Classical and quantum radiation reaction for linear acceleration

0.28 MB·English

by  Atsushi Higuchi

Classical and quantum radiation reaction for linear acceleration
Atsushi Higuchi1 and Giles D. R. Martin2
Department of Mathematics, University of York,
Heslington, York YO10 5DD, UK
email: [email protected], [email protected]
(Dated: January 5, 2005)
Weinvestigatetheeffectof radiation reaction onthemotion ofawavepacketofacharged scalar
particlelinearlyaccelerated inquantumelectrodynamics. Wegivethedetailsofthecalculationsfor
thecasewheretheparticleisacceleratedbyastaticpotentialthatwereoutlinedinPhys.Rev. D70
(2004)081701(R)andpresentsimilarresultsinthecaseofatime-dependentbutspace-independent
potential. In particular, we calculate the expectation value of the position of the charged particle
after theacceleration, to first order in the finestructure constant in the ~→0 limit, and find that
the change in the expectation value of the position (the position shift) due to radiation reaction
5
agrees exactly with the result obtained using the Lorentz-Dirac force in classical electrodynamics
0
forboth potentials. Wealso point outthat theone-loop correction tothepotentialmay contribute
0
tothe position change in this limit.
2

n PACSnumbers: 03.65.Sq,12.20.Ds,41.60.-m
a
J

6 I. INTRODUCTION

1 a
Achargedparticleradiateswhenitisaccelerated. This
v
simple discovery has been the basis of a multitude of
6
technological advances since. Most work, however, has
2
0 concentrated on the radiation itself. The effect of the
1 radiation, which carries energy and momentum, on the
0 particle itself is very small but still present. The overall
5 effectis achangeinthe equationofmotionforthe parti-
0
cle. The nature of this modified equation has been sub-
/ t
h ject to much controversy since the initial work of Abra-
p ham and Lorentz. The standardequationfor the change t0
- in the energy and momentum of the particle in classi-
t
n cal electrodynamics is the relativistic generalization by
a
Dirac of the work of Abraham and Lorentz and called FIG. 1: Preacceleration of a charged particle
u
the Lorentz-Dirac (or Abraham-Lorentz-Dirac)equation
q
: [1, 2, 3]. (See, e.g., Ref. [4] for a modern review.) Thus,
v
ifachargeewithmassmisacceleratedbyanexternal4-
Xi force Fµ , then its coordinates xµ(τ) at the proper time However,manyunphysicalandproblematiceffects occur
ext in the implementation. One example is the existence of
τ obey the following equation:
r run-away solutions for which a particle will continue to
a
accelerate under its own radiation reaction. This is cer-
d2xµ
m =Fµ +Fµ , (1.1) tainly an unphysical effect and thus a problem for the
dτ2 ext LD
theory. Alternative forms of implementing the Lorentz-
where the Lorentz-Dirac 4-force Fµ is given by Dirac force, including the integro-differential form, solve
LD
the problem of run-awaysbut create another problem in
2α d3xµ dxµ d2xν d2x the form of pre-acceleration. Thus, in this implementa-
FLµD ≡ 3c dτ3 + dτ dτ2 dτ2ν . (1.2) tion of the Lorentz-Dirac force the particle accelerates
(cid:20) (cid:18) (cid:19)(cid:21) before the external force causing the acceleration is ap-
We have let c = 1 and defined α e2/4π. Our metric plied;thisisanacausaleffectwhichshouldnotbepresent
c
is g = diag(+1, 1, 1, 1). Th≡is equation of motion in classical mechanics. This effect is represented in Fig.
µν
is fundamentally d−iffer−ent−in form from those which one 1,whichshowstheaccelerationoftheparticle,a,against
usuallyencountersinmechanicsasitisthird-order. This time, t, where a constant force is applied after time t0,
impliesthatathirdinitialconditionisneededinaddition i.e. in the shaded region.
tothepositionandvelocity. Hereinliesthecauseofmuch Additional conditions and procedures are therefore
ofthedebateastothephysicalcorrectnessofthistheory. necessaryin orderto treatthe Lorentz-Diractheory as a
Itshouldbenotedthattherearemanyderivationsofthe anormalcausalclassicaltheory,andtherearesuchproce-
Lorentz-Dirac force using a number of different methods dures (see, for example, the review [4] or the discussions
besides the original work of Dirac [3] (see, e.g., Ref. [5]). in [6] and [7]). One such procedure is the reduction of
2
order [8], in which the Lorentz-Dirac force is treated as
V
a perturbation order by order. Despite this, the debate V(z)=V0
overthe controversialaspects ofEq.(1.2) has continued,
its impact muted for a couple of reasons. Firstly, the
Acceleration
effect due to the third-derivative term, often called the
Schott term, is tiny for most physical situations; in fact
the time scale for the pre-acceleration is too small for
any known classical interactions for it to have a measur-
able effect. Secondly, the advent of quantum electrody-
V(z)=0
namics (QED), whichis the more fundamental theory of
electrodynamics, has rendered the problems concerning −Z1 −Z2 z
the classicalLorentz-Dirac force less urgent. The second
point,however,naturallyleadstothefollowingquestion:
“Iftheperturbativetreatmentofthe Lorentz-Diracforce
is causal and satisfactory, can it be derived in QED in FIG. 2: The potential V(z) and period of acceleration
the ~ 0 limit in perturbation theory?” This question
→
is the main motivation of this paper.
In fact it was found in Refs. [9, 10] (after an initial for z < Z and V(z)=0 for Z <z for some Z and
1 2 1
− −
claim to the contrary) that the change of the position, Z ,bothpositiveconstantswithZ >Z . Thus,thereis
2 1 2
calledthepositionshift,duetotheLorentz-Diracforceof non-zero acceleration only in the interval ( Z , Z ) as
1 2
− −
achargedparticlelinearlyacceleratedbyastaticexternal represented in Fig. 2 for the case with V >0.
