0.1 Klein-Gordon Equation 0.2 Dirac Equation

and a similar series for G with coefficients b p leads to the following recursion
relation
(s + p + 1)a p+1 − a p − βa p+1 − 1 ν b p − αb p+1 = 0
(s + p + 1)b p+1 − b p + βb p+1 − νa p + αa p+1 = 0
The series must terminate to avoid generating the exponentially increasing
solution. If p is the index of the highest power of ρ before termination, then
we can deduce that a p = −νb p from the fact that a p+1 and b p+1 are both zero
while a p and b p are non-zero. If we use this relation in the recursion relations
above with p → p − 1, we find that all the coefficients can be eliminated and
a quadratic equation for ν obtained:
The appropriate solution is positive:
αν 2 + 2(p + s)ν − α = 0.
ν = − p + s
α
√ (p ) 2 + s
+ + 1
α
It is then a matter of algebra to deduce from this that
E =
m
√
1 + α2
(p+s) 2
To make sense of this and to compare it to the NR result, we can expand it in
powers of α = e 2 /4π ≈ 1/137, noting the dependence of s = √ (j+ 1 2) 2 − α 2
on α. The final result is
E = m
(1 − α2
2n 2 − α4
2n 4 ( n
j+ 1 2
− 3 4
)
)
+ O(α 6 ) ,
in which n stands for p + j + 1 2. After the rest energy, the α 2 term is the
Rydberg energy calculated in NR QM and the α 4 term is the relativistic
correction. In all orders the energy depends only on j and n, so some but
not all of the ’accidental’ degeneracy of the H atom remains. DISCUSSION
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