steady Klein-Gordon equation with forcing by the spoton:

\begin{displaymath}
- \nabla^2\varphi^{\rm {p}}_{\rm {cl}}
= \frac{s_{\rm {p}}}{\epsilon_1} \psi_{\rm {p}}^*\psi_{\rm {p}}
\end{displaymath}

As a reminder that $\varphi^{\rm {p}}$ is a classical potential, not a quantum one, a subscript cl has been added. Also note that since this is a now a quantum description, the spoton sarge density $\sigma_{\rm {p}}$ has been identified as the spoton sarge $s_{\rm {p}}$ times the square magnitude of the spoton wave function $\vert\psi_{\rm {p}}\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {p}}^*\psi_{\rm {p}}$.

Now the classical potential is to be written in the form

\begin{displaymath}
\varphi^{\rm {p}}_{\rm {cl}}({\skew0\vec r})
= \sum_{{\r...
...vec k}} c_{{\vec k}}\, e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}
\end{displaymath}

To figure out the coefficients $c_{{\vec k}}$, plug it in the Poisson equation above. That gives

\begin{displaymath}
\sum_{{\rm all}\ {\vec k}} k^2 c_{{\vec k}}\, e^{{\rm i}{\...
...\frac{s_{\rm {p}}}{\epsilon_1} \psi_{\rm {p}}^*\psi_{\rm {p}}
\end{displaymath}

Note that in the left hand side each $\nabla$ produced a factor ${\rm i}{\vec k}$ for $\vphantom0\raisebox{1.5pt}{$-$}$$k^2$ total.

Now multiply this equation at both sides by some sample complex-conjugate momentum eigenfunction $e^{-{\rm i}\underline{\vec k}\cdot{\skew0\vec r}}$ and integrate over the entire volume ${\cal V}$ of the periodic box. In the left hand side, you only get something nonzero for the term in the sum where ${\vec k}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\underline{\vec k}$ because eigenfunctions are orthogonal. For that term, the exponentials multiply to 1. So the result is

\begin{displaymath}
{\underline k}^2 c_{\underline{\vec k}} {\cal V}= \frac{s_...
... r}} \psi_{\rm {p}}^*\psi_{\rm {p}} {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

Now in the right hand side, assume again that the spoton is almost exactly at the origin. In other words, assume that its wave function is zero except very close to the origin. In that case, the exponential in the integral is approximately 1 when the spoton wave function is not zero. Also, the square wave function integrates to 1. So the result is, after clean up,

\begin{displaymath}
c_{\underline{\vec k}} = \frac{s_{\rm {p}}}{\epsilon_1{\cal V}{\underline k}^2}
\end{displaymath}

This expression applies for any wave number vector $\underline{\vec k}$, so you can leave the underline away. It fully determines $\varphi^{\rm {p}}_{\rm {cl}}$ in terms of the momentum states:
\begin{displaymath}
\varphi^{\rm {p}}_{\rm {cl}}({\skew0\vec r}) = \sum_{{\rm ...
...psilon_1{\cal V}k^2} e^{{\rm i}{\vec k}\cdot{\skew0\vec r}} %
\end{displaymath} (A.118)

This solution is definitely one to remember. Note in particular that the coefficients of the momentum states are a constant divided by $k^2$. Recall also that for a unit value of $s_{\rm {p}}$$\raisebox{.5pt}{$/$}$$\epsilon_1$, this solution is the fundamental solution, or Green’s function, of the Poisson equation with point wise forcing at the origin.

If the requirement that the spoton wave function is completely at the origin is relaxed, the integral involving the spoton wave function stays:

\begin{displaymath}
\varphi^{\rm {p}}_{\rm {cl}}({\skew0\vec r}) = \sum_{{\rm ...
...m {p}}\big\rangle
e^{{\rm i}{\vec k}\cdot{\skew0\vec r}} %
\end{displaymath} (A.119)

where

\begin{displaymath}
\big\langle\psi_{\rm {p}}\big\vert e^{-{\rm i}{\vec k}\cdo...
...{\skew0\vec r}_{\rm {p}}) {\,\rm d}^3{\skew0\vec r}_{\rm {p}}
\end{displaymath}

Note that the integration variable over the spoton wave function has been renamed ${\skew0\vec r}_{\rm {p}}$ to avoid confusion with the position ${\skew0\vec r}$ at which the potential is evaluated. The above result is really better to work with in this subsection, since it does not suffer from some convergence issues that the Green’s function solution has. And it is exact for a spoton wave function that is somewhat spread out.

Now the objective is to reproduce this classical result using a proper quantum field theory. And to find the force when a selecton is added to the system.

To do so, consider initially a system of fotons and a single spoton. The spoton will be treated as a nonrelativistic particle. Then its wave function $\psi_{\rm {p}}$ describes exactly one spoton. The spoton wave function will be treated as given. Imagine something keeping the spoton in a ground state squeezed around the origin. Maxwell’s demon would work. He has not been doing much anyway after he failed his thermo test.

Next the fotons. Their description will be done based upon linear momentum states. Such a state corresponds to a single-foton wave function of the form $e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}$.

To keep it simple, for now only a single momentum state will be considered. In other words, only a single wave number vector ${\vec k}$ will be considered. But there might be multiple fotons in the state, or even uncertainty in the number of fotons.

Of course, at the end of the day the results must still be summed over all values of ${\vec k}$.

Some notations are needed now. A situation in which there are no fotons in the considered state will be indicated by the “Fock space ket” $\big\vert\big\rangle $. If there is one foton in the state, it is indicated by $\big\vert 1\big\rangle $, two by $\big\vert 2\big\rangle $, etcetera. In the mathematics of quantum field theory, kets are taken to be orthonormal, {A.15}:

\begin{displaymath}
\big\langle i_1\big\vert\big\vert i_2\big\rangle = \left\{...
...box{ if }i_1=i_2 \\ 0\mbox{ otherwise}\end{array}
\right. %
\end{displaymath} (A.120)

In words, the inner product of kets is 0 unless the numbers of fotons are equal. Then it is 1.

The ground state wave function for the combined spoton-fotons system is then assumed to be of the form

\begin{displaymath}
\psi_{\varphi\rm {p}} = C_0 \psi_{\rm {p}} \big\vert\big\r...
...vert^2 + \vert C_1\vert^2 + \vert C_2\vert^2 + \ldots = 1\; %
\end{displaymath} (A.121)

That is a linear combination of system states with 0, 1, 2, ... fotons. So it is assumed that there may be uncertainty about the number of fotons in the considered foton state. The normalization condition for the constants expresses that the total probability of finding the system in some state or the other is 1.

(It may be noted that in typical quantum field theories, a charged relativistic particle would also be described in terms of kets and some quantum field $\widehat\psi$. However, unlike for a photon, for a charged particle $\widehat\psi$ would normally be a complex quantum field. Then $\widehat\psi^*\widehat\psi$ or something along these lines provides a real probability for a photon to observe the particle. That resembles the Born interpretation of the nonrelativistic wave function somewhat, especially for a spinless particle. Compare [[17, pp. 49, 136, 144]]. The field $\widehat\psi$ will describe both the particle and its oppositely charged antiparticle. The spoton wave function $\psi_{\rm {p}}$ as used here represents some nonrelativistic limit in which the antiparticle has been approximated away from the field, [[17, pp. 41-45]]. Such a nonrelativistic limit simply does not exist for a real scalar field like the Koulomb one.)

Now, of course, the Hamiltonian is needed. The Hamiltonian determines the energy. It consists of three parts:

\begin{displaymath}
H = H_{\rm {p}} + H_\varphi + H_{\varphi\rm {p}}
\end{displaymath}

The first part is the Hamiltonian for the spoton in isolation. It consists of the kinetic energy of the spoton, as well as the potential provided by the fingers of the demon. By definition
\begin{displaymath}
H_{\rm {p}} \psi_{\rm {p}} = E_{\rm {p}} \psi_{\rm {p}} %
\end{displaymath} (A.122)

where $E_{\rm {p}}$ is the energy of the spoton in isolation.

The second part is the Hamiltonian of the free foton field. Each foton in the considered state should have an energy $\hbar\omega$ with $\omega$ $\vphantom0\raisebox{1.5pt}{$=$}$ $kc$. That is the energy that you get if you substitute the momentum eigenfunction $e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}$ into the Klein-Gordon eigenvalue problem (A.101) for a massless particle. And if one foton has an energy $\hbar\omega$, then $i$ of them should have energy $i\hbar\omega$, so

\begin{displaymath}
H_\varphi \big\vert i\big\rangle = i \hbar\omega \big\vert i\big\rangle %
\end{displaymath} (A.123)

Note that specifying what the Hamiltonian does to each separate ket tells you all you need to know about it. (Often there is an additional ground state energy shown in the above expression, but that does not make a difference here. It reflects the choice of the zero of energy.)

Finally, the third part of the total Hamiltonian is the interaction between the spoton and the foton field. This is the tricky one. First recall the classical expression for the interaction energy. According to the previous subsection, (A.111), it was $-s_{\rm {p}}\varphi_{\rm {p}}$. Here $\varphi_{\rm {p}}$ was the classical foton potential, evaluated at the position of the spoton.

In quantum field theory, the observable field $\varphi$ gets replaced by a quantum field $\widehat\varphi$. The interaction Hamiltonian then becomes

\begin{displaymath}
H_{\varphi\rm {p}} = - s_{\rm {p}} \widehat\varphi_{\rm {p}}
\end{displaymath} (A.124)

This Hamiltonian needs to operate on the wave function (A.121) involving the spoton wave function and Fock space kets for the fotons. The big question is now: what is that quantum field $\widehat\varphi$?

To answer that, first note that sarged particles can create and destroy fotons. The above interaction Hamiltonian must express that somehow. After all, it is the Hamiltonian that determines the time evolution of systems in quantum mechanics.

Now in quantum field theories, creation and destruction of particles are accounted for through creation and annihilation operators, {A.15}. A creation operator $\widehat a_{\vec k}$ creates a single particle in a momentum state $e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}$. An annihilation operator $\widehat a_{\vec k}$ annihilates a single particle from such a state. More precisely, the operators are defined as

\begin{displaymath}
\widehat a_{\vec k}\big\vert i\big\rangle = \sqrt{i} \big\...
...big\vert i{-}1\big\rangle = \sqrt{i} \big\vert i\big\rangle %
\end{displaymath} (A.125)

Here $\big\vert i\big\rangle $ is the Fock-space ket that indicates that there are $i$ fotons in the considered momentum state. Except for the numerical factor $\sqrt{i}$, the annihilation operator takes a foton out of the state. The creation operator puts it back in, adding another numerical factor $\sqrt{i}$.

Note incidentally that the foton field Hamiltonian given earlier can now be rewritten as

\begin{displaymath}
H_\varphi = \hbar\omega \widehat a^\dagger _{\vec k}\widehat a_{\vec k} %
\end{displaymath} (A.126)

That is because

\begin{displaymath}
\hbar\omega \widehat a^\dagger _{\vec k}\widehat a_{\vec k...
...\big\vert i\big\rangle
= H_\varphi \big\vert i\big\rangle
\end{displaymath}

In general this Hamiltonian will still need to be summed over all values of ${\vec k}$.

