Sub­sec­tions


D.36 Pho­ton wave func­tion de­riva­tions

The rules of en­gage­ment are as fol­lows:


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,

\begin{displaymath}
E = {\textstyle\frac{1}{2}}\epsilon_0 \int
\Bigg(\left\ver...
...igg(A_{i,t}^* A_{i,t} + c^2\sum_{j=1}^3 A_{i,j}^*A_{i,j}\bigg)
\end{displaymath}

How­ever, a bit more gen­eral ex­pres­sion is de­sir­able. If only the Lorenz con­di­tion is sat­is­fied, there may also be an elec­tro­sta­tic po­ten­tial $\varphi$. In that case, a more gen­eral ex­pres­sion for the en­ergy is:
$\parbox{400pt}{\hfill$\displaystyle
E = {\textstyle\frac{1}{2}}\epsilon_0 \big...
...hi_t^* \varphi_t
+ \sum_{j=1}^3 \varphi_j^* \varphi_j\bigg)\bigg]
$\hfill(1)}$
The mi­nus sign for the $\varphi$ 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

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
E = {\textstyle\frac{1}{2}}\e...
...nt
\bigg({\cal E}_i^*{\cal E}_i + c^2{\cal B}_i^*{\cal B}_i\bigg)
$\hfill(2)}$
Since $\skew3\vec{\cal E}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\partial\skew3\vec A/\partial{t}-\nabla\varphi$ and $\skew2\vec{\cal B}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\nabla$ $\times$ $\skew3\vec A$:

\begin{displaymath}
{\cal E}_i = - A_{i,t} -\varphi_i
\qquad
{\cal B}_i = A_{...
...th}}} - A_{{\overline{\imath}},{\overline{\overline{\imath}}}}
\end{displaymath}

From now on, it will be un­der­stood that there is a sum­ma­tion over $i$ and $j$ and that every­thing has a ${\textstyle\frac{1}{2}}\epsilon_0$. There­fore these will no longer be shown.

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

\begin{displaymath}
\int A_{i,t}^* A_{i,t} +
A_{i,t}^* \varphi_i + \varphi_i^* A_{i,t} + \varphi_i^* \varphi_i
\end{displaymath}

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

\begin{displaymath}
\int {} - A_{i,it}^* \varphi - \varphi^* A_{i,it} + \varphi_i^* \varphi_i
\end{displaymath}

How­ever, the di­ver­gence $A_{i,i}$ is ac­cord­ing to the Lorenz con­di­tion equal to $-\varphi_t$$\raisebox{.5pt}{$/$}$$c^2$, so

\begin{displaymath}
\int \frac{1}{c^2} \varphi_{tt}^* \varphi
+ \frac{1}{c^2} \varphi^* \varphi_{tt} + \varphi_i^* \varphi_i
\end{displaymath}

Us­ing the Klein-Gor­don equa­tion, $\varphi_{tt}$$\raisebox{.5pt}{$/$}$$c^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varphi_{ii}$, and then an­other in­te­gra­tion by parts on the first two terms and reno­tat­ing $i$ by $j$ gives the $\varphi_j^*\varphi_j$ terms in (1).

Now con­sider the in­te­gral of $\vert\skew2\vec{\cal B}\vert^2$ in (2). You get, mul­ti­ply­ing out,

\begin{displaymath}
c^2 \int \bigg(A_{{\overline{\overline{\imath}}},{\overline...
...* A_{{\overline{\imath}},{\overline{\overline{\imath}}}}\bigg)
\end{displaymath}

Now the first and last terms in the right hand side summed over $i$ pro­duce all terms $A_{i,j}^*A_{i,j}$ in (1) in which $i$ and $j$ are dif­fer­ent. That leaves the mid­dle terms. An in­te­gra­tion by parts yields

\begin{displaymath}
c^2 \int \bigg(A_{{\overline{\overline{\imath}}},{\overline...
...overline{\imath}}}}^* A_{{\overline{\overline{\imath}}}}\bigg)
\end{displaymath}

Reno­tate the in­dices cycli­cally to get

\begin{displaymath}
c^2 \int \bigg(A_{{\overline{\imath}},{\overline{\imath}}i}...
...rline{\imath}}},{\overline{\overline{\imath}}}i}^* A_{i}\bigg)
\end{displaymath}

