Electron self-energy - Apuntes 1 PDF

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Cálculo de la contribución del diagrama de Feynman...

Description

Following our discussion of loop corrections in QED, now we are going to look at the self-interaction of an electron by a photon. Such process is described by the following Feynman diagram: The diagram shows a fermion current that interacts twice with a photon. The amplitude is Z i((✁p + ✁q) + m) −iηµν d4 q 4 µ (−ieγν )u(p, s) M = (2π) u( ¯ p, s)(−ieγ ) (2π )4 (p + q )2 − m2 + iǫ q 2 + iǫ Both incoming and outgoing fermions are irrelevant since we are interested only in the loop. In a loop correction analysis, it is often considered the amputated diagrams only. The amplitude for such diagram is usually defined as −iΣ(p): Z d4 q µ i((✁p + q) ✁ + m) γ ν ηµν −iΣ(p) = −e2 γ (2π )4 (p + q )2 − m2 + iǫ q 2 + iǫ Since we will perform dimensional regularisation, it is convenient to set the spacetime dimension to an arbitrary dimension D and then recover by setting again D = 4 − ǫ. The above expression for −iΣ(p) can be simplified if we use the well-known Clifford’s algebra identity γ µ γ ν + γ ν γ µ = 2η µν : γ µ (✁p + q✁ + m)γ ν ηµν = γ µ ✁pγ ν ηµν + γ µ ✁qγ ν ηµν + γ µ mγ ν ηµν = γ µ γ α pα γ ν ηµν + γ µ γ α qα γ ν ηµν + γ µ mγ ν ηµν µ ν (2η µα − γ α γ µ )pα γ ν ηµν + (2η µα − γ α γ µ )qα γ ν ηµν + γ µ mγν ηµν = 2(✁p + q) ✁ + γ γ ηµν (m − ✁p − q) ✁

Since γ µ γµ = D: γ µ (✁p + q✁ + m)γ ν ηµν = 2(✁p + ✁q) + D(m − ✁p − q) ✁ +m·D ✁ = (2 − D)(✁p + q) Now we have:

dD q (2 − D)(p✁ + ✁q) + m · D (2π )D ((p + q )2 − m2 + iǫ)(q 2 + iǫ)

Z

−iΣ(p) = −e2

As we did with the loop correction of a photon, we use the Feynman identity so that: −iΣ(p) = −e2

= −e

2

Z

dD q (2π )D

Z

0

1

dx

dD q (2π )D

Z

Z

1

dx

0

(2 − D)(p✁ + ✁q) + m · D [x((p +

q)2

(2 − D)(✁p + q) ✁ +m·D [xp2

+

xq 2

+ 2xpq −

m2 x

+

q2

−

x2

+ iǫ]

2

− m2 + iǫ) + (1 − x)(q 2 + iǫ)]

= −e

2

Z

dD q (2π )D

Z

1

dx

0

2

=

(2 − D)(✁p + q) ✁ +m·D [(q + xp)2 − x(m2 − (1 − x)p2 ) + iǫ]

We define l ≡ q + xp and M 2 ≡ x(m2 − (1 − x)p2 ) so our expression becomes: −iΣ(p) = −e2

Z

dD l (2π )D

Z

1

dx

(2 − D)(l✄ + (1 − x)✁p) + m · D [l2 − M 2 + iǫ]

0

2

If we take into account the symmetry of the fraction in terms of ✄l, it’s clear that all terms proportional to ✄l vanish. So: Z 1 Z dD l (2 − D)(l✄ + (1 − x)✁p) + m · D −iΣ(p) = −e2 dx 2 (2π )D 0 [l2 − M 2 + iǫ] −iΣ(p) = −e2

Z

dD l 1 D 2 (2π ) [l − M 2 + iǫ]2

Z

1

dx [(2 − D)(1 − x)✁p + m · D]

0

The final result is: −iΣ(p) = −i

Z 1 Z       1 1 α 1 M2 M2 α 1 −γ− − γ − 1 − 2 dx (1 − x ) log dx log m − i ✁p 2 0 4π ε 4πµ2 4πµ2 π ε 0

2...