But surely, the creation and annihilation of particles should also depend on where the spoton is. Fotons in the considered state have a spatially varying wave function. That should be reflected in the quantum field $\widehat\varphi$ somehow. To find the correct expression, it is easiest to first perform a suitable normalization of the foton state. Now the full wave function corresponding to the single-foton momentum eigenstate in empty space is

\begin{displaymath}
\varphi_{\rm {f}} = C e^{-{\rm i}\omega t} e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}
\end{displaymath}

Here $C$ is some normalization constant to be chosen. The above wave function can be verified by putting it into the Klein-Gordon equation (A.102). The energy of the foton is given in terms of its wave function above as $\hbar\omega$. But the energy in the foton field is also related to the observable field $\varphi$; classical selectostatics gives that relation as (A.103). If you plug the foton wave function above into that classical expression, you do not normally get the correct energy $\hbar\omega$. There is no need for it; the foton wave function is not observable. However, it makes things simpler if you choose $C$ so that the classical energy does equal $\hbar\omega$. That gives a energy-normalized wave function
\begin{displaymath}
\varphi_{\vec k}= \frac{\varepsilon_k}{k} e^{{\rm i}{\vec ...
...on_k \equiv
\sqrt{\frac{\hbar\omega}{\epsilon_1{\cal V}}} %
\end{displaymath} (A.127)

In those terms, the needed quantum field turns out to be

\begin{displaymath}
\widehat\varphi
= \frac{1}{\sqrt{2}} (\varphi_{\vec k}\,...
...on_k \equiv
\sqrt{\frac{\hbar\omega}{\epsilon_1{\cal V}}} %
\end{displaymath} (A.128)

The first term in the right hand side is the normalized single-foton wave function at wave number ${\vec k}$ times the corresponding annihilation operator. The second term is the complex-conjugate foton wave function times the creation operator. There is also the usual factor 1$\raisebox{.5pt}{$/$}$$\sqrt{2}$ that appears when you take a linear combination of two states.

You might of course wonder about that second term. Mathematically it is needed to make the operator Hermitian. Recall that operators in quantum mechanics need to be Hermitian to ensure that observable quantities have real, rather than complex values, chapter 2.6. To check whether an operator is Hermitian, you need to check that it is unchanged when you take it to the other side of an inner product. Now the wave function is a numerical quantity that changes into its complex conjugate when taken to the other side. And $\widehat a$ changes into $\widehat a^\dagger $ and vice-versa when taken to the other side, {A.15.2}. So each term in $\widehat\varphi$ changes into the other one, leaving the sum unchanged. So the operator as shown is indeed Hermitian.

But what to make physically of the two terms? One way of thinking about it is that the observed field is real because it does not just involve an interaction with an $e^{{\rm i}({\vec k}\cdot{\skew0\vec r}-{\omega}t)}$ foton, but also with an $e^{-{\rm i}({\vec k}\cdot{\skew0\vec r}-{\omega}t)}$ antifoton.

In general, the quantum field above would still need to be summed over all wave numbers ${\vec k}$. (Or integrated over ${\vec k}$ in infinite space). It may be noted that for given ${\skew0\vec r}$ the sum of the creation operator terms over all ${\vec k}$ can be understood as a field operator that creates a particle at position ${\skew0\vec r}$, [34, p. 24]. That is a slightly different definition of the creation field operator than given in {A.15.9}, [42, pp. 22]. But for nonrelativistic particles (which have nonzero rest mass) it would not make a difference.

With the quantum field $\varphi$ now identified, the Hamiltonian of the spoton-fotons interaction becomes finally

\begin{displaymath}
H_{\varphi\rm {p}} = - s_{\rm {p}} \widehat\varphi_{\rm {p...
...on_k \equiv
\sqrt{\frac{\hbar\omega}{\epsilon_1{\cal V}}} %
\end{displaymath} (A.129)

Note that the spoton has uncertainty in position. The spoton position in the Hamiltonian above is just a possible spoton position. In usage it will still get multiplied by the square spoton wave function magnitude that gives the probability for that position. Still, at face value the interaction of the spoton with the field takes place at the location of the spoton. Interactions in quantum field theories are “local.” At least on macroscopic scales that is needed to satisfy the limitation of the speed of light.

Having a Hamiltonian allows quantum selectodynamics to be explored. That will be done to some detail for the case of the electromagnetic field in subsequent addenda. However, here the only question that will be addressed is whether classical selectostatics as described in the first subsection was correct. In particular, do equal sarges still attract in the quantum description?

Selectostatics of the spoton-fotons system should correspond to the ground state of the system. The ground state has the lowest possible energy. You can therefore find the ground state by finding the state of lowest possible system energy. That is the same trick as was used to find the ground states of the hydrogen molecular ion and the hydrogen molecule in chapters 4.6 and 5.2. The expectation value of the system energy is defined by the inner product

\begin{displaymath}
\big\langle E\big\rangle = \big\langle\psi_{\varphi\rm {p}...
..._\varphi+H_{\varphi\rm {p}})\psi_{\varphi\rm {p}}\big\rangle
\end{displaymath}

Here the Hamiltonians have already been described above.

Now to find the ground state, the lowest possible value of the expectation energy above is needed. To get that, the inner products between the kets in the factors $\psi_{\varphi\rm {p}}$ must be multiplied out. First apply the Hamiltonians (A.122), (A.123), and (A.129) on the wave function $\psi_{\varphi\rm {p}}$, (A.121), using (A.125). Then apply the orthogonality relations (A.120) of kets. Do not forget the complex conjugate on the left side of an inner product. That produces

\begin{eqnarray*}
\big\langle E\big\rangle & = & E_{\rm {p}} \\
&& \mbox{} ...
...g\rangle
\left(C_1^* C_0 + \sqrt{2}C_2^* C_1 + \ldots\right)
\end{eqnarray*}

Here the dots stand for terms involving the coefficients $C_3,C_4,\ldots$ of states with three or more fotons.

Note that the first term in the right hand side above is the energy $E_{\rm {p}}$ of the spoton by itself. That term is a given constant. The question is what foton states produce the lowest energy for the remaining terms. The answer is easy if the spoton sarge $s_{\rm {p}}$ is zero. Then the terms in the last two lines are zero. So the second line shows that $C_1,C_2,\ldots$ must all be zero. Then there are no fotons; only the state with zero fotons is then left in the system wave function (A.121).

If the spoton sarge is nonzero however, the interaction terms in the last two lines can lower the energy for suitable nonzero values of the constants $C_1$, $C_2$, .... To simplify matters, it will be assumed that the spoton sarge is nonzero but small. Then so will be the constants. In that case only the $C_1$ terms need to be considered; the other terms in the last two lines involve the product of two small constants, and those cannot compete. Further the normalization condition in (A.121) shows that $\vert C_0\vert$ will be approximately 1 since even $C_1$ is small. Then $C_0$ may be assumed to be 1, because any eigenfunction is indeterminate by a factor of magnitude 1 anyway.

Further, since any complex number may always be written as its magnitude times some exponential of magnitude 1, the second last line of the energy above can be written as

\begin{displaymath}
- \frac{s_{\rm {p}}\varepsilon_k}{\sqrt{2} k}
\big\langl...
...\Big\vert e^{{\rm i}\alpha}\; \vert C_1\vert e^{{\rm i}\beta}
\end{displaymath}

Replacing ${\rm i}$ everywhere by $\vphantom0\raisebox{1.5pt}{$-$}$${\rm i}$ gives the corresponding expression for the last line. The complete expression for the energy then becomes

\begin{eqnarray*}
E & = & E_{\rm {p}} + \vert C_1\vert^2 \hbar \omega \\
&&...
...g\rangle
\Big\vert \vert C_1\vert e^{-{\rm i}(\alpha+\beta)}
\end{eqnarray*}

In the last term, note that the sign of ${\rm i}$ inside an absolute value does not make a difference. Using the Euler formula (2.5) on the trailing exponentials gives

\begin{displaymath}
E = E_{\rm {p}} +
\vert C_1\vert^2 \hbar \omega - 2 \fra...
... {p}}\big\rangle
\Big\vert\vert C_1\vert\cos(\alpha+\beta)
\end{displaymath}

But in the ground state, the energy should be minimal. Clearly, that requires that the cosine is at its maximum value 1. So it requires taking $C_1$ so that $\beta$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$\alpha$.

That still leaves the magnitude $\vert C_1\vert$ to be found. Note that the final terms in expression above are now of the generic form

\begin{displaymath}
a \vert C_1\vert^2 - 2 b \vert C_1\vert
\end{displaymath}

where $a$ and $b$ are positive constants. That is a quadratic function of $\vert C_1\vert$. By differentiation, it is seen that the minimum occurs at $\vert C_1\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $b$$\raisebox{.5pt}{$/$}$$a$ and has a value $\vphantom0\raisebox{1.5pt}{$-$}$$b^2$$\raisebox{.5pt}{$/$}$$a$. Putting in what $a$ and $b$ are then gives

\begin{displaymath}
C_1 = \frac{s_{\rm {p}}\varepsilon_k}{\sqrt{2} k \hbar\ome...
...ot{\skew0\vec r}_{\rm {p}}} \psi_{\rm {p}}\big\rangle \vert^2
\end{displaymath}

The second part of the energy is the energy lowering achieved by having a small probability $\vert C_1\vert^2$ of a single foton in the considered momentum state.

This energy-lowering still needs to be summed over all states ${\vec k}$ to get the total:

\begin{displaymath}
E - E_{\rm {p}} = - \sum_{{\vec k}} \frac{s_{\rm {p}}^2}{2...
...\rm i}{\vec k}\cdot{\skew0\vec r}} \psi_{\rm {p}}\big\rangle
\end{displaymath}

Note that the final two inner products represent separate integrations. Therefore to avoid confusion, the subscript p was dropped from one integration variable. In the sum, the classical field (A.119) can now be recognized:

\begin{displaymath}
E - E_{\rm {p}} =
- \frac{s_{\rm {p}}}{2}
\big\langle\...
...c r})\psi_{\rm {p}}({\skew0\vec r}) {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

Ignoring the differences in notation, the energy lowering is exactly the same as (A.108) found in the classical analysis. The classical analysis, while not really justified, did give the right answer.

However now an actual picture of the quantum ground state has been obtained. It is a quantum superposition of system states. The most likely state is the one where there are no fotons at all. But there are also small probabilities for system states where there is a single foton in a single linear momentum foton state. This picture does assume that the spoton sarge is small. If that was not true, things would get much more difficult.

Another question is whether the observable values of the foton potential are the same as those obtained in the classical analysis. This is actually a trick question because even the classical foton potential is not observable. There is still an undetermined constant in it. What is observable are the derivatives of the potential: they give the observable selectic force per unit sarge on sarged particles.

Now, in terms of momentum modes, the derivatives of the classical potential can be found by differentiating (A.119). That gives

\begin{displaymath}
\varphi_i\strut_{\rm {p,cl}} = \sum_{{\rm all\ }{\vec k}}
...
...g\rangle
{\rm i}k_i e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}
\end{displaymath}

Recall again the convention introduced in the previous subsection that a subscript $i$ on $\varphi$ indicates the derivative $\partial$$\raisebox{.5pt}{$/$}$$\partial{r}_i$, where $r_1$, $r_2$, and $r_3$ correspond to $x$, $y$, and $z$ respectively. So the above expression gives the selectic force per unit sarge in the $x$, $y$, or $z$ direction, depending on whether $i$ is 1, 2, or 3.

The question is now whether the quantum analysis predicts the same observable forces. Unfortunately, the answer here is no. The observable forces have quantum uncertainty that the classical analysis missed. However, the Ehrenfest theorem of chapter 7.2.1 suggests that the expectation forces should still match the classical ones above.

