130_notes.dvi

We name thecreation and annihilation operators for the positron statesto bed(s)

†
~p and
d(~ps)and identify them to be.

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~′

The Dirac field and Hamiltonian can now berewritten.

ψ(~x,t) =

∑

~p

∑^2

s=1

√

mc^2

EV

(

b

(s)

~p u

(s)

~p e

i(~p·~x−Et)/ ̄h+d(s)†

~p v

(s)

~p 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 equations.

(iγμpμ+mc)u(~ps)= 0

(−iγμpμ+mc)v~p(s)= 0

Since we changed the sign of the momentum in our definition ofv(~ps), the momentum term in the
Dirac equation had to change sign.