D.16 Con­stant spher­i­cal po­ten­tials de­riva­tions

This note gives the de­riva­tions for con­stant po­ten­tials in spher­i­cal co­or­di­nates.

D.16.1 The eigen­func­tions

The de­riva­tion of the given spher­i­cal eigen­func­tion is al­most com­i­cally triv­ial com­pared to sim­i­lar prob­lems in quan­tum me­chan­ics.

Fol­low­ing the lines of the hy­dro­gen atom de­riva­tion, chap­ter 4.3.2, the ra­dial func­tions $R_{El}$ are found to sat­isfy the equa­tion

$$\frac{{\rm d}}{{\rm d}r} \left(r^2 \frac{{\rm d}R_{El}}{{\r... ...ft[\frac{p_{\rm {c}}^2}{\hbar^2}r^2 - l(l+1)\right] R_{El} = 0$$

To clean this up a bit more, de­fine new de­pen­dent and in­de­pen­dent vari­ables. In par­tic­u­lar, set $R_{El}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f_l$ and $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x\hbar$$\raisebox{.5pt}{$/$}$​$p_{\rm {c}}$. That pro­duces the spher­i­cal Bessel equa­tion

$$\frac{{\rm d}}{{\rm d}x}\left(x^2\frac{{\rm d}f_l}{{\rm d}x}\right) + \left[x^2 - l(l+1)\right] f_l = 0$$

It is now to be shown that the so­lu­tions $f_l$ to this equa­tion are the Han­kel and Bessel func­tions as given ear­lier.

To do so, make an­other change of de­pen­dent vari­able by set­ting $f_l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x^lg_l$. That gives for the $g_l$:

$$x \frac{{\rm d}^2 g_l}{{\rm d}x^2} + 2(l+1) \frac{{\rm d}g_l}{{\rm d}x} +x g_l = 0$$

Check, by sim­ply plug­ging it in, that $e^{{{\rm i}}x}$$\raisebox{.5pt}{$/$}$​$x$ is a so­lu­tion for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

Now make a fur­ther change in in­de­pen­dent vari­able from $x$ to $\xi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12x^2$ to give

$$2 \xi \frac{{\rm d}^2 g_l}{{\rm d}\xi^2} + 2(l+1)\frac{{\rm d}g_l}{{\rm d}\xi} + g_l = 0$$

Note that the equa­tion for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is ob­tained by dif­fer­en­ti­at­ing the one for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, (tak­ing $g_l'$ as the new un­known.). That im­plies that the $\xi$-​de­riv­a­tive of the so­lu­tion for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 above is a so­lu­tion for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. Keep dif­fer­en­ti­at­ing to get so­lu­tions for all val­ues of $l$. That pro­duces the spher­i­cal Han­kel func­tions of the first kind; the re­main­ing con­stant is just an ar­bi­trar­ily cho­sen nor­mal­iza­tion fac­tor.

Since the orig­i­nal dif­fer­en­tial equa­tion is real, the real and imag­i­nary parts of these Han­kel func­tions, as well as their com­plex con­ju­gates, must be so­lu­tions too. That gives the spher­i­cal Bessel func­tions and Han­kel func­tions of the sec­ond kind, re­spec­tively.

Note that all of them are just fi­nite sums of el­e­men­tary func­tions. And that physi­cists do not even dis­agree over their de­f­i­n­i­tion, just their names.

D.16.2 The Rayleigh for­mula

To de­rive the Rayleigh for­mula, con­vert the lin­ear mo­men­tum eigen­func­tion to spher­i­cal co­or­di­nates by set­ting $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r\cos\theta$. Also, for brevity set $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $p_{\infty}r$$\raisebox{.5pt}{$/$}$​$\hbar$. That turns the lin­ear mo­men­tum eigen­func­tion into

$$e^{{{\rm i}}x\cos\theta} = \sum_{{\underline l}=0}^\infty \frac{({{\rm i}}x\cos\theta)^{{\underline l}}}{{\underline l}!}$$

the lat­ter from Tay­lor se­ries ex­pan­sion of the ex­po­nen­tial.

Now this is an en­ergy eigen­func­tion. It can be writ­ten in terms of the spher­i­cal eigen­func­tions

$$\psi_{Elm} = j_l(x) Y_l^m(\theta,\phi)$$

with the same en­ergy be­cause the $\psi_{Elm}$ are com­plete. In ad­di­tion, the only eigen­func­tions needed are those with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. The rea­son is that the spher­i­cal har­mon­ics $Y_l^m$ are sim­ply Fourier modes in the $\phi$ di­rec­tion, {D.14} (D.5), and the lin­ear mo­men­tum eigen­func­tion above does not de­pend on $\phi$. There­fore

$$\sum_{{\underline l}=0}^\infty \frac{({{\rm i}}x\cos\theta)... ...!} = \sum_{l=0}^{\infty} c_{{\rm {w}},l} j_l(x) Y_l^0(\theta)$$

for suit­able co­ef­fi­cients $c_{{\rm {w}},l}$.

To find these co­ef­fi­cients, find the low­est power of $x$ in $j_l$ by writ­ing the sine in (A.19) as a Tay­lor se­ries and then switch­ing to $x^2$ as in­de­pen­dent vari­able. Sim­i­larly, find the high­est power of $\cos\theta$ in $Y_l^0$, {D.14} (D.5), by look­ing up the Ro­drigue’s for­mula for the Le­gendre poly­no­mial ap­pear­ing in it. That gives

$$\sum_{{\underline l}=0}^\infty \frac{({{\rm i}}x\cos\theta)... ...}} \left(\frac{(2l)!}{2^l(l!)^2}\cos^l \theta +\ldots \right)$$

Each co­ef­fi­cient $c_{{\rm {w}},l}$ must be cho­sen to match the term with ${\underline l}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l$ in the first sum, be­cause the terms for the other val­ues for $l$ do not have a low enough power of $x$ or a high enough power of the co­sine. That gives the Rayleigh val­ues of the co­ef­fi­cients as listed ear­lier.