QUANTIZED ELECTROMAGNETIC FIELD INTERACTING WITH A CLASSICAL SOURCE 4-1-1 Emission Probabilities in Visual Studio .NET

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4-1 QUANTIZED ELECTROMAGNETIC FIELD INTERACTING WITH A CLASSICAL SOURCE 4-1-1 Emission Probabilities
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We shall first consider the interaction of the quantized electromagnetic field with an external given source. The physical problem is thus the emission or absorption of photons by a classical conserved currentjll(x):
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Olljll(X) = 0
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From the electromagnetic action
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d4x (iFllvPV + jilAIl)
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we derive the equation of motion
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DA v - ovo A =j"
(4-3)
QUANTUM FIELD THEORY
As in the free case, we encounter a difficulty in the quantization of this field, which may be solved by the addition to the action of the term
d4X (O.A)Z
In this gauge, the equation of motion reduces to
DAJl=jIl
(4-4)
The operators AJl(x) act in a space with an indefinite metric and satisfy the same commutation relations as in the free case (4-5)
If we assume that the solution of (4-4) is defined on the same Fock space as the free field, the relation between the free and interacting fields amounts to a canonical transformation. In quantum mechanics with a finite number of degrees offreedom, this would be implemented by a unitary transformation. However, for an infinite number of degrees of freedom, there may be a difficulty. We illustrate this point with a simple model. Let a finite quantum system be composed of N half-integer spins at the vertices of a three-dimensional cubic lattice. They may carry, for instance, a magnetic moment. The observables are the 3N operators O"l(n), O"z(n), 0"3(n), n being the site index. The states are linear combinations of the eigenvectors of the 0"3, denoted by I ... ). They are generated by the action of the 0"upon the state 10) == 1+ + ... +). Consider now the rotated operators
r 1 (n)
= 0" 1 (n)
cos e - 0"3(n) sin e
rz(n) = O"z(n)
r3(n)
O"l(n) sin
e + 0"3(n) cos e
Clearly,
r,,(n) = exp [
-i ~ ptl O"z(p)]O",,(n) exp [i ~ ptl O"z(P)]
This means that the operators 0" and r which generate the same algebra are unitarily equivalent. Let Ie) denote the new ground state of the r:
Ie) = exp [ -
~ ptl O"z(P)J!O)
It is easy to compute the scalar product (Ole) = (cos e/2)N. Assume that in the limit N --+ 00, the Hilbert space of states is constructed from the ground state 10) == 1+ + .,. + ) by the action of a finite number of creation operators 0" _ and Cauchy completion. We may again rotate the operators 0" into r and define the state Ie) in analogy with 10). We may then seek a unitary transformation U(e) which would implement this rotation:
INTERACTION WITH AN EXTERNAL FIELD
Such a unitary operator clearly does not exist when N --+ 00. Any scalar product of a rotated state with a nonrotated one vanishes in this limit: for instance, <01 fJ) = (cos fJ/2t N-->d, O. The moral is that, in the case of an infinite number of degrees of freedom, physically equivalent observables, i.e., realizing the same algebra of commutation relations, etc., are not necessarily unitarily equivalent. We have to keep this in mind when we observe that our field AI' satisfies the free commutation relations (4-5). Returning to the field equation (4-4), we may write a 'particular c-number solution
A~(x) =
d4y G(x - yW(y)
expressed in terms of some Green function G(x. - y) of the d' Alembertian operator Ox:
OxG(x - y)
= c5 4 (x - y)
The general solution of Eq. (4-4) is therefore
AI'(x) =
A~O)(x) +
d4yG(x - y)jl'(Y)
(4-6)
where A(O) is a quantum free field. The precise form of Green function is specified by the boundary conditions. Assume that the current jl'(y) has been switched on adiabatically on a finite time interval. Using the advanced and retarded Green functions of Eq. (1-170) 1 ret Gadv (x) = - (2 n )4 we may write
AI'(x)
4 e- ' d P (Po _ 18 +')2
2nfJ xo)c5(x)
Afn(x)
d4y Gret(X - y)jll(y)
A~ut(x) + fd 4y Gadv(X -
y)jll(y)
(4-7)
The free fields Afn and A~ut describe the photon field before and after its interaction with the current j. In a formal sense, we have lim AI'(x) = Afn(x)
Xo--+ - 00
lim AI'(x)
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