Sub­sec­tions

### D.36 Pho­ton wave func­tion de­riva­tions

The rules of en­gage­ment are as fol­lows:

• The Carte­sian axes are num­bered us­ing an in­dex , with 1, 2, and 3 for , , and re­spec­tively.
• Also, in­di­cates the co­or­di­nate in the di­rec­tion, , , or .
• De­riv­a­tives with re­spect to a co­or­di­nate are in­di­cated by a sim­ple sub­script .
• If the quan­tity be­ing dif­fer­en­ti­ated is a vec­tor, a comma is used to sep­a­rate the vec­tor in­dex from dif­fer­en­ti­a­tion ones.
• In­dex is the num­ber im­me­di­ately fol­low­ing in the cyclic se­quence ...123123...and is the num­ber im­me­di­ately pre­ced­ing .
• Time de­riv­a­tives are in­di­cated by a sub­script t.
• A bare in­te­gral sign is as­sumed to be an in­te­gra­tion over all space, or over the en­tire box for par­ti­cles in a box. The is nor­mally omit­ted for brevity and to be un­der­stood.
• A su­per­script in­di­cates a com­plex con­ju­gate.

#### D.36.1 Rewrit­ing the en­ergy in­te­gral

As given in the text, the en­ergy in an elec­tro­mag­netic field in free space that sat­is­fies the Coulomb-Lorenz gauge is, writ­ing out the square mag­ni­tudes and in­di­vid­ual com­po­nents,

How­ever, a bit more gen­eral ex­pres­sion is de­sir­able. If only the Lorenz con­di­tion is sat­is­fied, there may also be an elec­tro­sta­tic po­ten­tial . In that case, a more gen­eral ex­pres­sion for the en­ergy is:
The mi­nus sign for the terms ap­pears be­cause this is re­ally a dot prod­uct of rel­a­tivis­tic four-vec­tors. The ze­roth com­po­nents in such a dot prod­uct ac­quire a mi­nus sign, chap­ter 1.2.4 and 1.3.2. In de­riva­tion {D.32} it was shown that each of the four in­te­grals is con­stant. That is be­cause each com­po­nent sat­is­fies the Klein-Gor­don equa­tion. So their sum is con­stant too.

The claim to ver­ify now is that the same en­ergy can be ob­tained from in­te­grat­ing the elec­tric and mag­netic fields as

Since and :

From now on, it will be un­der­stood that there is a sum­ma­tion over and and that every­thing has a . There­fore these will no longer be shown.

Start with the elec­tric field in­te­gral. It is, us­ing the above ex­pres­sions and mul­ti­ply­ing out,

The first term al­ready gives the vec­tor-po­ten­tial time de­riv­a­tives in (1). That leaves the fi­nal three terms. Per­form an in­te­gra­tion by parts on the first two. It will al­ways be as­sumed that the po­ten­tials van­ish at in­fin­ity or that the sys­tem is in a pe­ri­odic box. In that case there are no bound­ary terms in an in­te­gra­tion by parts. So the three terms be­come

How­ever, the di­ver­gence is ac­cord­ing to the Lorenz con­di­tion equal to , so

Us­ing the Klein-Gor­don equa­tion, , and then an­other in­te­gra­tion by parts on the first two terms and reno­tat­ing by gives the terms in (1).

Now con­sider the in­te­gral of in (2). You get, mul­ti­ply­ing out,

Now the first and last terms in the right hand side summed over pro­duce all terms in (1) in which and are dif­fer­ent. That leaves the mid­dle terms. An in­te­gra­tion by parts yields

Reno­tate the in­dices cycli­cally to get

(If you want, you can check that this is the same by writ­ing out all three terms in the sum.) This is equiv­a­lent to

as you can see from dif­fer­en­ti­at­ing and mul­ti­ply­ing out. The fi­nal term gives af­ter in­te­gra­tion by parts the terms in (1) in which and are equal. That leaves the first part. The term in paren­the­ses is the di­ver­gence , so the first part is

Per­form an in­te­gra­tion by parts

Rec­og­niz­ing once more the di­ver­gence, this gives the fi­nal term in (1)

#### D.36.2 An­gu­lar mo­men­tum states

The rules of en­gage­ment listed at the start of this sec­tion ap­ply. In ad­di­tion:

