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,,];,2 :b!(k)b,,(k) -
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dik)d!(k):
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(3-163)
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,,];,2 [b!(k)b,,(k) + d!(k)d,,(k)]
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leading to a sum of positive contributions to the energy of a quantum state. Let us elaborate the structure of the corresponding Fock space.
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3-3-2 Fock Space for Fermions
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We focus first on one-particle states. The necessary smearing in momentum space is implicitly understood. For a given four-momentum we observe that we now have a fourfold degeneracy. Denote the corresponding states la) (a = 1,2,3,4):
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= bI(k) 10) 12) = b1(k) 10) 13) = dI(k) 10) 14) = d1(k) 10) They satisfy Pilla) = kllla). We look for observables commuting with PIl in order
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to distinguish these states. One of these observables is the charge (normalized to 1 for a one-particle state)
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d 3 x /(x) =
d 3 x: l/Jt(x)l/J(x):
(3-164)
,,];,2 [b!(k)b,,(k) -
d!(k)d,,(k)]
QUANTIZATION-FREE FIELDS
When dealing with Dirac particles the notation dk will mean [d 3k/(2n)3](m/ko) instead of d3k/(2n)32ko as it is for bosons. Since the vacuum has zero charge and
[Q, bHk)]
we have
= bJ(k)
[Q, dJ(k)] = -dJ(k)
a = 1,2 a = 3,4
Qla)
{+ la)
-Ia)
In contradistinction to the "classical" case where we attempted to interpret (3-164) as a positive square norm it is seen that in the quantum case this is not so. The use of anticommutators has totaIIy reversed the situation; energy is now positive and charge an indefinite quantity. Dirac's quantized theory therefore describes particles of two types. In electrodynamics eQ wiII be the electric charge and 1/1 will annihilate electrons and create positrons. Since [Q, PI'] = 0, Q is time independent, and [Q, I/I(x)] = - I/I(x) and
[Q, ij/(x)] = ij/(x).
To characterize the one-particle states completely it remains to introduce spin. In analogy with (3-158) we have for Lorentz transformations
[JI'V, I/I(x)] = [ i(xl'oV - XVol')
+ (lI'V] 1/1 (x) 2
(3-165)
Since
blX(k)
f d 3x u(IX)(k) eik . x yOl/l(x)
dJ(k) = fd3xi/IX)(k)e-ik.xyOl/l(X)
we find (3-166)
Consider now the action of the Pauli-Lubanski operator W" = -ie"l'vpPvpp on a state la). The operator pP is replaced by its eigenvalue k P Let t == (1,0,0,0) be the time axis and
(t -
k m2
kot) Iki m
a normalized space-like four-vector orthogonal to k in the two-plane (t, k). Its space component is obviously along k. Choose the third axis along k. Then (Wo n/m) la) reduces to
-Ia) m
= J 12 1a)
QUANTUM FIELD THEORY
When acting on the state la) it is clear that the orbital contribution to J12 disappears, and since J121o) = 0 we are left with
w~n bZ(k) 10) =
~ utuJ)(k) (J~2 da)(k)b~(k)IO)
v(a).
To proceed we have to make a choice for the spinor basis d a ),
We set
with
0(=1 0(=2
The corresponding states are helicity states. Since the third axis was chosen along k we find
; ut<P)(k) (J~2 u(a)(k) = ;a [)aP
81 = 1,82 = -1
with the result that
a=1 a=2
Similarly,
(3-167)
Defining
v(a)
k _ - ~ +m ( 0) ( ) - J2m(kO + m) X(a)
with we obtain
0(=1 0(=2
81 = 1,82 = -1 and
a=3 a=4
(3-168)
Formulas (3-167) and (3-168) complete the characterization of states: II) and 14) have helicity i, 12) and 13) minus i.
QUANTIZATION-FREE FIELDS
We observe the typical inversion between the spinors u and v. It is the spinor V(4) with spin which corresponds to positive helicity, in agreement with the hole interpretation.
Consider now multiparticle states. Denote the creation operators by the collective symbol at, omitting therefore all indices of momentum, charge, or spin. A basis of Fock space is generated by the states
a t (1)'" at(n) 10)
From the anticommutation properties of the creation operators these states will be antisymmetric in the wave function arguments 1, ... , n and, in particular, will vanish if two of those coincide. We recover in this formalism Pauli's exclusion principle, valid for electrons and more generally for all identical fermions. Quantization using anticommutators has naturally generated Fermi-Dirac statistics.
The reader might enjoy discussing the free relativistic electron gas at thermal equilibrium. What happens if only the total charge is conserved and not electron and positron numbers separately
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