aa r X i v : . [ m a t h - ph ] F e b Quantum radiation from a classical point source
J. DimockDepartment of MathematicsSUNY at BuffaloFebruary 9, 2021
Abstract
We study the radiation of photons from a classical charged particle. Weparticularly consider a situation where the particle has a constant velocity in thedistant past, then is accelerated, and then has a constant velocity in the distantfuture. Starting with no photons in the distant past we seek to characterize thequantum state of the photon field in the distant future. Working in the Coulombgauge and in a C* algebra formulation, we give sharp conditions on whether thisstate is or is not in Fock space.
We study the radiation of the quantum electromagnetic field in the presence of a clas-sical source. With a specified four-current consisting of functions ( j , j ) satisfying ∂ µ j µ = 0, one seeks associated quantum field operators A = ( A , A ). In the Lorentzgauge ∂ µ A µ = 0 these are connected by the wave equation with source (cid:3) A µ = j µ .A textbook treatment of this problem can be found in Itzykson-Zuber [4] whose signconventions we follow. A mathematical treatment is due to Naudts-DeRoeck [6] . Acharacteristic phenomena is that if one starts in a state with no photons in the distantpast, the vacuum in Fock space, it may evolve into a state in the distant future withinfinitely many photons radiated and not in Fock space. This is sometimes known asthe infrared catastrophe.In this paper we are interested in characterizing exactly when this happens. Weparticularly study the case of a point source where the four-current has the form j ( x, t ) = δ ( x − x ( t )) j ( x, t ) = x ′ ( t ) δ ( x − x ( t )) (1)and x ( t ) is the trajectory of the charged particle in R . We suppose that the particlehas a velocity v in in the distant past, then accelerates for a finite amount of time, and1hen continues with a velocity v out in the distant future. We start with the Fock vacuumin the distant past considered as a state ω in on the C ∗ -algebra generated by the fields.It evolves to state ω out in the distant future. We establish the following results • If v out = v in = 0 and the trajectory x ( t ) is smooth then ω out is a Fock state. • If v out = v in = 0 and the velocity x ′ ( t ) has a discontinuity then ω out is not a Fockstate. • In general for a smooth trajectory ω out is a Fock state if and only if v in = v out .We work in the Coulomb gauge rather than the Lorentz gauge. The advantage isthat we avoid working with indefinite metric Hilbert spaces. The disadvantage is thatrelativistic properties like Lorentz invariance and locality are not manifest, althoughthey are still there. (See Skagerstam, Erikson, Rekdal [7] for a discussion of locality inthe Coulomb gauge.)The Coulomb gauge is defined by the condition ∇ · A = 0 on the magnetic potential.In this gauge Maxwells equations become − ∆ A = j ∂ A∂t − ∆ A + ∇ ∂∂t A = j (2)The first equation for the electrostatic potential A has the solution A ( x, t ) = Z V ( x − y ) j ( t, y ) dy (3)where V ( x ) = (4 π ) − | x | − is the Coulomb potential. In the second equation we callthe term ∇ ( ∂A /∂t ) the longitudinal current. Since ∂j /∂t + ∇ · j = 0 this can beexpressed as j L ( x, t ) = ∇ ( ∂A /∂t )( x, t ) = −∇ Z V ( x − y )( ∇ · j )( y, t ) dy (4)It satisfies ∇ · j L = ∇ · j . The transverse current is defined by j T = j − j L and itsatisfies ∇ · j T = 0. The second equation in (2) and the constraint can now be written ∂ A∂t − ∆ A = j T ∇ · A = 0 (5)or just (cid:3) A = j T with (cid:3) = ∂ t − ∆.The electrostatic potential A is given by (3) and is not quantized. We seek quantumfield operators A satisfying (5). 2 Preliminaries
