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PHY456H1F: Quantum Mechanics II. Lecture L23 (Taught by Prof J.E. Sipe). 3D Scattering.

Posted by peeterjoot on December 4, 2011

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Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

3D Scattering.

READING: section 20, and section 4.8 of our text [1].

We continue to consider scattering off of a positive potential as depicted in figure (\ref{fig:qmTwoL23:qmTwoL23fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL23fig1}
\caption{Radially bounded potential.}
\end{figure}

Here we have V(r) = 0 for r > r_0. The wave function

\begin{aligned}e^{i k \hat{\mathbf{n}} \cdot \mathbf{r}}\end{aligned} \hspace{\stretch{1}}(2.1)

is found to be a solution of the free particle Schr\”{o}dinger equation.

\begin{aligned}- \frac{\hbar^2}{2\mu} \boldsymbol{\nabla}^2e^{i k \hat{\mathbf{n}} \cdot \mathbf{r}} = \frac{\hbar^2 \mathbf{k}^2}{2 \mu}e^{i k \hat{\mathbf{n}} \cdot \mathbf{r}}\end{aligned} \hspace{\stretch{1}}(2.2)

Seeking a post scattering solution away from the potential

What other solutions can be found for r > r_0, where our potential V(r) = 0? We are looking for \Phi(\mathbf{r}) such that

\begin{aligned}- \frac{\hbar^2}{2\mu} \boldsymbol{\nabla}^2\Phi(r) = \frac{\hbar^2 \mathbf{k}^2}{2 \mu}\Phi(r)\end{aligned} \hspace{\stretch{1}}(2.3)

What can we find?

We split our Laplacian into radial and angular components as we did for the hydrogen atom

\begin{aligned}- \frac{\hbar^2}{2\mu} \frac{\partial^2 }{\partial {{r}}^2} (r \Phi(\mathbf{r})) +\frac{\mathcal{L}^2}{2 \mu r^2}\Phi(\mathbf{r})=E \Phi(\mathbf{r}),\end{aligned} \hspace{\stretch{1}}(2.4)

where

\begin{aligned}\mathcal{L}^2 = -\hbar^2 \left(\frac{\partial^2 }{\partial {{\theta}}^2}+ \frac{1}{{\tan\theta}} \PD{\theta}+ \frac{1}{{\sin^2\theta}} \frac{\partial^2 }{\partial {{\phi}}^2}\right)\end{aligned} \hspace{\stretch{1}}(2.5)

Assuming a solution of

\begin{aligned}\Phi(\mathbf{r}) = R(r) Y_l^m(\theta, \phi),\end{aligned} \hspace{\stretch{1}}(2.6)

and noting that

\begin{aligned}\mathcal{L}^2 Y_l^m(\theta, \phi) = \hbar^2 l (l+1) Y_l^m(\theta, \phi),\end{aligned} \hspace{\stretch{1}}(2.7)

we find that our radial equation becomes

\begin{aligned}- \frac{\hbar^2}{2 \mu r} \frac{\partial^2 }{\partial {{r}}^2} (r R(r))+\frac{\hbar^2 l (l+1)}{2 \mu r^2}R(r)=E R(r)=\frac{\hbar^2 k^2}{2\mu} R(r).\end{aligned} \hspace{\stretch{1}}(2.8)

Writing

\begin{aligned}R(r) = \frac{u(r)}{r},\end{aligned} \hspace{\stretch{1}}(2.9)

we have

\begin{aligned}- \frac{\hbar^2}{2 \mu r} \frac{\partial^2 {{u(r)}}}{\partial {{r}}^2}+\frac{\hbar^2 l (l+1)}{2 \mu r}u(r)=\frac{\hbar^2 k^2}{2\mu} \frac{u(r)}{r},\end{aligned} \hspace{\stretch{1}}(2.10)

or

\begin{aligned}\left( \frac{d^2}{dr^2} + k^2 -\frac{l (l+1) }{r^2} \right) u(r) = 0\end{aligned} \hspace{\stretch{1}}(2.11)

Writing \rho = k r, we have

\begin{aligned}\left( \frac{d^2}{d\rho^2} + 1 -\frac{l (l+1) }{\rho^2} \right) u(r) = 0\end{aligned} \hspace{\stretch{1}}(2.12)

The radial equation and its solution.