0
potentialisreproducedbythe~ 0limit ofone-photon The external4-forceinEq.(1.1)representingthe elec-
→
emission process in QED in the non-relativistic approxi- tric force corresponding to this potential is given as
mation. (See,e.g.,Refs.[12,13, 14]forotherapproaches
to arrive at the Lorentz-Dirac theory from QED.) Here Fetxt = −V′(z)dz/dτ,
we present in detail the generalization of this work to a Fz = V′(z)dt/dτ,
ext −
fully relativistic particle, which was outlined in [11]. At Fx = Fy =0. (2.1)
ext ext
the end of the paper we shall also show similar results
with a time-dependent but space-independent potential. The Lorentz-Dirac 4-force can be given in the following
The paper is organized as follows. In Secs. II and III form:
we outline the classical and quantum models that are
used in the paper. In Sec. IV we calculate expressions FLtD = FLDdz/dτ,
forthe positionexpectationvalue forboththehypothet- Fz = F dt/dτ,
LD LD
ical non-radiatingparticle (Subsection IVA) and the ra- Fx = Fy =0 (2.2)
LD LD
diating particle (Subsection IVB) before identifying the
expression for the position shift in terms of the emission with
amplitude. This amplitude is calculated in Sec. V. The
2α d
resultis usedin Sec. VI to show that the energy emitted F cγ (γ3z¨). (2.3)
LD
in the ~ 0 limit in QED is given by the classical Lar- ≡ 3 dt
→
mor formula and to compute the quantum position shift Adotindicatesthederivativewithrespecttot. Wehave
in Sec. VII. The comparison between the classical and defined γ (1 z˙2)−1/2 as usual.
≡ −
quantumpositionshiftsispresentedindetailinSec.VIII Suppose that this particle would be at z = 0 at time
and the time-dependent case is presented in Sec. IX be- t=0 if the Lorentz-Dirac force was absent. Thetrue po-
fore we conclude the paper in Sec. X. sition at t=0, which we denote δz and call the position
shift, can readily be found to lowest non-trivial order in
F by treatingthe Lorentz-Diracforceasperturbation.
LD
II. CLASSICAL MODEL The calculation can be facilitated by using the fact that
thetotalenergy,mdt/dτ+V(z),changesbythe amount
of work done by the Lorentz-Dirac force. Thus we find
We startbydescribing the modelto be investigatedin
thispaper. Considerachargedparticlewithchargeeand t d m
mass m moving in one space dimension under a poten- F (t′)z˙(t′)dt′ = δz˙+V′(z)δz
LD
dz˙√1 z˙2
tial. Letthismotionbeinthepositivez-directionandlet Z−∞ −
thepotentialbeafunctionofthespatialcoordinatez,viz d δz
= mγ3z˙2 , (2.4)
V =V(z). Wewishtoanalyzethechangeintheposition dt z˙
(cid:18) (cid:19)
of the particle due to radiation reaction. Thus we con-
where we have used
sider a model in which there has been a period of accel-
eration, i.e. non-constant potential, at some time in the d
particle’s history. We assume that V(z) = V = const. (mγz˙)=mγ3z¨= V′(z) (2.5)
0 dt −
3
to zeroth order, i.e. in the absence of the Lorentz-Dirac andchargee,thefieldA istheelectromagneticfield,and
µ
force. Rearrangingandintegrating,andtheninterchang- thefunctionV istheexternalpotentialwhichaccelerates
µ
ing the order of integration, we obtain the position shift the charged scalar particle and is given as V =V(z)δ
µ µ0
as for the static potential. The non-interacting quantum
electromagnetic field A (x) is expanded as
v 0 t 1 dz µ
δz = 0 dt′ F dt,
LD −m γ3(t′)[z˙(t′)]2 LDdt d3k
Z−∞(cid:18)Z0 (cid:19) A (x)= a (k)e−ik·x+a†(k)eik·x , (3.2)
(2.6) µ 2k(2π)3 µ µ
where v = z˙ is the final velocity in the absence of Z
0 t=0 (cid:2) (cid:3)
|
radiationreaction. We shall see that this result is repro- where k = k and the annihilation and creation opera-
duced in QED in the ~ 0 limit. tors, a (k)kankd a†(k), respectively, for the photons with
It is useful for later pu→rposes to find the change in the momenµta ~k in thµe Feynman gauge satisfy
position shift (2.6) caused by letting the particle be at
z = z0 6= 0 at t = 0 with the same final velocity v0 aµ(k),a†ν(k′) =−gµν(2π)32~kδ(k−k′). (3.3)
in the absence of the Lorentz-Dirac force. (We assume
Th(cid:2)e Fourier exp(cid:3)ansion of the non-interacting quan-
that z > Z .) The time this particle spends after the
0 2
− tumscalarfieldisnotasstraightforwardbecauseitsfield
acceleration and before getting to t = 0 (when the posi-
equation involves the potential V(z) as follows:
tion shift is “measured”) is lengthened by z /v . Hence
0 0
this position shift is obtained by using the same trajec-
~2[i∂ V(z)]2+~2 2 m2 ϕ=0. (3.4)
tory but defining it at t0 z0/v0. Hence, writing this t− ∇ −
≡
position shift as Positiv(cid:8)e-frequency mode functions c(cid:9)an be found in the
form
δz =δz +δz , (2.7)
class LD extra
i
we find Φ (t,x)=φ (z)exp (p x p t) , p >0,
p p ~ ⊥· ⊥− 0 0
(cid:20) (cid:21)
v t0 0 1 dz (3.5)
δzextra =−m0 Z−∞ Zt0 γ3(t′)[z˙(t′)]2dt′!FLDdtdt. zw-hceormepxo⊥nen≡t o(xf,tyh)eamndompe⊥ntu≡m(ppx,pyp)z. bWyethdeefirenleattiohne
(2.8) p2 = p2 +p2 +m2. We let p be ≡positive (negative) if
Noting thatz˙(t)=v for valuesoftbetween0andt we 0 ⊥
0 0 themode functioncorrespondstoawavecominginfrom
obtain
z = (z = + ). The function φ (z) with p = 0
p ⊥
z0 satisfi−e∞s ∞
δz = E (2.9)
extra −mγ3v2 em
0 0 d2
~2 φ (z)=[κ (z)]2φ (z), (3.6)
with γ (1 v2)−1/2, where E is the energy emitted − dz2 p p p
0 ≡ − 0 em
as radiation given by
where
0 dz
E = F dt κ (z) [p V(z)]2 m2. (3.7)
em LD p 0
− dt ≡ − −
Z−∞
= 2αc 0 (γ3z¨)2dt, (2.10) mIf2p+⊥p6=20.,Hthoewnevthere,mwpaesnseteedrmonmly2thhaesctaosebsewrietphlapced=by0
3 ⊥ ⊥
Z−∞ and p>0 in our calculations. The WKB approximation
which is the relativistic Larmor formula for one- for the function φ (z) with p > 0, which will be useful
p
dimensional motion. We shall see that the extra con- later, reads
tribution(2.9) tothe positionshiftis alsoreproducedby
QED in the ~ 0 limit. p z
φ (z)= exp κ (ζ)dζ , (3.8)
→ p p
κ (z)
r p (cid:20)Z0 (cid:21)
