This is the two component equation which is equivalent to the Dirac equation for energy eigenstates.
The one difference from our understanding of the Dirac equation is inthenormalization. We shall
see below that the normalization difference is small for non-relativistic electron states but needs to
be considered for atomic fine structure.
36.5.2 The Large and Small Components of the Dirac Wavefunction
Returning to the pair of equations inψAandψB. Note that forE≈mc^2 , that is non-relativistic
electrons,ψAis much bigger thanψB.
1
c(E+eA 0 +mc^2 )ψB=~σ·(
~p+e
cA~
)
ψAψB≈
c
2 mc^2~σ·(
~p+
e
cA~
)
ψA≈
pc
2 mc^2ψAIn the Hydrogen atom,ψBwould be of orderα 2 times smaller, so we callψAthe large component
andψBthe small component. When we include relativistic corrections for the fine structure of
Hydrogen, we must consider the effectψBhas on the normalization. Remember that the conserved
current indicates that the normalization condition for the four component Dirac spinor is.
j 0 =ψγ ̄ 4 ψ=ψ†γ 4 γ 4 ψ=ψ†ψ36.5.3 The Non-Relativistic Equation
Now we will calculate theprediction of the Dirac equation for the non-relativistic coulomb
problem, aiming to directly compare to what we have done with the Schr ̈odinger equation for
Hydrogen. As for previous Hydrogen solutions, we will setA~ = 0 but have a scalar potential
due to the nucleusφ=A 0. The energy we have been using in our non-relativistic formulation is
E(NR)=E−mc^2. We willwork with the equation for the large componentψA. Note that
A 0 is a function of the coordinates and the momentum operator will differentiate it.
~σ·(
~p+e
cA~
) c 2
(E+eA 0 +mc^2 )~σ·(
~p+e
cA~
)
ψA = (E+eA 0 −mc^2 )ψA~σ·~pc^2
(E+eA 0 +mc^2 )~σ·~pψA = (E(NR)+eA 0 )ψAExpand the energy term on the left of the equation for the non-relativistic case.
c^2
E+eA 0 +mc^2=
1
2 m(
2 mc^2
mc^2 +E(NR)+eA 0 +mc^2)
=
1
2 m(
2 mc^2
2 mc^2 +E(NR)+eA 0)
=
1
2 m(
1
1 +E
(NR)+eA 0
2 mc^2)
≈
1
2 m(
1 −
E(NR)+eA 0
2 mc^2+
(
E(NR)+eA 0
2 mc^2