Lorentz-Dirac force from quantum electrodynamics

Lorentz-Dirac force from quantumelectrodynamics

Atsushi Higuchi, Giles Martin and Phil WalkerUniversity of York, UK

CAPRA Meeting, 27 June, 2008

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Content

We derive the Lorentz-Dirac force from quantumelectrodynamics (QED) in the ~ → 0 limit.

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Content

Our derivation is a straightforward exercise in QED

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Content

Our derivation is a straightforward exercise in QED with noextra assumptions, such as a finite size of electrons,

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Content

Our derivation is a straightforward exercise in QED with noextra assumptions, such as a finite size of electrons, and wetreat the electrons as quantum fields.

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Content

Our derivation is a straightforward exercise in QED with noextra assumptions, such as a finite size of electrons, and wetreat the electrons as quantum fields. This was not donepreviously, except possibly by VS Krivitskiı and VN Tsytovich,Sov. Phys. Usp. 34, 250 (1991).

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Content

Our derivation is a straightforward exercise in QED with noextra assumptions, such as a finite size of electrons, and wetreat the electrons as quantum fields. This was not donepreviously, except possibly by VS Krivitskiı and VN Tsytovich,Sov. Phys. Usp. 34, 250 (1991).

The infinite mass of the charged particle is subtracted by themass counterterm in QED. (Not discussed.)

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Outline

Motivation

Which “straightforward exercise” should we do?

Quantum position shift

Equality of classical and quantum position shifts

Summary and outlook

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Motivation

I Classical electrodynamics is an approximation to QED.The radiation-reaction force should ultimately be ofquantum origin.

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Motivation

I Derivation of the Lorentz-Dirac force in perturbative QEDwill be another justification for the “reduction of order”.

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Motivation

I Derivation of the Lorentz-Dirac force in perturbative QEDwill be another justification for the “reduction of order”.

I There might be a one-loop quantum correction bigger thanthe radiation-reaction force.

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Outline

Motivation

Summary and outlook

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How do we compare quantum and classical forces?

The “force” is not easy to define in quantum mechanics.

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The Lorentz-Dirac force changes the position of a chargedparticle.

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Can we reproduce this position change in QED?

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Can we reproduce this position change in QED?

We first discuss the general formula for the position change dueto a perturbative force.

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Position shift by a perturbative force in general

Consider the motion of a particle described by a HamiltonianH(x,p):

xi =∂H∂pi

, pi = −∂H∂xi

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xi =∂H∂pi

, pi = −∂H∂xi

Let (Xi(t),Pi(t)) be a solution.

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xi =∂H∂pi

, pi = −∂H∂xi

+ ∆Fi(t).

Now add a perturbative force ∆Fi(t).

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xi =∂H∂pi

, pi = −∂H∂xi

+ ∆Fi(t).

The solution will change to (Xi(t) + ∆Xi(t),Pi(t) + ∆Pi(t)).

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xi =∂H∂pi

, pi = −∂H∂xi

+ ∆Fi(t).

Expression for the retarded solution ∆Xi(t) in terms of ∆Fi(t)?

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xi =∂H∂pi

, pi = −∂H∂xi

+ ∆Fi(t).

Expression for the retarded solution ∆Xi(t) in terms of ∆Fi(t)?

H =√

(p − eA(x, t)))2 + m2 + eA0(x, t) (the Lorentz force)∆F (t) (the Lorentz-Dirac force),for example.

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We linearize the zeroth-order Hamilton equations about asolution (Xi(t),Pi(t)).

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Let (∆xi;k (t ; s),∆pi;k (t ; s)) be a set of solutions to the linearizedHamilton equations labelled by k = 1,2,3 and s ∈ R

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Let (∆xi;k (t ; s),∆pi;k (t ; s)) be a set of solutions to the linearizedHamilton equations labelled by k = 1,2,3 and s ∈ R satisfying

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∆xi;k (s; s) = 0, ∆pi;k (s; s) = δik .

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∆xi;k (s; s) = 0, ∆pi;k (s; s) = δik .

Then the retarded position shift, ∆Xi(t), due to the perturbation∆Fi(t) is

∆Xi(t) =

−∞

∆xi;k (t ; s)∆Fk (s) ds.

