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## Does solving the Dirac equation give the right energy spectrum in S4?

We want to consider the potential $A_\mu = ( Ze/r, 0,0,0)$ and solve on $S^4(1/h)$ the Dirac equation:

$(D - i\frac{eA}{c\hbar} -m_0c^2) \psi = 0$

Using the calculations of Camperosi and Higuchi, we know that in geodesic polar coordinates $(\theta, \Omega)$ the eigenspinors have the factor

$\phi_{nl}(\theta) = \cos(\theta/2)^{l+1}\sin(\theta/2)^{l} P(\cos(\theta)$

where $n$ is the eigenvalue number and $l \le n$.  Let’s approximate $1/r$ by $1/\sin(\theta)$ because we will consider the large radius $1/h$.  Now we are integrating over $S^4$ for which the volume form in polar coordinates will have a factor $\sin^3(\theta) d\theta$.  If we were doing the calculation in latex $S^3$ then the factor of interest in the volume form would be $\sin^2(\theta) d\theta$ which multiplied by $1/\sin(\theta)$ produces an odd function and for $l=0$  we have even functions in $\phi_{nl}$ and we have zero integral.  However this vanishing does not hold for $S^4$.  These integrals would correspond to the inner product $\langle F, \Phi_n\rangle$ for eigenspinors $\Phi_n$ where $F = Z e/\sin(\theta)$.

Now let us take a slightly more abstract approach.   We want to calculate $\beta_{nl} = \langle F, \Phi_n \rangle$ in order to solve the Dirac equation in eigenspinor expansions.  We can essentially bring in the Dirac-squared on the right side and by dividing by $\lambda_n^2$ and then use the self-adjointness and the Lichnerowicz formula:

$\beta_{nl} \lambda_n^2 = \langle \Delta(F) + \frac{1}{4}R F, \Phi_n \rangle$

where $R$ is the scalar curvature.  Since the scalar curvature is constant, we have:

(*) $\beta_{nl} ( \lambda_n^2 - \frac{1}{4} R ) = \langle \Delta(F), \Phi_n \rangle$

For the right side we can use the formula for the Laplacian in coordinates:

$\Delta_{S^4} f(t,\xi) = \sin^{-3}(t) \partial_t ( \sin^3(t) \partial_t f ) + \sin^{-2}\Delta_{\xi}f$

A small calculation gives $\Delta_{S^4} (1/\sin(t)) = \sin^{-3}(t) [ 1 + 2 \sin^2(t)]$ which we can then plug in with $\Phi$ in the integrand, use the fact that the volume form contains a $\sin^3(t)$.  The right hand side of (*) is then the sum of $\int_0^{2\pi} \Phi(t) dt$ and $\int_0^{2\pi} \sin^2 (t) \Phi(t) dt$.  Let’s call the right side $C_{nl}$ and all of these will be finite. Then we have the expression

$\beta_{nl} = \frac{C_{nl}}{\lambda_n^2 - \frac{1}{4}R}$

We can get some asymptotic approximation for $C_{nl}$ by using the formula here:

$P_n^{(a,b)}(\cos(t)) = n^{-1/2} k(t) \cos( Nt + \gamma)$

and we focus on the $k(t)$ term

$k(t) = \pi^{-1/2} \sin^{-a-1/2}(t/2)\cos^{-b-1/2}(t/2)$

Once we absorb the cosine and sine terms in the eigenspinor formula, we need only worry about $a=1$ and $b=2$ in our case.  Then use the double angle formula for $\sin^2(t) = 4 \sin^2(t/2)\cos^2(t/2)$ to get approximations for integrals of $\Phi_n(t)$ and $\sin^2(t)\Phi_n(t)$.

Now let us return to the Dirac equation and its inner product with eigenspinor $\Phi_n$ to examine the linear equation in eigenspaces.  We have either

$\lambda_n - \beta_{nl} = m_0 c^2$

or the coefficient of eigenspinor $\Phi_n$ in the solution expansion is zero.  Since $\beta_{nl}$ coefficients can be calculated before attempting to solve the Dirac operator (these are the effect of the multiplication by $F$ on the eigenspaces), in principle, we have a complete solution of the Dirac equation: for each $n,l$, check whether $\lambda_n – \beta_{nl}$ is equal to $m_0c^2$.  If so, then all the eigenfunctions in that eigenspace produce solutions to the Dirac equation; if not, then none of the eigenfunctions in the eigenspace enter into the solution to the Dirac equation.

EXPLICIT FORMULA

Gradsheyn and Ryzhik formulae in 7.391 allow us to get exact expressions for both $\int_{-1}^1 \Phi_n(t) dt$ and $\int_{-1}^1 \sin^2(t)\Phi_n(t) dt$.  For both, use the change of variables $x = \cos t$ in the latter using $\sin(t) = (1+x)^{1/2}(1-x)^{1/2}$.  The general formula is (see GradshteynTableIntegrals):

$\int_{-1}^1 (1-x)^{\rho}(1+x)^{\sigma} P_n^{(a,b)}(x) dx = A B$

where

$A = \frac{2^{\rho+\sigma+1} \Gamma(\rho+1)\Gamma(\sigma+1)\Gamma(n+1+\alpha)}{n! \Gamma(\rho+\sigma+2)\Gamma(n+\alpha)}$

and

$B = {}_3F_2(-n, \alpha+\beta+n+1, \rho+1; \alpha + 1, \rho+\sigma+2; 1)$

We use $\alpha=1$ and $\beta=2$ for integrals of both $\Phi(t)$ and $\sin^2(t)\Phi(t)$.  For the former, set $\rho = \sigma = 0$ and for the latter $\rho=\sigma= 1/2$.  So we have computable expressions for the right hand side of (*).