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## Numerical calculations for S4 hydrogen

The exact integral formula (see 7.391 of Gradshteyn) for us will be:

$\int_{-1}^1 (1-x)^\rho(1+x)^\sigma P^{(a,b)}_n(x) dx = \frac{2^{\rho+\sigma+1}\Gamma(\rho+1)\Gamma(\sigma+1)\Gamma(n+1+a)}{n!\Gamma(\rho+\sigma+2)\Gamma(1+a)} \times 3F2( -n, a+b+n+1, \rho+1; a+1, \rho+\sigma+2; 1)$

Let the right hand side above be denoted by $Q(\rho,\sigma, a, b, n)$.  Then we can write for $C_{nl}$ introduced here by

$C_{nl} = Q( l/2, l/2, l+1, l, n) + 2 Q(1/2+2,l/2+2, l+1,l,n )$

A SUBTLETY
The Dirac operator $D$ has no harmonic spinors on $S^4$ and so it is sensible to map $n=3$ to the lowest energy shell of hydrogen.  This is not ‘cheating’ and is in fact justified also because the denominator of for $\lambda_n$ is not $1/\lambda_n^2$ but instead $1/(\lambda_n^2-1/4 R)$ and the scalar curvature is going to be equal to 12.  This is crucial for the Lamb shift issue as well as we will see shortly that there are no degeneracies among energy levels if we consider for example $n\ge 2$ for $C_{nl}$.

We can proceed now with numerical estimation using python libraries mpmath and math with simple code.  We want to map \$l

from mpmath import hyp3f2
from math import log, gamma, exp

def Q(r,s,a,b,n):
g1 = gamma(1+r) * gamma(1+s) / gamma(r+s+2)
g2 = (gamma(n+1+a) / gamma(1+n)) / gamma(1+a)
f = hyp3f2( -n, a+b+n+1, r+1, a+1, r+s+2, 1.0)
c = exp(log(2)*(r+s+1))
return c * g1 * g2 * f

def Cnl( n, l ):
return Q(l/2,l/2,l+1,l, n) + 2*Q(l/2+2,l/2+2,l+1,l,n)

print Cnl(3,0)
print Cnl(3,1)
print Cnl(3,2)
print Cnl(3,3)
print Cnl(4,0)
print Cnl(4,1)
print Cnl(4,2)
print Cnl(4,3)
print Cnl(4,4)

The output of this code is the following:

–> n = 3 for Dirac (lowest order shell for hydrogen)

0.271428571428571
1.33333333333333
0.666666666666667
1.5047619047619

–> n=4 for Dirac (n=1 shell for hydrogen)
0.4
2.0
0.857142857142857
2.59393939393939
1.58135198135198