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QUANTIZATIONFREE FIELDS 131 in .NET framework
QUANTIZATIONFREE FIELDS 131 Scanning PDF417 In VS .NET Using Barcode Control SDK for Visual Studio .NET Control to generate, create, read, scan barcode image in .NET applications. PDF417 2d Barcode Creation In Visual Studio .NET Using Barcode generation for Visual Studio .NET Control to generate, create PDF 417 image in .NET framework applications. only involves longitudinal and scalar polarizations: Decode PDF417 2d Barcode In VS .NET Using Barcode reader for VS .NET Control to read, scan read, scan image in VS .NET applications. Make Barcode In Visual Studio .NET Using Barcode drawer for VS .NET Control to generate, create bar code image in Visual Studio .NET applications. iO'A(+)=Jdke ik ' X
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(3118) n<4>n l4>n> =
Hence
If n = 0 such a state therefore has a zero norm. A general solution is of the form (3118) with 2 <4>14 = IC o l 20 the coefficients C; remaining arbitrary and each of the l4>n> fulfilling (3117). It remains to show that the arbitrariness in 14 does not affect the physical observables. Consider the energy, for instance. The hamiltonian reads = dk
f f w{t
d x :nl'AI'  2': = 1 2' d x: ~l [Al
",3. + (VAi)2]  A~  (VA O)2: (3120) d1.)t(k)d1.)(k)  dO)t(k)dO)(k)] QUANTUM FJELD THEORY
When acting with H on a state in :Yt'1 we find in general a contribution from the unphysical part 14 , but it vanishes when we take the mean value since <1/11 H 11/1) <1/111/1)  <I/ITI dk Wk ).];,2 d).)t(k)d).)(k) II/IT) <I/ITII/IT) (3121) by virtue of (3117). As far as momentum is concerned, the formula is analogous with Wk replaced by k. Insofar as only average values are observed, we see that not only have negative probabilities disappeared when restricting ourselves to :Yt'1 but also scalar and longitudinal photons do not contribute. Only the two physical transverse polarization states manifest themselves. It is nice to realize that the arbitrariness on 14 is reflected by the fact that the mean value of A" in such a state is a pure gradient. Let us compute it as an exercise. We readily derive from (3118) that <4>1 A"(x) 14 c~c <4>01
dk e ik x
[e~3)(k)a(3)(k) + e~O)(k)dO)(k)] 14>,) + complex conjugate (cc) 14>0) can be taken
equal
All other components 14>.) do not contribute since A" changes n by a unit; to 10), and 14>,) is necessarily of the form 14>,) = dq J(q) [a(3)t(q)  do>t(q)] 10 ) Choosing real polarization vectors we have
We recall that e(O)(k) n and by its definition e(3)(k) = k  n(k' n) (k'n) Therefore, Hence, provided the integral is meaningful, (3122) where the scalar cnumber function A(x) is a solution of DA(x) = 0, and may be chosen at will by adjusting the vector 14 . This is another proof that the arbitrariness of 14 reflects the gauge arbitrariness of A" without practical consequence. We may select in the equivalence class of state vectors of the form II/IT) 14 a convenient representative by imposing that 14 == 10), this again with reference to a choice of basis polarization vectors. Instead of taking the transverse ones, real and orthogonal, we frequently choose the complex combinations (8(1) i8(2))!.j2 to represent the two helicities of the photon. QUANTIZATIONFREE FIELDS
We shall not elaborate in detail the proof of Lorentz invariance. It is clear that .Yt\ is Lorentz invariant as well as the equivalence class representing a given ph)!sical state. The reader is invited to construct explicitly the generators M"' of Lorentz transformations. He or she will verify that photons are not only massless but carry an helicity 1. It is sometimes said that the photon has spin 1. In spite of the fact that only vectors in 1 receive a physical interpretation it is worth emphasizing that the full indefinite norm Fock space is necessary to preserve the locality properties. They appear generally in complete sums over intermediate states. As an exercise one can easily extend the discussion of Sec. 315 to recover the blackbody spectrum within the present formalism.