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∆Xi(t) =

−∞

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∆Xi(t) =

−∞

The fact that (∆xi,k (t ; s),∆pi,k (t ; s)) are solutions to alinearized Hamiltonian system implies

∆xi;k (t ; s) = −∆xk ;i(s; t).

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∆Xi(t) =

−∞

∆xi;k (t ; s) = −∆xk ;i(s; t).

∆Xi(t) = −

−∞

∆xk ;i(s; t)∆Fk (s) ds,

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∆Xi(t) =

−∞

∆xi;k (t ; s) = −∆xk ;i(s; t).

∆Xi(t) = −

−∞

∆xk ;i(s; t)∆Fk (s) ds,

or by letting t = 0

∆Xi(0) = −

−∞

∆Fk (t)∆xk ;i(t ; 0) dt .

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∆Xi(0) = −

−∞

∆Fk (t)∆xk ;i(t ; 0) dt .

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∆Xi(0) = −

−∞

∆Fk (t)∆xk ;i(t ; 0) dt .

Interpretation of ∆xk ;i(t ; 0)?

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∆Xi(0) = −

−∞

∆Fk (t)∆xk ;i(t ; 0) dt .

Interpretation of ∆xk ;i(t ; 0)? Suppose the unperturbed solutiongoes through the spacetime origin, i.e. Xk (0) = 0 and considera set of solutions (xk(p)(t),pk(p)(t)) labelled by p such that

xk(p)(0) = 0, pk(p)(0) = pk .

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∆Xi(0) = −

−∞

∆Fk (t)∆xk ;i(t ; 0) dt .

xk(p)(0) = 0, pk(p)(0) = pk .

∂xk(p)(0)

∂pi= 0,

∂pk(p)(0)

∂pi= δki .

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∆Xi(0) = −

−∞

∆Fk (t)∆xk ;i(t ; 0) dt .

xk(p)(0) = 0, pk(p)(0) = pk .

∂xk(p)(0)

∂pi= 0,

∂pk(p)(0)

∂pi= δki .

Hence∂xk(p)(t)

∂pi= ∆xk ;i(t ; 0).

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Therefore, the position shift at t = 0 due to a perturbative force∆Fi(t) is

∆Xi(0) = −

−∞

∆Fk (t)∂xk(p)(t)

∂pidt ,

where xk(p)(t) is the position of the unperturbed particle whichat t = 0 is at the orgin and has momentum p.

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Therefore, the position shift at t = 0 due to a perturbative force∆Fi(t) is

∆Xi(0) = −

−∞

∂pidt ,

where xk(p)(t) is the position of the unperturbed particle whichat t = 0 is at the orgin and has momentum p.

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Outline

Motivation

Summary and outlook

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Position expectation value of a wave packet

Aim: to reproduce the position-shift formula with ∆Fi(t) beingthe Lorentz-Dirac force in the ~ → 0 limit of QED.

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The model: massive scalar field theory with externalelectromagnetic vector potential A(t) which is nonzero only in−T1 < t < −T2 < 0.

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We calculate the shift in the position expectation value of awave packet at t = 0 due to radiation reaction.

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We calculate the shift in the position expectation value of awave packet at t = 0 due to radiation reaction.

The position expectation value can be identified with the centreof the charge distribution: particle creation is absent to allorders in ~ if A(t) is smooth.

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The charge-density operator is ρ(t ,x) = i~

: ϕ†↔

∂t ϕ : , whereϕ(t ,x) is the scalar field operator.

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: ϕ†↔

For the state |ψ〉 the position expectation value at t = 0 is〈xi〉 =

d3x xi〈ψ|ρ(0,x)|ψ〉.

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: ϕ†↔

With e = 0, if the corresponding one-particle wave function is

ϕ(t ,x) =

(2π~)3√

2p0f (p)e−i(p0t−p·x)/~,

where p0 =√

p2 + m2 (c = 1), we have

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: ϕ†↔

With e = 0, if the corresponding one-particle wave function is

ϕ(t ,x) =

(2π~)3√

2p0f (p)e−i(p0t−p·x)/~,

where p0 =√

p2 + m2 (c = 1), we have

〈xi〉0 = i~

d3p(2π~)3 f ∗(p)

∂pi f (p).