III. THE QED MODEL where we have required φp(0)=1.
The non-interacting quantum charged scalar field is
expanded in terms of these mode functions as
We now turn our attention to the quantum field theo-
retic model. The corresponding Lagrangiandensity is
d3p
ϕ(x)=~ A(p)Φ (x)+B†(p)Φ∗(x) .
L = [(D1µ+ieAµ)ϕ1]†[(Dµ+ieAµ)ϕ]−(m/~)2ϕ†ϕ Z 2p0(2π~)3 (cid:2) p p ((cid:3)3.9)
FµνFµν (∂µAµ)2, (3.1) (This expansion needs to be modified if there are modes
−4 − 2
which are totally reflected by the potential — this is the
where F ∂ A ∂ A and D ∂ +iV /~. The case if V = 0 — or if there are modes which fail to
µν µ ν ν µ µ µ µ 0
≡ − ≡ 6
field ϕ describes a charged scalar particle with mass m reach z = . However, such modifications will not
| | ∞
4
be relevant since the affected modes will not enter our whichisautomaticallysatisfiedaslongasκ (z)andκ′(z)
p p
calculations.) The operators A(p) and B(p) satisfy can be regarded as quantities of order ~0.
The normalization condition i i =1 implies
A(p),A†(p′) = B(p),B†(p′) =(2π~)32p0δ(p p′), h | i
−
(3.10) d3p
(cid:2) (cid:3) (cid:2) (cid:3) f(p)2 =1 (3.17)
if one normalizesthe mode functions appropriately,with (2π~)3| |
all other commutators vanishing. The WKB mode func- Z
because of the commutation relations (3.10). This equa-
tions Φ (t,x)with φ (z)givenby Eq.(3.8)arecorrectly
p p
tion shows that the function f(p) can heuristically be
normalizedsince it satisfies the usualnormalizationcon-
regarded as the one-particle wave function in the mo-
dition Φ (t,x) = 1 in the region ( Z ,+ ) where
p 2
| | − ∞ mentumrepresentation. InRef. [11]itwasassumedthat
V(z)=0.
f(p) was real for simplicity. This assumption will be
Now, in the interaction picture, if there is one scalar
dropped in this paper.
particle with momentum p in the initial state, then the
final state to lowest non-trivialorder in e is a superposi-
tionofastateproportionalto theinitialstate andstates
IV. POSITION EXPECTATION VALUE
with one scalarparticle and one photon. Thus, the state
A†(p)0 evolves as follows:
| i Inthissectionwederiveaformulaforthepositionshift
A†(p)0 [1+i (p)]A†(p)0 in terms of the emission amplitude µ(p,k). We closely
| i → F | i A
i d3k follow Ref. [10], correcting a few errors. In particular,
+~ 2k(2π)3Aµ(p,k)a†µ(k)A†(P)|0i. the fourth term in Eq. (4.32) was missing in Ref. [10].
Z
(3.11)
A. Non-radiating particle
The (p) is the forward-scattering amplitude of order
F
e2 coming from the one-loop diagram, which we do not
evaluateexplicitly. Theemissionamplitude (p,k)will Define the charge density by
µ
A
playacentralroleinourcalculations. Notethatthemo- i
mentum ~k of the photon is of order~ because the wave ρ(x)≡ ~ :ϕ†∂tϕ−∂tϕ†·ϕ:, (4.1)
number k rather than the momentum has the classical
where:...:denotes normalordering. The operatorρ(x)
limit. The momentum P of the charged particle in the
is the t-component of the conserved current
final state is determined using conservationof the trans-
verse momentum and energy: i
Jµ(x) :ϕ†∂µϕ ∂µϕ† ϕ: . (4.2)
≡ ~ − ·
p = P +~k , (3.12)
⊥ ⊥ ⊥
We let the potential satisfy V < 2m, thus precluding
p0 = P0+~k. (3.13) the possibility of scalar-parti|cle0|pair creation. Then, the
The z-component P Pz is determined by the on-shell chargedensityρ(x)coincideswiththeprobabilitydensity
condition ≡ fortheparticleifthereisonlyonechargedparticleinthe
state. Hence the expectation value of z is
P2 =P2+P2 +m2. (3.14)
0 ⊥
z = d3xz ρ(t,x) . (4.3)
We shall consider the initial state i given by h i h i
| i Z
d3p We shallfirstneedto considerthe expectationvalue ofz
|ii= √2p (2π~)3f(p)A†(p)|0i, (3.15) forthehypotheticalscalarparticlethatinteractwiththe
Z 0 externalpotentialbutnotwiththequantumelectromag-
where the function f(p) is sharply peaked about a mo- netic field. This willbe the benchmark againstwhich we
mentuminthepositivez-directionwithwidthoforder~. candefinethe positionshiftdue totheradiation. Hence,
We also require that the p-derivative of f(p) has width wemustevaluatetheexpectationvalueoftheprobability
oforder~. This wavepacketcorrespondsto the classical density in the absence of radiation and use Eq. (4.3) to
particle considered in the previous section. We make a find z . The initial and final states in this case are the
h i
further assumption that the WKB approximation (3.8) same. Thus, using the commutation relations (3.10) for
is valid for the momenta p where the function f(p) is the expectation value of the charge density (4.1) in the
not negligibly small. The WKB approximationis known state i given by (3.15), we find
| i
to be applicable if the wavelength stays approximately
ρ(t,x)
constant over many periods (see, e.g. Ref. [15]). Since h i
the “time-dependentwavelength”is 2π~[κp(z)]−1,where =i~ d3p′ d3p f∗(p′)f(p)
κp(z) is defined by Eq. (3.7), this condition is Z 2p′0(2π~)3 Z √2p0(2π~)3
~ d 1 1, (3.16) × Φ∗p′(tp,x)∂tΦp(t,x)−∂tΦ∗p′(t,x)·Φp(t,x) .