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

\begin{displaymath}
c^2 \int \bigg(A_{i,i}^* + A_{{\overline{\imath}},{\overlin...
...line{\overline{\imath}}}}^*\bigg)_i A_{i}
- A_{i,i i}^* A_{i}
\end{displaymath}

as you can see from dif­fer­en­ti­at­ing and mul­ti­ply­ing out. The fi­nal term gives af­ter in­te­gra­tion by parts the $A_{i,j}^*A_{i,j}$ terms in (1) in which $i$ and $j$ are equal. That leaves the first part. The term in paren­the­ses is the di­ver­gence $-\varphi_t$$\raisebox{.5pt}{$/$}$$c^2$, so the first part is

\begin{displaymath}
\int - \varphi_{it}^* A_{i}
\end{displaymath}

Per­form an in­te­gra­tion by parts

\begin{displaymath}
\int \varphi_t^* A_{i,i}
\end{displaymath}

Rec­og­niz­ing once more the di­ver­gence, this gives the fi­nal $-\varphi_t^*\varphi_t$$\raisebox{.5pt}{$/$}$$c^2$ term in (1)


D.36.2 An­gu­lar mo­men­tum states

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


D.36.2.1 About the scalar modes

The scalar modes are the $jY$.

It will be as­sumed that the $j$ are zero at the large ra­dius $r_{\rm {max}}$ at which the do­main is as­sumed to ter­mi­nate. That makes the scalar modes a com­plete set; any scalar func­tion $f$ 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 $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r_{\rm {max}}$ makes the Lapla­cian Her­mit­ian. This will not be ex­plic­itly proved since it is very stan­dard.)

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

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
r^2 j'' + 2 r j' = l(l+1) j - k^2r^2 j
$\hfill(3)}$
The fol­low­ing in­te­gral is needed (note that $j$ is real):
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\int_0^{r_{\rm max}} j^2r^2{\,\rm d}r \sim \frac{r_{\rm{max}}}{2k^2}
$\hfill(4)}$
This is valid for large $kr_{\rm {max}}$, which ap­plies since $r_{\rm {max}}$ is large and the $k$ val­ues of in­ter­est are fi­nite. The above re­sult comes from the in­te­gral of the square two-di­men­sion­al Bessel func­tions $J$, and a re­cur­rence re­la­tion, [40, 27.18,88], us­ing $j_l(kr)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $J_{l+\frac12}(kr)\sqrt{\pi/2kr}$, [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 $j_l$ is zero at $r_{\rm {max}}$ be­cause of the bound­ary con­di­tion, the one in $j_{l+1}$ has mag­ni­tude 1.

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

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\int \bar Y Y{\,\rm d}\Omega = \delta_{\bar{l}l}\delta_{\bar{m}m}
$\hfill(5)}$
In other words, the in­te­gral is only 1 if $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\bar{l}$ and $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\bar{m}$ and oth­er­wise it is zero. Fur­ther
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\int \left(\bar Y_\theta Y_\t...
...right) {\,\rm d}\Omega = l(l+1) \delta_{\bar{l}l}\delta_{\bar{m}m}
$\hfill(6)}$


D.36.2.2 Ba­sic ob­ser­va­tions and eigen­value prob­lem

For any func­tion $f$

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
{\skew0\vec r}\times \nabla f = - \nabla \times ({\skew0\vec r}f)
$\hfill(7)}$
This fol­lows from writ­ing out the right hand side

\begin{displaymath}
- (r_{\overline{\overline{\imath}}}f)_{\overline{\imath}}+ ...
...imath}}+ r_{\overline{\imath}}f_{\overline{\overline{\imath}}}
\end{displaymath}

the lat­ter since ${\overline{\imath}}$ and ${\overline{\overline{\imath}}}$ are dif­fer­ent in­dices.