The quantum expectation force per unit sarge in the $i$-​direction is given by

\begin{displaymath}
\big\langle\varphi_i\big\rangle =
\big\langle\Psi_{\varp...
...\varphi\rm {p}} = e^{-{\rm i}E t/\hbar} \psi_{\varphi\rm {p}}
\end{displaymath}

Here $E$ is the ground state energy. Note that in this case the full, time-dependent wave function $\Psi_{\varphi\rm {p}}$ is used. That is done because in principle an observed field could vary in time as well as in space. Substituting in the $r_i$-​derivative of the quantum field (A.128) gives

\begin{displaymath}
\big\langle\varphi_i\big\rangle = \frac{\varepsilon_k}{\sq...
...widehat a^\dagger _{\vec k})\psi_{\varphi\rm {p}}\big\rangle
\end{displaymath}

Note that here ${\skew0\vec r}$ is not a possible position of the spoton, but a given position at which the selectic force per unit sarge is to be found. Also note that the time dependent exponentials have dropped out against each other; the expectation forces are steady like for the classical field.

The above expression can be multiplied out as before. Using the obtained expression for $C_1$, and the fact that $\big\langle\psi_{\rm {p}}\big\vert\psi_{\rm {p}}\big\rangle $ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 because wave functions are normalized, that gives.

\begin{displaymath}
\big\langle\varphi_i\big\rangle =
\frac{s_{\rm {p}}}{2\e...
...\rangle
{\rm i}k_i e^{-{\rm i}{\vec k}\cdot{\skew0\vec r}}
\end{displaymath}

Summed over all ${\vec k}$, the two terms in the right hand side produce the same result, because opposite values of ${\vec k}$ appear equally in the summation. In other words, for every ${\vec k}$ term in the first sum, there is a $\vphantom0\raisebox{1.5pt}{$-$}$${\vec k}$ term in the second sum that produces the same value. And that then means that the expectation selectic forces are the same as the classical ones. The classical analysis got that right, too.

To see that there really is quantum uncertainty in the forces, it suffices to look at the expectation square forces. If there was no uncertainty in the forces, the expectation square forces would be just the square of the expectation forces. To see that that is not true, it is sufficient to simply take the spoton sarge zero. Then the expectation field is zero too. But the expectation square field is given by

\begin{displaymath}
\big\langle\varphi_i^2\big\rangle = \frac{\varepsilon_k^2}...
... \psi_{\varphi\rm {p}} = \psi_{\rm {p}} \big\vert\big\rangle
\end{displaymath}

Multiplying this out gives

\begin{displaymath}
\big\langle\varphi_i^2\big\rangle = \frac{\varepsilon_k^2k...
...}{2 k^2} =
\frac{\hbar\omega k_i^2}{2\epsilon_1{\cal V}k^2}
\end{displaymath}

Since Planck’s constant is not zero, this is not zero either. So even without the spoton, a selectic force measurement will give a random, but nonzero value. The average of a large number of such force measurements will be zero, but not the individual measurements.

The above expression can be compared with the corresponding $\vert\varphi_{{\vec k},i}\vert^2$ of a single foton, as given by (A.127). That comparison indicates that even in the ground state in empty space, there is still half a foton of random field energy left. Recall now the Hamiltonian (A.126) for the foton field. Usually, this Hamiltonian would be defined as

\begin{displaymath}
H_\varphi = \sum_{{\vec k}} \hbar\omega (\widehat a^\dagger _{\vec k}\widehat a_{\vec k}+{\textstyle\frac{1}{2}})
\end{displaymath}

The additional ${\textstyle\frac{1}{2}}$ expresses the half foton of energy left in the ground state. The ground state energy does not change the dynamics. However, it is physically reflected in random nonzero values if the selectic field is measured in vacuum.

The bad news is that if you sum these ground state energies over all values of ${\vec k}$, you get infinite energy. The exact same thing happens for the photons of the electromagnetic field. Quantum field theories are plagued by infinite results; this “vacuum energy” is just a simple example. What it really means physically is as yet not known either. More on this can be found in {A.23.4}.

The final issue to be addressed is the attraction between a spoton and a selecton. That can be answered by simply adding the selecton to the spoton-fotons analysis above, {D.37.2}. The answer is that the spoton-selecton interaction energy is the same as found in the classical analysis.

So equal sarges still attract.


A.22.4 Poincaré and Einstein try to save the universe

The Koulomb universe is a grim place. In selectodynamics, particles with the same sarge attract. So all selectons clump together into one gigantic ball. Assuming that spotons have the opposite sarge, they clump together into another big ball at the other end of the universe. But actually there is no justification to assume that spotons would have a different sarge from selectons. That then means that all matter clumps together into a single gigantic satom. A satom like that will form one gigantic, obscene, black hole. It is hardly conductive to the development of life as we know it.

Unfortunately, the Koulomb force is based on highly plausible, apparently pretty unavoidable assumptions. The resulting force simply makes sense. None of these things can be said about the Coulomb force.

But maybe, just maybe, the Koulomb juggernaut can be tripped up by some legal technicality. Things like that have happened before.

Now in a time not really that very long ago, there lived a revolutionary of mathematics called Poincaré. Poincaré dreamt of countless shining stars that would sweep through a gigantic, otherwise dark universe. And around these stars there would be planets populated by living beings called observers. But if the stars all moved in random directions, with random speeds, then which star would be the one at rest? Which star would be the king around which the other stars danced? Poincaré thought long and hard about that problem. No! he thundered eventually; “It shall not be. I hereby proclaim that all stars are created equal. Any observer at any star can say at any given time that its star is at rest and that the other stars are moving. On penalty of dead, nothing in physics may indicate that observer to be wrong.”

Now nearby lived a young physicist called Einstein who was very lazy. For example, he almost never bothered to write the proper summation symbols in his formulae. Of course, that made it difficult for him to find a well paying job in some laboratory where they smash spotons into each other. Einstein ended up working in some patent office for little pay. But, fortunate for our story, working in a patent office did give Einstein a fine insight in legal technicalities.

First Einstein noted that the Proclamation of Poincaré meant that observers at different stars had to disagree seriously about the locations and times of events. However, it would not be complete chaos. The locations and times of events as perceived by different observers would still be related. The relation would be a transformation that the famous physicist Lorentz had written down earlier, chapter 1.2.1 (1.6).

And the Proclamation of Poincaré also implied that different observers had to agree about the same laws of physics. So the laws of physics should remain the same when you change them from one observer to the next using the Lorentz transformation. Nowadays we would say that the laws of physics should be “Lorentz invariant.” But at the time, Einstein did not want to use the name of Lorentz in vain.

Recall now the classical action principle of subsection A.22.2. The so-called action integral had to be unchanged under small deviations from the correct physics. The Proclamation of Poincaré demands that all observers must agree that the action is unchanged. If the action is unchanged for an observer at one star, but not for one at another star, then not all stars are created equal.

To see what that means requires a few fundamental facts about special relativity, the theory of systems in relative motion.

The Lorentz transformation badly mixes up spatial positions and times of events as seen by different observers. To deal with that efficiently, it is convenient to combine the three spatial coordinates and time into a single four-di­men­sion­al vector, a four-vector, chapter 1.2.4. Time becomes the “zeroth coordinate” that joins the three spatial coordinates. In various notations, the four-vector looks like

\begin{displaymath}
\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hsp...
...trut x^3\end{array}\right)
\equiv
\{x^\mu\}
\to
x^\mu
\end{displaymath}

First of all, note that the zeroth coordinate receives an additional factor $c$, the speed of light. That is to ensure that it has units of length just like the other components. It has already been noted before that the spatial coordinates $x$, $y$, and $z$ are indicated by $r_1$, $r_2$, and $r_3$ in this addendum. That allows a generic component to be indicated by $r_i$ for $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, or 3. Note also that curly brackets are a standard mathematical way of indicating a set or collection. So $\{r_i\}$ stands for the set of all three $r_i$ values; in other words, it stands for the complete position vector ${\skew0\vec r}$. That is the primary notation that will be used in this addendum.

However, in virtually any quantum field book, you will find four-vectors indicated by $x^\mu$. Here $\mu$ is an index that can have the values 0, 1, 2, or 3. (Except that some books make time the fourth component instead of the zeroth.) An $x^\mu$ by itself probably really means $\{x^\mu\}$, in other words, the complete four-vector. Physicists have trouble typing curly brackets, so they leave them away. When more than one index is needed, another Greek symbol will be used, like $x^\nu$. However, $x^i$ would stand for just the spatial components, so for the position vector $\{r_i\}$. The give-away is here that a roman superscript is used. Roman superscript $j$ would mean the same thing as $i$; the spatial components only.

There are similar notations for the derivatives of a function $f$:

\begin{displaymath}
\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hsp...
...}\right)
\equiv
\{\partial_\mu f\}
\to
\partial_\mu f
\end{displaymath}

Note again that time derivatives in this addendum are indicated by a subscript $t$ and spatial derivatives by a subscript $i$ for $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, or 3.

Quantum field books use $\partial_{\mu}f$ for derivatives. They still have problems with typing curly brackets, so $\partial_{\mu}f$ by itself probably means the set of all four derivatives. Similarly $\partial_if$ would probably mean the spatial gradient $\nabla{f}$.

The final key fact to remember about special relativity is:

In dot products between four-vectors, the product of the zeroth components gets a minus sign.
Dot products between four-vectors are very important because all observers agree about the values of these dot products. They are Lorentz invariant. (In nonrelativistic mechanics, all observers agree about the usual dot products between spatial vectors. That is no longer true at relativistic speeds.)

One warning. In almost all modern quantum field books, the products of the spatial components get the minus sign instead of the time components. The purpose is to make the relativistic dot product incompatible with the nonrelativistic one. After all, backward compatibility is so, well, backward. (One source that does use the compatible dot product is [48]. This is a truly excellent book written by a Nobel Prize winning pioneer in quantum field theory. It may well be the best book on the subject available. Unfortunately it is also very mathematical and the entire thing spans three volumes. Then again, you could certainly live without supersymmetry.)

One other convention might be mentioned. Some books put a factor ${\rm i}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{-1}$ in the zeroth components of four-vectors. That takes care of the minus sign in dot products automatically. But modern quantum field books do not this.

Armed with this knowledge about special relativity, the Koulomb force can now be checked. Action is defined as

\begin{displaymath}
{\cal S}\equiv \int_{t_1}^{t_2} {\cal L}{\,\rm d}t
\end{displaymath}

Here the time range from $t_1$ to $t_2$ should be chosen to include the times of interest. Further ${\cal L}$ is the so-called Lagrangian.

If all observers agree about the value of the action in selectodynamics, then selectodynamics is Lorentz invariant. Now the Lagrangian of classical selectodynamics was of the form, subsection A.22.2,

\begin{displaymath}
\Lag_{\varphi\rm {p}} = \int \pounds _\varphi {\,\rm d}^3{...
..._{\rm {p}} \vec v_{\rm {p}}^2 + \varphi_{\rm {p}} s_{\rm {p}}
\end{displaymath}

Here the Lagrangian density of the foton field $\varphi$ was

\begin{displaymath}
\pounds _\varphi = - \frac{\epsilon_1}{2}
\left(-\frac{1}{c^2}\varphi_t^2 + \varphi_i^2\right)
\end{displaymath}

To this very day, a summation symbol may not be used to reveal to nonphysicists that the last term needs to be summed over all three values of $i$. That is in honor of the lazy young physicist, who tried to save the universe.

Note that the parenthetical term in the Lagrangian density is simply the square of the four-vector of derivatives of $\varphi$. Indeed, the relativistic dot product puts the minus sign in front of the product of the time derivatives. Since all observers agree about dot products, they all agree about the values of the Lagrangian density. It is Lorentz invariant.

To be sure, it is the action and not the Lagrangian density that must be Lorentz invariant. But note that in the action, the Lagrangian density gets integrated over both space and time. Such integrals are the same for any two observers. You can easily check that from the Lorentz transformation chapter 1.2.1 (1.6) by computing the Jacobian of the ${\rm d}{x}{\rm d}{t}$ integration between observers.