With a last substitution of u(r) = U( k r ) = U(\rho), and introducing an explicit l suffix on our eigenfunction U(\rho) we have

\begin{aligned}\left( -\frac{d^2}{d\rho^2} +\frac{l (l+1) }{\rho^2} \right) U_l(\rho) = U_l(\rho).\end{aligned} \hspace{\stretch{1}}(2.13)

We’d not have done this before with the hydrogen atom since we had only finite E = \hbar^2 k^2/2 \mu. Now this can be anything.

Making one final substitution, U_l(\rho) = \rho f_l(\rho) we can rewrite 2.13 as

\begin{aligned}\left( \rho^2 \frac{d^2}{d\rho^2} + 2 \rho \frac{d}{d\rho} + (\rho^2 - l(l+1)) \right) f_l = 0.\end{aligned} \hspace{\stretch{1}}(2.14)

This is the spherical Bessel equation of order l and has solutions called the Bessel and Neumann functions of order l, which are

\begin{subequations}

\begin{aligned}j_l(\rho) = (-\rho)^l \left( \frac{1}{{\rho}} \frac{d}{d\rho} \right)^l \left( \frac{\sin\rho}{\rho} \right)\end{aligned} \hspace{\stretch{1}}(2.15a)

\begin{aligned}n_l(\rho) = (-\rho)^l \left( \frac{1}{{\rho}} \frac{d}{d\rho} \right)^l \left( -\frac{\cos\rho}{\rho} \right).\end{aligned} \hspace{\stretch{1}}(2.15b)

\end{subequations}

We can easily calculate

\begin{aligned}U_0(\rho) &= \rho j_0(\rho) = \sin\rho \\ U_1(\rho) &= \rho j_1(\rho) = -\cos\rho + \frac{\sin\rho}{\rho}\end{aligned} \hspace{\stretch{1}}(2.16)

and can plug these into 2.13 to verify that they are a solution. A more general proof looks a bit trickier.

Observe that the Neumann functions are less well behaved at the origin. To calculate the first few Bessel and Neumann functions we first compute

\begin{aligned}\frac{1}{{\rho}} \frac{d}{d\rho} \frac{\sin\rho}{\rho}&= \frac{1}{{\rho}} \left(\frac{\cos\rho}{\rho}-\frac{\sin\rho}{\rho^2}\right) \\ &=\frac{\cos\rho}{\rho^2}-\frac{\sin\rho}{\rho^3}\end{aligned}

\begin{aligned}\left( \frac{1}{{\rho}} \frac{d}{d\rho} \right)^2 \frac{\sin\rho}{\rho}&= \frac{1}{{\rho}} \left(-\frac{\sin\rho}{\rho^2}-2\frac{\cos\rho}{\rho^3}-\frac{\cos\rho}{\rho^3}+3\frac{\sin\rho}{\rho^4}\right) \\ &=\sin\rho\left(-\frac{1}{\rho^3}+\frac{3}{\rho^5}\right)-3\frac{\cos\rho}{\rho^4}\end{aligned}

and

\begin{aligned}\frac{1}{{\rho}} \frac{d}{d\rho} -\frac{\cos\rho}{\rho}&= \frac{1}{{\rho}} \left(\frac{\sin\rho}{\rho}+\frac{\cos\rho}{\rho^2}\right) \\ &=\frac{\sin\rho}{\rho^2}+\frac{\cos\rho}{\rho^3}\end{aligned}

\begin{aligned}\left( \frac{1}{{\rho}} \frac{d}{d\rho} \right)^2 -\frac{\cos\rho}{\rho}&= \frac{1}{{\rho}} \left(\frac{\cos\rho}{\rho^2}-2\frac{\sin\rho}{\rho^3}-\frac{\sin\rho}{\rho^3}-3\frac{\cos\rho}{\rho^4}\right) \\ &=\cos\rho\left(\frac{1}{\rho^3}-\frac{3}{\rho^5}\right)-3\frac{\sin\rho}{\rho^4}\end{aligned}

so we find

\begin{aligned}\begin{array}{l l l l}j_0(\rho) &= \frac{\sin\rho}{\rho} 					& n_0(\rho) &= -\frac{\cos\rho}{\rho} 	\\ j_1(\rho) &= \frac{\sin\rho}{\rho^2} -\frac{\cos\rho}{\rho} 		& n_1(\rho) &= -\frac{\cos\rho}{\rho^2} -\frac{\sin\rho}{\rho} \\ j_2(\rho) &= \sin\rho \left(-\frac{1}{\rho} + \frac{3}{\rho^3} \right) +\cos\rho \left(-\frac{3}{\rho^2} \right)& n_2(\rho) &= \cos\rho \left(\frac{1}{\rho} - \frac{3}{\rho^3} \right) +\sin\rho \left(-\frac{3}{\rho^2} \right)\end{array}\end{aligned} \hspace{\stretch{1}}(2.18)

