charge mom. Energy Sz ~j(EM) ~v spinor
spin up, positive energy electron −e pˆz +
√
p^2 c^2 +m^2 c^4 + ̄h 2 −zˆ +ˆz u(1)(~p)
spin up, positive energy positron +e pˆz +
√
p^2 c^2 +m^2 c^4 + ̄h 2 +ˆz +ˆz v(1)(~p)
(spin down “negative energy” hole)
spin down, “negative energy” electron −e −pˆz −
√
p^2 c^2 +m^2 c^4 − ̄h 2 −zˆ +ˆz −u(4)(−~p)
We havedefined the positron spinorv(1)to be the one with positive momentum and spin up.
Note that the minus sign onu(4)is conventional and will come from our future definition of the
charge conjugation operator.
Similarly we can make a table starting from a spin down electron.
charge mom. Energy Sz ~j(EM) ~v spinor
spin down, positive energy electron −e pˆz +
√
p^2 c^2 +m^2 c^4 − ̄h 2 −zˆ +ˆz u(2)(~p)
spin down, positive energy positron +e pˆz +
√
p^2 c^2 +m^2 c^4 − ̄h 2 +ˆz +ˆz v(2)(~p)
(spin up “negative energy” hole)
spin up, “negative energy” electron −e −pˆz −
√
p^2 c^2 +m^2 c^4 + ̄h 2 −zˆ +ˆz u(3)(−~p)
We have now alsodefined the spinor,v(2), for the spin down positron.
36.18Charge Conjugate Waves
Assume that, in addition to rotation, boost and parity symmetry, the Dirac equation also has a
symmetry undercharge conjugation. We wish to write the Dirac equation in a way that makes
the symmetry between electron and positron clear. Start from the Dirac equation and include the
coupling to the EM field with the substitution that~p→
(
~p+ecA~
)
.
∂
∂xμ
γμψ+
mc
̄h
ψ= 0
(
∂
∂xμ
+
ie
̄hc
Aμ
)
γμψ+
mc
̄h
ψ= 0
The strategy is to try to write the charge conjugate of this equation then show that it is equivalent
to the Dirac equation with the right choice of charge conjugation operator forψ. First of all, the
sign ofeAμis expected to change in the charge conjugate equation. (Assumethe equation, including
the constanteis the same but the sign of the EM fieldAμchanges.) Second assume, for now, that
theDirac spinor is transformed to its charge conjugateby the operation
ψC=SCψ∗
where we are motivated by complex scalar field experience. SCis a 4 by 4 matrix. Thecharge
conjugate equationthen is
(
∂
∂xμ
−
ie
̄hc
Aμ
)
γμSCψ∗+
mc
̄h
SCψ∗= 0.