We review some facts about solutions of the wave equation with source. At first this isjust classical so we seek functions A satisfying (cid:3) A = j . We are interested in solutionsof the form A = G ± ∗ j where G ± are retarded and advanced fundamental solutionswhich we define shortly. We want to allow j to be certain distributions so G ± ∗ j isdefined as < G ± ∗ j, φ > = < j, G ∓ ∗ φ > for φ in (vector-valued) S ( R ), the Schwartzspace of smooth rapidly decreasing functions. However we also want to work at sharptime so the test functions φ will be replaced by f ⊗ δ t with f in (vector-valued) S ( R ).The upshot is that we want to define functions G ± ∗ ( f ⊗ δ t ), or without the convolution G ± ( f ⊗ δ t ) ≡ G ± ( f, t ).The G ± are given by G ± ( x, t ) = − π ) Z p ∈ R ± iǫ e i ( − p t + px ) p − | p | dp dp (6)For f ∈ S ( R ) this is to be understood as for G ± ( f, t ) = R f ( x ) G ± ( x, t ) dx given by G ± ( f, t ) = − (2 π ) − Z p ∈ R ± iǫ p − | p | e − ip t ˜ f ( − p ) dp dp (7)where ˜ f ( p ) = (2 π ) − R e − ipx f ( x ) dx is the Fourier transform of f .Consider G + ( f, t ). We have | e − ip t | = e Im( p ) t so if t < p -contourin the upper half plane and get zero. On the other hand If t > p = ± ω where ω ( p ) = | p | . Similarly for G − ( f, t ). We obtain G ± ( f, t ) = ± (2 π ) − θ ( ± t ) i Z (cid:16) e − iωt − e iωt (cid:17) ˜ f ( − p ) dp ω (8)where θ is the Heaviside function.The convolution is defined by ( G ± ∗ ( f ⊗ δ t ))( x, t ) = G ± (cid:16) f ( x − · ) , δ t ( t − · ) (cid:17) . Butthe Fourier transform of f ( x − · ) at − p is e ipx ˜ f ( p ) and δ t ( t − · ) = δ t − t . Thus we have( G ± ∗ ( f ⊗ δ t ))( x, t ) = ± (2 π ) − θ ( ± ( t − t )) i Z (cid:16) e − iω ( t − t ) − e iω ( t − t ) (cid:17) e ipx ˜ f ( p ) dp ω (9)We also will want to consider the propagator G = G + − G − . In this expression theHeaviside functions θ disappear and we have( G ∗ ( f ⊗ δ t ))( x, t ) = u f ( x, t − t ) (10)where u f ( x, t ) = (2 π ) − i Z (cid:16) e − iωt − e iωt (cid:17) e ipx ˜ f ( p ) dp ω (11)3s a smooth solution of the wave equation with data at t = 0 u f ( x,
0) = 0 ∂u f ∂t ( x,
0) = f ( x ) (12) We review the mathematics of the quantized free field in the Coulomb gauge. In thiscase we can take A = 0, and study the field equation ∂ A∂t − ∆ A = 0 (13)We want to define a quantum field operator satisfying the equation, the constraint ∇ · A = 0, and the canonical commutation relations.It is convenient to use the relativistic Hilbert space which is (vector valued) H = L ( R , (2 ω ) − dp ) (14)again with ω ( p ) = | p | . On this space define a projection operator.( P h ) j ( p ) = X k =1 (cid:16) δ jk − p j p k | p | (cid:17) h k ( p ) (15)which is the orthogonal projection onto the space of transverse functions satisfying p · h ( p ) = 0. The one photon Hilbert space is H T = P H , the n-photon Hilbert space H nT is the n-fold symmetric tensor product of H T with itself, and the Fock space over H T is F ( H T ) = ∞ M n =0 H nT (16)For h ∈ H T let a ∗ ( h ) and a ( h ) be the creation and annihilation operators on theFock space, with a ( h ) anti-linear in h and a ∗ ( h ) linear in h . These operators (and alloperators in this section) are defined on the dense subspace D consisting elements ofFock space with a finite number of particles. They satisfy [ a ( h ) , a ∗ ( h )] = ( h , h ). Wealso define momentum space fields A ◦ ( h ) and Π ◦ ( h ) = A ◦ ( ih ) by A ◦ ( h ) = 1 √ (cid:16) a ∗ ( h ) + a ( h ) (cid:17) Π ◦ ( h ) = i √ (cid:16) a ∗ ( h ) − a ( h ) (cid:17) (17)These are real linear in h , symmetric, and satisfy [ A ◦ ( h ) , A ◦ ( h )] = i Im( h , h ).The field operator is defined as a distribution. For a real test function f ∈ S ( R )define A ( f, t ) (formally R f ( x ) A ( x, t ) dx ) by A ( f, t ) = √ A ◦ (cid:16) e iωt ˜ f T (cid:17) = a ∗ ( e iωt ˜ f T ) + a ( e iωt ˜ f T ) (18)4here ˜ f T ≡ P ˜ f . This satisfies the field equation ∂ A∂t ( f, t ) − A (∆ f, t ) = 0 (19)It also satisfies ∇ · A = 0 since for scalar g ∈ S ( R ) we have ( ∇ · A )( g, t ) ≡ − A ( ∇ g, t )and ( ∇ g ) ∼ T ( p ) = iP ( p ) p ˜ g ( p ) = 0. We have the commutation relations[ A ( f , t ) , A ( f , t )] = 2Im( ˜ f ,T , ˜ f ,T ) = 0 (20)The vanishing follows since for f real ˜ f ( p ) = ˜ f ( − p ) so ( ˜ f ,T , ˜ f ,T ) is real.The field has a conjugate momentum Π( h, t ) = ∂/∂t A ( h, t ) given byΠ( f, t ) = √ ◦ (cid:16) e iωt ω ˜ f T (cid:17) = i (cid:16) a ∗ ( e iωt ω ˜ f T ) − a ( e iωt ω ˜ f T ) (cid:17) (21)The Π( f, t ) commute with each other and satisfy the