Observe that our radial functions R(r) are proportional to these Bessel and Neumann functions

\begin{aligned}R(r)&= \frac{u(r)}{r}  \\ &= \frac{U(kr)}{r}  \\ &=\left\{\begin{array}{l}\frac{j_l(\rho) \rho}{r} \\ \frac{n_l(\rho) \rho}{r}\end{array}\right. \\ &=\left\{\begin{array}{l}\frac{j_l(\rho) k \not{{r}}}{\not{{r}}} \\ \frac{n_l(\rho) k \not{{r}}}{\not{{r}}}\end{array}\right.\end{aligned}

Or

\begin{aligned}R(r) \sim j_l(\rho), n_l(\rho).\end{aligned} \hspace{\stretch{1}}(2.19)

Limits of spherical Bessel and Neumann functions

With n!! denoting the double factorial, like factorial but skipping every other term

\begin{aligned}n!! = n(n-2)(n-4) \cdots,\end{aligned} \hspace{\stretch{1}}(2.20)

we can show that in the limit as \rho \rightarrow 0 we have

\begin{subequations}

\begin{aligned}j_l(\rho) \rightarrow \frac{\rho^l}{(2 l + 1)!!} \end{aligned} \hspace{\stretch{1}}(2.21a)

\begin{aligned}n_l(\rho) \rightarrow -\frac{(2 l - 1)!!}{\rho^{(l+1)}},\end{aligned} \hspace{\stretch{1}}(2.21b)

\end{subequations}

(for the l=0 case, note that (-1)!! = 1 by definition).

Comparing this to our explicit expansion for j_1(\rho) in 2.18 where we appear to have a 1/\rho dependence for small \rho it is not obvious that this would be the case. To compute this we need to start with a power series expansion for \sin\rho/\rho, which is well behaved at \rho =0 and then the result follows (done later).

It is apparently also possible to show that as \rho \rightarrow \infty we have

\begin{subequations}

\begin{aligned}j_l(\rho) \rightarrow \frac{1}{{\rho}} \sin\left( \rho - \frac{l \pi}{2} \right) \end{aligned} \hspace{\stretch{1}}(2.22a)

\begin{aligned}n_l(\rho) \rightarrow -\frac{1}{{\rho}} \cos\left( \rho - \frac{l \pi}{2} \right).\end{aligned} \hspace{\stretch{1}}(2.22b)

\end{subequations}

Back to our problem.

For r > r_0 we can construct (for fixed k) a superposition of the spherical functions

\begin{aligned}\sum_l \sum_m \left( A_l j_l( k r ) + B_l n_l(k r) \right) Y_l^m(\theta, \phi)\end{aligned} \hspace{\stretch{1}}(2.23)

we want outgoing waves, and as r \rightarrow \infty, we have

\begin{subequations}

\begin{aligned}j_l(k r) \rightarrow \frac{\sin\left(kr - \frac{l \pi}{2}\right)}{k r} \end{aligned} \hspace{\stretch{1}}(2.24a)

\begin{aligned}n_l(k r) \rightarrow -\frac{\cos\left(kr - \frac{l \pi}{2}\right)}{k r}\end{aligned} \hspace{\stretch{1}}(2.24b)

\end{subequations}

Put A_l/B_l = -i for a given l we have

\begin{aligned}\frac{1}{{k r}} \left( -i\frac{\sin\left(kr - \frac{l \pi}{2}\right)}{k r}-\frac{\cos\left(kr - \frac{l \pi}{2}\right)}{k r} \right)\sim \frac{1}{{k r}} e^{i (k r - \pi l/2)}\end{aligned} \hspace{\stretch{1}}(2.25)

