How do you use spherical Bessel functions of the first and second kind to solve the following ordinary differential equation?

#(d^2R(r))/(dr^2) + 2/r (dR(r))/(dr) + [(2mE)/(ℏ^2) - (l(l+1))/(r^2)]R(r) = 0#

where #R(r)# is to be a radial wave function that vanishes at the boundary #r = R#.

This is the differential equation I solved for when I was working with an infinite spherically symmetric well with a bound electron. By multiplying through by #r^2#, this is a spherical Bessel differential equation:

#r^2(d^2R(r))/(dr^2) + 2r (dR(r))/(dr) + [(2mE)/(ℏ^2)r^2 - l(l+1)]R(r) = 0#

with #k^2 = (2mE)/(ℏ^2)#.

Another reference is:
http://farside.ph.utexas.edu/teaching/qmech/Quantum/node81.html

1 Answer
Truong-Son N.
Jan 16, 2018

Using the substitution that #k^2 = (2mE)/(ℏ^2)#, the common way to solve this seems to be to let #rho = kr#, so that:

  • #(rho/k)^2(d^2R(r))/(d(rho//k)^2) = rho^2(d^2R(r))/(drho^2)#

  • #2(rho/k) (dR(r))/(d(rho//k)) = 2rho (dR(r))/(drho)#

This would then lead to

#r^2 (d^2R(r))/(dr^2) + 2r(dR(r))/(dr) + [(2mE)/(ℏ^2)r^2 - l(l+1)]R(r) = 0#

becoming:

#rho^2(d^2R(rho))/(drho^2) + 2rho (dR(r))/(drho) + [rho^2 - l(l+1)]R(r) = 0#

or

#(d^2R(rho))/(drho^2) + 2/rho (dR(r))/(drho) + [1 - (l(l+1))/rho^2]R(r) = 0#

From the second reference, the solutions to this are some linear combination of spherical Bessel functions of the first and second kind, #j_l(rho)# and #y_l(rho)#, where #l# is the angular momentum of the state.

These are given by:

#j_l(rho) = rho^l (-1/rho d/(drho))^l ((sin rho)/rho)#

#y_l(rho) = -rho^l (-1/rho d/(drho))^l ((cos rho)/rho)#

However, #y_l(rho)# is not square-integrable at #rho = 0#:

graph{(cosx)/x [-10, 10, -5, 5]}

So, we can choose only #j_l(rho)#. In this case, I was asked to only look for #s# states, so #l = 0# and the unnormalized wave function is:

#j_0(rho) = rho^0 (-1/rho d/(drho))^0 ((sin rho)/rho)#

#= (sin rho)/rho#

Applying the boundary condition that #R(r) = 0# at #r = R# (the sphere radius),

#j_0(kR) = (sin (kR))/(kR) = 0#

and this is only true when #kR -= rho_(nl) = npi#, where #n = 1, 2, 3, . . . #.

So, #rho_(nl) = npi = sqrt((2mE)/ℏ^2)R#. This allows us to solve for the energies of the #1s, 2s, 3s, . . . # states in terms of the roots of #j_0(rho)#:

#color(blue)(E_(n0) = (ℏ^2rho_(nl)^2)/(2mR^2))#

This also leads to the unnormalized radial wave function:

#R(r) = (sin (npir//R))/(npir//R)#

Normalizing this in spherical coordinates, one finds:

#1 = N^2 int_(0)^(2pi) 1^2 d phi int_(0)^(pi) sintheta d theta int_(0)^(R) R^"*"(r) R(r) r^2 dr#

#1 = N^2 int_(0)^(2pi) d phi int_(0)^(pi) sintheta d theta int_(0)^(R) (sin^2 (npir//R))/(npir//R)^2r^2dr#

#= 4pi N^2 int_(0)^(R) (sin^2 (npir//R))/(n^2pi^2//R^2cancel(r^2))cancel(r^2)dr#

#= (4piR^2)/(n^2pi^2) cdot N^2 int_(0)^(R) sin^2 (npir//R)dr#

Using the identity #sin^2u = 1/2(1 - cos2u)#, one would get:

#= (4piR^2)/(2n^2pi^2) cdot N^2 (R - cancel(R/(2npi)sin((2npiR)/R))^(0))#

Therefore:

#N = sqrt((2n^2pi^2)/(4piR^3))#

#= 1/sqrt(4pi)sqrt(2/R)(npi)/R#

#= 1/sqrt(2piR)(npi)/R#

As a result, the normalized wave function would be:

#color(blue)(R(r)) = 1/sqrt(2piR) cancel((npi)/R) sin(npir//R)/(cancel(npi)r//cancel(R))#

#= color(blue)(1/sqrt(2piR) sin(npir//R)/r)#

And this looks something like this:

graph{sin(x)/x [-0.5, 20, -1, 1.2]}

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