dz κ (z) ≪ (4.4)
(cid:12) (cid:20) p (cid:21)(cid:12) (cid:2) (cid:3)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
5
The t-dependence of the mode functions is given by B. Radiating particle
Φ (t,x) e−ip0t/~. Hence, we find
p
∝
1 d3p′ d3p Nowweconsiderthecaseoftheradiatingparticle. The
ρ(t,x) = f∗(p′)f(p) finalstateresultingfromtheinitialstate i canbefound
h i 2 (2π~)3 (2π~)3 | i
Z Z from Eq. (3.11) as
p p′
× rp0′0 +sp00!Φ∗p′(t,x)Φp(t,x). |fi = √2pd(32pπ~)3F(p)A†(p)|0i
(4.5) Z 0
i d3k d3p
+
We arrange the wave packet so that at time t = 0 it is ~ 2k(2π)3 √2p (2π~)3
Z Z 0
located far into the region where the potential vanishes.
Gµ(p,k)a†(k)A†(P)0 , (4.12)
Thenwemayapproximatethemodefunctionsasfollows: × µ | i
where P = p ~k and P = p ~k as before. We
⊥ ⊥ ⊥ 0 0
− −
Φ (0,x) e−i(pz+p⊥·x⊥)/~. (4.6) have defined
p
≈
After substituting Eq. (4.5) in the position expectation F(p) [1+i (p)]f(p), (4.13)
≡ F
formula (4.3) we use the approximation(4.6) to write Gµ(p,k) µ(p,k)f(p). (4.14)
≡A
∂
zΦp(0,x) i~ Φp(0,x). (4.7) One can heuristically regard the function F(p) as the
≈− ∂p
one-particle wave function in the zero-photon sector in
We can now use integration by parts for the variable p the p-representation and the function Gµ(p,k) as that
in Eq. (4.3) to obtain the expectation value of z for the in the one-photon sector with a photon with momentum
hypothetical non-radiating particle, which we denote by ~kintheP-representation. Byintroducingthedefinition
z , as
0
i~ d3p′ d3p i d3p
z0 = 2 (2π~)3 (2π~)3 C†µ(k)= ~ √2p (2π~)3Aµ(p,k)f(p)A†(P), (4.15)
Z Z Z 0
∂ p p′
0 + 0 f∗(p′)f(p) the final state can be written
× ∂p" rp′0 sp0! # d3p
f = F(p)A†(p)0
× d3xΦ∗p′(0,x)Φp(0,x). (4.8) | i Z √2p0(2π~)3 | i
Z d3k
Substitutingtheapproximation(4.6)forΦp givesadelta + 2k(2π)3C†µ(k)a†µ(k)|0i. (4.16)
function: Z
i~ d3p′ d3p Wefirstderivetherelationbetweentheimaginarypart
z0 = 2 (2π~)3 (2π~)3f∗(p′) ofF(p)andthescatteringprobabilitywhichresultsfrom
Z Z unitarity. Recalling the definition (4.13) and using the
∂ p p′ unitarity of time evolution, we find
0 + 0 f(p)
× ∂p" rp′0 sp0! # d3p
(2π~)3δ(p p′). (4.9) 1=hf|fi= (2π)3(1−2ImF(p))|f(p)|2+Pem,
× − Z
Performingthep′-integrationafterexplicitlyworkingout (4.17)
to first order in e2, where the emission probability
the p-derivative, we find Pem
is given by
d3p ∂
z0 = i~ (2π~)3f∗(p)∂pf(p) (4.10) = ~ d3k 0C (k′)C†µ(k)0
Z Pem − 2k(2π)3h | µ | i
or, by integration by parts, Z
1 d3k d3p
z0 = i2~ (2dπ3~p)3f∗(p) ↔∂p f(p), (4.11) = −~Z 2k(2π)3 Z √2p0(2π~)3f∗(p)A∗µ(p,k)
Z d3p′
f(p′) µ(p′,k)
wthheereexp↔∂epc=ta→∂tipon−v←a∂lpu.e Tofhitshreespuolstitciaonn boepeirnatteorprrie~t∂edp ains ×Z0Ap(P2p)′0A(†2(πP~′))30 . A (4.18)
the p-representation of the one-particle wave function. ×h | | i
An example of a function f(p) satisfying Eq. (4.11) is To perform the p′-integration recall first that
f(p)=f (p)e−iz0p/~ withthefunctionf (p)beingreal
R R
though it is not necessary to make this assumption. 0A(P)A†(P′)0 =(2π~)32P δ(P P′). (4.19)
0
h | | i −
6
We need to integrate this delta function with respect to WeconverttheintegrationvariablefromPtopbyusing
p′ by relating it to δ(p p′). We note that d3P=(∂P/∂p)d3pandtheP-derivativetoap-derivative
−
by using ∂ =(∂p/∂P)∂ to find
δ(P P′ )=δ(p p′ ) (4.20) P p
⊥− ⊥ ⊥− ⊥
because P⊥ =p⊥−~k⊥, and hziem = −2i (2dπ3~p)3f∗(p) ↔∂p f(p)
dp Z
δ(P P′)= δ(p p′). (4.21) d3k P dp 2
− dP − µ∗(p,k) (p,k) 0
× 2k(2π)3A Aµ p dP
By using these formulae in Eq. (4.19) we find Z 0 (cid:18) (cid:19)
i d3p
P dp f(p)2
0A(P)A†(P′)0 = 0 (2π~)32p δ(p p′). (4.22) −2 (2π~)3| |
h | | i p dP 0 − Z
0 d3k ↔
By substituting this formula in Eq. (4.18) we obtain × 2k(2π)3Aµ∗(p,k) ∂p Aµ(p,k). (4.30)
Z
1 d3p
= f(p)2 We have dropped the factor (P /p )(dp/dP)2 from the
Pem −~ (2π~)3| | 0 0
Z second term for the following reason. It will turn out
× 2kd(32kπ)3A∗µ(p,k)Aµ(p′,k)Pp0ddPp . alarteerboththatotfhoerdemeris~s0io.nTahmuspltihtuedseecAonµdantedrmitsinp-dtheerivaabtoivvee
Z 0 (4.23) expression for z em is of order ~0. Since P0 = p0 and
dp/dP =1 to ohrdier ~0, we can dropthe factors of P /p
0 0
Eq. (4.17) must hold for any function f(p), so we must and dp/dP from the second term. However, one needs
have to keep these factors in the first term, which is of order
1 d3k P dp ~−1.
