At this point we take the difference between the two equations to get one condition.
(
(s+nr−κ)
√
k 1
k 2+Zα)
anr+(
Zα√
k 1
k 2−(s+nr+κ))
bnr= 0
(
−(s+nr−κ)√
k 1
k 2−Zα)√
k 2
k 1bnr+(
Zα√
k 1
k 2−(s+nr+κ))
bnr= 0−(s+nr−κ)−Zα√
k 2
k 1+Zα√
k 1
k 2−(s+nr+κ) = 0−(s+nr−κ)√
k 1 k 2 −Zαk 2 +Zαk 1 −(s+nr+κ)√
k 1 k 2 = 0
−2(s+nr)√
k 1 k 2 +Zα(k 1 −k 2 ) = 0
2(s+nr)√
k 1 k 2 =Zα(k 1 −k 2 )
2(s+nr)√
m^2 c^4 −E^2 = 2ZαE
(s+nr)√
m^2 c^4 −E^2 =ZαE
(s+nr)^2 (m^2 c^4 −E^2 ) =Z^2 α^2 E^2
(s+nr)^2 (m^2 c^4 ) = (Z^2 α^2 + (s+nr)^2 )E^2
(s+nr)^2
((s+nr)^2 +Z^2 α^2 )(m^2 c^4 ) =E^2E^2 =
m^2 c^4
(1 + Z(^2) α 2
(s+nr)^2 )
E=
mc^2
√
1 + Z
(^2) α 2
(nr+s)^2
E=
mc^2
√
1 + Z
(^2) α 2
(nr+
√
κ^2 −Z^2 α^2 )^2
E=
mc^2
√
1 + Z
(^2) α 2
(
nr+
√
(j+^12 )^2 −Z^2 α^2) 2
Using the quantum numbers from four mutually commuting operators, we have solved the radial
equation in a similar way as for the non-relativistic case yielding the exactenergy relationfor
relativistic Quantum Mechanics.
E=
mc^2
√
1 + Z
(^2 α^2
nr+√
(j+^12 )^2 −Z^2 α^2) 2
We can identify the standard principle quantum number in this case asn=nr+j+^12. This
result gives the same answer as our non-relativistic calculation to orderα^4 but is alsocorrect to
higher order. It is anexact solution to the quantum mechanics problemposed but does
not include the effects offield theory, such as the Lamb shift and the anomalous magnetic moment
of the electron.