canonical commutation relations(CCR) [ A ( f , t ) , Π( f , t )] = i ( ˜ f ,T , ω ˜ f ,T ) H = i ( ˜ f ,T , ˜ f ,T ) L ( R ,dp ) (22) C ∗ algebra and coherent states We want to allow states for this system which are not Fock states, and for this we focuson the algebra of the fields. See for example Bratteli-Robinson [1] for more details onthis standard material. We consider operators A ◦ ( h ) for h in a dense subspace S ⊂ H T .The operator is essentially self-adjoint on the domain D ⊂ F ( H T ), and so we can define W ◦ ( h ) = e iA ◦ ( h ) h ∈ S (23)This satisfies W ◦ ( h ) W ◦ ( h ) = e − i Im( h ,h ) W ◦ ( h + h ) (24)This is the Weyl form of the CCR. The operators W ◦ ( h ) generate a C ∗ algebra, denoted A , which is the closure of the set of operators which are finite sums P n c n W ◦ ( h n ) in theBanach space of bounded operators on F ( H T ). Different choices of the dense subspace S give different algebras, and we will make a specific choice later.A state ω on the algebra is a positive linear functional of norm one. Positivity isthe condition that ω ( B ∗ B ) ≥ B ∈ A . Any normalized state in Fock spacedefines a state on the algebra. In particular the Fock vacuum Ω defines a state by ω ( · ) = (Ω , [ · ]Ω ). It satisfies a ( h )Ω = 0 and is characterized by ω ( W ◦ ( h )) = (Ω , e iA ( h ) Ω ) = (Ω , e ia ∗ ( h ) / √ e ia ( h ) / √ Ω ) e − k h k = e − k h k (25)Another class of states are the coherent states. Given J ∈ H T and the Fock vacuumΩ we define the coherent stateΩ J = e − k J k e a ∗ ( J ) Ω = e − k J k ∞ X n =0 n ! ( a ∗ ( J )) n Ω (26)5he series converges in Fock space since k ( a ∗ ( J )) n Ω k = √ n ! k J k n . It is normalized to k Ω J k = 1 and is an eigenstate of the annihilation operator: a ( h )Ω J = ( h, J )Ω J (27)As a state on the algebra ω h = (Ω h , [ · ]Ω h ) it satisfies ω J ( W ◦ ( h )) =(Ω J , e iA ( h ) Ω J )=(Ω J , e ia ∗ ( h ) / √ e ia ( h ) / √ Ω J ) e − k h k = e i ( h,J ) / √ e i ( h,J ) / √ e − k h k = e − k h k + i √ J,h ) (28)More generally we define a coherent state on A to be a state satisfying ω ( W ◦ ( h )) = e − k h k + i √ L ( h ) (29)for some linear function L on the dense domain S . It can also be written ω ( W ◦ ( h )) = ω ( W ◦ ( h )) e i √ L ( h ) (30)The positivity condition is satisfied since for any finite sequence of complex numbers c n and h n ∈ S and B = P n c n W ◦ ( h n ) we have ω ( B ∗ B ) = X n,m ¯ c m c n ω (cid:16) W ◦ ( h n − h m ) (cid:17) e i Im( h n ,h m ) = X n,m (cid:16) ¯ c m e − i √ L ( f m ) (cid:17)(cid:16) c n e i √ L ( f n ) (cid:17) ω (cid:16) W ◦ ( h n − h m ) (cid:17) e i Im( h n ,h m ) = ω ( B ∗ B ) ≥ B = X n c n e i √ L ( f n ) W ◦ ( h n ) (32)The coherent state ω defined by L is known to be a pure state (Honegger-Rapp [3]).Furthermore ω is a Fock state if and only if L is continuous on the pre-Hilbert space S (Honegger-Rieckers [2]). This is the criterion we will use in the following.6 Regular source
To warm up we first consider our problem with a regular source. We take j ∈ C ∞ ( R )and seek quantum field operators which solve (cid:3) A = j T and ∇ · A = 0. Note thatpartial Fourier transform ˜ j ( p, t ) is bounded and rapidly decreasing in p as is ˜ j T ( p, t ) = P ( p )˜ j ( p, t ) The inverse Fourier transform j T ( x, t ) then has compact support in t , andfor each t is bounded and continuous in x (since ˜ j ( p, t ) is integrable in p ) and squareintegrable in x (since ˜ j ( p, t ) is square integrable in p ). The localization of j T ( x, t ) in x is considerably weaker than that of j ( x, t ).We want a solution which is a free field in the distant past and the distant future.Therefore we take A in ( f, t ) to be a free field defined from creation and annihilationoperators a ∗ in ( h ) , a in ( h ) as in (18). Then for real f ∈ S ( R ) we define A ( f, t ) = A in ( f, t ) + ( G + ∗ j T )( f, t ) (33)The expression G + ∗ j T can be analyzed pointwise. Indeed we have explicitly( G ± ∗ j T )( x, t ) = ± (2 π ) − i Z θ ( ± ( t − s )) i (cid:16) e − iω ( t − s ) − e iω ( t − s ) t (cid:17) e ipx ˜ j T ( p, s ) dp ω ds (34)Then ( G + ∗ j T )( x, t ) vanishes in