A.21 Pho­ton type 2 wave func­tion

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

A.21.1 The wave func­tion

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

$$\mbox{electron:} \quad \skew{-1}\vec\Psi({\skew0\vec r};t)... ...0\vec r};t) \\ \Psi_-({\skew0\vec r};t) \end{array} \right)$$

This wave func­tion takes on two dif­fer­ent mean­ings

1.

It gives the prob­a­bil­ity per unit vol­ume of find­ing the elec­tron at a given po­si­tion with a given spin. For ex­am­ple, $\vert\Psi_+({\skew0\vec r};t)\vert^2{ \rm d}^3{\skew0\vec r}$ gives the prob­a­bil­ity of find­ing the elec­tron with spin-up in an vicin­ity of in­fin­i­tes­i­mal vol­ume ${ \rm d}^3{\skew0\vec r}$ around po­si­tion ${\skew0\vec r}$. That is the Born sta­tis­ti­cal in­ter­pre­ta­tion.

2.

It is the un­ob­serv­able func­tion that na­ture seems to use to do its quan­tum com­pu­ta­tions of how physics be­haves.

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

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

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

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

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

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

But quan­tum me­chan­ics uses po­ten­tials, not forces. For ex­am­ple, the so­lu­tion of the hy­dro­gen atom of chap­ter 4.3 used a po­ten­tial en­ergy of the elec­tron $V$. In elec­tro­sta­t­ics, this po­ten­tial en­ergy is writ­ten as $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-e\varphi$ where $\vphantom{0}\raisebox{1.5pt}{$-$}$$e$ is the charge of the elec­tron and $\varphi$ is called the elec­tro­sta­tic po­ten­tial. This po­ten­tial is not di­rectly ob­serv­able nor unique; you can add any con­stant to it with­out chang­ing the ob­served physics.

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

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

$$\skew3\vec{\cal E}= - \nabla \varphi - \frac{\partial \skew... ...l t} \qquad \skew2\vec{\cal B}= \nabla \times \skew3\vec A %$$ (A.86)

Here the op­er­a­tor

$$\nabla = {\hat\imath}\frac{\partial}{\partial x} + {\hat\j... ...ac{\partial}{\partial y} + {\hat k}\frac{\partial}{\partial z}$$

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

$${\cal E}_z = - \frac{\partial\varphi}{\partial z} - \frac{\... ...ac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}$$

When both po­ten­tials are al­lowed for, the nonunique­ness be­comes much larger. In par­tic­u­lar, for any ar­bi­trary func­tion $\chi$ of po­si­tion and time, you can find two dif­fer­ent po­ten­tials $\varphi'$ and $\skew3\vec A'$ that pro­duce the ex­act same elec­tric and mag­netic fields as $\varphi$ and $\skew3\vec A$. These po­ten­tials are given by

$$\varphi' = \varphi - \frac{\partial\chi}{\partial t} \qquad \skew3\vec A' = \skew3\vec A+ \nabla \chi %$$ (A.87)

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

Fi­nally, it turns out that clas­si­cal rel­a­tivis­tic me­chan­ics likes to com­bine the four scalar po­ten­tials in a four-di­men­sion­al vec­tor, or four-vec­tor, chap­ter 1.3.2:

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

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

The pho­ton wave func­tion ${\buildrel\raisebox{-1.5pt}[0pt][0pt] {\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}_{\kern-1pt\gamma}$ should not be con­fused with the clas­si­cal four-po­ten­tial ${\buildrel\raisebox{-1.5pt}[0pt][0pt] {\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}$.

Wave func­tions are in gen­eral com­plex. The clas­si­cal four-po­ten­tial, and es­pe­cially its phys­i­cally ob­serv­able de­riv­a­tives, the elec­tric and mag­netic fields, must be real. In­deed, ac­cord­ing to quan­tum me­chan­ics, ob­serv­able quan­ti­ties cor­re­spond to eigen­val­ues of Her­mit­ian op­er­a­tors, not to wave func­tions. What the op­er­a­tors of the ob­serv­able elec­tric and mag­netic fields are will be dis­cussed in ad­den­dum {A.23}.

