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The physical meaning of negative energy solutions has yet to be clarified. Also, our construction of plane wave states does not make sense for zero-mass particles. Those will be studied in detail in Sec. 2-4-3. To characterize the remaining degeneracy of the plane wave solutions u and v, we construct the projectors onto states of definite polarization. For any spacelike normalized four-vector n (n 2 = -1) orthogonal to k, we have, from (2-21), (2-45)
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(0,0,0,1), we see that the solutions (2-36) are eigenstates of - W n(3)/m = W 3/m, with eigenvalues +i (spin-up) for U(l) and v(1), and -i (spin-down) for d 2 ) and V(2). The projector onto u(1)(m, 0) and v(2)(m, 0) may thus be written
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n(3)
I+ Y5~(3) = ~ (I +
1-0"3
After a Lorentz transformation, the spinors da)(k), v(a)(k) are eigenstates of - W n/m, where n is now the transform of n(3) :
THE DIRAC EQUATION
W~ n U<a)(k) = ~ Y5Yzu(a)(k) = ~ u<a)(k)
Won - -;;:;- v(a)(k)
-"2 Y5 Yzv(a)(k) = "2 v(a)(k)
(2-46)
The plus sign refers to a = 1, the minus sign to a = 2. The projector onto
U<1)(k) and v(2)(k) reads P(n) = i(I
+ Y5Yz)
(2-47)
This expression remains valid for an arbitrary normalized vector n, orthogonal to k; P(n) projects onto the state which in its rest frame has a spin (f n/2 = i for a positive energy solution, and a spin (f n/2 = -i for a negative energy one. (Note the signs! !) We shall denote u(p, n) and v(p, n) the (positive and negative energy respectively) eigenvectors of P(n):
P(n)u(k, n) P(n)v(k, n)
= u(k, n)
v(k, n)
(2-48)
The projector P(n) has the following properties:
[A (k), P(n)]
A+(k)P(n)
+ A_(k)P(n) + A+(k)P( -
+ A-(k)P( -
tr A (k)P( n) = 1 Relaxing the condition on the norm of n,
p(n)
1 + Y5P 2
-1 < n 2 < 0
may be interpreted as a spin density matrix tr p
tr p2 < 4
There exists a particular choice of n, such that n is proportional to k in the reference frame. Let nk be equal to
(2-49)
This definition of polarization is called helicity, and is such that
(2-50)
Therefore P(nk) projects over positive helicity, positive energy and negative helicity, negative energy states.
60 QUANTUM FIELD THEORY
In the ultrarelativistic limit m/ko -> 0,
Ikl/ko -> 1, nC -> k"/m, and
1 + Ys 2~
P( nk)A+(k) ->
(2-51)
2-2-2 Wave Packets
Let us proceed to the construction of normalizable wave packets. We would like to superpose only positive energy plane waves, the only physically sensible ones at this stage. However, we shall be led 'to inconsistencies and will have to abandon this requirement. Let l/J(+)(x) be
l/J(+)(x)
d ~ (2n) 3 E
a= 1,2
b(p,a)u<a)(p)e- ip . x
(2-52)
The factor [lj(2nn(mjE) is designed to make the normalization condition simple:
d xf+)o(t, x) =
f(~:;6 ff
d p dV
;~, ~, b*(p, a)b(p', a')
x u(a)t(p)u(a')(p') ei(E-E')t-i(p-p')'x
(2-53)
where we observe that d 3 pjE is a Lorentz invariant measure. We also compute the total current
J(+)
d x j<+)(t, x) =
f(g~t ff
d p dV
;~, ~ b*(p, a)b(p', a')
x u(a)t(p)au<a')(p') ei(E-E')t-i(p-p')'x
We need here the Gordon identity which states that for any two positive energy solutions u<a)(p) and U<P)(q) of the Dirac equation, we have
1 ij(a)(p)yllu(P)(q) = 2m ij(a)(p) [(p
+ q)" + i(JIlV(p -
q)v] u(P)(q)
(2-54)
Indeed, as a consequence of the Dirac equation
0= ij(a)(p) [p(~ - m)
+ (p -
m)p]u(P)(q)
2mii(a)(p) u(P)(q)
+ ii(a)(p) ({1> ; fl,,b} + [1> ; fl,,b J)U<P)(q)
THE DIRAC EQUATION
where a is an arbitrary four-vector. Equation (2-54) follows by differentiation with respect to all" Using this identity together with Eq. (2-39), we may write
J(+) =
"f 7
d p m Ib( )1 2 ~ = \E (21ll E p, a E
(2-55)
Therefore the total current for a superposition of positive energy solutions is just the group velocity. This is analogous to what happens in the Schrodinger theory, and seems satisfactory. However, there is an inconsistency in the assumption of superposition of positive energy solutions only. To illustrate this point, let us consider the time evolution of a wave packet given at time t = by a gaussian distribution of half width d:
_ 1 _X2/2d 2 '1'(0, x) - (nd 2 /4 e w
(2-56)
where w is some fixed spinor, say
(~).
The corresponding (normalizable) solution of the Dirac equation has the form
If;(t, x) =
f(~:~3 ~ ~ f
d3 x
[b(p, a)u(a)(p) e- ip ' x
+ d*(p, a)v(a)(p) eip . X]
(2-57)
Since the Fourier transform of a gaussian is a gaussian
e-x2/2d2_ip'x =
(2nd 2 )3/2 e-p2d2/2
we may write
(4nd 2 )3/4 e-p2d2/2 w = ~ ~ [b(P, a)u<a)(p)
+ d*(p, a)v(a)(p)]
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