The elec­tric modes

\begin{displaymath}
\nabla\times{\skew0\vec r}\times\nabla f
\end{displaymath}

are so­le­noidal be­cause $\nabla\cdot\nabla$ $\times$ $\ldots$ gives zero. The mag­netic modes

\begin{displaymath}
{\skew0\vec r}\times\nabla f
\end{displaymath}

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

\begin{displaymath}
(r_{\overline{\imath}}f_{\overline{\overline{\imath}}}- r_{...
...}}}} - 2 f_{{\overline{\overline{\imath}}}{\overline{\imath}}}
\end{displaymath}

and the fi­nal two terms can­cel. And the Lapla­cian also com­mutes with the ad­di­tional $\nabla\times$ 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 $-\nabla^2jY$ $\vphantom0\raisebox{1.5pt}{$=$}$ $k^2jY$. In ad­di­tion,

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\nabla\times\nabla\times{\skew0\vec r}\times\nabla f
= k^2 {\skew0\vec r}\times\nabla f
$\hfill(8)}$
be­cause $\nabla$ $\times$ $\nabla\times$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$\nabla^2$ for a so­le­noidal func­tion, (D.1).


D.36.2.3 Spher­i­cal form and net an­gu­lar mo­men­tum

In spher­i­cal co­or­di­nates, the mag­netic mode is

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\skew3\vec A^{\rm M} = {\skew...
...Y_\theta - {\hat\imath}_\theta
\frac{1}{\sin\theta} Y_\phi\right]
$\hfill(9)}$
and then the elec­tric mode is
$\parbox{400pt}{\hspace{15pt}\hfill$\displaystyle
\skew3\vec A^{\rm E} = \nabla...
... Y_\theta + {\hat\imath}_\phi
\frac{1}{\sin\theta} Y_\phi\right]
$\hfill(10)}$
from [40, 20.74,76,82] and for the $r$ com­po­nent of $\skew3\vec A^{\rm {E}}$ the eigen­value prob­lem of chap­ter 4.2.3.

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

\begin{displaymath}
{\widehat J}_z {\skew0\vec r}\times\nabla f = {\skew0\vec r...
...}\times\nabla f = \nabla\times{\skew0\vec r}\times\nabla L_z f
\end{displaymath}

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

At the cut-off $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r_{\rm {max}}$, $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, which gives:

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\mbox{At $r_{\rm{max}}$:} \qq...
... Y_\theta + {\hat\imath}_\phi
\frac{1}{\sin\theta} Y_\phi\right]
$\hfill(11)}$
Also needed is, dif­fer­en­ti­at­ing (10):
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\mbox{At $r_{\rm{max}}$:} \qq...
... Y_\theta + {\hat\imath}_\phi
\frac{1}{\sin\theta} Y_\phi\right]
$\hfill(12)}$
which used (3) to get rid of the sec­ond or­der de­riv­a­tive of $j$.


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 $r_{\rm {max}}$.

In gen­eral the im­por­tant re­la­tions here

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\begin{array}{l}
\displaysty...
... A_{i,j})
\frac{\partial r_j}{\partial r} {\,\rm d}S
\end{array} $\hfill(13)}$
where $S$ is the sur­face of the sphere $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r_{\rm {max}}$. 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 $k$ val­ues, but or­thog­o­nal­ity should be shown any­way.). For un­equal $k$ 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:

\begin{displaymath}
\int (r_{\overline{\imath}}\bar f_{\overline{\overline{\ima...
...{\overline{\imath}}}\bar f A_{i,{\overline{\imath}}}^{\rm {E}}
\end{displaymath}

The first two terms in the right hand side can be in­te­grated in the ${\overline{\overline{\imath}}}$, re­spec­tively ${\overline{\imath}}$ di­rec­tion and are then zero be­cause $\bar{f}$ is zero on the spher­i­cal sur­face $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r_{\rm {max}}$. The fi­nal two terms summed over $i$ can be reno­tated by shift­ing the sum­ma­tion in­dex one unit down, re­spec­tively up in the cyclic se­quence to give

\begin{displaymath}
\sum_i - \int \bar f r_i(A_{{\overline{\overline{\imath}}},...
... \int \bar f {\skew0\vec r}\cdot {\skew0\vec r}\times \nabla f
\end{displaymath}

the lat­ter be­cause of the form of $\skew3\vec A^{\rm {E}}$, the fact that $\nabla$ $\times$ $\nabla\times$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$\nabla^2$ for a so­le­noidal vec­tor, and the en­ergy eigen­value prob­lem es­tab­lished for $\skew3\vec A^{\rm {M}}$. The fi­nal term is zero be­cause ${\skew0\vec r}\cdot{\skew0\vec r}\times$ is.

