# <111 IlIz> = exp in VS .NET Making PDF417 in VS .NET <111 IlIz> = exp

<111 IlIz> = exp
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[f dk lI!(k)IIZ(k)]
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Therefore two such states are in general not orthogonal and the set of coherent states is an overcomplete one. If we denote 111) as the normalized state corresponding to 111>:
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The vacuum state corresponds to 11 = 0, and more generally a(k) 111 > = lI(k) 111 >, so that we have diagonalized the annihilation (positive frequency) part of the field
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Coherent states remain coherent under time evolution. Indeed,
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e- iH , 111 > = exp
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{f dk lI(k) [e- iH' at(k) e+ iH'] }IO>
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[f dk lI(k) e- ikot at(k)] 0 >
= exp = 111,> where IIt(k)
(3-61)
e- iko' lI(k).
QUANTIZATION-FREE FIELDS
It is also interesting to construct the unitarity operator D(1]) which realizes the transformation
The reader may convince himself that
D(1]) = exp {-
dk [1]*(k)a(k) -1](k)a t (k)]}
does the job. Since <p(x) is an operator-valued distribution let us smear it with a test function
<PI =
d4x f(x)<p(x)
(3-62)
When f approaches <5(t)g(x) or <5'(t)g(x) we recover a space average of <p(0, x) or n(O, x). We may study the probability distribution of <PI in a state corresponding to a fixed number r of particles
p,(a) = <r 1<5(<PI - a) 1 = r>
da. a . - e- .. <rl e"''''f Ir> 2n
(3-63)
To compute this quantity we can use the coherent states as generating functions so the problem is reduced to the evaluation of (3-64) where 1 is the Fourier transform of f:
Observe that only the mass shell components of 1 contribute to Eq. (3-64). Leaving the general case as an exercise to the reader, we derive the vacuum distribution of the field as
po(a) =
f [f
da exp [ - iaa 2n
dk Il(k)!2
a -"2
f -- ]
dk 1f(k) 12
] [ a2 exp - 2 dk Il(k)!2
(3-65)
Not surprisingly we obtain a gaussian distribution with mean square fluctuation
around a mean value zero. The field at a fixed space-time point corresponding to 1= 1 has an infinite fluctuation and is therefore unobservable. But smeared operators are meaningful even when the function f has a support restricted to a fixed time. We note that the vacuum is not a zero field state and we shall soon show that vacuum fluctuations are observable. Instead of considering <p( + )(x) we may also wish to construct a complete set of tested values of <p(0, x):
(3-66)
corresponding to fn(x) = <5(x )Fn(x) in the previous notation, with
(3-67)
QUANTUM FIELD THEORY
normalized according to
dk Fn* (k) Fn,(k) =
Fn( -k)
bn,n' Fn*(k)
(3-68)
Fn(k)F~(k')
== (2nj32wkb3(k - k')
<{In,,
An eigenstate of the operators <{In with eigenvalues eigenstate of the field <(J(O, x) with eigenvalue
J,(x) =
may, by abuse of language, be called an
~ <(In,,
dk e- ik ' x F~(k)
(3-69)
such that
Let us denote by I<{J,) the corresponding state, To obtain its components in the Fock basis it is sufficient to know its scalar product with a coherent state 11]). This is obtained by solving
(3-70)
for every value of n, If the normalization is such that
(3-71)
we find that
from which we may study the physical content of these states.
3-1-3 Charged Scalar Field
The above description of the hermitian scalar field does not allow a distinction between particles and antiparticles. Particles and antiparticles have to carry some opposite-charge quantum number, whatever the nature of this charge may be. Classically, the minimal coupling prescription requires at least a complex field. In the quantum case let us therefore introduce a doublet of hermitian fields CP1 and CP2 represented by the complex quantity
CP1 + iCP2 cP=
and its hermitian conjugate. The total lagrangian 2'tot will be the sum of two identicallagrangians of the form (3-7) pertaining to CP1 and CP2 : 2 2'tot = 2'(CP1) + 2'(CP2) = (Ol'cp)t(iJl'CP) - m cptcp (3-73) The quadratic 2' is invariant under any rotation in the internal space characterized by the indices 1 and 2, or equivalently when cP is multiplied by a phase and cpt
QUANTIZATION-FREE FIELDS
by the conjugate one. The invariance group is U(l) or 0(2). Quantization is performed independently for <PI and <Pz, with the states labeled by the numbers nl and nz of quanta of type 1 or 2 corresponding to the operators Ni
f dk at(k)ai(k)
= 1,2
(3-74)
In terms of the fields <p and <p t and the conjugate momenta n=<p= the hamiltonian reads H = f d 3x :ntn + V<ptV<p + mZ<pt<p: (3-76)
CPl - icpz
t n=<p=
+ icpz
(3-75)
Note the absence of factor i in Eqs. (3-73) and (3-76) as compared to the hermitian case. The commutation rules [<p(x), <pt(y)] = id(x - y) reduce at equal times to [<p(t, x), n(t, y)] = [<pt(t, x), nt(t, y)] = ic5\x - y) The field itself may be written <p(x) = fdk[a(k)e-ik.X
(3-77)
+ bt(k)e ik ' x ]
<pt(x) = f dk [b(k) e- ik ' x + at(k) e ik . X] a(k) = al(k) at(k) =
+ iaz(k)
a~(k) - ia~(k)