• The quan­tum num­bers and will be reno­tated by and , while stays . That is eas­ier to type.
• The quan­tum num­bers are not shown un­less needed. For ex­am­ple, stands for .
• A bar on a quan­tity, like in , means the com­plex con­ju­gate. In ad­di­tion, the (un­listed) quan­tum num­bers of spher­i­cal har­monic may in gen­eral be dif­fer­ent from those of and are in­di­cated by bars too.
• The sym­bols and are used as generic scalar func­tions. They of­ten stand in par­tic­u­lar for the scalar modes.
• An in­te­gral in spher­i­cal co­or­di­nates takes the form where .

##### D.36.2.1 About the scalar modes

The scalar modes are the .

It will be as­sumed that the are zero at the large ra­dius at which the do­main is as­sumed to ter­mi­nate. That makes the scalar modes a com­plete set; any scalar func­tion can be writ­ten as a com­bi­na­tion of them. (That is be­cause they are the eigen­func­tions of the Lapla­cian in­side the sphere, and the zero bound­ary con­di­tion on the sphere sur­face makes the Lapla­cian Her­mit­ian. This will not be ex­plic­itly proved since it is very stan­dard.)

The Bessel func­tion of the scalar modes sat­isfy the or­di­nary dif­fer­en­tial equa­tion, {A.6}

The fol­low­ing in­te­gral is needed (note that is real):
This is valid for large , which ap­plies since is large and the val­ues of in­ter­est are fi­nite. The above re­sult comes from the in­te­gral of the square two-di­men­sion­al Bessel func­tions , and a re­cur­rence re­la­tion, [40, 27.18,88], us­ing , [1, p 437, 10.1.1], and the as­ymp­totic be­hav­ior of the Bessel func­tion you get from {A.6} (A.19). To get the lead­ing as­ymp­totic term, each time you have to dif­fer­en­ti­ate the trigono­met­ric func­tion. And where the trigono­met­ric func­tion in is zero at be­cause of the bound­ary con­di­tion, the one in has mag­ni­tude 1.

The spher­i­cal har­mon­ics are or­tho­nor­mal on the unit sphere, {D.14.4}

In other words, the in­te­gral is only 1 if and and oth­er­wise it is zero. Fur­ther

##### D.36.2.2 Ba­sic ob­ser­va­tions and eigen­value prob­lem

For any func­tion

This fol­lows from writ­ing out the right hand side

the lat­ter since and are dif­fer­ent in­dices.

The elec­tric modes

are so­le­noidal be­cause gives zero. The mag­netic modes

are so­le­noidal for the same rea­son, af­ter not­ing (7) above.

The Lapla­cian com­mutes with the op­er­a­tors in front of the scalar func­tions in the elec­tric and mag­netic modes. That can be seen for the mag­netic ones from

and the fi­nal two terms can­cel. And the Lapla­cian also com­mutes with the ad­di­tional in the elec­tric modes since dif­fer­en­ti­a­tions com­mute.

From this it fol­lows that the en­ergy eigen­value prob­lem is sat­is­fied be­cause by de­f­i­n­i­tion of the scalar modes . In ad­di­tion,

be­cause for a so­le­noidal func­tion, (D.1).

##### D.36.2.3 Spher­i­cal form and net an­gu­lar mo­men­tum

In spher­i­cal co­or­di­nates, the mag­netic mode is

and then the elec­tric mode is
from [40, 20.74,76,82] and for the com­po­nent of the eigen­value prob­lem of chap­ter 4.2.3.

Now note that the de­pen­dence of is through a sim­ple fac­tor , chap­ter 4.2.3. There­fore it is seen that if the co­or­di­nate sys­tem is ro­tated over an an­gle around the -​axis, it pro­duces a fac­tor in the vec­tors. First of all that means that the az­imuthal quan­tum num­ber of net an­gu­lar mo­men­tum is , {A.19}. But it also means that, {A.19},

be­cause ei­ther way the vec­tor gets mul­ti­plied by for the modes. And if it is true for all the modes, then it is true for any func­tion . Since the -​axis is not spe­cial for gen­eral , the same must hold for the and an­gu­lar mo­men­tum op­er­a­tors. From that it fol­lows that the modes are also eigen­func­tions of net square an­gu­lar mo­men­tum, with az­imuthal quan­tum num­ber .