the distant past since j T ( p, s ) does, and hence A ( f, t ) = A in ( f, t ) in the distant past. We have (cid:3) ( G ∗ j T ) = j T , also in the sense of distributions,and hence (cid:3) A = j T . Furthermore ∇ · ( G ∗ j T ), also in the sense of distributions, andhence ∇ · A = 0Next we define A out ( f, t ) by A ( f, t ) = A out ( f, t ) + ( G − ∗ j T )( f, t ) (35)Then A ( f, t ) = A out ( f, t ) in the distant future since ( G − ∗ j T )( f, t ) vanishes in thedistant future. Combining (33) and (35) and identifying G = G + − G − gives A out ( f, t ) = A in ( f, t ) + ( G ∗ j T )( f, t ) (36)where now( G ∗ j T )( x, t ) = (2 π ) − i Z (cid:16) e − iω ( t − s ) − e iω ( t − s ) (cid:17) e ipx ˜ j T ( p, s ) dp ω ds = (2 π ) − i Z (cid:16) e − iωt ˆ j T ( ω, p ) − e iωt ˆ j T ( − ω, p ) (cid:17) e ipx dp ω (37)where ˆ j is the four dimensional Fourier transform with Lorentz inner product:ˆ j ( p , p ) = (2 π ) − Z e i ( p t − px ) j ( x, t ) dxdt = (2 π ) − Z e ip t ˜ j ( p, t ) dt (38)Then G ∗ j T satisfies (cid:3) ( G ∗ j T ) = 0 and ∇· ( G ∗ j T ) = 0, and hence so does A out ( f, t ) Fur-thermore together with Π out ( f, t ) = ∂A out ( f, t ) /∂t it satisfies the CCR. Thus A out ( f, t )is a free field. 7n the second term in (37) make the change of variable p → − p . Since j is real wehave ˆ j T ( ω, p ) = ˆ j T ( − ω, − p ) and the second term is identified as the complex conjugateof the first term. Thus if we define J ( p ) = −√ π i ˆ j ( ω, p ) (39)and J T ( p ) = P ( p ) J ( p ) then we have( G ∗ j T )( x, t ) = − (2 π ) − Z e − i ( ωt − px ) J T ( p ) dp ω (40)This gives( G ∗ j T )( f, t ) = − Z ˜ f ( − p ) e − iωt J T ( p ) dp ω = − e iωt ˜ f T , J T ) (41)Now we have A out ( f, t ) = A in ( f, t ) − e iωt ˜ f T , J T ) (42)We define creation and annihilation operators by taking what amounts to the pos-itive and negative frequency parts of the last expression. We define for h ∈ H T a ∗ out ( h ) = a ∗ in ( h ) − ( J T , h ) a out ( h ) = a in ( h ) − ( h, J T ) (43)These satisfy the standard [ a out ( h ) , a ∗ out ( h )] = ( h , h ). Then a ∗ out ( e iωt ˜ f T ) = a ∗ in ( e iωt ˜ f T ) − ( J T , e iωt ˜ f T ) a out ( e iωt ˜ f T ) = a in ( e iωt ˜ f T ) − ( e iωt ˜ f T , J T ) (44)If we add these we get A in ( f, t ) − e iωt ˜ f T , J T ) = A out ( f, t ) on the right. Thus wemust get the same on the left which says A out ( f, t ) = a ∗ out ( e iωt ˜ f T ) + a out ( e iωt ˜ f T ) (45)just as for A in ( f, t ) .The out vacuum in Fock space should satisfy a out ( h )Ω out = 0. Note that j ∈ C ∞ ( R )implies that ˆ j is bounded and rapidly decreasing, hence J T ∈ H T . Therefore we canconstruct the coherent stateΩ out = Ω J T = e − k J T k exp( a ∗ in ( J T ))Ω in (46)This gives the the desired a out ( h )Ω out = (cid:16) a in ( h ) − ( h, J T ) (cid:17) Ω out = 0 (47)We summarize: 8 roposition 1. Let j ∈ C ∞ ( R ) . Then J T ( p ) = −√ π i ˆ j T ( ω, p ) is in H T and the outvacuum Ω out exists in Fock space. Remarks .1. To obtain this result can certainly relax the condition that j ∈ C ∞ ( R ). The keycondition is is J T ∈ H T which follows from J ∈ H which says Z | ˆ j ( ω, p ) | dp ω < ∞ (48)This is similar to a condition that Naudts - De Roeck [6] found in the Lorentzgauge.2. We sketch some further developments of the scattering theory associated with aregular source. As in (17) define Π ◦ , in ( h ) = i √ (cid:16) a ∗ in ( h ) − a in ( h ) (cid:17) . This is essentiallyself-adjoint on D and we define a unitary operator by S = exp (cid:16) i √ ◦ , in ( J T ) (cid:17) = exp (cid:16) − a ∗ in ( J T ) + a in ( J T ) (cid:17) (49)We have S − Ω in = exp (cid:16) a ∗ in ( J T ) − a in ( J T ) (cid:17) Ω in = e − k J T k exp (cid:16) a ∗ in ( J T ) (cid:17) exp (cid:16) − a in ( J T ) (cid:17) Ω in = e − k J T k exp (cid:16) a ∗ in ( J T ) (cid:17) Ω in = Ω out (50)and S − a in ( h ) S = a in ( h ) + [ a ∗ in ( J T ) , a in ( h )] = a in ( h ) − ( h, J T ) = a out ( h ) (51)So the in and out fields are unitarily equivalent.One can form other asymptotic states by applying creation operators a ∗ out ( h ) toΩ out and then form scattering amplitudes like (cid:16) a ∗ out ( h ′ ) · · · a ∗ out ( h ′ m )Ω out , a ∗ in ( h ) · · · a ∗ in ( h n )Ω in (cid:17) (52)This can also be written (cid:16) a ∗ in ( h ′ ) · · · a ∗ in ( h ′ m )Ω in , S a ∗ in ( h ) · · · a ∗ in ( h n )Ω in (cid:17) (53)and S is revealed as a scattering operator.9 Point source - I
