Pictures Paths Particles Processes

November 7, 2013 51
(
¯hm ∂
∂ ¯J + ∂ 2 )
e¯h2 ∂ ¯J∂H − J Z( ¯J, J, H) = 0 ,
)
∂H + e
∂2
∂ ¯J∂J − H Z( ¯J, J, H) = 0 . (1.113)
(
¯hµ ∂
The field-generating functions (the ‘field functions’) are, of course, each a function
of J, ¯J and H, and are given by
so that
ψ = ¯h ∂
∂ ¯J log Z , ∂
∂
¯ψ = ¯h log Z , A = ¯h log Z , (1.114)
∂J ∂H
¯h ∂
∂
Z = ψ Z , ¯h
∂ ¯J ∂J Z = ¯ψ Z , ¯h ∂
∂H Z = A Z , (1.115)
and Eq.(1.112) can be written as
ψ = 1 m J − e (
A ψ + ¯h ∂ )
m ∂H ψ ,
¯ψ = 1 m ¯J − e (
¯ψ A + ¯h ∂ )
m ∂H ¯ψ ,
A = 1 µ H − e (
¯ψ ψ + ¯h ∂ )
µ ∂J ψ . (1.116)
Incidentally, note that we could rewrite these SDe’s since
∂
∂H ψ = ∂
∂ ¯J A , ∂
∂H ¯ψ = ∂
∂J H , ∂
∂J ψ = ∂
∂ ¯J ¯ψ . (1.117)
The Feynman rules are, for this action, as follows :
ψ ψ ↔ ¯h m ,
↔ ¯h µ ,
↔ − ē h . (1.118)
A few things are of interest here. In the first place, all diagrams have a symmetry
factor of unity. In the second place, the ϕ propagator links two different fields
(ϕ to ¯ϕ) and must therefore carry an orientation (an arrow on the line is usually
employed). In the third place, in the action we find the two terms ¯Jϕ and ¯ϕJ,
which would suggest that J is the source in the SDe of ¯ψ, and ¯J is the source in
the SDe for ψ ; but it is actually the other way around ! What is the source for a
given field function is seen by taking the derivative of the action, and inspecting
which field then occurs as a linear term, and which source term is left by itself
after the differentiation.