2ImF(p)=−~ 2k(2π)3A∗µ(p,k)Aµ(p,k)p0dP . Fromtheexpressionsforthefinalstate(4.16)itisclear
Z 0 that the contribution to z from the process without
(4.24) h i
photonemissionis obtainedfromEq.(4.11)byreplacing
The contribution of the one-photon-emission term to
f(p) with F(p) [1+i (p)]f(p). Thus
ρ(t,x) is ≡ F
h i
d3k i~ d3p
hρ(t,x)iem =−~ 2k(2π)3h0|Cµ(k)ρ(t,x)Cµ†(k)|0i. hzifor = 2 (2π~)3 2i|f(p)|2∂pReF(p)
Z Z
(4.25) (cid:8) ↔
Theintegrandhereisobtainedfrom ρ(t,x) forthenon- +[1−2ImF(p)]f∗(p) ∂p f(p) (4.31)
h i
radiating particle if we replace the state i in Eq. (3.15) o
by | i to first order in e2. (Recall that the forward-scattering
amplitude (p) is of order e2.) We can now add the
~1/2C†(k)0 =i d3P g (P)A†(P)0 , two contribFutions z em and z for together to find the
µ | i √2P (2π~)3 µ | i position expectatiohnivalue: h i
Z 0
(4.26)
where i~ d3p ↔
hzi = 2 (2π~)3f∗(p) ∂p f(p)
P dp Z
gµ(P)≡~−1/2Aµ(p,k)f(p)sp00 dP . (4.27) −~ (2dπ3~p)3|f(p)|2∂pReF(p)
Z
Hence we conclude from Eq. (4.11) that i d3p
f(p)2
−2 (2π~)3| |
~ d3xz 0Cµ(k)ρ(t,x)C†(k)0 Z
h | µ | i d3k ↔
Zi~ d3P ↔ × 2k(2π)3Aµ∗(p,k) ∂p Aµ(p,k)
= gµ∗(P)∂ g (P). (4.28) Z
2 (2π)3 P µ i d3p ↔
Z −2 (2π~)3f∗(p) ∂p f(p)
This implies that the one-photon-emission contribution
Z
to z is d3k
h i × 2k(2π)3Aµ∗(p,k)Aµ(p,k)
i d3k d3P Z
z =
h iem −2 2k(2π)3 (2π~)3 P dp dp
Z Z 0 1 . (4.32)
[f∗(p) µ∗(p,k)] ↔∂P [f(p) µ(p,k)] ×p0 dP (cid:18)dP − (cid:19)
× A A
P0 dp 2 The unitarity relation (4.24) has been used to eliminate
. (4.29)
×p dP 2Im (p). Each term in Eq. (4.32) can be interpreted
0 (cid:18) (cid:19) F
7
as follows. The first term is the position expectation V. EMISSION AMPLITUDE
value z for the non-radiating particle. The second term
0
is the contribution from what can be regarded as the The one-photon-emission part of the evolution of the
one-loopquantumcorrectiontothepotential. Therefore, state was represented earlier in Eq. (3.11) by
we identify the sum of the third and forth terms as the
position shift to be compared with the classical position i d3k
A†(p)0 + µ(p,k)a†(k)A†(P)0 .
shift δzclass given by Eq. (2.7). Using Eq. (4.11) and | i→··· ~ 2k(2π)3A µ | i
the assumptionthat the function f(p) is sharply peaked Z (5.1)
about a momentum in the positive z-direction with the The evolution in perturbation theory in the interaction
width of order ~, we find these two terms in the ~ 0 picture is generated by the interaction Lagrangian den-
→
limit as sity as follows:
δz =δz +δz , (4.33)
q q1 q2 i
A†(p)0 + d4x (x)A†(p)0 . (5.2)
where | i→··· ~ LI | i
Z
i d3k
δz = We can take the inner product of the two expressions
q1 −2Z 2k(2π)3 (5.1) and (5.2) with the state 0aν(k)A(p′) and equate
×Aµ∗(p,k) ↔∂p Aµ(p,k), (4.34) cthedemureinleoarddsertoto find the amplhitu|de Aµ(p,k). This pro-
z d3k
0
δz =
q2 − ~ 2k(2π)3 2~p′(2π~)3δ(p′ P) (p,k)
Z − 0 − Aν
dp
×Aµ∗(p,k)Aµ(p,k) dP −1 ,(4.35) = d4xh0|aν(k)A(p′)LI(x)A†(p)|0i. (5.3)
(cid:18) (cid:19) Z
where the momentum at which the function f(p) is Integrating over the variable p′, with the appropriate
peaked is now denoted simply by p. We have used
measure,andrearranging,we find that (p,k) is given
the fact that dp/dP 1 is of order ~ to drop the fac- Aµ
− by
tor (P /p )(dp/dP) in δz . We shall demonstrate that
0 0 q2
the quantum position shift δzq in the ~ 0 limit is 1 d3p′
identical with the classical counterpart δzc→lass by show- Aµ(p,k) = −~ 2p′(2π~)3
ing δz =δz and δz =δz . Z 0
q1 LD q2 extra
Let us first examine the latter equality. To this end d4x 0a (k)A(p′) (x)A†(p)0 .