(OK, the limits of integration are not really the same for different observers. One simple way to get around that is to assume that the field vanishes at large negative and positive times. Then you can integrate over all space-time. A more sophisticated argument can be given based on the derivation of the action principle {A.1.5}. From that derivation it can be seen that it suffices to consider small deviations from the correct physics that are localized in both space and time. It implies that the limits of integration in the action integral are physically irrelevant.)

(Note that this subsection does no longer mention periodic boxes. In relativity periodicity is not independent of the observer, so the current arguments really need to be done in infinite space.)

The bottom line is that the first, integral, term in the Lagrangian produces a Lorentz-invariant action. The second term in the Lagrangian is the nonrelativistic kinetic energy of the spoton. Obviously the action produced by this term will not be Lorentz invariant. But you can easily fix that up by substituting the corresponding relativistic term as given in chapter 1.3.2. So the lack of Lorentz invariance of this term will simply be ignored in this addendum. If you want, you can consider the spoton mass to be the moving mass in the resulting equations of motion.

The final term in the Lagrangian is the problem. It represents the spoton-fotons interaction. The term by itself would be Lorentz invariant, but it gets integrated with respect to time. Now in relativity time intervals ${\rm d}{t}$ are not the same for different observers. So the action for this term is not Lorentz invariant. Selectodynamics cannot be correct. The Koulomb juggernaut has been stopped by a small legal technicality.

(To be sure, any good lawyer would have pointed out that there is no problem if the spoton sarge density, instead of the spoton sarge $s_{\rm {p}}$, is the same for different observers. But the Koulomb force was so sure about its invincibility that it never bothered to seek competent legal counsel.)

The question is now of course how to fix this up. That will hopefully produce a more appealing universe. One in which particles like protons and electrons have charges $q$ rather than sarges $s$. Where these charges allow them to interact with the photons of the electromagnetic field. And where these photons assure that particles with like charges repel, rather than attract.

Consider the form of the problem term in the Koulomb action:

\begin{displaymath}
\int_{t_1}^{t_2} \varphi_{\rm {p}} s_{\rm {p}} {\,\rm d}t
\end{displaymath}

It seems logical to try to write this in relativistic terms, like

\begin{displaymath}
\int_{t_1}^{t_2} \Big(\frac{\varphi_{\rm {p}}}{c}\Big)
\...
...{c}\Big)
\Big(s_{\rm {p}} {\,\rm d}r_{\rm {p}}\strut_0\Big)
\end{displaymath}

Here ${\rm d}{r}_{\rm {p}}\strut_0$ is the zeroth component of the change in spoton four-vector ${\rm d}\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over r}
\kern-1.3pt_{\rm {p}}$. The product of the two parenthetical factors is definitely not Lorentz invariant. But suppose that you turn each of the factors into a complete four-vector? Dot products are Lorentz invariant. And the four-vector corresponding to ${\rm d}{r}_{\rm {p}}\strut_0$ is clearly ${\rm d}\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over r}
\kern-1.3pt_{\rm {p}}$.

But the photon potential must also become a four-vector, instead of a scalar. That is what it takes to achieve Lorentz invariance. So electrodynamics defines a four-vector of potentials of the form

\begin{displaymath}
{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt...
...trut A^3\end{array}\right)
\equiv
\{A^\mu\}
\to
A^\mu
\end{displaymath}

Here $\skew3\vec A$ is the so-called vector potential while $\varphi$ is now the electrostatic potential.

The interaction term in the action now becomes, replacing the spoton sarge $s_{\rm {p}}$ by minus the proton charge $q_{\rm {p}}$,

\begin{displaymath}
\int_{t_1}^{t_2} {\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\h...
...ce{0pt}}}\over r}
\kern-1.3pt_{\rm {p}}}{{\rm d}t} {\,\rm d}t
\end{displaymath}

In writing out the dot product, note that the spatial components of ${\rm d}\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over r}
\kern-1.3pt_{\rm {p}}$$\raisebox{.5pt}{$/$}$${\rm d}{t}$ are simply the proton velocity components $v_{\rm {p}}\strut_j$. That gives the interaction term in the action as

\begin{displaymath}
\int_{t_1}^{t_2} \left(-\varphi_{\rm {p}} q_{\rm {p}}
+ ...
..._{\rm {p}} q_{\rm {p}} v_{\rm {p}}\strut_j \right) {\,\rm d}t
\end{displaymath}

Once again nonphysicists may not be told that the second term in parentheses must be summed over all three values of $j$ since $j$ appears twice.

The integrand above is the interaction term in the electromagnetic Lagrangian,

\begin{displaymath}
\Lag_{\rm int} = -\varphi_{\rm {p}} q_{\rm {p}}
+ A_j\strut_{\rm {p}} q_{\rm {p}} v_{\rm {p}}\strut_j
\end{displaymath}

For now at least.

The Lagrangian density of the photon field is also needed. Since the photon field is a four-vector rather than a scalar, the self-evident electromagnetic density is

\begin{displaymath}
\pounds _{\rm seem} = - \frac{\epsilon_0}{2}
\left(
- ...
...\strut_i^2 + \frac{1}{c^2}\varphi_t^2 - \varphi_i^2
\right)
\end{displaymath}

Here the constant $\epsilon_0$ is called the “permittivity of space.” Note that the second term in parentheses must be summed over both $i$ and $j$. The curious sign pattern for the parenthetical terms arises because it involves two dot products: one from the square four-gradient (derivatives), and one from the square four-potential. Simply put, having electrostatic potentials is worth a minus sign, and having time derivatives is too.

It might be noted that in principle the proper Lagrangian density could be minus the above expression. But a minus sign in a Lagrangian does not change the motion. The convention is to choose the sign so that the corresponding Hamiltonian describes energies that can be increased by arbitrarily large amounts, not lowered by arbitrarily large amounts. Particles can have unlimited amounts of positive kinetic energy, not negative kinetic energy.

Still, it does seem worrisome that the proper sign of the Lagrangian density is not self-evident. But that issue will have to wait until the next subsection.

Collecting things together, the self-evident Lagrangian for electromagnetic field plus proton is

\begin{displaymath}
\Lag_{\rm seem+p} = \int \pounds _{\rm seem} {\,\rm d}^3{\...
...rm {p}} + A_j\strut_{\rm {p}} q_{\rm {p}} v_{\rm {p}}\strut_j
\end{displaymath}

Here $\pounds _{\rm {seem}}$ was given above.

The first thing to check now is the equation of motion for the proton. Following subsection A.22.2, it can be found from

\begin{displaymath}
\frac{{\rm d}}{{\rm d}t}
\left(\frac{\partial{\cal L}}{\...
...(\frac{\partial{\cal L}}{\partial r_{\rm {p}}\strut_i}\right)
\end{displaymath}

Substituting in the Lagrangian above gives

\begin{displaymath}
\frac{{\rm d}}{{\rm d}t}
\left(m_{\rm {p}} v_{\rm {p}}\s...
...
+ {{A_j}_i}\strut_{\rm {p}} q_{\rm {p}}v_{\rm {p}}\strut_j
\end{displaymath}

This can be cleaned up, {D.6}. In short, first an electric” field $\skew3\vec{\cal E}$ and a “magnetic field $\skew2\vec{\cal B}$ are defined as, in vector notation,

\begin{displaymath}
\skew3\vec{\cal E}= - \nabla \varphi - \frac{\partial\skew...
... t}
\qquad \skew2\vec{\cal B}= \nabla \times \skew3\vec A %
\end{displaymath} (A.130)

The individual components are
\begin{displaymath}
{\cal E}_i = - \varphi_i - A_i\strut_t \qquad
{\cal B}_i...
... A_{\overline{\imath}}\strut_{\overline{\overline{\imath}}} %
\end{displaymath} (A.131)

Here $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, or 3 corresponds to the $x$, $y$, or $z$ components respectively. Also ${\overline{\imath}}$ follows $i$ in the periodic sequence $\ldots123123\ldots$ and ${\overline{\overline{\imath}}}$ precedes $i$. In these terms, the simplified equation of motion of the proton becomes, in vector notation,
\begin{displaymath}
\frac{{\rm d}m_{\rm {p}}\vec v_{\rm {p}}}{{\rm d}t} = q_{\...
...}+\vec v_{\rm {p}}\times\skew2\vec{\cal B}_{\rm {p}}\right) %
\end{displaymath} (A.132)

The left hand side is mass times acceleration. Relativistically speaking, the mass should really be the moving mass here, but OK. The right hand side is known as the Lorentz force.

Note that there are 4 potentials with 4 derivatives each, for a total of 16 derivatives. But matter does not observe all 16 individually. Only the 3 components of the electric field and the 3 of the magnetic field are actually observed. That suggests that there may be changes to the fields that can be made that are not observable. Such changes are called gage (or gauge) changes. The name arises from the fact that a gage is a measuring device. You and I would then of course say that these changes should be called nongage changes. They are not measurable. But gage is really shorthand for “Take that, you stupid gage.”

Consider the most general form of such gage changes. Given potentials $\varphi$ and $\skew3\vec A$, equivalent potentials can be created as

\begin{displaymath}
\varphi' = \varphi - \chi_t \qquad \skew3\vec A' = \skew3\vec A+ \nabla\chi
\end{displaymath}

Here $\chi$ can be any function of space and time that you want.

The potentials $\varphi'$ and $\skew3\vec A'$ give the exact same electric and magnetic fields as $\varphi$ and $\skew3\vec A$. (These claims are easily checked using a bit of vector calculus. Use Stokes to show that they are the most general changes possible.)

The fact that you can make unmeasurable changes to the potentials like that is called the gage (or gauge) property of the electromagnetic field. Nonphysicists probably think it is something you read off from a voltage gage. Hilarious, isn’t it?

Observable or not, the evolution equations of the four potentials are also needed. To find them it is convenient to spread the proton charge out a bit. That is the same trick as was used in subsection A.22.2. For the spread-out charge, a “charge density” $\rho_{\rm {p}}$ can be defined as the charge per unit volume. It is also convenient to define a “current density” $\vec\jmath_{\rm {p}}$ as the charge density times its velocity. Then the proton-photons interaction terms in the Lagrangian are:

\begin{displaymath}
\int\Big(- \varphi \rho_{\rm {p}} + A_j \jmath_{\rm {p}}\s...
...p}}
+ A_j\strut_{\rm {p}} q_{\rm {p}} v_{\rm {p}}\strut_j %
\end{displaymath} (A.133)

Here the right hand side is an approximation if the proton charge is almost concentrated at a single point, or exact for a point charge.

The interaction terms can now be included in the Lagrangian density to give the total Lagrangian

\begin{displaymath}
\Lag_{\rm seem+p} = \int \Big(\pounds _{\rm seem} + \pound...
...+ {\textstyle\frac{1}{2}} m_{\rm {p}} v_{\rm {p}}\strut_j^2 %
\end{displaymath} (A.134)


\begin{displaymath}
\pounds _{\rm {seem}} = - \frac{\epsilon_0}{2}
\left(
...
...nt} = - \varphi \rho_{\rm {p}} + A_j \jmath_{\rm {p}}\strut_j
\end{displaymath}

If there are more charged particles than just a proton, their charge and current densities will combine into a net $\rho$ and $\vec\jmath$.

The field equations now follow similarly as in subsection A.22.2. For the electrostatic potential:

\begin{displaymath}
\frac{\partial}{\partial t}
\left(\frac{\partial\pounds ...
...varphi_i}\right) =
\frac{\partial\pounds }{\partial\varphi}
\end{displaymath}

where $\pounds $ is the combined Lagrangian density. Worked out and converted to vector notation, that gives
\begin{displaymath}
\frac{1}{c^2}\frac{\partial^2\varphi}{\partial t^2} - \nabla^2\varphi
= \frac{\rho}{\epsilon_0} %
\end{displaymath} (A.135)

This is the same equation as for the Koulomb potential earlier.