A.21.2 Sim­pli­fy­ing the wave func­tion

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

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

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

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

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

$$- \frac{1}{c^2} \frac{\partial^2\chi}{\partial t^2} + \nabl... ...rphi_\gamma'/c}{\partial t} + \nabla\cdot\skew3\vec A_\gamma'$$

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

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

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

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

Maxwell’s clas­si­cal elec­tro­mag­net­ics pro­vides ad­di­tional sup­port for that idea. There the Klein-Gor­don equa­tion for the po­ten­tials also re­quires that the Lorenz con­di­tion is sat­is­fied, {A.37}.

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

And so the fully sim­pli­fied pho­ton wave func­tion be­comes:

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

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

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

A.21.3 Pho­ton spin

Now that the pho­ton wave func­tion has been sim­pli­fied the pho­ton spin can be de­ter­mined. Re­call that for the elec­tron, the two com­po­nents of the wave func­tion cor­re­spond to its two pos­si­ble val­ues of the spin an­gu­lar mo­men­tum ${\widehat S}_z$ in the cho­sen $z$-​di­rec­tion. In par­tic­u­lar, $\Psi_+$ cor­re­sponds to ${\widehat S}_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\hbar$, and $\Psi_-$ to ${\widehat S}_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\frac12\hbar$. Since the wave func­tion of the pho­ton is a four-di­men­sion­al vec­tor, at first it might there­fore look like the pho­ton should have spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. That would make ${\widehat S}_z$ one of $\frac32\hbar$, $\frac12\hbar$, $-\frac12\hbar$, or $-\frac32\hbar$. But that is not true.

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

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

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

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

A.21.4 En­ergy eigen­states

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

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

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

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

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

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

A.21.5 Nor­mal­iza­tion of the wave func­tion

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

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

$$\int_{\rm all} \left( \left\vert \frac{\partial\skew3\vec... ...ight) { \rm d}^3{\skew0\vec r}= \mbox{the same for all time}$$

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

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

$$\int_{\rm all} \left(\left\vert\skew3\vec{\cal E}\right\ver... ...ight) { \rm d}^3{\skew0\vec r}= \mbox{the same for all time}$$

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

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

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

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

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

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

$$\skew3\vec{\cal E}_\gamma^{\rm n}= {\rm i}k c \skew3\vec A_... ...l B}_\gamma^{\rm n}= \nabla \times \skew3\vec A_\gamma^{\rm n}$$

(To be sure, clas­si­cal physics would take $\skew3\vec{\cal E}$ to be mi­nus the time de­riv­a­tive of the po­ten­tial $\skew3\vec A$. But for an en­ergy eigen­state, the time de­riv­a­tive gives a sim­ple fac­tor $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm i}\omega$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$${{\rm i}}kc$.) The func­tions $\skew3\vec{\cal E}_\gamma^{\rm n}$ and $\skew2\vec{\cal B}_\gamma^{\rm n}$ will be re­ferred to as “un­ob­serv­able fields” to avoid con­fu­sion with the ob­serv­able elec­tric and mag­netic fields.

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

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

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

A.21.6 States of def­i­nite lin­ear mo­men­tum

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

In that case, the pho­ton wave func­tion takes the form

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

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

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

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

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

There is an­other use­ful way to write the wave func­tion:

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

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

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

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

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

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

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

$$\skew3\vec A_\gamma^{\rm {e}} = {\hat k}e^{{\rm i}k y}$$

The nor­mal­ized eigen­func­tion and un­ob­serv­able fields are in that case

$$\fbox{$\displaystyle \skew3\vec A_\gamma^{\rm n}= \frac{\v... ...psilon_k = \sqrt{\frac{\hbar\omega}{\epsilon_0{\cal V}}} $} %$$ (A.95)

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

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

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

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

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

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

A.21.7 States of def­i­nite an­gu­lar mo­men­tum

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(The ex­pres­sion for the mag­netic field arises be­cause for a so­le­noidal vec­tor $\nabla$ $\times$ $\nabla\times$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$\nabla^2$, and that pro­duces a fac­tor $k^2$ ac­cord­ing to the en­ergy eigen­value prob­lem.)

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

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

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

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

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

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

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

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

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