Again we study the equations (cid:3) A = j T and ∇ · A = 0 and look for solutions which arefree in the distant past and future. We now specialize to a point source with current j ( t, x ) = x ′ ( t ) δ ( x − x ( t )). We assume in this section that x ′ ( t ) has compact support.So the particle is at rest in the distant past, accelerates for finite amount of time, andthen stops in the distant future.Again we take A in ( f, t ) to be a free field as defined in (18) for real f ∈ S ( R ). Thenwe define A ( f, t ) = A in ( f, t ) + ( G + ∗ j T )( f, t ) (54)Now we do not define G + ∗ j T pointwise, but treat it as a distribution and proceedby throwing everything onto the test function f . Since G + ( x, t ) = G − ( − x, − t ) weinterpret ( G + ∗ j T )( f, t ) as D G + ∗ j T , f ⊗ δ t E = D j T , G − ∗ ( f ⊗ δ t ) E (55)Then formally we can transfer the transverse projection from j and to f and get D j T , G − ∗ ( f ⊗ δ t ) E = D j, G − ∗ ( f T ⊗ δ t ) E (56)The latter is well-defined for our current since G − ∗ ( f T ⊗ δ t ) as given by (9) is a boundedfunction continuous in the spatial variable. With this interpretation (54) becomes A ( f, t ) = A in ( f, t ) + D j, G − ∗ ( f T ⊗ δ t ) E (57)Since j has compact support and G − ∗ ( f T ⊗ δ t ) vanishes to the future of t , we havethat < j, G − ∗ ( f T ⊗ δ t ) > vanishes for t sufficiently negative and so A ( f, t ) = A in ( f, t )in the distant past. Since A in ( f, t ) is free and since G − is a fundamental solution wehave (cid:3) A out ( f, t ) = ∂ A out ∂t ( f, t ) − A out (∆ f, t )= D j, ∂ ∂t G − ∗ ( f T ⊗ δ t ) − G − ∗ ((∆ f ) T ⊗ δ t ) E = D j, (cid:3) ( G − ∗ ( f T ⊗ δ t )) E = D j, f T ⊗ δ t E ≡ D j T , f ⊗ δ t E (58)So A satisfies (cid:3) A = j T in the sense of distributions. Similarly ∇ · A = 0.Next we define A out ( f, t ) by A ( f, t ) = A out ( f, t ) + G − ∗ j T ( f, t ) (59)which is interpreted as A ( f, t ) = A out ( f, t ) + D j, G + ∗ ( f T ⊗ δ t ) E (60)10hen A ( f, t ) = A out ( f, t ) in the distant future.Combining (57) and (60) gives A out ( f, t ) = A in ( f, t ) − D j, G ∗ ( f T ⊗ δ t ) E (61)Then A out satisfies (cid:3) A out = 0 and ∇ · A out = 0 in the sense of distributions and is afree field.Now take the expression ( G ∗ ( f T ⊗ δ t ))( x, s ) = u f T ( x, s − t ) from (10) and find D j, G ∗ ( f T ⊗ δ t ) E = Z x ′ ( s ) δ ( x − x ( s )) u f T ( x, s − t ) dxds = Z x ′ ( s ) u f T ( x ( s ) , s − t ) ds =(2 π ) − i Z x ′ ( s ) (cid:16) e − iω ( s − t ) − e iω ( s − t ) (cid:17) e ipx ( s ) ˜ f T ( p ) dp ω ds =(2 π ) − Z x ′ ( s ) h ie − iω ( s − t )+ ipx ( s ) ˜ f T ( p ) − ie iω ( s − t ) − ipx ( s ) ˜ f T ( − p ) i dp ω ds =(2 π ) − Z x ′ ( s ) h − ie iω ( s − t ) − ipx ( s ) ˜ f T ( − p ) i dp ω ds (62)Now change the order of integration (easily justified), and define (new definition) J ( p ) = − i (2 π ) − Z x ′ ( s ) e iωs − ipx ( s ) ds (63)Then D j, G ∗ ( f T ⊗ δ t ) E = 2 Re Z e − iωt ˜ f T ( − p ) J ( p ) dp ω = 2 Re ( e iωt ˜ f T , J T ) (64)Then integral converges since J T ( p ) is bounded and ˜ f T ( p ) is bounded and rapidlydecreasing. We have written it as an inner product in H but are not asserting that J T ∈ H T , at least not yet. Now we have as before A out ( f, t ) = A in ( f, t ) − J T , e iωt ˜ f T ) (65)We consider the linear function h → ( J T , h ) for h in a dense domain S ⊂ H T containing the functions e iωt ˜ f T for f ∈ S ( R ). It is the domain of of rapidly decreasingfunctions: S = { h ∈ H T : for k ≥ C so | h (( p ) | ≤ C (1 + | p | ) − k } (66)For h ∈ S we define a ∗ out ( h ) = a ∗ in ( h ) − ( J T , h ) a out ( h ) = a in ( h ) − ( h, J T ) (67)11nd then as in (43) - (45) A out ( f, t ) = a ∗ out ( e iωt ˜ f T ) + a out ( e iωt ˜ f T ) (68)just as for A in ( f, t ).Now we can write A in / out ( f, t ) = √ A ◦ , in / out ( e iωt ˜ f T ) where