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

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

To clean this up a bit more, define new dependent and independent variables. In particular, set $R_{El}=f_l$ and $r=x\hbar/p_{\rm {c}}$. 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 $f_l=x^lg_l$. 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 $e^{{{\rm i}}x}/x$ is a solution for $l=0$.

Now make a further change in independent variable from $x$ to $\xi=\frac12x^2$ 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 $l=1$ is obtained by differentiating the one for $l=0$, (taking $g_l'$ as the new unknown.). That implies that the $\xi$-derivative of the solution for $l=0$ above is a solution for $l=1$. 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 $z=r\cos\theta$. Also, for brevity set $x=p_{\infty}r/\hbar$. That turns the linear momentum eigenfunction into

$$e^{{{\rm i}}x\cos\theta} = \sum_{{\underline l}=0}^\inft... ...rac{({{\rm i}}x\cos\theta)^{{\underline l}}}{{\underline l}!}$$

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 $m=0$. 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.18) 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 ${\underline l}=l$ 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.