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〈xi〉0 = i~

d3p(2π~)3 f ∗(p)

∂ pi f (p),

where f (p) is the momentum-space representation of theone-particle wave function.

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〈xi〉0 = i~

d3p(2π~)3 f ∗(p)

∂ pi f (p),

where f (p) is the momentum-space representation of theone-particle wave function.

With e 6= 0 (neglecting terms which vanish as ~ → 0 in the end)we have (with k ≡ ‖k‖)

f (p) → [1 + iF(p)] f (p)

f (p) ⊗

d3k(2π)32k

Aµ(p,k)a†µ(k)|0〉,

F(p): forward scattering amplitude of order e2,Aµ(p,k): one-photon emission amplitude of order e.

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F (p) ≡ [1 + iF(p)] f (p)

Gµ(p,k) ≡ Aµ(p,k)f (p) .

We find the position expectation value to be

〈xi〉e =i~2

d3p(2π~)3 F ∗(p)

∂ pi F (p)

d3k(2π)32k

d3p(2π~)3 G∗

µ(p,k)↔

∂ pi Gµ(p,k).

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F (p) ≡ [1 + iF(p)] f (p)

Gµ(p,k) ≡ Aµ(p,k)f (p) .

We find the position expectation value to be

〈xi〉e =i~2

d3p(2π~)3 F ∗(p)

∂ pi F (p)

d3k(2π)32k

d3p(2π~)3 G∗

µ(p,k)↔

∂ pi Gµ(p,k).

We let f (p) be sharply peaked with width of order ~.

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Then, neglecting the terms which will vanish as ~ → 0 in theend,

∆QXi(0) ≡ 〈xi〉e − 〈xi〉0

= −i~∂

∂piReF(p)

d3k(2π)32k

A∗µ(p,k)

∂ pi Aµ(p,k),

where p is now the expectation value of the momentum of thecharged scalar particle at t = 0 (after the acceleration).

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∆QXi(0) ≡ 〈xi〉e − 〈xi〉0

= −i~∂

∂piReF(p)

d3k(2π)32k

A∗µ(p,k)

∂ pi Aµ(p,k),

where p is now the expectation value of the momentum of thecharged scalar particle at t = 0 (after the acceleration). Theone-loop contribution ReF(p) (naturally of order e2/~2) iscancelled by the mass counterterm in QED.

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∆QXi(0) ≡ 〈xi〉e − 〈xi〉0

= −i~∂

∂piReF(p)

d3k(2π)32k

A∗µ(p,k)

∂ pi Aµ(p,k),

where p is now the expectation value of the momentum of thecharged scalar particle at t = 0 (after the acceleration). Theone-loop contribution ReF(p) (naturally of order e2/~2) iscancelled by the mass counterterm in QED.

The one-photon emission term?

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Outline

Motivation

Summary and outlook

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One-photon emission amplitude

We now have

∆Xi(0) = −

−∞

∂pidt

∆QXi(0) = −i2

d3k(2π)32k

A∗µ(p,k)

∂ pi Aµ(p,k).

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We now have

∆Xi(0) = −

−∞

∂pidt

∆QXi(0) = −i2

d3k(2π)32k

A∗µ(p,k)

∂ pi Aµ(p,k).

∆QXi(0) = ∆Xi(0), with ∆Fi(t) = Lorentz-Dirac force?

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We now have

∆Xi(0) = −

−∞

∂pidt

∆QXi(0) = −i2

d3k(2π)32k

A∗µ(p,k)

∂ pi Aµ(p,k).

∆QXi(0) = ∆Xi(0), with ∆Fi(t) = Lorentz-Dirac force?

To lowest order in ~ in the WKB approximation, one finds

Aµ(p,k) = −e∫

d4xeik ·x jµ(p)(x),

where j(p)(x) is the 4-current of the classical particleaccelerated by A(t), passing through the origin withmomentum p at t = 0.

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∆QXi(0) in terms of classical field

Remark: The current jµ(p)(x) is smoothly cut off for large |t | asjµ(p)(x)χ(t), where χ(t) = 0 for large enough |t |.