the momentum spacefields are A ◦ , in / out ( h ) = √ ( a ∗ in / out ( h ) + a in / out ( h )). In terms of these fields the definingrelation (67) can be written A ◦ , out ( h ) = A ◦ , in ( h ) − √ J T , h ) (69)With this identity we pass to the C ∗ algebra. Define W out ( h ) = e iA ◦ , out ( h ) W in ( h ) = e iA ◦ , in ( h ) (70)and then W out ( h ) = W in ( h ) e − i √ J T ,h ) (71)These are elements of C ∗ algebra A generated by W in ( h ), h ∈ S . In the Fock vacuum ω in = (Ω ,in , [ · . ]Ω ,in ) we have ω in ( W out ( h )) = ω in ( W in ( h )) e − i √ J T ,h ) (72)But W out ( h ) is again a representation of the CCR and generates the same C ∗ algebra A as W in ( h ). Then there is a unique ∗ -automorphism α on A such that α ( W out ( h )) = W in ( h ) (see for example [1]). This satisfies α ( W in ( h )) = W in ( h ) e i √ J T ,h ) . If we definea new state by ω out ≡ ω in ◦ α then ω out ( W in ( h )) = ω in ( W in ( h )) e i √ J T ,h ) (73)Instead of changing operators we have changed states. The form of the equation showsthat ω out is a coherent state by our C ∗ algebra definition.The question is now whether ω out is a Fock state. Before answering the question wesharpen our criterion for when this is the case. Proposition 2. ω out is a Fock state if and only if J T ∈ H T . Proof . We have already noted that ω out is a Fock state if and only if the linear function h → ( J T , h ) is continuous on the pre-Hilbert space S . We now claim this is equivalentto J T ∈ H T . If J T ∈ H T then continuity is immediate. If the continuity holds thenthere is J ′ ∈ H T such that ( J T , h ) = ( J ′ , h ) for h ∈ S . If h rapidly decreasing then h T ∈ S and ( J T − J ′ , h ) = ( J T − J ′ , h T ) = 0. Then ( J T − J ′ , h ) = 0 for h ∈ S ( R ).Hence ( J T − J ′ ) / ω as distributions and hence almost everywhere as functions. Thus J T = J ′ almost everywhere and so J T ∈ H T . This completes the proof.In the following we assume that the velocity of the charged particle is less than thespeed of light which is here one. 12 roposition 3. Let x ( t ) be C with | x ′ ( t ) | ≤ v < , and suppose x ′ ( t ) has compactsupport. Then J T ∈ H T and ω out is the Fock state defined by Ω out = Ω J T . Proof . It suffices to show J ∈ H . Ignoring the multiplicative constants in (63) wemight as well assume that J ( p ) = Z x ′ ( t ) e iψ ( p,t ) dt ψ ( p, t ) = ωt − p · x ( t ) (74)Since this is a bounded continuous function of p , to decide whether it is in H we need tostudy asymptotics as p → ∞ . Our first thought might be the stationary phase method.So we look for points where ψ t ( p, t ) = ω − p · x ′ ( t ) = 0 (75)However | p · x ′ ( t ) | < | p | v = ωv and ω (1 − v ) ≤ | ψ t ( p, t ) | ≤ ω (1 + v ) (76)There is no point of stationary phase for p = 0. Instead we can integrate by parts andwrite with ψ tt = − p · x ′′ ( t ) J ( p ) = Z x ′ ( t ) 1 iψ t ∂∂t e iψ ( p,t ) dt = 1 i Z (cid:20) ψ tt ψ t x ′ ( t ) − ψ t x ′′ ( t ) (cid:21) e iψ ( p,t ) dt (77)Now x ′ , x ′′ are bounded and ψ − t and ψ tt /ψ t are O ( ω − ) for | p | ≥
1. Hence J ( p ) = O ( ω − ). This is not quite enough for convergence, but we can repeat the last step andget J ( p ) = 1 i Z (cid:20) ψ tt ψ t x ′ ( t ) − ψ t x ′′ ( t ) (cid:21) iψ t ∂∂t e iψ ( p,t ) dt = Z (cid:20)(cid:16) ψ ttt ψ t − ψ tt ψ t (cid:17) x ′ ( t ) + 3 ψ tt ψ t x ′′ ( t ) − ψ t x ′′′ ( t ) (cid:21) e iψ ( p,t ) dt (78)Now x ′ , x ′′ , x ′′′ are all bounded and the coefficients are all O ( ω − ) for | p | ≥
1. Thus J ( p ) = O ( ω − ) and we have Z | p |≥ | J ( p ) | dp ω ≤ const Z | p |≥ ω − dp < ∞ (79)Hence J ∈ H and the proof is complete. Remark . We can allow jump discontinuities in x ′′′ ( t ) without changing the result.Suppose we allow a jump discontinuity in x ′′ ( t ), say at t = 0. Then in the integrationby parts in (78) we get a boundary term which is (cid:20) p · ( x ′′ (0 − ) − x ′′ (0 + )) x ′ (0) ψ t ( p,
0) + x ′′ (0 − ) − x ′′ (0 + ) ψ t ( p, (cid:21) e iψ ( p, (80)13his is still O ( ω − ) and our results still hold. However a discontinuity in the velocity x ′ ( t ) is more serious as we now show. Proposition 4.