µ I
we need to find an expression for dp/dP in terms of p × h | L | i
Z
and k. The energy conservation equation p P = ~k (5.4)
0 0
−
gives a one-to-one relation between p and P for a given
k after letting P2 = p2 = 0 because these are of order The I(x) is obtainedfromthe Lagrangiandensity (3.1)
⊥ ⊥ L
~2. Then we find as follows:
dp m2 ie
=1 ~k. (4.36) (x)= A : ϕ†Dµϕ (Dµϕ)†ϕ : . (5.5)
dP − p2p LI −~ µ −
0
(cid:2) (cid:3)
By using this formula in Eq. (4.35) we obtain Recall that D = ∂ +iV , where V = δ V(z). By
µ µ µ µ µ0
m2z substituting Eq.(5.5)inEq.(5.4)andusing the commu-
0
δz = , (4.37) tation relations (3.3) and (3.10) we find
q2 −p2p Eem
0
where d3p′
(p,k) = ie~ d4xeik·x
d3k Aµ − 2p′(2π~)3
k µ∗(p,k) (p,k) (4.38) Z 0 Z
Eem ≡−Z 2k(2π)3 A Aµ × Φ∗p′(x)DµΦp(x)−(DµΦp′(x))†Φp(x) ,
is the expectation value of the energy emitted as radia- h (5i.6)
tion. By comparing Eq. (4.37) with Eq. (2.9), it can be
seen that the equality δz =δz will hold if
q2 extra where the mode functions Φp(x) are given by Eq. (3.5).
2α 0 As we mentioned before, we assume that the WKB ap-
= c (γ3z¨)2dt, (4.39) proximation (3.8) is valid for the initial mode function.
em
E 3 Z−∞ Since the momentum ~k of the emitted photon is of or-
which is identical to the relativistic generalizationof the der ~, the WKB approximation can also be used for the
classical Larmor formula. To show this equality (and mode functions for the final state.
also δz = δz ) we need to find the ~ 0 limit of Since the transverse momentum p is assumed to be
q1 LD ⊥
theone-photon-emissionamplitude (p,k)→towhichwe of order ~, the final transverse momentum p ~k
µ ⊥ ⊥
now turn. A is also of order ~. This means that we do not n−eed to
8
considerthex-andy-components (p,k)becausethey momentum. Hence, v (z) κ (z)/[p V(z)] is the ve-
µ p p 0
are smaller by a factor of ~ compAared to the t- and z- locity of this classical part≡icle. Thus, −
components. Using D =∂ +iV(z)/~ we obtain
0 t ~k
κ (z) κ (z)= . (5.14)
d3p′ p − P v (z)
(p,k) = e d4x p
At − 2p′(2π~)3
Z 0 Z By substituting this formula in Eq. (5.12) and letting
× φ∗p′(z)φp(z)[p0+p′0−2V(z)] P =p and κP(z)=κp(z), because the differences P p
eik·xe−i[(p′0−p0)t−(p′⊥−p⊥)·x⊥]/~. (5.7) and κP(z) κp(z) are of order ~, we find −
× −
Integration with respect to t, x and y gives φ∗p′(z)φp(z) = κ p(z)exp ik z vd(zz)
dp′ pp (cid:20) Z0 p (cid:21)
At(p,k) = −e 2p′ dz = κ (z)eikt, (5.15)
Z 0 Z p
×[p0+p′0−2V(z)]φ∗p′(z)φp(z)e−ikzz where the time t is defined by dz/dt = v (z) with the
×δ(p′0+~k−p0). (5.8) condition t=0 at z =0. Hence p
Wtheatcodnpv′/epr′0tt=hedpp′0′-/inp′teagnrdatfiionndtop′0-integrationbynoting At(p,k) = −eZ−+∞∞ vpd(zz)eikt−ikzz
+∞
(p,k)= e dz P0+p0−V(z)φ∗(z)φ (z)e−ikzz, = e dteikt−ikzz. (5.16)
At − Z 2P P p − Z−∞
(5.9)
where P =p ~k with P P2 m2 as before. We It is convenient to define the variable ξ by
0 0− ≡ 0 −
letP =p andP =pinEq.(5.9)becausethedifferences
P 0p an0d P p are of orderp~. Thus, we obtain ξ ≡ t−zcosθ, (5.17)
0− 0 − cosθ kz/k. (5.18)
≡
At(p,k)=−e dz p0−pV(z)φ∗P(z)φp(z)e−ikzz. Then
Z
(5.10) +∞ dt
Proceeding similarly for the z-component (p,k) we (p,k)= e dξ eikξ. (5.19)
Az At − dξ
have Z−∞
ie~ Next we turn our attention to the z-componentgivenby
z(p,k) = dz Eq.(5.11), We note that, to lowestorderin~, wecanlet
A −2p
Z
φ∗(z)dφp(z) dφ∗P(z)φ (z) e−ikzz.
