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44 November 7, 2013
Such subdominant solutions to the classical field equations are called instantons.
Their contribution to Green’s functions do, as we see, not have a series expansion
around ¯h = 0. Such nonpertubative effects are therefore not accessible using
Feynman diagrams. This is not to say that they are irrelevant. Indeed, we
usually have a finite value for ¯h ; more dramatically, if we let J vary as a
parameter, ϕ (1)
c , say, may for some value of J take over from ϕ (0)
c as the true
maximum position of the probability density, causing a sudden shift in the value
of φ c (J) from ϕ (0)
c to ϕ (1)
c .
1.5 The effective action
1.5.1 The effective action as a Legendre transform
Since perturbation theory presumes that higher orders in the loop expansion
are small compared to lower orders, the following question suggests itself : is
it possible to find, for a given action S(ϕ), another action, called the effective
action, with the property that its tree approximation reproduces the full field
function of the original action S ? If such an effective action, denoted by Γ(φ),
exists, we must have
Γ ′ (φ) = J , (1.81)
where φ(J) is the full solution to the SDe belonging with S(ϕ). We can use
partial integration to find
∫
∫
Γ(φ) = J dφ = J φ − φ dJ = J φ − ¯hW , (1.82)
where J is now to be interpreted as a function of φ. The transition from W (J)
to Γ(φ) is called the Legrendre transform. In classical mechanics, we have the
same situation : there, ¯hW would be the Lagrangian with J as the velocity and
φ as the momentum, and then the effective action would turn out to be the
Hamiltonian.
An important fact to be noted about the effective action can be inferred as
follows. Let us consider the derivative of φ(J). If we denote the probability
density (including the sources) of the quantum field ϕ by P J (ϕ), that is,
P J (ϕ) =
A(ϕ) ∫
dϕA(ϕ)
, A(ϕ) = exp
we can write this derivative as
1
¯h φ′ (J) = 1¯h
d
dJ
=
(∫ )
PJ (ϕ) ϕ dϕ
∫
PJ (ϕ) dϕ
∫
PJ (ϕ) ϕ 2 dϕ
∫
PJ (ϕ) dϕ
−
(− 1¯h (S(ϕ) − Jϕ) )
(∫
PJ (ϕ) ϕ dϕ ) 2
(∫
PJ (ϕ) dϕ ) 2
, (1.83)
=
∫
PJ (ϕ 1 )P J (ϕ 2 ) (ϕ 1 2 − ϕ 1 ϕ 2 ) dϕ 1 dϕ 2
(∫
PJ (ϕ) dϕ ) 2
. (1.84)