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In the second expression of (2-124), the sum S d3pl(2n) 3 has been replaced by a t sum over the final states in a finite space volume V: 11 V .f...J na Isates kfoCI. ; then the wave (mIVE)1/2 e- ip x u<a)(p) describes a particle of velocity piE and polarization a in the volume V. Therefore, the transition amplitude between the state
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THE DIRAC EQUATION
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and an analogous state !/Jf(x) is
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Sfi= -ie V(Ei;f)1 /2
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d4 zii(a)(Pf)J(z)e i(p/-p,l.zu<P)(p;)
(2-125a)
For the Coulomb problem, A == (Ao, 0) and A o = - Ze/4nr; therefore, (2-126)
It is gratifying to recover the energy conservation. We finally need the Fourier
transform of the potential
d3 r
r~q r = 1:1
q== Pf - Pi
We are using stationary plane waves instead of wave packets; no wonder, then, that the amplitude (2-126) cannot be squared. This is remedied if according to Fermi's golden rule we consider a finite time interval and replace the E> function by
2 sin [T(Ef - Ei)/2]
Ef-Ei
The square of this expression behaves for large T as T2nE>(Ef - Ei). The transition probability between states i and f per unit time and per incident particle is then
dP _ dt fi
1iZIX
3 m 4n -(a) (P) 12 E> ) d pf (EfEY/2 U (Pf)Y u (Pi) 2n (E f - Ei V (2n)3
Iql2
(2-127)
The summation runs over all possible final states, the number of which in the momentum volume element d3pf is V d 3 pf/{2n)3. Dividing by the incident flux (1/V)(lpd/Ei), we obtain the differential cross section
We use
Ipd = IPfl =
Pf and Pf dpf = Ef dE f to perform the trivial Pf integration 4Z 2IX 2m 2 d<Jfi = lii(a\Pf)yOU(fJ)(PiW dOf (2-128)
Iql4
In the nonrelativistic limit, ii(a)yOu(P) is proportional to E>aP If we do not observe . the final polarization, a sum over IX has to be performed, whereas for an un-
QUANTUM FIELD THEORY
polarized incident state we average over the two equally probable polarizations f3: Z mZ d(JJi! = 4Z C(: L i a(<<)(PJ)y OU(/l)(Pi)j2 dO. unpolarized iqi 2 /l
_4ZzC(zmZ! (OPi+m OPJ+ m ) iqi 4 2 tr y 2m y 2m
(2-129)
Once again, the expression (2-40) has been used. We now need identities for traces of y matrices. For any product of an odd number of y matrices, the traces vanish. For an even number, the following identity may be proved by induction: tr ( l Z ... Zn) = al " az tr ( 3 Zn) - al " a3 tr ( Z 4" Zn) Here this reduces to tr yOpiyOJ6J = 4(E iEJ - Pi" PJ tr yOyO = 4 We also need the kinematical relations
+ ...
+ E;E J)
Ei = EJ = E
where f3 == vic =
Pi" PJ = E Z - pZ cos e = m Z + 2E z f3 z sin z ~
ipilE is the incoming (or outgoing) velocity and iqi 4= 16ipi4 sin 4~ = 16f3zEzpz sin4~
d(J dO.
The final expression of the unpolarized cross section (Mott cross section) is
unpolarized
ZZC(z ( 1 - f3 z Z e) Z Z 4 sm4p f3 sm e 12 2
(2-130)
When f3 -+ 0, this reduces to the Rutherford formula. Also notice that the relativistic correction (1 - f3z sin z eI2)(1 - f3 Z)-l affects predominantly the backward scattering. This result has been derived for incident electrons. Let us discuss briefly the scattering of positrons in the same Coulomb field. The attractive Coulomb force is now replaced by a repulsive one. In classical nonrelativistic mechanics, this leads to the same Rutherford formula (this remarkable result is a peculiar feature of the Coulomb field). In our quantum treatment, we know that the theory is invariant under charge conjugation. The scattering of an electron off a charge - Ze is the same as that of a positron off a charge + Ze. On the other hand, to lowest order, the cross section is an even function of Z. Therefore, the Mott cross section is also valid for positrons. We may verify this by a direct calculation; we shall rather use the hole theory interpretation. An outgoing positron of four-momentum PJ and polarization C( is represented by an "incoming" negative energy solution running back-
THE DIRAC EQUATION
ward in time:
> 0
We thus return to Eq. (2-116), using this time the boundary conditions
I{I(+)(tl,X) = 0
I{I(-)(t2,X)
t1 = - 00
XO =
eiPJ xv(-a)(Pf)
t2 =
Repeating the steps which led us from (2-122) to (2-125a), with due attention paid to the signs, we get
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