For

\begin{aligned}\sum_l\sum_m B_l\frac{1}{{k r}} e^{i (k r - \pi l/2)} Y_l^m(\theta, \phi).\end{aligned} \hspace{\stretch{1}}(2.26)

Making this choice to achieve outgoing waves (and factoring a (-i)^l out of B_l for some reason, we have another wave function that satisfies our Hamiltonian equation

\begin{aligned}\frac{e^{i k r}}{k r}\sum_l\sum_m(-1)^lB_lY_l^m(\theta, \phi).\end{aligned} \hspace{\stretch{1}}(2.27)

The B_l coefficients will depend on V(r) for the incident wave e^{i \mathbf{k} \cdot \mathbf{r}}. Suppose we encapsulate that dependence in a helper function f_\mathbf{k}(\theta, \phi) and write

\begin{aligned}\frac{e^{i k r}}{r} f_\mathbf{k}(\theta, \phi)\end{aligned} \hspace{\stretch{1}}(2.28)

We seek a solution \psi_\mathbf{k}(\mathbf{r})

\begin{aligned}\left( - \frac{\hbar^2}{2\mu} \boldsymbol{\nabla}^2+ V(\mathbf{r})\right)\psi_\mathbf{k}(\mathbf{r}) = \frac{\hbar^2 \mathbf{k}^2}{2 \mu}\psi_\mathbf{k}(\mathbf{r}),\end{aligned} \hspace{\stretch{1}}(2.29)

where as r \rightarrow \infty

\begin{aligned}\psi_\mathbf{k}(\mathbf{r}) \rightarrow e^{i \mathbf{k} \cdot \mathbf{r}} + \frac{e^{i k r}}{r} f_\mathbf{k}(\theta, \phi).\end{aligned} \hspace{\stretch{1}}(2.30)

Note that for r < r_0 in general for finite r, \psi_k(\mathbf{r}), is much more complicated. This is the analogue of the plane wave result

\begin{aligned}\psi(x) = e^{i k x} + \beta_k e^{-i k x}\end{aligned} \hspace{\stretch{1}}(2.31)

Scattering geometry and nomenclature.

We can think classically first, and imagine a scattering of a stream of particles barraging a target as in
figure (\ref{fig:qmTwoL23:qmTwoL23fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.3\textheight]{qmTwoL23fig2}
\caption{Scattering cross section.}
\end{figure}

Here we assume that d\Omega is far enough away that it includes no non-scattering particles.

Write P for the number density

\begin{aligned}P = \frac{\text{number of particles}}{\text{unit volume}},\end{aligned} \hspace{\stretch{1}}(3.32)

and

\begin{aligned}J = P v_0 =\frac{\text{Number of particles flowing through}}{\text{a unit area in unit time}}\end{aligned} \hspace{\stretch{1}}(3.33)

We want to count the rate of particles per unit time dN through this solid angle d\Omega and write

\begin{aligned}dN = J \left( \frac{d \sigma(\Omega)}{d\Omega} \right) d\Omega.\end{aligned} \hspace{\stretch{1}}(3.34)

The factor

\begin{aligned}\frac{d \sigma(\Omega)}{d\Omega},\end{aligned} \hspace{\stretch{1}}(3.35)

is called the differential cross section, and has “units” of

\begin{aligned}\frac{\text{area}}{\text{steradians}}\end{aligned} \hspace{\stretch{1}}(3.36)

(recalling that steradians are radian like measures of solid angle [2]).

The total number of particles through the volume per unit time is then

\begin{aligned}\int J \frac{d \sigma(\Omega)}{d\Omega} d\Omega= J \int \frac{d \sigma(\Omega)}{d\Omega} d\Omega= J \sigma\end{aligned} \hspace{\stretch{1}}(3.37)

where \sigma is the total cross section and has units of area. The cross section \sigma his the effective size of the area required to collect all particles, and characterizes the scattering, but isn’t necessarily entirely geometrical. For example, in photon scattering we may have frequency matching with atomic resonance, finding \sigma \sim \lambda^2, something that can be much bigger than the actual total area involved.

Appendix

Q: Are Bessel and Neumann functions orthogonal?

Answer: There is an orthogonality relation, but it is not one of plain old multiplication.