Next con­sider the or­thog­o­nal­ity of the mag­netic modes for dif­fer­ent quan­tum num­bers. For $\bar{l}$ $\raisebox{.2pt}{$\ne$}$ $l$ or $\bar{m}$ $\raisebox{.2pt}{$\ne$}$ $m$, the or­thog­o­nal­ity fol­lows from (9) and (6). For $\bar{k}$ $\raisebox{.2pt}{$\ne$}$ $k$, the or­thog­o­nal­ity fol­lows from the fi­nal line in (13) since the mag­netic modes are zero at $r_{\rm {max}}$, (11).

Fi­nally the elec­tric modes. For $\bar{l}$ $\raisebox{.2pt}{$\ne$}$ $l$ or $\bar{m}$ $\raisebox{.2pt}{$\ne$}$ $m$, the or­thog­o­nal­ity fol­lows from (10), (5), and (6). For $\bar{k}$ $\raisebox{.2pt}{$\ne$}$ $k$, the or­thog­o­nal­ity fol­lows from the fi­nal line in (13). To see that, rec­og­nize that $A_{i,j}\partial{r}_j$$\raisebox{.5pt}{$/$}$$\partial{r}$ is the ra­dial de­riv­a­tive of $\skew3\vec A$; 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),

\begin{displaymath}
\int \skew3\vec A^{{\rm {M}}*}\cdot \skew3\vec A^{\rm {M}}
= l(l+1) \frac{r_{\rm {max}}}{2k^2}
\end{displaymath}

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

\begin{displaymath}
l^2(l+1)^2 \int_0^{r_{\rm max}} j^2 {\,\rm d}r
+ l(l+1) \int_0^{r_{\rm max}} (rj)'(rj)' {\,\rm d}r
\end{displaymath}

Ap­ply an in­te­gra­tion by parts on the sec­ond in­te­gral,

\begin{displaymath}
l^2(l+1)^2 \int_0^{r_{\rm max}} j^2 {\,\rm d}r
- l(l+1) \int_0^{r_{\rm max}} jr(rj)'' {\,\rm d}r
\end{displaymath}

and then use (3) to get

\begin{displaymath}
\int \skew3\vec A^{{\rm {M}}*}\cdot \skew3\vec A^{\rm {M}}
= k^2 l(l+1) \frac{r_{\rm {max}}}{2k^2}
\end{displaymath}

The nor­mal­iza­tions given in the text fol­low.


D.36.2.5 Com­plete­ness

Be­cause of the con­di­tion $\nabla\cdot\skew3\vec A$ $\vphantom0\raisebox{1.5pt}{$=$}$ 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 $\nabla\cdot\skew3\vec A$ $\vphantom0\raisebox{1.5pt}{$=$}$ 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 $jY$ sat­is­fy­ing a zero bound­ary con­di­tion at $r_{\rm {max}}$. 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

\begin{displaymath}
\skew3\vec A= {\skew0\vec r}\times\nabla f + \nabla\times{\skew0\vec r}\times\nabla g
\end{displaymath}

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

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

\begin{displaymath}
\vec v = {\skew0\vec r}\times\nabla f + {\hat\imath}_r\times{\skew0\vec r}\times\nabla g
\end{displaymath}

That is suf­fi­cient be­cause the Fourier trans­form of $\skew3\vec A$ does not have a ra­dial com­po­nent, so it will be of this form. And the in­verse Fourier trans­form of $\vec{v}$ is of the form $\skew3\vec A$, com­pare any book on Fourier trans­forms and (7).

The proof that $\vec{v}$ 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 $r$

\begin{displaymath}
- \frac{\partial f}{\partial\phi}
- \sin\theta \frac{\part...
...\theta}
- \frac{\partial g}{\partial\phi}
= v_\phi\sin\theta
\end{displaymath}

Now de­com­pose this in Fourier modes $e^{{{\rm i}}m\phi}$ in the $\phi$ di­rec­tion:

\begin{displaymath}
- {\rm i}m f_m
- \sin\theta \frac{\partial g_m}{\partial\t...
...l f_m}{\partial\theta}
- {\rm i}m g_m
= v_{\phi m}\sin\theta
\end{displaymath}

For $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, $f_0$ and $g_0$ fol­low by in­te­gra­tion. Note that the in­te­grands are pe­ri­odic of pe­riod $2\pi$ and an­ti­sym­met­ric about the $z$-​axis. That makes $f$ and $g$ pe­ri­odic of pe­riod $2\pi$ too,