Suppose x ( t ) is continuous on R and C on ( −∞ , and [0 , ∞ ) with x ′ (0 − ) − x ′ (0 + ) = 0 (81) Suppose also x ′ ( t ) has compact support and | x ′ ( t ) | ≤ v < . Then J T / ∈ H T and ω out is not a Fock state Proof . Since J ( p ) is still bounded, it again suffices to consider asymptotics as p → ∞ .In the integration by parts in (77) there is now a boundary term and we have J ( p ) = Z x ′ ( t ) 1 iψ t ∂∂t e iψ ( p,t ) dt = 1 i Z (cid:20) ψ tt ψ t x ′ ( t ) − ψ t x ′′ ( t ) (cid:21) e iψ ( p,t ) dt + 1 i ∆( p ) e iψ ( p, (82)where ∆( p ) = x ′ (0 − ) ω − p · x ′ (0 − ) − x ′ (0 + ) ω − p · x ′ (0 + ) (83)The first term in (82) is analyzed as in the previous proposition, unaffected by possiblediscontinuities in x ′′ , x ′′′ , and the transverse part is in H T . Thus to prove our result itsuffices to show that ∆ T is not in H T .Let v − = x ′ (0 − ) and v + = x ′ (0 + ). Then this can be written∆( p ) = ω ( v − − v + ) + ( v + p · v − − v − p · v + )( ω − p · v − )( ω − p · v + ) (84)Now v + p · v − − v − p · v + = p × ( v + × v − ). If we define δv = v − − v + = 0 and n p = p/ω = p/ | p | then ∆( p ) = ∆ ( p ) + ∆ ( p ) where∆ ( p ) = 1 ω (cid:20) δv (1 − n p · v − )(1 − n p · v + ) (cid:21) ∆ ( p ) = 1 ω (cid:20) n p × ( v + × v − )(1 − n p · v − )(1 − n p · v + ) (cid:21) (85)Note that the denominators here are bounded below since | v ± | <
1. Now ∆ ( p ) isalready transverse, and the transverse part of ∆ ( p ) is with n δv = δv/ | δv | ∆ ,T ( p ) = 1 ω (cid:20) δv − n p ( n p · δv )(1 − n p · v − )(1 − n p · v + ) (cid:21) = | δv | ω (cid:20) n δv − n p ( n p · n δv )(1 − n p · v − )(1 − n p · v + ) (cid:21) (86)We have now ∆ T ( p ) = ∆ ,T ( p ) + ∆ ( p ). 14ow we divide into two cases. Either v ± are colinear or not. If they are colinearthen ∆ ( p ) = 0, and we compute Z | p |≥ | ∆ T ( p ) | dp ω ≥ Z | p |≥ , | n p · n δv |≤ | ∆ ,T ( p ) | dp ω ≥ const | δv | Z | p |≥ , | n p · n δv |≤ ω dp = ∞ (87)If v ± are not colinear then v × ≡ v + × v − = 0. We look at the component of ∆ T ( p )along v × . Again ∆ ( p ) does not contribute, nor does the n δv term in ∆ ,T ( p ). We have n v × · ∆ T ( p ) = | δv | ω (cid:20) − ( n p · n v × )( n p · n δv )(1 − n p · v − )(1 − n p · v + ) (cid:21) (88)Then Z | p |≥ | ∆ T ( p ) | dp ω ≥ Z | p |≥ , | n p · n v × |≥ , | n p · n δv |≥ | n v × · ∆ T ( p ) | dp ω ≥ const | δv | Z | p |≥ , | n p · n v × |≥ , | n p · n δv |≥ ω dp = ∞ (89)The last step follows since δv orthogonal to v × implies the cones | n v × · n p | ≥ and | n p · n δv | ≥ have a non-empty intersection. This completes the proof. Now we consider a trajectory x ( t ) which has constant velocity v in in the distant past,and constant velocity v out in the distant future, and is otherwise smooth. So we nolonger have compact support for x ′ ( t ), but we do have compact support for x ′′ ( t ).There is now a problem with the convergence of integrals like (63). To avoid thiswe approximate the current j by j R ( x, t ) = ( x ′ ( t ) δ ( x − x ( t )) | t | ≤ R R → ∞ . The analysis proceeds as in the the previous sectionand we have A out ( f, t ) = A in ( f, t ) − D j R , G ∗ ( f T ⊗ δ t ) E where D j R , G ∗ ( f T ⊗ δ t ) E =(2 π ) − Z R − R x ′ ( s ) Z h − ie iω ( s − t ) − ipx ( s ) ˜ f T ( − p ) i dp ω ds (91)15 roposition 5. Let x ( t ) be C with | x ′ ( t ) | ≤ v < , and suppose x ′′ ( t ) has compactsupport. Then the limit R → ∞ exists on the right side of (91) and defines the left sideat R = ∞ . We have as in (64) D j, G ∗ ( f T ⊗ δ t ) E = 2 Re Z e − iωt ˜ f T ( − p ) J ( p ) dp ω = 2 Re ( e iωt ˜ f T , J T ) (92) where up to a multiplicative constant J ( p ) = 1 i Z (cid:20) ψ tt ψ t x ′ ( t ) − ψ t x ′′ ( t ) (cid:21) e iψ ( p,t ) dt (93) as in (77), and the representation (78) for J ( p ) holds as well. Proof . Note that since ψ tt = − p · x ′′ ( t ) the integrand on the right side of (93) hascompact support in t and that integral exists. Furthermore J ( p ) is O ( ω − ) everywhereand ˜ f T ( p ) is bounded and rapidly decreasing so the integral (92) converges. Once (93)is established (78) holds by a further integration by parts.To derive (92) it suffices to consider the case t = 0. Then in (91) with ψ ( p, s ) = ωs − p · x ( s ) we need to find an R → ∞ limit for Z R − R x ′ ( s ) (cid:20)Z e iψ ( p,s ) ˜ f T ( − p ) dp ω (cid:21) ds = Z R − R x ′ ( s ) dds (cid:20)Z iψ s e iψ ( p,s ) ˜ f T ( − p ) dp ω (cid:21) ds + Z R − R x ′ ( s ) (cid:20)Z ψ ss iψ s e iψ ( p,s ) ˜ f T ( − p ) dp ω (cid:21) ds = 1 i Z R − R (cid:20)Z (cid:18) ψ ss ψ s x ′ ( s ) − ψ s x ′′ ( s ) (cid:19) e iψ ( p,s ) ˜ f T ( − p ) dp ω (cid:21) ds + (cid:20) x ′ ( s ) Z iψ s e iψ ( p,s ) ˜ f T ( − p ) dp ω (cid:21) s = Rs = − R (94)In the double integral the R → ∞ limit exists by the support of x ′′ ( t ). The integrandis O ( ω − ) as | p | → | p | → ∞ , so the integral is is abso-lutely convergent. Changing the order of integration gives the result (92), provided theendpoint term goes to zero.For the endpoint term we need to show R ψ − t e iψ ( t,p ) ˜ f T ( − p ) dp/ ω goes to zero as | t | → ∞ . As t → ∞ we have x ′ ( t ) → v out so it suffices to show Z ω − p · v out e i ( ω − p · v out ) t ˜ f T ( − p ) dp ω (95)goes to zero as t → ∞ . Let p → − p and then write this in polar coordinates ( | p | , θ, φ )with v out as the z -axis. Then ω + p · v out = ω (1 + cos θ v out ) and (1 + cos θ v out ) isbounded above and below. The integral is now Z θ v out ) h Z e i | p | (1+cos θ v out ) t ˜ f T ( | p | , θ, φ ) d | p | i sin θdθdφ (96)16ow the interior integral goes to zero by the Riemann- Lebesgue lemma, and the overallintegral goes to zero by dominated convergence. The limit t → −∞ is entirely similar.This completes the proof.Having established (92), (93) we proceed as in (65) - (73) and again find ω out ( W in ( h )) = ω in ( W in ( h )) e i √ J T ,h ) (97) Proposition 6.
