Current

• Renormalization

• Electric Charge

• Interpretation

2010年12月30日木曜日

Renormalization

= + +1PR 1PI 1PR 1PR

Γµ0,1PR = Γµ

0,1PI +−Π0(q2)

1 + Π0(q2)

�gµν −

qµqν

q2

�Γν

0,1PI

+δZ3

e0Z3

�−q2gµν + qµqν

� −i

q2(1 + Π0(q2))(ie0)Γν

0,1PI

= Γµ0,1PI +

�−Π0(q2)

1 + Π0(q2)− δZ3

Z3

11 + Π0(q2)

� �gµν −

qµqν

q2

�Γν

0,1PI

Renormalized to zero at q2=0 thanks to

�p�| (−Jµphys(x)) |p� = �p�| (−Jµ

N(x)) |p� +δZ3

Z3

1e0

�p�| ∂νF νµ0 |p�

generated by Noether current

Z3 =1

1 + Π0(0)

=1Z3

11 + Π0(q2)

− 1 q2→0−→ 0

2010年12月30日木曜日

Electric charge

�e−(p)|QN|e−(p)� =�

d3x�e(p)|J0N(x)|e(p)�

= limq→0

�d3x e−iqx�e(p)|J0

N(x)|e(p)�

= limq→0

(2π)3δ(3)(�p− �p− �q)e−iq0x0�e(p)|J0

N(0)|e(p)�

= (2π)3δ(3)(�0)× u(p)�

(−1)Z2Γ00,1PI(p, p)

�u(p)

�1 +

�1

1 + Π0(0)− 1

��

= (2π)3δ(3)(�0)×�

(−1)Z2F01 (q2 = 0) 2p0

�×

�1

1 + Π0(0)

�

�e−(p)|Qphys|e−(p)� =�

d3x�e(p)|�

J0N(x)− δZ3

Z3

1e0

∂νF ν00

�|e(p)�

= (2π)3δ(3)(�0 )×�

(−1)Z2F01 (0) 2p0

�×

�1

1 + Π0(0)− δZ3

Z3

11 + Π0(0)

�

1PI 1PR

1PR

= 1= 1

Two vertex diagrams for Noether current insertion

Counter term for physical current

�e±(p)|Qphys|e±(p)� = (±1) × (2π)32p0δ(3)(0)

= (±1) × �e±(p)|e±(p)�

2010年12月30日木曜日

Interpretation

1PI 1PR

1PI Noether current insertion

�e−(p)|QN|e−(p)� = (2π)3δ(3)(�0)× u(p)�

(−1)Z2Γ00,1PI(p, p)

�u(p)

�1 +

�1

1 + Π0(0)− 1

��

1PR

1PR Noether current insertion 1PR Counter term

1PR Noether current insertion

• 1PR Noether current insertion can be understood as, the charge operator QN itself creates the e+ e- pair in vacuum and see it as charge of an object to be measured.•We could define renormalized Noether charge by subtracting the IV divergence of 1PR-diagram in MSbar. But such a renormalization leaves 1PR contribution at q2=0. This leads strange phenomena; for instance, electron Noether charge is non-zero even for muon.•Form factor <p’|JN|p> defined by Noether current insertion receive 1PR contribution, which can be understood as running factor of QED effective charge. •Thus 1PR diagram must be completely subtracted at q2=0, which is the physical charge defined here.•The physical charge is UV finite and the one which satisfies Gauss’s law.

2010年12月30日木曜日