Similarly, for the components of the vector potential

\begin{displaymath}
\frac{\partial}{\partial t}
\left(\frac{\partial\pounds ...
...\strut_i}\right) =
\frac{\partial\pounds }{\partial\varphi}
\end{displaymath}

That gives
\begin{displaymath}
\frac{\partial^2 \skew3\vec A}{\partial t^2} - c^2 \nabla^2 \skew3\vec A
= \frac{\vec\jmath}{\epsilon_0} %
\end{displaymath} (A.136)

The above equations are again Klein-Gordon equations, so they respect the speed of light. And since the action is now Lorentz invariant, all observers agree with the evolution. That seems very encouraging.

Consider now the steady case, with no charge motion. The current density $\vec\jmath$ is then zero. That leads to zero vector potentials. Then there is no magnetic field either, (A.130).

The steady equation (A.135) for the electrostatic field $\varphi$ is exactly the same as the one for the Koulomb potential. But note that the electric force per unit charge is now minus the gradient of the electrostatic potential, (A.130) and (A.132). And that means that like charges repel, not attract. All protons in the universe no longer clump together into one big ball. And neither do electrons. That sounds great.

But wait a second. How come that apparently protons suddenly manage to create fields that are repulsive to protons? What happened to energy minimization? It seems that all is not yet well in the universe.


A.22.5 Lorenz saves the universe

The previous subsection derived the self-evident equations of electromagnetics. But there were some worrisome aspects. A look at the Hamiltonian can clarify the problem.

Given the Lagrangian (A.134) of the previous subsection, the Hamiltonian can be found as, {A.1.5}:

\begin{displaymath}
H_{\rm seem+p} =
\int \left(\frac{\partial\pounds }{\par...
...}{\partial v_{\rm {p}}\strut_j} v_{\rm {p}}\strut_j - {\cal L}
\end{displaymath}

That gives
\begin{displaymath}
H_{\rm seem+p} = \frac{\epsilon_0}{2} \int
\left(
A_j\...
...j^2 +
\int \varphi\rho_{\rm {p}}{\,\rm d}^3{\skew0\vec r} %
\end{displaymath} (A.137)

(This would normally still need to be rewritten in terms of canonical momenta, but that is not important here.)

Note that the electrostatic potential $\varphi$ produces negative electromagnetic energy. That means that the electromagnetic energy can have arbitrarily large negative values for large enough $\varphi$.

That then answers the question of the previous subsection: “How come a proton produces an electrostatic field that repels it? What happened to energy minimization?” There is no such thing as energy minimization here. If there is no lowest energy, then there is no ground state. Instead the universe should evolve towards larger and larger electrostatic fields. That would release infinite amounts of energy. It should blow life as we know it to smithereens. (The so-called second law of thermodynamics says, simply put, that thermal energy is easier to put into particles than to take out again. See chapter 11.)

In fact, the Koulomb force would also produce repulsion between equal sarges, if its field energy was negative instead of positive. Just change the sign of the constant $\epsilon_1$ in classical selectodynamics. Then its universe should blow up too. Unlike what you will read elsewhere, the difference between the Koulomb force, (or its more widely known sibling, the Yukawa force of {A.41}), and the Coulomb force is not simply that the photon wave function is a four-vector. It is whether negative field energy appears in the most straightforward formulation.

As the previous subsection noted, you might assume that the electrodynamic Lagrangian, and hence the Hamiltonian, would have the opposite sign. But that does not help. In that case the vector potentials $A_j$ would produce the negative energies. Reversing the sign of the Hamiltonian is like reversing the direction of time. In either direction, the universe gets blown to smithereens.

To be sure, it is not completely sure that the universe will be blown to smithereens. A negative field energy only says that it is in theory possible to extract limitless amounts of energy out of the field. But you would still need some actual mechanism to do so. There might not be one. Nature might be carefully constrained so that there is no dynamic mechanism to extract the energy.

In that case, there might then be some mathematical expression for the constraint. As another way to look at that, suppose that you would indeed have a highly unstable system. And suppose that there is still something recognizable left at the end of the first day. Then surely you would expect that whatever is left is special in some way. That it obeys some special mathematical condition.

So presumably, the electromagnetic field that we observe obeys some special condition, some constraint. What could this constraint be? Since this is very basic physics, you would guess it to be relatively simple. Certainly it must be Lorentz invariant. The simplest condition that meets this requirement is that the dot product of the four-gradient $\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over\nabla}
\kern-1.3pt$ with the four-potential ${\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}$ is zero. Written out that produces the so-called “Lorenz condition:”

\begin{displaymath}
\fbox{$\displaystyle
\frac{1}{c}\frac{\partial\varphi/c}{\partial t} + \nabla\cdot\skew3\vec A= 0
$} %
\end{displaymath} (A.138)

This condition implies that only a very special subset of possible solutions of the Klein-Gordon equations given in the previous subsection is actually observed in nature.

Please note that the Lorenz condition is named after the Danish physicist Ludvig Lorenz, not the Dutch physicist Hendrik Lorentz. Almost all my sources mislabel it the Lorentz condition. The savior of the universe deserves more respect. Always remember: the Lorenz condition is Lorentz invariant.

(You might wonder why the first term in the Lorenz condition does not have the minus sign of dot products. One way of thinking about it is that the four-gradient in its natural condition already has a minus sign on the time derivative. Physicists call it a covariant four-vector rather than a contravariant one. A better way to see it is to grind it out; you can use the Lorentz transform (1.6) of chapter 1.2.1 to show directly that the above form is the same for different observers. But those familiar with index notation will recognize immediately that the Lorenz condition is Lorentz invariant from the fact that it equals $\partial_{\mu}A^{\mu}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, and that has $\mu$ as both subscript and superscript. See chapter 1.2.5 for more.)

To be sure, the Lorenz condition can only be true if the interaction with matter does not produce violations. To check that, the evolution equation for the Lorenz condition quantity can be obtained from the Klein-Gordon equations of the previous subsection. In particular, in vector notation take $\partial$$\raisebox{.5pt}{$/$}$$\partial{t}$ (A.135) plus $\nabla$ (A.136) to get

\begin{displaymath}
\left[\frac{\partial^2}{\partial t^2} - c^2 \nabla^2\right...
...{\partial\rho}{\partial t}
+ \nabla\cdot\vec\jmath\right) %
\end{displaymath} (A.139)

The parenthetical expression in the left hand side should be zero according to the Lorenz condition. But that is only possible if the left hand side is zero too, so

\begin{displaymath}
\frac{\partial\rho}{\partial t} = - \nabla\cdot\vec\jmath
\end{displaymath}

This important result is known as “Maxwell’s continuity equation.” It expresses conservation of charge. (To see that, take any arbitrary volume. Integrate both sides of the continuity equation over that volume. The left hand side then becomes the time derivative of the charge inside the volume. The right hand side becomes, using the [divergence] [Gauss] [Ostrogradsky] theorem, the net inflow of charge. And if the charge inside can only change due to inflow or outflow, then no charge can be created out of nothing or destroyed.) So charge conservation can be seen as a consequence of the need to maintain the Lorenz condition.

Note that the Lorenz condition (A.138) looks mathematically just like the continuity equation. It produces conservation of the integrated electrostatic potential. In subsection A.22.7 it will be verified that it is indeed enough to produce a stable electromagnetic field. One with meaningfully defined energies that do not run off to minus infinity.

Note that charge conservation by itself is not quite enough to ensure that the Lorenz condition is satisfied. However, if in addition the Lorenz quantity and its time derivative are zero at just a single time, it is OK. Then (A.139) ensures that the Lorenz condition remains true for all time.


A.22.6 Gupta-Bleuler condition

The ideas of the previous subsection provide one way to quantize the electromagnetic field, [[17, 6]].

As already seen in subsection A.22.3 (A.128), in quantum field theory the potentials become quantum fields, i.e. operator fields. For electromagnetics the quantum field four-vector is a bit more messy

\begin{displaymath}
\widehat{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspa...
...}\cdot{\skew0\vec r}}\widehat a^\dagger _{{\vec k}\nu}\right)
\end{displaymath}

Since a four-vector has four components, a general four-vector can be written as a linear combination of four chosen basis four-vectors $\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over e}
\kern-1.3pt_{{\vec k}}^{\,0}$, $\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over e}
\kern-1.3pt_{{\vec k}}^{\,1}$, $\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over e}
\kern-1.3pt_{{\vec k}}^{\,2}$, and $\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over e}
\kern-1.3pt_{{\vec k}}^{\,3}$. (That is much like a general vector in three dimensions can be written as a linear combination of ${\hat\imath}$, ${\hat\jmath}$, and ${\hat k}$.) The four basis vectors physically represent different possible polarizations of the electromagnetic field. That is why they are typically aligned with the momentum of the wave rather than with some Cartesian axis system and its time axis. Note that each polarization vector has its own annihilation operator $\widehat a_{{\vec k}\nu}$ and creation operator $\widehat a^\dagger _{{\vec k}\nu}$. These annihilate respectively create photons with that wave number vector ${\vec k}$ and polarization.

(Electromagnetic waves in empty space are special; for them only two independent polarizations are possible. Or to be precise, even in empty space the Klein-Gordon equations with Lorenz condition allow a third polarization. But these waves produce no electric and magnetic fields and contain no net electromagnetic energy. So they are physically irrelevant. You can call them “gage equivalent to the vacuum.” That sounds better than irrelevant.)

The Lorenz condition of the previous subsection is again needed to get rid of negative energy states. The question is now exactly how to phrase the Lorenz condition in quantum terms.

(There is an epidemic among my, highly authorative, sources that come up with negative norm states without Lorenz condition. Now the present author himself is far from an expert on quantum field theories. But he knows one thing: norms cannot be negative. If you come up with negative norms, it tells you nothing about the physics. It tells you that you are doing the mahematics wrong. I believe the correct argument goes something like this: “Suppose that we can do our usual stupid canonical quantization tricks for this system. Blah Blah. That gives negative norm states. Norms cannot be negative. Ergo: we cannot do our usual stupid canonical quantization tricks for this system.” If you properly define the creation and annihilation operators to put photons in negative energy states, there is no mathematical problem. The commutator relation for the negative energy states picks up a minus sign and the norms are positive as they should. Now the mathematics is sound and you can start worrying about problems in the physics. Like that there are negative energy states. And maybe lack of Lorentz invariance, although the original system is Lorentz invariant, and I do not see what would not be Lorentz invariant about putting particles in the negative energy states.)

The simplest idea would be to require that the quantum field above satisfies the Lorenz condition. But the quantum field determines the dynamics. Like in the classical case, you do not want to change the dynamics. Instead you want to throw certain solutions away. That means that you want to throw certain wave functions $\big\vert\Psi\big\rangle $ away.

The strict condition would be to require (in the Heisenberg picture {A.12})

\begin{displaymath}
\bigg(\frac{1}{c}\frac{\partial\widetilde\varphi/c}{\parti...
...w4\widetilde{\skew3\vec A}\bigg) \big\vert\Psi\big\rangle = 0
\end{displaymath}

for all physically observable states $\big\vert\Psi\big\rangle $. Here the parenthetical expression is the operator of the Lorenz quantity that must be zero. The above requirement makes $\big\vert\Psi\big\rangle $ an eigenvector of the Lorenz quantity with eigenvalue zero. Then according to the rules of quantum mechanics, chapter 3.4, the only measurable value of the Lorenz quantity is zero.