Curious about this, I find an orthogonality condition in [3]

\begin{aligned}\int_0^\infty J_\alpha(z) J_\beta(z) \frac{dz}{z} = \frac{2}{\pi} \frac{\sin\left(\frac{\pi}{2}\left( \alpha - \beta\right) \right) }{\alpha^2 - \beta^2},\end{aligned} \hspace{\stretch{1}}(4.38)

from which we find for the spherical Bessel functions

\begin{aligned}\int_0^\infty j_l(\rho) j_m(\rho) d\rho =\frac{\sin\left(\frac{\pi}{2}\left( l - m \right) \right) }{(l+ 1/2)^2 - (m + 1/2)^2}.\end{aligned} \hspace{\stretch{1}}(4.39)

Is this a satisfactory orthogonality integral? At a glance it doesn’t appear to be well behaved for l = m, but perhaps the limit can be taken?

Deriving the large limit Bessel and Neumann function approximations.

For 2.22 we are referred to any “good book on electromagnetism” for details. I thought that perhaps the weighty [4] would be to be such a book, but it also leaves out the details. In section 16.1 the spherical Bessel and Neumann functions are related to the plain old Bessel functions with

\begin{subequations}

\begin{aligned}j_l(x) = \sqrt{\frac{\pi}{2x} } J_{l+1/2}(x) \end{aligned} \hspace{\stretch{1}}(4.40a)

\begin{aligned}n_l(x) = \sqrt{\frac{\pi}{2x} } N_{l+1/2}(x)\end{aligned} \hspace{\stretch{1}}(4.40b)

\end{subequations}

Referring back to section 3.7 of that text where the limiting forms of the Bessel functions are given

\begin{subequations}

\begin{aligned}J_\nu(x) \rightarrow \sqrt{\frac{2}{\pi x}} \cos\left(x - \frac{\nu\pi}{2} - \frac{\pi}{4} \right) \end{aligned} \hspace{\stretch{1}}(4.41a)

\begin{aligned}N_\nu(x) \rightarrow \sqrt{\frac{2}{\pi x}} \sin\left(x - \frac{\nu\pi}{2} - \frac{\pi}{4} \right)\end{aligned} \hspace{\stretch{1}}(4.41b)

\end{subequations}

This does give us our desired identities, but there’s no hint in the text how one would derive 4.41 from the power series that was computed by solving the Bessel equation.

Deriving the small limit Bessel and Neumann function approximations.

Writing the \text{sinc} function in series form

\begin{aligned}\frac{\sin x}{x} = \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k + 1)!},\end{aligned} \hspace{\stretch{1}}(4.42)

we can differentiate easily

\begin{aligned}\begin{aligned}\frac{1}{{x}} \frac{d}{dx} \frac{\sin x}{x}&= \sum_{k=1}^\infty (-1)^k (2k) \frac{x^{2k-2}}{(2k + 1)!} \\ &= (-1) \sum_{k=0}^\infty (-1)^k (2k + 2) \frac{x^{2k}}{(2k + 3)!} \\ &= (-1) \sum_{k=0}^\infty (-1)^k \frac{1}{{2k + 3}} \frac{x^{2k}}{(2k + 1)!} \\ \end{aligned}\end{aligned} \hspace{\stretch{1}}(4.43)

Performing the derivative operation a second time we find

\begin{aligned}\begin{aligned}\left(\frac{1}{{x}} \frac{d}{dx}\right)^2 \frac{\sin x}{x}&= (-1) \sum_{k=1}^\infty (-1)^k \frac{1}{{2k + 3}} (2k) \frac{x^{2k-2}}{(2k + 1)!} \\ &= \sum_{k=0}^\infty (-1)^k \frac{1}{{2k + 5}} \frac{1}{{2k + 3}} \frac{x^{2k}}{(2k + 1)!}\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.44)

It appears reasonable to form the inductive hypotheses

\begin{aligned}\left(\frac{1}{{x}} \frac{d}{dx}\right)^l \frac{\sin x}{x}= (-1)^l\sum_{k=0}^\infty (-1)^k \frac{(2k+1)!!}{(2(k + l) + 1)!!}\frac{x^{2k}}{(2k + 1)!},\end{aligned} \hspace{\stretch{1}}(4.45)

and this proves to be correct. We find then that the spherical Bessel function has the power series expansion of

\begin{aligned}j_l(x) =\sum_{k=0}^\infty (-1)^k \frac{(2k+1)!!}{(2(k + l) + 1)!!}\frac{x^{2k + l}}{(2k + 1)!}\end{aligned} \hspace{\stretch{1}}(4.46)

and from this the Bessel function limit of 2.21a follows immediately.

