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THE DIRAC EQUATION
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1. (r a = I. 2. For any ra(ra #- rs = I), there exists a r b such that 3. Thus the trace of all ra, except r s, vanishes:
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ra(r b = - tr rbrar b= - tr (rb ra = 0 4. For any couple (ra, r b), a #- b, there exists rc #- r S = I such that rar b= rc up to a factor lor i. 5. From these properties, we deduce the linear independence of our set {ra}.
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tr Suppose that
Multiplying successively by all the for all a. 6. We note the following identities:
yfl-yfl-= 4 yfl-yvypYfl- = 4gVP
and taking the trace leads to A. a = 0
yfl- yV fl- = - 2yv y yfl-y"yPy"y fl- = - 2y"yP yV
(2-32)
These and other useful identities have been listed in the Appendix. Using this basis, we now give the properties of the corresponding bilinears
Ij/ AIjI under proper Lorentz or parity transformations: s: 1j/'(x)IjI'(x') = Ij/(x) ljI(x) scalar
Ij/'(x')yfl- 1jI'(x') = Afl- v Ij/(x)y" ljI(x) 1j/'(x')(Jfl-vljI'(x')
vector antisymmetric tensor pseudo vector pseudo scalar
(2-33)
Afl-pA""Ij/(x)(JP"IjI(x)
1j/'(x')ysyfl-IjI'(x') = det(A)Afl-vlj/(x)YsyvljI(x) Ij/'(x')ys 1jI'(x')
det (A)ljJ(x)ys ljI(x)
The prefix "pseudo" refers to parity and x' stands for x'fl- = Afl-vxv.
2-2 PHYSICAL CONTENT 2-2-1 Plane Wave Solutions and Projectors
We seek plane wave solutions of the free Dirac equation (2-9), i.e., solutions of the form 1jI(+)(x) = e-ik,xu(k) positive energy
1jI( - )(x) = eik . x v(k)
negative energy
(2-34)
QUANTUM FIELD THEORY
with the condition that kO is positive. To verify the Klein-Gordon equation, we also must have P = m 2 The positive time-like four-vector kt< is nothing but the energy momentum of the particle (with h set equal to one). The Dirac equation implies
- m)u(k)
+ m)v(k) = 0
(2-35)
Let us assume that the particle is massive, m = O. In the rest frame of the particle, kt< = (m,O), and Eqs. (2-35) reduce to
(yO _ 1)u(m, 0) (yO
+ 1)v(m, 0) =
There are clearly two linearly independent u solutions, and two v's. In the usual representation (2-10), we denote them as follows:
um(mo)~m
We could now boost these solutions from rest up to a velocity v = !k!/ko by a pure Lorentz transformation, using Eq. (2-19). It is simpler to observe that
(~ - m)(~
+ m) =
k 2 - m2 = 0
so that we may write E
u(a)(k) =
+ m)1/2
J2m(m
u(a)(m, O) =
+ E)
[2m(m
(10k + E)] 1/2 X a (m, 0)
X (m,O)
v(a)(k)
- ~ +m
,j2m(m
+ E)
V(a)(m, O)
= [(
2m m + E 1/2
(10k
(2-37)
(E 2m m)1/2 +
Here E denotes the positive quantity: E == kO = (k 2 + m2)1/2 ; the two component spinors qJ and Xare the non vanishing components of u(m, 0) and v(m, 0) respectively. For the conjugate spinors we find
u(a)(k)
= u(a)(m, 0)
~ +m
J2m(m
+ E)
v(a)(k) = v(a)(m, O)
- ~+m .j2m(m + E)
(2-38)
THE DIRAC EQUATION
The normalization factors have been chosen in order that
u(a)(k)u(P)(k) = (jap v(a)(k)v(P)(k)
u(a)(k)v(P)(k) = 0 v(a)(k)u(P)(k)
(jap
(2-39)
Consider now the matrices
A+(k) ==
a= 1,2
u(a)(k) u(a)(k)
1 1 + yO 2m(m + E) (~+ m) ~2-(~
+ m)
(2-40)
where use has been made of the identity valid for k 2
m2 :
m)yO(~
+ m) =
2E(~
+ m)
Similarly, let
A_(k) == -
a= 1,2
v(a)(k) v(a)(k)
1 1 _ yO 2m(m + E) (~- m) ~2- (~- m)
-~+m
(2-41)
The operators A+ and A_ project over the positive and negative energy states respectively. They satisfy
Ai(k) = A (k)
tr A (k) = 2
A+(k)
(2-42)
+ A-(k) =
The normalization in (2-39) is Lorentz invariant. However, the positive definite density per unit volume is p = l(k) = If/(k)y !/l(k). Let us compute it for our plane wave functions
If/( + )(a)(x)yO !/I (+ )(P)(x) =
u(a)(k)yO u(P)(k)
= u(a)(k) {~, yO} U<P)(k)
(2-43a)
QUANTUM FIELD THEORY
for positive energy solutions, and
1f/(-)(a)(x)yOl/l(-)(P)(x) = v(a)(k)yOv(P)(k) = -v(a)(k)
{~, yO}
v(P)(k)
(2-43b)
for negative ones. The spinors have been normalized at rest; since density times volume has to remain constant, when the latter is reduced by the contraction factor E/m the former must increase by the same amount. Positive and negative energy states are mutually orthogonal, if we consider states with opposite energies but the same three-momentum:
I/I(+)(a)(x) = e-i(kOxO-k.x)u(a)(k) I/I(-)(P)(x) = ei(kOxO+k x)v(P)(k)
with k == (kO, -k) (2-44)
If/( -)(P)(x)I/I( + )(a)(x) =
r2ikoxo v(P)(k)yOu(a)(k)
=1-e-2ikoxo v(P)(k) ( _
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