130_notes.dvi

with an operator to create a positron state with the right momentum and spin.

d(1)~p = −b(4)

†

~p

d(2)~p = b(3)

†

~p

Theseanti-commutewith everything else with the exception that

{d(~ps),d(s

′)†

p~′ }=δss′δ~pp~′

Now rewrite the fields and Hamiltonian.

ψ(~x,t) =

∑

~p

∑^2

s=1

√

mc^2

EV

(

b(~ps)u(~ps)ei(~p·~x−Et)/ ̄h+d(~ps)†v(~ps)e−i(~p·~x−Et)/ ̄h

)

ψ†(~x,t) =

∑

~p

∑^2

s=1

√

mc^2

EV

(

b(~ps)†u(~ps)†e−i(~p·~x−Et)/ ̄h+d(~ps)v(~ps)†ei(~p·~x−Et)/ ̄h

)

H =

∑

~p

∑^2

s=1

E

(

b(s)

†

~p b

(s)

~p −d

(s)

~p d

(s)†

~p

)

=

∑

~p

∑^2

s=1

E

(

b(s)

†

~p b

(s)

~p +d

(s)†

~p d

(s)

~p −^1

)

All theenergies of these states are positive.

There is an (infinite) constant energy, similar but of opposite sign tothe one for the quantized
EM field, which we must add to make the vacuum state have zero energy. Note that, had we
used commuting operators (Bose-Einstein) instead of anti-commuting, there would have been no
lowest energy ground state so this Energy subtraction would not have been possible.Fermi-Dirac
statistics are required for particles satisfying the Diracequation.

Since theoperators creating fermion states anti-commute, fermion states must be antisym-
metric under interchange. Assumeb†randbrare the creation and annihilation operators for fermions
and that they anti-commute.
{b†r,b†r′}= 0

Thestates are then antisymmetric under interchangeof pairs of fermions.

b†rb†r′| 0 〉=−b†r′b†r| 0 〉

Its not hard to show that theoccupation number for fermion states is either zero or one.

Note that thespinors satisfy the following slightly different equations.

(iγμpμ+mc)u(~ps)= 0

(−iγμpμ+mc)v~p(s)= 0