Finding the matching induction series for the Neumann functions is a bit harder. It’s not really any more difficult to write it, but it is harder to put it in a tidy form that is.

We find

\begin{aligned}-\frac{\cos x}{x} &= - \sum_{k=0}^\infty (-1)^k \frac{x^{2k-1}}{(2k)!} \\ \frac{1}{{x}} \frac{d}{dx}-\frac{\cos x}{x} &= - \sum_{k=0}^\infty (-1)^k \frac{2k-1}{2k} \frac{x^{2k-3}}{(2k-2)!} \\ \left( \frac{1}{{x}} \frac{d}{dx} \right)^2-\frac{\cos x}{x} &= - \sum_{k=0}^\infty (-1)^k \frac{(2k-1)(2k -3)}{2k(2k -2)} \frac{x^{2k-3}}{(2k-4)!}\end{aligned} \hspace{\stretch{1}}(4.47)

The general expression, after a bit of messing around (and I got it wrong the first time), can be found to be

\begin{aligned}\begin{aligned}\left( \frac{1}{{x}} \frac{d}{dx} \right)^l-\frac{\cos x}{x} &= (-1)^{l+1}\sum_{k=0}^{l-1} \prod_{j=0}^{l-1}  {\left\lvert{ 2(k-j)-1}\right\rvert} \frac{x^{2(k-l)-1}}{(2k)!} \\ &\quad +(-1)^{l+1}\sum_{k=0}^\infty (-1)^k \frac{(2(k+l)-1)!!}{(2k - 1)!!}\frac{x^{2k-1}}{(2(k + l)!}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.50)

We really only need the lowest order term (which dominates for small x) to confirm the small limit 2.21b of the Neumann function, and this follows immediately.

For completeness, we note that the series expansion of the Neumann function is

\begin{aligned}\begin{aligned}n_l(x)&= -\sum_{k=0}^{l-1} \prod_{j=0}^{l-1}  {\left\lvert{ 2(k-j)-1}\right\rvert} \frac{x^{2 k -l -1}}{(2k)!} \\ &\quad -\sum_{k=0}^\infty (-1)^k \frac{(2 k + 3 l - 1)!!}{(2k - 1)!!}\frac{x^{2k-1}}{(2(k + l)!}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.51)

Verifying the solution to the spherical Bessel equation.

One way to verify that 2.15a is a solution to the Bessel equation 2.14 as claimed should be to substitute the series expression and verify that we get zero. Another way is to solve this equation directly. We have a regular singular point at the origin, so we look for solutions of the form

\begin{aligned}f = x^r \sum_{k=0}^\infty a_k x^k\end{aligned} \hspace{\stretch{1}}(4.52)

Writing our differential operator as

\begin{aligned}L = x^2 \frac{d^2}{dx^2} + 2 x \frac{d}{dx} + x^2 - l(l+1),\end{aligned} \hspace{\stretch{1}}(4.53)

we get

\begin{aligned}0 &= L f \\ &= \sum_{k=0}^\infty a_k \Bigl( (k+r)(k+r-1) + 2 (k + r) - l (l+1) \Bigr) x^{k + r} + a_k x^{k + r + 2} \\ &= a_0 \Bigl( r( r + 1) - l(l + 1)\Bigr) x^r \\ &+a_1 \Bigl( (r+ 1)( r + 2) - l(l + 1)\Bigr) x^{r+1} \\ &+\sum_{k=2}^\infty a_k \Bigl( (k+r)(k+r-1) + 2 (k + r) - l (l+1) + a_{k-2} \Bigr) x^{k + r} \end{aligned}

Since we require this to be zero for all x including non-zero values, we must have constraints on r. Assuming first that a_0 is non-zero we must then have

\begin{aligned}0 = r( r + 1) - l(l + 1).\end{aligned} \hspace{\stretch{1}}(4.54)

One solution is obviously r = l. Assuming we have another solution r = l + k for some integer k we find that r = -l-1 is also a solution. Restricting attention first to r = l, we must have a_1 = 0 since for non-negative l we have (l+1)(l+2) - l(l+1) = 2(l+1) \ne 0. Thus for non-zero a_0 we find that our function is of the form

\begin{aligned}f = \sum_k a_{2k} x^{2k + l}.\end{aligned} \hspace{\stretch{1}}(4.55)

