130_notes.dvi

(Frankie) #1

This“Schr ̈odinger equation”, derived from the Dirac equation, agrees well with the one
we used to understand the fine structure of Hydrogen. The firsttwo terms are the kinetic and
potential energy terms for the unperturbed Hydrogen Hamiltonian. The third term is the relativistic
correction to the kinetic energy. The fourth term is the correctspin-orbit interaction,
including theThomas Precessioneffect that we did not take the time to understand when we did
the NR fine structure. The fifth term is the so calledDarwin termwhich we said would come from
the Dirac equation; and now it has.


For a free particle, each component of the Dirac spinor satisfies the Klein-Gordon equation.


ψ~p=u~pei(~p·~x−Et)/ ̄h

This is consistent with the relativistic energy relation.


The four normalized solutions for a Dirac particle at rest are.


ψ(1)=ψE=+mc (^2) ,+ ̄h/ 2 =


1


V




1

0

0

0



e−imc

(^2) t/ ̄h
ψ(2)=ψE=+mc (^2) ,− ̄h/ 2 =


1


V




0

1

0

0



e−imc

(^2) t/ ̄h
ψ(3)=ψE=−mc (^2) ,+ ̄h/ 2 =


1


V




0

0

1

0



e+imc

(^2) t/ ̄h
ψ(4)=ψE=−mc (^2) ,− ̄h/ 2 =


1


V




0

0

0

1



e+imc

(^2) t/ ̄h
Thefirst and third have spin upwhile the second and fourth have spin down. The first and second
are positive energy solutions while thethird and fourth are “negative energy solutions”, which
we still need to understand.
The next step is to find the solutions with definite momentum. The four plane wave solutions to
the Dirac equation are
ψ
(r)
~p ≡



mc^2
|E|V
u
(r)
~p e

i(~p·~x−Et)/ ̄h

where the four spinors are given by.


u(1)~p =


E+mc^2
2 mc^2





1

0

pzc
E+mc^2
(px+ipy)c
E+mc^2




 u

(2)
~p =


E+mc^2
2 mc^2





0

1

(px−ipy)c
E+mc^2
−pzc
E+mc^2





u(3)~p =


−E+mc^2
2 mc^2





−pzc
−E+mc^2
−(px+ipy)c
−E+mc^2
1
0




 u

(4)
~p =


−E+mc^2
2 mc^2





−(px−ipy)c
−Ep+zmcc^2
−E+mc^2
0
1




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