× P dz − dz p dφp(z) κp(z)
(cid:20) (cid:21) (5.11) dz =i ~ φp(z). (5.20)
Hence
Tosimplifytheexpressionfortheemissionamplitude
µ
A
fbuyrtEhqe.r,(3w.8e)n.eWedetnootuesefirtshte explicit form of φp(z) given Az(p,k) = e +∞dzκp2+pκPφ∗P(z)φp(z)e−ikzz
Z−∞
+∞
φ∗(z)φ (z) = Pp = e dzeikt−ikzz
P p sκP(z)κp(z) Z−∞
+∞ dz
i z = e dξ eikξ, (5.21)
×exp ~ [κp(η)−κP(η)]dη . Z−∞ dξ
(cid:26) Z0 (cid:27)
(5.12) where we have let κ (z) = κ (z) as before and used
P p
Eq.(5.15). ThisformulaandEq.(5.19)canbecombined
Now, to lowest order in ~ we have as
κp(z) κP(z)= ∂κp(z)(p0 P0)= p0−V(z)~k. µ(p,k)= e +∞dξ dxµ eikξ, (5.22)
− ∂p0 − κp(z) (5.13) A − Z−∞ dξ
Note that κ (z) and p V(z) are the z- and t- wherexµ istheclassicaltrajectorywithfinalmomentum
p 0
components of mdxµ/dτ, −respectively, of the classical p which passes through (t,z) = (0,0). (The minus sign
particlewithfinalmomentumpandvanishingtransverse for the z-component is due to the raised index.) This
9
emission amplitude is identical with that for a classical By substituting dξ/dt=1 z˙cosθ we find
−
point charge passing through (0,0) [11].
The expression (5.22) is ill-defined since dxµ/dξ re- d2z z¨
= , (6.6)
mains finite as ξ . Therefore, we introduce dξ2 (1 z˙cosθ)3
→ ±∞ −
a smooth cut-off function χ(ξ) which takes the value d2t d2z
one while the acceleration is non-zero with the property = cosθ, (6.7)
dξ2 dξ2
lim χ(ξ)=0. Thus,
ξ→±∞
and hence
+∞ dxµ
µ(p,k)= e dξ χ(ξ)eikξ. (5.23) d2xµd2x z¨2sin2θ dt
A − Z−∞ dξ dξ2 dξ2µ =−(1 z˙cosθ)5 dξ . (6.8)
In the end we take the limit χ(ξ) 1 at any given ξ in −
such a way that +∞dξ[χ′(ξ)]2 →0. By substituting this formula in Eq. (6.4) we obtain
−∞ → Eq. (4.39), thus demonstrating the equality δz =
q2
R δz .
extra
VI. DERIVATION OF THE LARMOR
FORMULA
VII. QUANTUM POSITION SHIFT
In this section we derive the relativistic generaliza-
tion of the Larmor formula for one-dimensional motion, Our next task is to show that δzq1 = δzLD. This will
Eq. (4.39), thus completing the demonstration of the establish that the classical position shift δzclass is equal
eloqwuianligtyfoδrmzq2of=thδezeexmtrais.siIotniasmcopnlivtuendieenotbttaoinuesdebthyeinftoel-- toItnhethqeuapnrtoudmuctoneµδ∗z(qp,ink)t∂hpe ~µ→(p,0kl)iminit.Eq. (4.34) we
gration by parts: useEq.(6.1)for µA∗andEq.(5A.23)for∂p µ. Proceeding
A A
in a way similar to that led to Eq. (6.4) we find
ie +∞ d2xµ dxµ
µ(p,k)= dξ + χ′(ξ) eikξ,
A −k Z−∞ (cid:20) dξ2 dξ (cid:21) (6.1) δzq1 = −4απc dΩ dξ dd2ξx2µ∂∂p ddxξµ
Z Z (cid:26) (cid:18) (cid:19)
where we have used the condition that χ(ξ) = 1 if 1 ∂ dxµdx d
d2xµ/dξ2 =0. Bysubstituting thisformulainEq.(4.38) + µ [χ(ξ)]2 .(7.1)
6 4∂p dξ dξ dξ
we obtain (cid:18) (cid:19) (cid:27)
∞ dk +∞ +∞ Next we demonstrate that the second term, which still
Eem = −e2 dΩ 16π3 dξ′ dξ contains the cut-off function, vanishes when integrated
Z Z0 Z−∞ Z−∞ over the solid angle and ξ. Noting that
d2xµ dxµ
× dξ′2 + dξ′ χ′(ξ′) dxµdxµ dτ 2 dxµdxµ dτ 2
(cid:20) (cid:21) = = , (7.2)
d2xµ + dxµχ′(ξ) eik(ξ−ξ′), (6.2) dξ dξ (cid:18)dξ(cid:19) dτ dτ (cid:18)dξ(cid:19)
× dξ2 dξ we find by integration by parts that the contribution of
(cid:20) (cid:21)
wheredΩisthesolidangleinthek-space. Weextendthe this term is proportional to the following integral:
integration range for k from [0,+ ) to ( ,+ ) and
dividebytwo. Thenusing +∞eik∞(ξ−ξ′)dk−=∞2πδ∞(ξ ξ′), I dΩ +∞dξd ∂ dτ 2 [χ(ξ)]2. (7.3)
we find −∞ − ≡Z Z−∞ dξ (∂p(cid:18)dξ(cid:19) )
R
e2 +∞ Since ∂/∂pis takenwithξ fixed, the ξ-andp-derivatives
= dΩ dξ
Eem −16π2 commute. Hence this integral is equal to
Z Z−∞
d2x d2xµ dx dxµ
µ + µ [χ′(ξ)]2 . (6.3) ∂ +∞ d dτ 2
× dξ2 dξ2 dξ dξ I = dΩ dξ ,
(cid:26) (cid:27) ∂p dξ dξ
Z Z−∞ (cid:18) (cid:19)
The secondterm tends to zeroin the limit χ(ξ) 1 due
tinotthhiesrleimquitirement R−+∞∞[χ′(ξ)]2dξ → 0. Hence,→we have = ∂∂p"Z dΩ(cid:18)ddτξ(cid:19)2#ξξ==−+∞∞ , (7.4)
α +∞ d2x d2xµ
= c dξ dΩ µ , (6.4) where we have used the fact that the ξ-derivative of
Eem −4π Z−∞ Z dξ2 dξ2 (dτ/dξ)2 is non-zero only if the acceleration is non-zero
where α e2/4π as before. Now, one can readily show and,therefore,onlywhenthecut-offfunctionχ(ξ)equals
c
that ≡ one. Since
d2xµ dt 3 dξd2xµ d2ξdxµ dτ 2 4π
= . (6.5) dΩ = =4π, (7.5)
dξ2 dξ dt dt2 − dt2 dt dξ (dt/dτ)2 (dz/dτ)2
(cid:18) (cid:19) (cid:20) (cid:21) Z (cid:18) (cid:19) −
10
wehaveI =0,andthusthecontributionfromthesecond Using this expressioninthe positionshift(7.10) wehave
terminEq.(7.1)vanishes. Theremainingtermgivesthe
α 0 ∂z
main contribution to the position shift due to radiation δz = c dΩ dt
reaction in QED in the ~→0 limit as q1 −4π Z Z−∞ (cid:18)∂p(cid:19)t
δz = αc dΩ dξd2xµ ∂ dxµ . (7.6) dz¨ sin2θ +3z¨2 sin2θcosθ .
