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= A~ut(x)
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The precise mathematical meaning of these expressions depends on the source j. We may require at least that the matrix elements of some local average of the fields between normalized states satisfy these relations. This weak limit relies crucially on the assumption of an adiabatic vanishing of the source as Itl--+ 00.
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QUANTUM FIELD THEORY
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This is often not actually realized and therefore Eq. (4-8) will have to be taken with a grain of salt. The construction takes place in a given Hilbert space, the Fock space of incoming photons, for instance. The vacuum is then the state annihilated by the operators ai:)(k), and we try to find the canonical transformation represented by a unitary operator S which connects the in- and out-fields, that is,
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A~ut(x) = S-l Afn(x)S
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and in- and out-states according to lout> = S-l lin> = or lin>
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st lin>
(4-10)
S lout>
Suppose that at time t = - 00 the system is in a definite state, the vacuum for instance, i.e., contains no (physical) photon. The final state has some computable probability to contain zero, one, two, etc., emitted photons. For example, the probability amplitude to remain in the ground state is <0 outlO in> = <0 inlSIO in> = <0 outlSIO out>
For this interpretation to make sense, we should verify that the probability = 1<0 outlO in>12 is less than one. Theprobabilitiespl' P2,"" the final polarizations, and the angular distributions are computable in an analogous way. Therefore, the operator S contains all the information about the final state. We now turn to its determination. Equation (4-7) implies that
A~ut(x) =
Afn(x)
d4 y [Gret(X - y) - Gadv(x - y)]P'(y)
(4-11) The second term of the right-hand side of (4-11) is, of course, a solution of the homogeneous equation, and is nothing but the classical field Ag1 radiated by the current j (1-206). The combination G(-) == Gret - Gady has already been encountered in Chap. 1 [see Eq. (1-173)] :
G(-)(x)
= Gret(x) - Gadv(x) =
(2~)3
d4p e- ip ' x e(po)c5(p2)
= 2n e(xO)c5(x 2) = - d(x)
It coincides up to a sign with the commutator d of scalar massless free fields (3-56). This enables us to rewrite (4-11) as
A~ut(x) =
S-l Afn(x)S = Afn(x) - i
d4 y [Afn(x), Ain(y) j(y)]
(4-12)
INTERACTION WITH AN EXTERNAL FIELD
This is reminiscent of the formula (2-81):
Indeed, only the first two terms of the right-hand side do not vanish, since A and B represent free fields, the commutator of which is a c number. Therefore, we may write S = exp
[-i f
d4 x Ain(X).j(X)]
exp [ - i
(4-13)
d x Aout(x) j(X)]
This form does satisfy all the conditions, including unitarity in the indefinite metric space. Only a c-number phase, depending possibly on j, is still arbitrary in S; the latter does not affect the physical quantities. It is convenient to rewrite S in normal order. We decompose Afn as a sum of annihilation Af~+) and creation Af~-) operators. We observe that aIlAf~+) = aIlA~~i), and therefore S must leave the positive metric physical subspace invariant. The commutator of AIl(-) and AV(+) is a c-number function: (4-14) Therefore we may use the identity
eA e B
eA+B+[A, Bl/2
(4-15)
which is valid whenever [A, [A, B]] = [B, [A, B]] = 0, and write the normal form ofS: S = exp [ - i
exp [ - i x exp
d4 y Ain(y) j(y)]
d4 y Al:)(y) j(y)] exp [ - i
d4 y
Al: )(y). j(y)]
(4-16)
{i ff
d4 x d4 y [Al:)(x) j(x),
Al: )(y). j(y)] }
We introduce the Fourier transform of the current j(x):
JIl(k) =
d4 xjll(x) e- ik x
(4-17)
The real character of j and its conservation law are expressed by
jll( -k) = jll*(k)
(4-18)
168 QUANTUM FIELD THEORY
The exponent of the last term in Eq. (4-16) is
~ ffd4x d4y [A~:)(x) oj(x), A~:)(y) oj(y)] = ~ f 2k~;;n J*(k)
jt'(k)
J(k)Iko=lkl
As expected, only the Fourier components of the source corresponding to lightlike arguments do contribute. Moreover, for P = 0, we may decompose
kt< It(k)
+ Jt'r(k)
(4-19)
where J1(k) is a number and Jfr(k) is a space-like vector orthogonal to k. For instance, if k = (kO, k), we introduce the space-like four-vectors G1 = (0, e1), G2 = (0, e2) with ei = e~ = 1, e1 e2 = e1 k = e2 k = 0. We may then choose Jfr(k) = - Li= 1,2 Ji(k)Gf(k) with Ji(k) = Gi J(k). Using this decomposition, it is easy to see that
0 0 0 0
In other words, only transverse components contribute to the last term of (4- i6):
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