At the cut-off , 0, which gives:

Also needed is, dif­fer­en­ti­at­ing (10):
which used (3) to get rid of the sec­ond or­der de­riv­a­tive of .

##### D.36.2.4 Or­thog­o­nal­ity and nor­mal­iza­tion

Whether the modes are or­thog­o­nal, and whether the Lapla­cian is Her­mit­ian, is not ob­vi­ous be­cause of the weird bound­ary con­di­tions at .

In gen­eral the im­por­tant re­la­tions here

where is the sur­face of the sphere . The sec­ond last line can be ver­i­fied by dif­fer­en­ti­at­ing out and the last line is the di­ver­gence the­o­rem.

The first and sec­ond line in (13) show that the Lapla­cian is Her­mit­ian if all un­equal modes are or­thog­o­nal (or have equal val­ues, but or­thog­o­nal­ity should be shown any­way.). For un­equal val­ues or­thog­o­nal­ity may be shown by show­ing that the fi­nal sur­face in­te­gral is zero.

It is con­ve­nient to show right away that the elec­tric and mag­netic modes are al­ways mu­tu­ally or­thog­o­nal:

The first two terms in the right hand side can be in­te­grated in the , re­spec­tively di­rec­tion and are then zero be­cause is zero on the spher­i­cal sur­face . The fi­nal two terms summed over can be reno­tated by shift­ing the sum­ma­tion in­dex one unit down, re­spec­tively up in the cyclic se­quence to give

the lat­ter be­cause of the form of , the fact that for a so­le­noidal vec­tor, and the en­ergy eigen­value prob­lem es­tab­lished for . The fi­nal term is zero be­cause is.

Next con­sider the or­thog­o­nal­ity of the mag­netic modes for dif­fer­ent quan­tum num­bers. For or , the or­thog­o­nal­ity fol­lows from (9) and (6). For , the or­thog­o­nal­ity fol­lows from the fi­nal line in (13) since the mag­netic modes are zero at , (11).

Fi­nally the elec­tric modes. For or , the or­thog­o­nal­ity fol­lows from (10), (5), and (6). For , the or­thog­o­nal­ity fol­lows from the fi­nal line in (13). To see that, rec­og­nize that is the ra­dial de­riv­a­tive of ; there­fore us­ing (11) and (12), the in­te­grand van­ishes.

The in­te­gral of the ab­solute square in­te­gral of a mag­netic mode is, us­ing (9), (6), and (4),

The in­te­gral of the ab­solute square in­te­gral of an elec­tric mode is, us­ing (10), (5), and (6),

Ap­ply an in­te­gra­tion by parts on the sec­ond in­te­gral,

and then use (3) to get

The nor­mal­iza­tions given in the text fol­low.

##### D.36.2.5 Com­plete­ness

Be­cause of the con­di­tion 0, you would gen­er­ally speak­ing ex­pect two dif­fer­ent types of modes de­scribed by scalar func­tions. The elec­tric and mag­netic modes seem to fit that bill. But that does not mean that there could not be say a few more spe­cial modes. What is needed is to show com­plete­ness. That means to show that any smooth vec­tor field sat­is­fy­ing 0 can be writ­ten as a sum of the elec­tric and mag­netic modes, and noth­ing else.

This au­thor does not know any sim­ple way to do that. It would be au­to­matic with­out the so­le­noidal con­di­tion; you would just take each Carte­sian com­po­nent to be a com­bi­na­tion of the scalar modes sat­is­fy­ing a zero bound­ary con­di­tion at . Then com­plete­ness would fol­low from the fact that they are eigen­func­tions of the Her­mit­ian Lapla­cian. Or from more rig­or­ous ar­gu­ments that you can find in math­e­mat­i­cal books on par­tial dif­fer­en­tial equa­tions. But how to do some­thing sim­i­lar here is not ob­vi­ous, at least not to this au­thor.

What will be done is show that any rea­son­able so­le­noidal vec­tor can be writ­ten in the form

where and are scalar func­tions. Com­plete­ness then fol­lows since the modes pro­vide a com­plete de­scrip­tion of any ar­bi­trary func­tion and .