q1 −4π dξ2 ∂p dξ ×"dt (1 z˙cosθ)4 (1 z˙cosθ)5#
Z Z (cid:18) (cid:19) − −
To compare δz given by this equation with δz (7.14)
q1 LD
given by Eq. (2.6) we need to find an expression of δz
q1 We can now perform the integrationover the solid angle
in terms of t rather than ξ. Thus, we need to eliminate
with the following result:
thevariableξ usingitsdefinitionξ =t zcosθ. Thetwo
components of the second derivative d−2xµ/dξ2 are given 2α 0 dz¨ ∂z
δz = c dt γ4 +3γ6z¨2z˙
byEqs.(6.6)and(6.7). Letusconsiderthesecondfactor q1 − 3 dt ∂p
of the integrand of (7.6), i.e. the p-derivative of dxµ/dξ. Z−∞ (cid:20) (cid:21)(cid:18) (cid:19)t
0 2α d ∂z
By interchanging the order of integration we obtain = dt cγ γ3z¨
− 3 dt ∂p
∂ dxµ = dt d ∂xµ , (7.7) Z−0∞ (cid:20) ∂z (cid:0) (cid:1)(cid:21)(cid:18) (cid:19)t
∂p dξ dξdt ∂p = dtF , (7.15)
(cid:18) (cid:19) (cid:18) (cid:19)ξ − LD ∂p
Z−∞ (cid:18) (cid:19)t
wherethesubscriptξindicatesthatthepartialderivative
where we have used the expression of F given by
with respect to p is taken with ξ fixed. Then by differ- LD
Eq. (2.3). This formula will be used in the next section
entiating the equation t = zcosθ +ξ with respect to p
to show that δz =δz . We note here that Eq. (7.15)
with ξ fixed, we find q1 LD
is valid for an external force of any form provided that
∂t ∂z the photon emission amplitude is given by Eq. (5.22).
= cosθ. (7.8)
∂p ∂p
(cid:18) (cid:19)ξ (cid:18) (cid:19)ξ
CombiningthisequationwithEqs.(6.6)and(6.7),which VIII. COMPARISON OF CLASSICAL AND
gives d2xµ/dξ2, we can write the integrand in Eq. (7.6) QUANTUM POSITION SHIFTS
as
Comparison between Eqs. (7.15) and (2.6) shows that
d2xµ ∂ dx z¨ d ∂z
µ = sin2θ. the equality δzq1 =δzLD follows if
dξ2 ∂p dξ −(1 z˙cosθ)4dt ∂p
(cid:18) (cid:19) − (cid:18) (cid:19)ξ (7.9) ∂z = v0z˙(t) t dt′ dt′. (8.1)
We substitute this expression in the quantum position (cid:18)∂p(cid:19)t m Z0 γ3(t′)[z˙(t′)]2
shift (7.6) and change the integration variable from ξ to
This equationcan be demonstratedas follows. Since the
t and integrate by parts. The result is
energy is conserved, we have
α 0 d z¨
c
δz = dΩ dt
q1 −4π dt (1 z˙cosθ)3 m2(dz/dτ)2+m2+V(z)= p2+m2, (8.2)
Z Z−∞ (cid:20) − (cid:21)
∂z sin2θ, (7.10) and,qhence, p
× ∂p
(cid:18) (cid:19)ξ dz −2 1/2
= 1 m2 p2+m2 V(z) . (8.3)
where we have also changed the integration range from
dt − −
( ,+ ) to ( ,0] because z¨ = 0 for [0,+ ). Fi- (cid:20) (cid:16)p (cid:17) (cid:21)
n−al∞ly, w∞e need t−o∞relate (∂z/∂p) to (∂z/∂p) .∞By re- By differentiating both sides with respect to p with t
ξ t
garding z as a function of t and p we have fixed, and noting that
∂z p/ p2+m2 = v , (8.4)
0
dz =z˙dt+ dp. (7.11)
(cid:18)∂p(cid:19)t p2+mp2 V(z) = mγ, (8.5)
−
By substituting dt = dξ +cosθdz in this equation and we obtain p
rearranging,we obtain
d ∂z 1 ∂z
= v V′(z) . (8.6)
z˙ 1 ∂z dt ∂p mγ3z˙ 0− ∂p
dz = dξ+ dp. (7.12) (cid:18) (cid:19)t (cid:20) (cid:18) (cid:19)t(cid:21)
1 z˙cosθ 1 z˙cosθ ∂p
− − (cid:18) (cid:19)t By substituting the formula V′(z) = mγ3z¨ [see
−
Thus, Eq. (2.5)] in Eq. (8.6) we find
∂z 1 ∂z d 1 ∂z v
= . (7.13) = 0 . (8.7)
∂p 1 z˙cosθ ∂p dt z˙ ∂p mγ3z˙2
(cid:18) (cid:19)ξ − (cid:18) (cid:19)t (cid:20) (cid:18) (cid:19)t(cid:21)