For $m$ $\raisebox{.2pt}{$\ne$}$ 0 make a co­or­di­nate trans­form from $\theta$ to

\begin{displaymath}
t = \int{\rm d}\theta/\sin\theta = \ln\tan{\textstyle\frac{1}{2}}\theta
\end{displaymath}

Note that $\vphantom{0}\raisebox{1.5pt}{$-$}$$\infty$ $\raisebox{.3pt}{$<$}$ $t$ $\raisebox{.3pt}{$<$}$ $\infty$. If any­body is ac­tu­ally read­ing this, send me an email. The sys­tem be­comes af­ter clean­ing up

\begin{displaymath}
- {\rm i}m f_m
- \frac{\partial g_m}{\partial t}
= v_{\th...
...partial f_m}{\partial t}
- {\rm i}m g_m
= v_{\phi m}/\cosh t
\end{displaymath}

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 $f_m$ must sat­isfy

\begin{displaymath}
\frac{\partial^2 f_m}{\partial t^2} - m^2f_m
= \frac{-2m{\rm i}v_{\theta m}}{e^t+e^{-t}}
\end{displaymath}

and ${{\rm i}}mg_m$ must the de­riv­a­tive of $f_m$. The so­lu­tion sat­is­fy­ing the re­quired reg­u­lar­ity at $\pm\infty$ is, [40, 19.8],

\begin{displaymath}
f_m =
\int_t^{\infty} {\rm i}v_{\theta m}\frac{e^{-m(\tau-...
..._{\theta m}\frac{e^{m(\tau-t)}}{e^\tau+e^{-\tau}}{\,\rm d}\tau
\end{displaymath}

That fin­ishes the con­struc­tion, but you may won­der about po­ten­tial non­ex­po­nen­tial terms in the first in­te­gral at $\vphantom{0}\raisebox{1.5pt}{$-$}$$\infty$ and the sec­ond in­te­gral at $\infty$. Those would pro­duce weak log­a­rith­mic sin­gu­lar­i­ties in the phys­i­cal $f$ and $g$. 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 $z$-​axis. If you choose a dif­fer­ent axis, it will show no sin­gu­lar­i­ties at the old $z$-​axis, and the so­lu­tion is unique.

For more con­fi­dence, you can check the can­cel­la­tion ex­plic­itly for the lead­ing or­der, $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 terms. But there is a bet­ter way. If the right hand sides are zero within a nonzero an­gu­lar dis­tance $\Delta\theta_1$ from the $z$-​axis, there are no sin­gu­lar­i­ties. And it is easy to split off a part of $\vec{v}$ that is zero within $\Delta\theta_1$ of the axis and then changes smoothly to the cor­rect $\vec{v}$ in an an­gu­lar range from $\Delta\theta_1$ to $\Delta\theta_2$ from the axis. The re­main­der of $\vec{v}$ can than be han­dled by us­ing say the $x$-​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 $\sin(kr)$$\raisebox{.5pt}{$/$}$$r$ or $\cos(kr)$$\raisebox{.5pt}{$/$}$$r$. Ei­ther way, the ze­ros are spaced $\Delta{k}\,r_{\rm {max}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$ apart. So there is one state $\Delta{N}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 in an in­ter­val $\Delta{E}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\Delta{k}c$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar{\pi}c$$\raisebox{.5pt}{$/$}$$r_{\rm {max}}$. 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 ${\skew0\vec r}$ by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. The par­ity of some­thing is 1 or even if it does not change, and $\vphantom{0}\raisebox{1.5pt}{$-$}$1 or odd if it changes sign. Un­der a par­ity trans­for­ma­tion, the op­er­a­tors ${\skew0\vec r}\times$ and $\nabla\times$ flip over the par­ity of what they act on. On the other hand, $\nabla{j}_jY_j^{m_j}$ has the same par­ity as $j_jY_j^{m_j}$; the spa­tial com­po­nents flip over, but so do the unit vec­tors that mul­ti­ply them. And the par­ity of $j_jY_j^{m_j}$ is even if $j$ is even and odd if $j$ 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 $j$ and def­i­nite spin 1 may in­volve or­bital an­gu­lar mo­men­tum $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j-1$, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j$ and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j-1$, 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, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j$ for mag­netic states and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j-1$ or $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j+1$ for elec­tric ones.

woof.