It doesn’t matter that we started with a_0 \ne 0. If we instead start with a_1 \ne 0 we find that we must have r = l-1, -l-2, so end up with exactly the same functional form as 4.55. It ends up slightly simpler if we start with 4.55 instead, since we now know that we don’t have any odd powered a_k‘s to deal with. Doing so we find

\begin{aligned}0 &= L f \\ &= \sum_{k=0}^\infty a_{2k} \Bigl((2k + l)(2k + l - 1) + 2(2k + l) - l (l+1)\Bigr) x^{2k + l} + a_{2k} x^{2k + l + 2} \\ &=\sum_{k=1}^\infty \Bigl(a_{2k} 2k (2 (k+l) + 1) + a_{2(k-1)}\Bigr) x^{2k + l} \end{aligned}

We find

\begin{aligned}\frac{a_{2k} }{a_{2(k-1)}}= \frac{-1}{2k (2 (k+l) + 1) }.\end{aligned} \hspace{\stretch{1}}(4.56)

Proceeding recursively, we find

\begin{aligned}f = a_0 (2 l + 1)!! \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!! (2 (k+l) + 1)!!} x^{2k + l}.\end{aligned} \hspace{\stretch{1}}(4.57)

With a_0 = 1/(2l + 1)!! and the observation that

\begin{aligned}\frac{1}{{(2k)!!}} = \frac{(2k + 1)!!}{(2k+1)!},\end{aligned} \hspace{\stretch{1}}(4.58)

we have f = j_l(x) as given in 4.46.

If we do the same for the r = -l-1 case, we find

\begin{aligned}\frac{a_{2k} }{a_{2(k-1)}}= \frac{-1}{2k (2 (k-l) - 1) },\end{aligned} \hspace{\stretch{1}}(4.59)

and find

\begin{aligned}\frac{a_{2k}}{a_0} = \frac{(-1)^k}{(2k)!! (2(k-l) -1)(2(k-l)-3)\cdots(-2l + 1)}.\end{aligned} \hspace{\stretch{1}}(4.60)

Flipping signs around, we can rewrite this as

\begin{aligned}\frac{a_{2k} }{a_0}= \frac{1}{(2k)!! (2(l-k) + 1) (2(l-k) + 3) \cdots (2 l - 1)}.\end{aligned} \hspace{\stretch{1}}(4.61)

For those values of l > k we can write this as

\begin{aligned}\frac{a_{2k} }{a_0}= \frac{(2(l-k)-1)!!}{(2k)!! (2 l - 1)!!}.\end{aligned} \hspace{\stretch{1}}(4.62)

Comparing to the small limit 2.21b, the k=0 term, we find that we must have

\begin{aligned}\frac{a_0}{(2 l - 1)!!} = -1.\end{aligned} \hspace{\stretch{1}}(4.63)

After some play we find

\begin{aligned}a_{2k}= \left\{\begin{array}{l l}-\frac{(2(l-k)-1)!!}{ (2k)!!  } & \quad \mbox{if latex l \ge k$} \\ \frac{(-1)^{k-l+1}}{ (2k)!! (2 (k-l) -1)!! } & \quad \mbox{if l \le k} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(4.64)$

Putting this all together we have

\begin{aligned}n_l(x) =-\sum_{0 \le k \le l}(2(l-k)-1)!!\frac{x^{2k -l -1}}{(2k)!!}-\sum_{l < k}\frac{(-1)^{k-l}} { (2 (k-l) -1)!! }\frac{x^{2k -l -1}}{(2k)!!}\end{aligned} \hspace{\stretch{1}}(4.65)

FIXME: check that this matches the series calculated earlier 4.51.

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] Wikipedia. Steradian — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 4-December-2011]. http://en.wikipedia.org/w/index.php?title=Steradian&oldid=462086182.

[3] Wikipedia. Bessel function — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 4-December-2011]. http://en.wikipedia.org/w/index.php?title=Bessel_function&oldid=461096228.

[4] JD Jackson. Classical Electrodynamics Wiley. John Wiley and Sons, 2nd edition, 1975.

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