But the above strict condition is too restrictive. Not even the vacuum state with no photons would be physically observable. That is because the creation operators in $\widehat\varphi$ and $\skew6\widehat{\skew3\vec A}$ will create nonzero photon states when applied on the vacuum state. That suggests that only the annihilation terms should be included. That then gives the “Gupta-Bleuler condition:”

\begin{displaymath}
\bigg(\frac{1}{c}\frac{\partial\widetilde\varphi^+\!/c}{\p...
...widetilde{\skew3\vec A}}^+\bigg) \big\vert\Psi\big\rangle = 0
\end{displaymath}

for physically observable states $\big\vert\Psi\big\rangle $. Here the superscript $+$ on the quantum fields means that only the $\widehat a_{{\vec k}\nu}$ annihilation operator terms are included.

You might of course wonder why the annihilation terms are indicated by a plus sign, instead of the creation terms. After all, it are the creation operators that create more photons. But the plus sign actually stands for the fact that the annihilation terms are associated with an $e^{-{\rm i}{\omega}t}$ time dependence instead of $e^{{\rm i}{\omega}t}$. Yes true, $e^{-{\rm i}{\omega}t}$ has a minus sign, not a plus sign. But $e^{-{\rm i}{\omega}t}$ has the normal sign, and normal is represented by a plus sign. Is not addition more normal than subtraction? Please do not pull at your hair like that, there are less drastic ways to save on professional hair care.

Simply dropping the creation terms may seem completely arbitrary. But it actually has some physical logic to it. Consider the inner product

\begin{displaymath}
\big\langle\Psi'\big\vert
\bigg(\frac{1}{c}\frac{\partia...
...\widetilde{\skew3\vec A}\,\bigg) \big\vert\Psi\big\rangle = 0
\end{displaymath}

This is the amount of state $\big\vert\Psi'\big\rangle $ produced by applying the Lorenz quantity on the physically observable state $\big\vert\Psi\big\rangle $. The strict condition is equivalent to saying that this inner product must always be zero; no amount of any state may be produced. For the Gupta-Bleuler condition, the above inner product remains zero if $\big\vert\Psi'\big\rangle $ is a physically observable state. (The reason is that the creation terms can be taken to the other side of the inner product as annihilation terms. Then they produce zero if $\big\vert\Psi'\big\rangle $ is physically observable.) So the Gupta-Bleuler condition implies that no amount of any physically observable state may be produced by the Lorenz quantity.

There are other ways to do quantization of the electromagnetic field. The quantization following Fermi, as discussed in subsection A.22.8, can be converted into a modern quantum field theory. But that turns out to be a very messy process indeed, [[17, 6]]. The derivation is essentially to mess around at length until you more or less prove that you can use the Lorenz condition result instead. You might as well start there.

It does turns out that the so-called path-integral formulation of quantum mechanics does a very nice job here, [52, pp. 30ff]. It avoids many of the contortions of canonical quantization like the ones above.

In fact, a popular quantum field textbook, [34, p. 79], refuses to do canonical quantization of the electromagnetic field at all, calling it an awkward subject. This book is typically used during the second year of graduate study in physics, so it is not that its readers are unsophisticated.


A.22.7 The conventional Lagrangian

Returning to the classical electromagnetic field, it still needs to be examined whether the Lorenz condition has made the universe safe for life as we know it.

The answer depends on the Lagrangian, because the Lagrangian determines the evolution of a system. So far, the Lagrangian has been written in terms of the four potentials $\varphi$ and $A_j$ (with $j$ = 1, 2, and 3) of the electromagnetic field. But recall that matter does not observe the four potentials directly. It only notices the electric field $\skew3\vec{\cal E}$ and the magnetic field $\skew2\vec{\cal B}$. So it may help to reformulate the Lagrangian in terms of the electric and magnetic fields. Concentrating on the observed fields is likely to show up more clearly what is actually observed.

With a bit of mathematical manipulation, {D.37.3}, the self-evident electromagnetic Lagrangian density can be written as:

\begin{displaymath}
\pounds _{\rm seem} = \frac{\epsilon_0}{2} \bigg({\cal E}^...
...\partial t}
+\nabla\cdot\skew3\vec A\bigg\}^2\,\bigg) + ...
\end{displaymath}

Here the dots stand for terms that do not affect the motion. (Since in the action, Lagrangian densities get integrated over space and time, terms that are pure spatial or time derivatives integrate away. The quantities relevant to the action principle vanish at the limits of integration.)

The term inside the curly brackets is zero according to the Lorenz condition (A.138). Therefore, it too does not affect the motion. (To be precise, the term does not affect the motion because it is squared. By itself it would affect the motion. In the formal way in which the Lagrangian is differentiated, one power is lost.)

The conventional Lagrangian density is found by disregarding the terms that do not change the motion:

\begin{displaymath}
\pounds _{\rm conem} = \frac{\epsilon_0}{2} \Big({\cal E}^2 - c^2 {\cal B}^2\Big)
\end{displaymath}

So the conventional Lagrangian density of the electromagnetic field is completely in terms of the observable fields.

As an aside, it might be noted that physicists find the above expression too intuitive. So you will find it in quantum field books in relativistic index notation as:

\begin{displaymath}
\pounds _{\rm conem} = - \frac{\epsilon_0}{4} F_{\mu\nu}F^{\mu\nu}
\end{displaymath}

Here the “field strength tensor” is defined by

\begin{displaymath}
F_{\mu\nu} = c \left(\partial_\mu A_\nu - \partial_\nu A_\mu\right)
\qquad \mu=0,1,2,3 \quad \nu=0,1,2,3
\end{displaymath}

Note that the indices on each $A$ are subscripts instead of superscripts as they should be. That means that you must add a minus sign whenever the index on an $A$ is 0. If you do that correctly, you will find that from the 16 $F_{\mu\nu}$ values, some are zero, while the rest are components of the electric or magnetic fields. To go from $F_{\mu\nu}$ to $F^{\mu\nu}$, you must raise both indices, so add a minus sign for each index that is zero. If you do all that the same Lagrangian density as before results.

Because the conventional Lagrangian density is different from the self-evident one, the field equations (A.135) and (A.136) for the potentials pick up a few additional terms. To find them, repeat the analysis of subsection A.22.4 but use the conventional density above in (A.134). Note that you will need to write the electric and magnetic fields in terms of the potentials using (A.131). (Using the field strength tensor is actually somewhat simpler in converting to the potentials. If you can get all the blasted sign changes right, that is.)

Then the conventional field equations become:

\begin{displaymath}
\frac{1}{c^2}\frac{\partial^2\varphi}{\partial t^2} - \nab...
...a\cdot\skew3\vec A}{\partial t}
= \frac{\rho}{\epsilon_0} %
\end{displaymath} (A.140)


\begin{displaymath}
\frac{\partial^2 \skew3\vec A}{\partial t^2} - c^2 \nabla^...
...\nabla\cdot\skew3\vec A)
= \frac{\vec\jmath}{\epsilon_0} %
\end{displaymath} (A.141)

Here $\rho$ is again the charge density and $\vec\jmath$ the current density of the charges that are around,

The additional terms in each equation above are the two before the equals signs. Note that these additional terms are zero on account of the Lorenz condition. So they do not change the solution.

The conventional field equations above are obviously more messy than the original ones. Even if you cancel the second order time derivatives in (A.140). However, they do have one advantage. If you use these conventional equations, you do not have to worry about satisfying the Lorenz condition. Any solution to the equations will give you the right electric and magnetic fields and so the right motion of the charged particles.

To be sure, the potentials will be different if you do not satisfy the Lorenz condition. But the potentials have no meaning of their own. At least not in classical electromagnetics.

To verify that the Lorenz condition is no longer needed, first recall the indeterminacy in the potentials. As subsection A.22.4 discussed, more than one set of potentials can produce the same electric and magnetic fields. In particular, given potentials $\varphi$ and $\skew3\vec A$, you can create equivalent potentials as

\begin{displaymath}
\varphi' = \varphi - \chi_t \qquad \skew3\vec A' = \skew3\vec A+ \nabla\chi
\end{displaymath}

Here $\chi$ can be any function of space and time that you want. The potentials $\varphi'$ and $\skew3\vec A'$ give the exact same electric and magnetic fields as $\varphi$ and $\skew3\vec A$. Such a transformation of potentials is called a gage transform.

Now suppose that you have a solution $\varphi$ and $\skew3\vec A$ of the conventional field equations, but it does not satisfy the Lorenz condition. In that case, simply apply a gage transform as above to get new fields $\varphi'$ and $\skew3\vec A'$ that do satisfy the Lorenz condition. To do so, write out the Lorenz condition for the new potentials,

\begin{displaymath}
\frac{1}{c^2}\frac{\partial\varphi'}{\partial t} + \nabla\...
...chi}{\partial t^2}
+ \nabla\cdot\skew3\vec A+ \nabla^2 \chi
\end{displaymath}

You can always choose the function $\chi$ to make this quantity zero. (Note that that gives an inhomogeneous Klein-Gordon equation for $\chi$.)

Now it turns out that the new potentials $\varphi'$ and $\skew3\vec A'$ still satisfy the conventional equations. That can be seen by straight substitution of the expressions for the new potentials in the conventional equations. So the new potentials are perfectly OK: they satisfy both the Lorenz condition and the conventional equations. But the original potentials $\varphi$ and $\skew3\vec A$ produced the exact same electric and magnetic fields. So the original potentials were OK too.

The evolution equation (A.140) for the electrostatic field is worth a second look. Because of the definition of the electric field (A.130), it can be written as

\begin{displaymath}
\nabla\cdot\skew3\vec{\cal E}= \frac{\rho}{\epsilon_0} %
\end{displaymath} (A.142)

That is called Maxwell’s first equation, chapter 13.2. It ties the charge density to the electric field quite rigidly.

Maxwell’s first equation is a consequence of the Lorenz condition. It would not be required for the original Klein-Gordon equations without Lorenz condition. In particular, it is the Lorenz condition that allows the additional two terms in the evolution equation (A.140) for the electrostatic potential. These then eliminate the second order time derivative from the equation. That then turns the equation from a normal evolution equation into a restrictive spatial condition on the electric field.

It may be noted that the other evolution equation (A.141) is Maxwell’s fourth equation. Just rewrite it in terms of the electric and magnetic fields. The other two Maxwell equations follow trivially from the definitions (A.130) of the electric and magnetic fields in terms of the potentials.

Since there is no Lorenz condition for the conventional equations, it becomes interesting to find the corresponding Hamiltonian. That should allow the stability of electromagnetics to be examined more easily.

The Hamiltonian for electromagnetic field plus a proton may be found the same way as (A.137) in subsection A.22.5, {A.1.5}. Just use the conventional Lagrangian density instead. That gives

\begin{displaymath}
H_{\rm conem+p} = \int\Big( \frac{\epsilon_0}{2}
({\cal ...
...
+ {\textstyle\frac{1}{2}} m_{\rm {p}} v_{\rm {p}}\strut_j^2
\end{displaymath}

But the proton charge density $\rho_{\rm {p}}$ may eliminated using Maxwell’s first equation above. An additional integration by parts of that term then causes it to drop away against the previous term. That gives the conventional energy as
\begin{displaymath}
E_{\rm conem+p} = \frac{\epsilon_0}{2} \int({\cal E}^2 + c...
... + {\textstyle\frac{1}{2}}m_{\rm {p}}\vec v_{\rm {p}}^{\,2} %
\end{displaymath} (A.143)

The first term is the energy in the observable fields and the final term is the kinetic energy of the proton.

The simplified energy above is no longer really a Hamiltonian; you cannot write Hamilton’s equations based on it as in {A.1.5}. But it does still give the energy that is conserved.