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The retarded field Aµ−(p)(x) from the current ejµ(p)(x) is

Aµ−(p)(x) = −

d3k(2π)32k

Aµ(p,k)e−ik ·x −Aµ∗(p,k)eik ·x]

for large t such that jµ(p)(x) (which is smoothly cut off) vanishes

there.

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The retarded field Aµ−(p)(x) from the current ejµ(p)(x) is

Aµ−(p)(x) = −

d3k(2π)32k

Aµ(p,k)e−ik ·x −Aµ∗(p,k)eik ·x]

for large t such that jµ(p)(x) (which is smoothly cut off) vanishes

there. This equation allows us to write ∆QXi(0) in terms of theretarded field A−(p)(x).

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The result is

∆QXi(0) = −12

t=Td3x(∂pi A

µ−(p))

∂ t A−(p) µ,

where jµ−(p)(x) = 0 at t = T due to the cut off.

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The result is

∆QXi(0) = −12

t=Td3x(∂pi A

µ−(p))

∂ t A−(p) µ,

Then using∫

t=Td3x G−

µα(x − y)

∂ t G−µβ(x − z) = −2GRαβ(y − z) ,

where GR = G− − G+, if y0, z0 < T ,

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The result is

∆QXi(0) = −12

t=Td3x(∂pi A

µ−(p))

∂ t A−(p) µ,

Then using∫

t=Td3x G−

µα(x − y)

∂ t G−µβ(x − z) = −2GRαβ(y − z) ,

where GR = G− − G+, if y0, z0 < T , we obtain

∆QXi(0) = e2∫

d4yd4z ∂pi j(p)µ(y)GµνR (y − z)j(p)ν(z)

= e∫

d4y ∂pi jµ(p)(y)AR(y) µ(y) .

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∆QXi(0) = e∫

d4x ∂pi jµ(p)(x)AR(p) µ(x).

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∆QXi(0) = e∫

jµ(p)(x) =dxµ

dtδ(3)(x − x(p)(t))χ(t),

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∆QXi(0) = e∫

jµ(p)(x) =dxµ

dtδ(3)(x − x(p)(t))χ(t),

∆QXi(0) = e∫ 0

−∞

dt FRkν

dxν(p)

∂xk(p)

∂pi.

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∆QXi(0) = e∫

jµ(p)(x) =dxµ

dtδ(3)(x − x(p)(t))χ(t),

∆QXi(0) = e∫ 0

−∞

dt FRkν

dxν(p)

∂xk(p)

∂pi.

This agrees with

∆Xi(0) = −

−∞

∆Fk (t)∂xk (t ; p)

∂pidt

with ∆Fk (t) = −eFRkνdxν

dt , which is the Lorentz-Dirac force.(The minus sign is due to index lowering.)

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Outline

Motivation

Summary and outlook

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Summary and comments on the one-loop contribution

I We showed that the change in the position of a chargedparticle due to the Lorentz-Dirac force can be reproducedin QED (at least) if the charge is accelerated by atime-dependent vector potential A(t).

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I The one-loop correction to the accelerating potential canbe of lower order in ~ than the Lorentz-Dirac force.

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I The one-loop correction to the accelerating potential canbe of lower order in ~ than the Lorentz-Dirac force.However, so far there are no examples of large one-loopcorrections of this type in physically relevant theories. (Notdiscussed in this talk.)

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Outlook

I It should be investigated how unique the Lorentz-Diracforce is, given the position shift formula.

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Outlook

I The generalization to arbitrary accelerating forces will beinteresting. We have studied only the cases where thepotential depends on one coordinate.

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Outlook

I It will be interesting to reproduce radiation reaction force incurved spacetime and for gravitational radiation. The tailterm will appear in the one-loop diagram.

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Outlook

I The structure of the one-loop diagram needs to beinvestigated further. It would be very interesting if there is a“large” one-loop correction in physically relevant situations.

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Outlook

I The structure of the one-loop diagram needs to beinvestigated further. It would be very interesting if there is a“large” one-loop correction in physically relevant situations.

See, Giles Martin, arXiv:0805.0666.

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Lorentz-Dirac force from quantum electrodynamics

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