Let x ( t ) be C with | x ′ ( t ) | ≤ v < , and suppose x ′ ( t ) = v in in thedistant past and x ′ ( t ) = v out in the distant future so x ′′ ( t ) has compact support. Then ω out is a Fock state if and only if v in = v out . Proof . We must show that J T ∈ H T if and only if v in = v out . We can use therepresentation (78) as before to get convergence for | p | ≥
1. So the issue is whether thethe integral for | p | ≤ | p | ≤ J ( p ) is O ( ω − ), as in J T ( p ). We show that in case v in = v out this is actually sharp and gives a logarithmic divergence.In the representation (93) for | p | ≤ e iψ ( p,t ) = 1+ O ( ω ) and correspondingly J ( p ) = F ( p ) + O (1). The O (1) term is in H and the transverse part in in H T , so itsuffices to consider F ( p ) and show that F T ( p ) either is or is not square integrable for | p | ≤ ω ) − dp .We have up to a multiplicative constant F ( p ) = Z (cid:20) x ′ ( p · x ′′ ) ω − p · x ′ ( t )) + x ′′ ( ω − p · x ′ ( t )) (cid:21) dt = Z ddt (cid:20) x ′ ( t )( ω − p · x ′ ( t )) (cid:21) dt = v out ( ω − p · v out ) − v in ( ω − p · v in )= ω ( v out − v in ) + ( v in p · v out − v out p · v in )( ω − p · v out )( ω − p · v in ) (98)where δv = v out − v in . We see that if v in = v out then F = 0 and F T = 0 and hence J ∈ H and J T ∈ H T and ω out is Fock.We proceed with the case v out = v in . We are now almost exactly in the situationthat occurred in the proof of Proposition 4. The main difference is that we consider | p | ≤ | p | ≥
1. We write F T ( p ) = F ,T ( p ) + F ( p ) where with δv ≡ v out − v in and v × = v in × v out F ,T ( p ) = | δv | ω (cid:20) n δv − n p ( n p · n δv )(1 − n p · v out )(1 − n p · v in ) (cid:21) F ( p ) = 1 ω (cid:20) n p × n v × (1 − n p · v out )(1 − n p · v in )) (cid:21) (99)17f v in and v out are colinear (which includes the case where one of them is zero), thenonly F ,T contributes and we have as in (87) Z | p |≤ | F T ( p ) | dp ω ≥ const | δv | Z | p |≤ , | n p · n δv |≤ ω dp = ∞ (100)If v in and v out are not colinear we look at the component along v × and find as in (88) n v × · F T ( p ) = | δv | ω (cid:20) − ( n p · n v × )( n p · n δv )(1 − n p · v out )(1 − n p · v in ) (cid:21) (101)Then as in (89) Z | p |≤ | F T ( p ) | dp ω ≥ const | δv | Z | p |≤ , | n p · n v × |≥ , | n p · n δv |≥ ω dp = ∞ (102)Thus the F T condition fails, and ω out is not Fock. Remarks .1. Note that if the particle has constant velocity then x ′′ ( t ) = 0, hence J ( p ) = 0,and ω out = ω in . No photons are radiated.2. The result does not depend on the complete history of the trajectory, only on theinitial and final states. Except in a the special circumstance v in = v out we alwaysleave Fock space.3. It is noteworthy that the departure from Fock space comes from a UV divergencein the case of of discontinuities in the velocity (proposition 4), and from an IRdivergence in the case v in = v out (proposition 6).4. It might be of interest to obtain results of this kind for the radiation of gravitonsby a classical source in linearized quantum gravity. See Skagerstam, Eriksson,Rekdal [8], for a formulation of this model in an analog of the Coulomb gauge.In this connection we also mention the work of Kegeles, Oriti, Tomlin [5] whosuggest non-Fock coherent states of the type used here as appropriate for a groupfield theory model of quantum gravity.18 eferenceseferences