The energy above is always positive. So it can no longer be lowered by arbitrary amounts. The system will not blow up. And that then means that the original Klein-Gordon equations (A.135) and (A.136) for the fields are stable too as long as the Lorenz condition is satisfied. They produce the same evolution. And they satisfy the speed of light restriction and are Lorentz invariant. Lorenz did it!

Note also the remarkable result that the interaction energy between proton charge and field has disappeared. The proton can no longer minimize any energy of interaction between itself and the field it creates. Maxwell’s first equation is too restrictive. All the proton can try to do is minimize the energy in the electric and magnetic fields.


A.22.8 Quantization following Fermi

Quantizing the electromagnetic field is not easy. The previous subsection showed a couple of problems. The gage property implies that the electromagnetic potentials $\varphi$ and $\skew3\vec A$ are indeterminate. Also, taking the Lorenz condition into account, the second order time derivative is lost in the Klein-Gordon equation for the electrostatic potential $\varphi$. The equation turns into Maxwell’s first equation,

\begin{displaymath}
\nabla\cdot\skew3\vec{\cal E}= \frac{\rho}{\epsilon_0}
\end{displaymath}

That is not an evolution equation but a spatial constraint for the electric field $\skew3\vec{\cal E}$ in terms of the charge density $\rho$.

Various ways to deal with that have been developed. The quantization procedure discussed in this subsection is a simplified version of the one found in Bethe’s book, [6, pp. 255-271]. It is due to Fermi, based on earlier work by Dirac and Heisenberg & Pauli. This derivation was a great achievement at the time, and fundamental to more advanced quantum field approaches, [6, p. 266]. Note that all five mentioned physicists received a Nobel Prize in physics at one time or the other.

The starting point in this discussion will be the original potentials $\varphi$ and $\skew3\vec A$ of subsection A.22.4. The ones that satisfied the Klein-Gordon equations (A.135) and (A.136) as well as the Lorenz condition (A.138).

It was Fermi who recognized that you can make things a lot simpler for yourself if you write the potentials as sums of exponentials of the form $e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}$:

\begin{displaymath}
\varphi = \sum_{{\rm all}\ {\vec k}} c_{\vec k}e^{{\rm i}{...
...vec k}} \vec d_{\vec k}e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}
\end{displaymath}

That is the same trick as was used in quantizing the Koulomb potential in subsection A.22.3. However, in classical mechanics you do not call these exponentials momentum eigenstates. You call them Fourier modes. The principle is the same. The constant vector ${\vec k}$ that characterizes each exponential is still called the wave number vector. Since the potentials considered here vary with time, the coefficients $c_{\vec k}$ and $\vec{d}_{\vec k}$ are functions of time.

Note that the coefficients $\vec{d}_{\vec k}$ are vectors. These will have three independent components. So the vector potential can be written more explicitly as

\begin{displaymath}
\skew3\vec A= \sum_{{\rm all}\ {\vec k}} d_{1,{\vec k}\,} ...
... \vec e_{3,{\vec k}\,} e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}
\end{displaymath}

where $\vec{e}_{1,{\vec k}}$, $\vec{e}_{2,{\vec k}}$, and $\vec{e}_{3,{\vec k}}$ are unit vectors. Fermi proposed that the smart thing to do is to take the first of these unit vectors in the same direction as the wave number vector ${\vec k}$. The corresponding electromagnetic waves are called longitudinal. The other two unit vectors should be orthogonal to the first component and to each other. That still leaves a bit choice in direction. Fortunately, in practice it does not really make a difference exactly how you take them. The corresponding electromagnetic waves are called transverse.

In short, the fields can be written as

\begin{displaymath}
\varphi = \sum_{{\rm all}\ {\vec k}} c_{{\vec k}\,} e^{{\r...
...vec e_{3,{\vec k}\,} e^{{\rm i}{\vec k}\cdot{\skew0\vec r}} %
\end{displaymath} (A.144)

where

\begin{displaymath}
\vec e_{1,{\vec k}\,} = \frac{{\vec k}}{k} \qquad \vec e_{...
...vec k}= \vec e_{2,{\vec k}\,} \cdot \vec e_{3,{\vec k}\,} = 0
\end{displaymath}

From those expressions, and the directions of the unit vectors, it can be checked by straight substitution that the curl of the longitudinal potential is zero:

\begin{displaymath}
\mathop{\rm curl}\nolimits \skew3\vec A_\parallel \equiv
...
...a\times\skew3\vec A_\parallel = 0 \quad \mbox{(irrotational)}
\end{displaymath}

A vector field with zero curl is called “irrotational.” (The term can be understood from fluid mechanics; there the curl of the fluid velocity field gives the local average angular velocity of the fluid.)

The same way, it turns out that that the divergence of the transverse potential is zero

\begin{displaymath}
\div\skew3\vec A_\perp \equiv \nabla\cdot\skew3\vec A_\perp = 0 \quad \mbox{(solenoidal)}
\end{displaymath}

A field with zero divergence is called “solenoidal.” (This term can be understood from magnetostatics; a magnetic field, like the one produced by a solenoid, an electromagnet, has zero divergence.)

To be fair, Fermi did not really discover that it can be smart to take vector fields apart into irrotational and solenoidal parts. That is an old trick known as the “Helmholtz decomposition.”

Since the transverse potential has no divergence, the longitudinal potential is solely responsible for the Lorenz condition (A.138). The transverse potential can do whatever it wants.

The real problem is therefore with the longitudinal potential $\skew3\vec A_\parallel$ and the electrostatic potential $\varphi$. Bethe [6] deals with these in terms of the Fourier modes. However, that requires some fairly sophisticated analysis. It is actually easier to return to the potentials themselves now.

Reconsider the expressions (A.130) for the electric and magnetic fields in terms of the potentials. They show that the electrostatic potential produces no magnetic field. And neither does the longitudinal potential because it is irrotational.

They do produce a combined electric field $\skew3\vec{\cal E}_{\varphi\parallel}$. But this electric field is irrotational, because the longitudinal potential is, and the gradient $\nabla$ of any scalar function is. That helps, because then the Stokes theorem of calculus implies that the electric field $\skew3\vec{\cal E}_{\varphi\parallel}$ is minus the gradient of some scalar potential:

\begin{displaymath}
\skew3\vec{\cal E}_{\varphi\parallel} = - \nabla\varphi_{\rm {C}}
\end{displaymath}

Note that normally $\varphi_{\rm {C}}$ is not the same as the electrostatic potential $\varphi$, since there is also the longitudinal potential. To keep them apart, $\varphi_{\rm {C}}$ will be called the Coulomb potential.

As far as the divergence of the electric field $\skew3\vec{\cal E}_{\varphi\parallel}$ is concerned, it is the same as the divergence of the complete electric field. The reason is that the transverse field has no divergence. And the divergence of the complete electric field is given by Maxwell’s first equation. Together these observations give

\begin{displaymath}
\skew3\vec{\cal E}_{\varphi\parallel} = - \nabla\varphi_{\...
...l} =
- \nabla^2 \varphi_{\rm {C}} = \frac{\rho}{\epsilon_0}
\end{displaymath}

Note that the final equation is a Poisson equation for the Coulomb potential.

Now suppose that you replaced the electrostatic field $\varphi$ with the Coulomb potential $\varphi_{\rm {C}}$ and had no longitudinal field $\skew3\vec A_\parallel$ at all. It would give the same electric and magnetic fields. And they are the only ones that are observable. They give the forces on the particles. The potentials are just mathematical tools in classical electromagnetics.

So why not? To be sure, the combination of the Coulomb potential $\varphi_{\rm {C}}$ and remaining vector potential $\skew3\vec A_\perp$ will no longer satisfy the Lorenz condition. But who cares?

Instead of the Lorenz condition, the combination of Coulomb potential plus transverse potential satisfies the so-called “Coulomb condition:”

\begin{displaymath}
\fbox{$\displaystyle
\nabla\cdot\skew3\vec A= 0
$}
\end{displaymath} (A.145)

The reason is that now $\skew3\vec A$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\skew3\vec A_\perp$ and the transverse vector potential has no divergence. Physicists like to say that the original potentials used the Lorenz gage, while the new ones use the Coulomb gage.

Because the potentials $\varphi_{\rm {C}}$ and $\skew3\vec A_\perp$ do no longer satisfy the Lorenz condition, the Klein-Gordon equations (A.135) and (A.136) do no longer apply. But the conventional equations (A.140) and (A.141) do still apply; they do not need the Lorenz condition.

Now consider the Coulomb potential somewhat closer. As noted above it satisfies the Poisson equation

\begin{displaymath}
- \nabla^2 \varphi_{\rm {C}} = \frac{\rho}{\epsilon_0}
\end{displaymath}

The solution to this equation was already found in the first subsection, (A.107). If the charge distribution $\rho$ consists of a total of $I$ point charges, it is
\begin{displaymath}
\varphi_{\rm {C}}({\skew0\vec r};t) = \sum_{i=1}^I \frac{q_i}{4\pi\epsilon_0\vert{\skew0\vec r}-{\skew0\vec r}_i\vert}
\end{displaymath} (A.146)

Here $q_i$ is the charge of point charge number $i$, and ${\skew0\vec r}_i$ its position.

If the charge distribution $\rho$ is smoothly distributed, simply take it apart in small point charges $\rho({\underline{\skew0\vec r}};t){\rm d}^3{\underline{\skew0\vec r}}$. That gives

\begin{displaymath}
\varphi_{\rm {C}}({\skew0\vec r};t) = \int_{{\rm all\ }{\u...
...ine{\skew0\vec r}}\vert}{\,\rm d}^3{\underline{\skew0\vec r}}
\end{displaymath} (A.147)

The key point to note here is that the Coulomb potential has no life of its own. It is rigidly tied to the positions of the charges. That then provides the most detailed answer to the question: “What happened to energy minimization?” Charged particles have no option of minimizing any energy of interaction with the field. Maxwell’s first equation, the Poisson equation above, forces them to create a Coulomb field that is repulsive to them. Whether they like it or not.

Note further that all the mechanics associated with the Coulomb field is quasi-steady. The Poisson equation does not depend on how fast the charged particles evolve. The Coulomb electric field is minus the spatial gradient of the potential, so that does not depend on the speed of evolution either. And the Coulomb force on the charged particles is merely the electric field times the charge.

It is still not obvious how to quantize the Coulomb potential, even though there is no longer a longitudinal field. But who cares about the Coulomb potential in the first place? The important thing is how the charged particles are affected by it. And the forces on the particles caused by the Coulomb potential can be computed using the electrostatic potential energy, {D.37.4},

\begin{displaymath}
V_{\rm C} = {\textstyle\frac{1}{2}} \sum_{i=1}^I\sum_{\tex...
...\vert{\skew0\vec r}_i-{\skew0\vec r}_{{\underline i}}\vert} %
\end{displaymath} (A.148)

For example, this is the Coulomb potential energy that was used to find the energy levels of the hydrogen atom in chapter 4.3. It can still be used in unsteady motion because everything associated with the Coulomb potential is quasi-steady. Sure, it is due to the interaction of the particles with the electromagnetic field. But where in the above mathematical expression does it say electromagnetic field? All it contains are the coordinates of the charged particles. So what difference does it make where the potential energy comes from? Just add the energy above to the Hamiltonian and then pretend that there are no electrostatic and longitudinal fields.

Incidentally, note the required omission of the terms with ${\underline i}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i$ in the potential energy above. Otherwise you would get infinite energy. In fact, a point charge in classical electromagnetics does have infinite Coulomb energy. Just take any of the point charges and mentally chop it up into two equal parts sitting at the same position. The interaction energy between the halves is infinite.