But to show the above does not seem easy ei­ther, so what will be ac­tu­ally shown is that any vec­tor with­out ra­dial com­po­nent can be writ­ten in the form

That is suf­fi­cient be­cause the Fourier trans­form of does not have a ra­dial com­po­nent, so it will be of this form. And the in­verse Fourier trans­form of is of the form , com­pare any book on Fourier trans­forms and (7).

The proof that must be of the stated form is by con­struc­tion. Note that au­to­mat­i­cally, the ra­dial com­po­nent of the two terms is zero. Writ­ing out the gra­di­ents in spher­i­cal co­or­di­nates, [40, 20.74,82], mul­ti­ply­ing out the cross prod­ucts and equat­ing com­po­nents gives at any ar­bi­trary ra­dius

Now de­com­pose this in Fourier modes in the di­rec­tion:

For 0, and fol­low by in­te­gra­tion. Note that the in­te­grands are pe­ri­odic of pe­riod and an­ti­sym­met­ric about the -​axis. That makes and pe­ri­odic of pe­riod too,

For 0 make a co­or­di­nate trans­form from to

Note that . If any­body is ac­tu­ally read­ing this, send me an email. The sys­tem be­comes af­ter clean­ing up

It is now eas­i­est to solve the above equa­tions for each of the two right hand sides sep­a­rately. Here the first right hand side will be done, the sec­ond right hand side goes sim­i­larly.

From the two equa­tions it is seen that must sat­isfy

and must the de­riv­a­tive of . The so­lu­tion sat­is­fy­ing the re­quired reg­u­lar­ity at is, [40, 19.8],

That fin­ishes the con­struc­tion, but you may won­der about po­ten­tial non­ex­po­nen­tial terms in the first in­te­gral at and the sec­ond in­te­gral at . Those would pro­duce weak log­a­rith­mic sin­gu­lar­i­ties in the phys­i­cal and . You could sim­ply guess that the two right hand sides will com­bine so that these terms drop out. Af­ter all, there is noth­ing spe­cial about the cho­sen di­rec­tion of the -​axis. If you choose a dif­fer­ent axis, it will show no sin­gu­lar­i­ties at the old -​axis, and the so­lu­tion is unique.

For more con­fi­dence, you can check the can­cel­la­tion ex­plic­itly for the lead­ing or­der, 1 terms. But there is a bet­ter way. If the right hand sides are zero within a nonzero an­gu­lar dis­tance from the -​axis, there are no sin­gu­lar­i­ties. And it is easy to split off a part of that is zero within of the axis and then changes smoothly to the cor­rect in an an­gu­lar range from to from the axis. The re­main­der of can than be han­dled by us­ing say the -​axis as the axis of the spher­i­cal co­or­di­nate sys­tem.

##### D.36.2.6 Den­sity of states

The spher­i­cal Bessel func­tion is for large ar­gu­ments pro­por­tional to or . Ei­ther way, the ze­ros are spaced apart. So there is one state 1 in an in­ter­val . The ra­tio gives the stated den­sity of states.

##### D.36.2.7 Par­ity

Par­ity is what hap­pens to the sign of the wave func­tion un­der a par­ity trans­for­ma­tion. A par­ity trans­for­ma­tion in­verts the pos­i­tive di­rec­tion of all three Carte­sian axes, re­plac­ing any po­si­tion vec­tor by . The par­ity of some­thing is 1 or even if it does not change, and 1 or odd if it changes sign. Un­der a par­ity trans­for­ma­tion, the op­er­a­tors and flip over the par­ity of what they act on. On the other hand, has the same par­ity as ; the spa­tial com­po­nents flip over, but so do the unit vec­tors that mul­ti­ply them. And the par­ity of is even if is even and odd if is odd. The stated par­i­ties fol­low.

##### D.36.2.8 Or­bital an­gu­lar mo­men­tum of the states

In prin­ci­ple a state of def­i­nite net an­gu­lar mo­men­tum and def­i­nite spin 1 may in­volve or­bital an­gu­lar mo­men­tum , and , chap­ter 7.4.2. But states of def­i­nite par­ity re­strict that to ei­ther only odd val­ues or only even val­ues, {A.20}. To get the stated par­i­ties, for mag­netic states and or for elec­tric ones.

woof.