The issue does not exist if the charge is smoothly distributed. In that case the Coulomb potential energy is, {D.37.4},

\begin{displaymath}
V_{\rm C} = {\textstyle\frac{1}{2}} \int_{{\rm all\ }{\ske...
...\,\rm d}^3{\skew0\vec r}{\rm d}^3{\underline{\skew0\vec r}} %
\end{displaymath} (A.149)

While the integrand is infinite at ${\underline{\skew0\vec r}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}$, the integral remains finite.

So the big idea is to throw away the electrostatic and longitudinal potentials and replace them with the Coulomb energy $V_{\rm {C}}$, origin unknown. Now it is mainly a matter of working out the details.

First, consider the Fermi Lagrangian. It is found by throwing out the electrostatic and longitudinal potentials from the earlier Lagrangian (A.134) and subtracting $V_{\rm {C}}$. That gives, using the point charge approximation (A.133) and in vector notation,

\begin{displaymath}
\Lag_{\rm F} = \frac{\epsilon_0}{2}
\int \bigg[\left\ver...
... {\textstyle\frac{1}{2}} m_i \vec v_i^{\,2} \Big] - V_{\rm C}
\end{displaymath} (A.150)

Note that it is now assumed that there are $I$ particles instead of just the single proton in (A.134). Because $i$ is already used to index the particles, ${\underline j}$ is used to index the three directions of spatial differentiation. The Coulomb energy $V_{\rm {C}}$ was already given in (A.148). The velocity of particle $i$ is $\vec{v}_i$, while $q_i$ is its charge and $m_i$ its mass. The subscript $i$ on the transverse potential in the interaction term indicates that it is evaluated at the location of particle $i$.

You may wonder how you can achieve that only the transverse potential $\skew3\vec A_\perp$ is left. That would indeed be difficult to do if you work in terms of spatial coordinates. The simplest way to handle it is to work in terms of the transverse waves (A.144). They are transverse by construction.

The unknowns are now no longer the values of the potential at the infinitely many possible positions. Instead the unknowns are now the coefficients $d_{2,{\vec k}}$ and $d_{3,{\vec k}}$ of the transverse waves. Do take into account that since the field is real,

\begin{displaymath}
d_{2,-{\vec k}} = d_{2,{\vec k}}^* \qquad d_{3,-{\vec k}} = d_{3,{\vec k}}^*
\end{displaymath}

So the number of independent variables is half of what it seems. The most straightforward way of handling this is to take the unknowns as the real and imaginary parts of the $d_{2,{\vec k}}$ and $d_{3,{\vec k}}$ for half of the ${\vec k}$ values. For example, you could restrict the ${\vec k}$ values to those for which the first nonzero component is positive. The corresponding unknowns must then describe both the ${\vec k}$ and $\vphantom0\raisebox{1.5pt}{$-$}$${\vec k}$ waves.

(The ${\vec k}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 terms are awkward. One way to deal with it is to take an adjacent periodic box and reverse the sign of all the charges and fields in it. Then take the two boxes together to be a new bigger periodic box. The net effect of this is to shift the mesh of ${\vec k}$-​values figure 6.17 by half an interval. That means that the ${\vec k}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 terms are gone. And other problems that may arise if you sum over all boxes, like to find the total Coulomb potential, are gone too. Since the change in ${\vec k}$ values becomes zero in the limit of infinite box size, all this really amounts to is simply ignoring the ${\vec k}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 terms.)

The Hamiltonian can be obtained just like the earlier one (A.137), {A.1.5}. (Or make that {A.1.4}, since the unknowns, $d_{2,{\vec k}}$ and $d_{3,{\vec k}}$, are now indexed by the discrete values of the wave number ${\vec k}$.) But this time it really needs to be done right, because this Hamiltonian is supposed to be actually used. It is best done in terms of the components of the potential and velocity vectors. Using $j$ to index the components, the Lagrangian becomes

\begin{displaymath}
\Lag_{\rm F} = \frac{\epsilon_0}{2}
\int \sum_{j=1}^3 \b...
...+ {\textstyle\frac{1}{2}} m_i v_i\strut_j^2 \Big] - V_{\rm C}
\end{displaymath}

Now Hamiltonians should not be in terms of particle velocities, despite what (A.137) said. Hamiltonians should be in terms of canonical momenta, {A.1.4}. The canonical momentum corresponding to the velocity component $v_i\strut_j$ of a particle $i$ is defined as

\begin{displaymath}
p^{\rm {c}}_i\strut_j \equiv \frac{\partial {\cal L}}{\partial v_i\strut_j}
\end{displaymath}

Differentiating the Lagrangian above gives

\begin{displaymath}
p^{\rm {c}}_i\strut_j = m_i v_i\strut_j + q_i A_j\strut_i
\end{displaymath}

It is this canonical momentum that in quantum mechanics gets replaced by the operator $\hbar\partial$$\raisebox{.5pt}{$/$}$${\rm i}\partial{r}_j$. That is important since, as the above expression shows, canonical momentum is not just linear momentum in the presence of an electromagnetic field.

The time derivatives of the real and imaginary parts of the coefficients $d_{2,{\vec k}}$ and $d_{3,{\vec k}}$ should be replaced by similarly defined canonical momenta. However, that turns out to be a mere rescaling of these time derivatives.

The Hamiltonian then becomes, following {A.1.4} and in vector notation,

 $\displaystyle H_{\rm F}$ $\textstyle =$ $\displaystyle \frac{\epsilon_0}{2}
\int \bigg[\left\vert\skew3\vec A_\perp\stru...
...\vec A_\perp\strut_{\underline j}\right\vert^2 \bigg] {\,\rm d}^3{\skew0\vec r}$   
     $\displaystyle \mbox{} + \sum_{i=1}^I \frac{({\skew0\vec p}^{\,\rm {c}}_i - q_i ...
...}}{4\pi\epsilon_0\vert{\skew0\vec r}_i-{\skew0\vec r}_{{\underline i}}\vert}%
$  (A.151)

Note in particular that the center term is the kinetic energy of the particles, but in terms of their canonical momenta.

In terms of the waves (A.144), the integral falls apart in separate contributions from each $d_{2,{\vec k}}$ and $d_{3,{\vec k}}$ mode. That is a consequence of the orthogonality of the exponentials, compare the Parseval identity in {A.26}. (Since the exponentials are complex, the absolute values in the integral are now required.) As a result, the equations for different coefficients are only indirectly coupled by the interaction with the charged particles. In particular, it turns out that each coefficient satisfies its own harmonic oscillator equation with forcing by the charged particles, {A.1.4},

\begin{displaymath}
\epsilon_0 {\cal V}(\ddot d_{j,{\vec k}} + k^2c^2 d_{j,{\v...
...vec k}\cdot{\skew0\vec r}_i}
\quad\mbox{for $j$\ = 2 and 3}
\end{displaymath}

If the speed of the particle gets comparable to the speed of light, you may want to use the relativistic energy (1.2);

\begin{displaymath}
\frac{({\skew0\vec p}^{\,\rm {c}}_i - q_i \skew3\vec A_\pe...
...vec p}^{\,\rm {c}}_i - q_i \skew3\vec A_\perp\strut_i)^2 c^2}
\end{displaymath}

Sometimes, it is convenient to assume that the system under consideration also experiences an external electromagnetic field. For example, you might consider an atom or atomic nucleus in the magnetic field produced by an electromagnet. You probably do not want to include every electron in the wires of the electromagnet in your model. That would be something else. Instead you can simply add the vector potential $\skew3\vec A_{\rm {ext}}\strut_i$ that they produce to $\skew3\vec A_\perp\strut_i$ in the Hamiltonian. If there is also an external electrostatic potential, add a separate term $q_i{\varphi_{\rm {ext}}}_i$ to the Hamiltonian for each particle $i$. The external fields will be solutions of the homogeneous evolution equations (A.140) and (A.141), (i.e. the equations without charge and current densities). However, the external fields will not vanish at infinity; that is why they can be nonzero without charge and current densities.

Note that the entire external vector potential is needed, not just the transverse part. The longitudinal part is not included in $V_{\rm {C}}$. Bethe [6, p. 266] also notes that the external field should satisfy the Lorenz condition. No further details are given. However, at least in various simple cases, a gage transform that kills off the Lorenz condition may be applied. See for example the gage property for a pure external field {A.19.5}. In the classical case a gage transform of the external fields does not make a difference either, because it does not change either the Lagrangian equations for the transverse field nor those for the particles. Using the Lorenz condition cannot hurt, anyway.

Particle spin, if any, is not included in the above Hamiltonian. At nonrelativistic speeds, its energy can be described as a dot produce with the local magnetic field, chapter 13.4.

So far all this was classical electrodynamics. But the interaction between the charges and the transverse waves can readily be quantized using essentially the same procedure as used for the Koulomb potential in subsection A.22.3. The details are worked out in addendum {A.23} for the fields. It allows a relativistic description of the emission of electromagnetic radiation by atoms and nuclei, {A.24} and {A.25}.

While the transverse field must be quantized, the Coulomb potential can be taken unchanged into quantum mechanics. That was done, for example, for the nonrelativistic hydrogen atom in chapter 4.3 and for the relativistic one in addendum {D.82}.

Finally, any external fields are assumed to be given; they are not quantized either.

Note that the Fermi quantization is not fully relativistic. In a fully relativistic theory, the particles too should be described by quantum fields. The Fermi quantization does not do that. So even the relativistic hydrogen atom is not quite exact, even though it is orders of magnitude more accurate than the already very accurate nonrelativistic one. The energy levels are still wrong by the so-called Lamb shift, {A.38} But this is an extremely tiny effect. Little in life is perfect, isn’t it?


A.22.9 The Coulomb potential and the speed of light

The Coulomb potential

\begin{displaymath}
\varphi_{\rm {C}}({\skew0\vec r};t) =
\sum_{i=1}^I \frac{q_i}{4\pi\epsilon_0\vert{\skew0\vec r}- {\skew0\vec r}_i\vert}
\end{displaymath}

does not respect the speed of light $c$. Move a charge, and the Coulomb potential immediately changes everywhere in space. However, special relativity says that an event may not affect events elsewhere unless these events are reachable by the speed of light. Something else must prevent the use of the Coulomb potential to transmit observable effects at a speed greater than that of light.

To understand what is going on, assume that at time zero some charges at the origin are given a well-deserved kick. As mentioned earlier, the Klein-Gordon equations respect the speed of light. Therefore the original potentials $\varphi$ and $\skew3\vec A$, the ones that satisfied the Klein-Gordon equations and Lorenz condition, are unaffected by the kick beyond a distance $ct$ from the origin. The original potentials do respect the speed of light.

The Coulomb potential above, however, includes the longitudinal part $\skew3\vec A_\parallel$ of the vector potential $\skew3\vec A$. As the Coulomb potential reflects, $\skew3\vec A_\parallel$ does change immediately all the way up to infinity. But the transverse part $\skew3\vec A_\perp$ also changes immediately all the way up to infinity. Beyond the limit dictated by the speed of light, the two parts of the potential exactly cancel each other. As a result, beyond the speed of light limit, the net vector potential $\skew3\vec A$ does not change.

The bottom line is

The mathematics of the Helmholtz decomposition of $\skew3\vec A$ into $\skew3\vec A_\parallel$ and $\skew3\vec A_\perp$ hides, but of course does not change, the limitation imposed by the speed of light.
The limitation is still there, it is just much more difficult to see. The change in current density $\vec\jmath$ caused by kicking the charges near the origin is restricted to the immediate vicinity of the origin. But both the longitudinal part $\vec\jmath_\parallel$ and the transverse part $\vec\jmath_\perp$ extend all the way to infinity. And then so do the longitudinal and transverse potentials. It is only when you add the two that you see that the sum is zero beyond the speed of light limit.