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November 7, 2013 71<br />

= ∑ [ 1<br />

2 m2 ϕ(x) 2 + 1 ]<br />

2 ϕ′ (x) 2 ∆<br />

n<br />

∫ [ 1<br />

=<br />

2 m2 ϕ(x) 2 + 1 ]<br />

2 ϕ′ (x) 2 dx . (2.29)<br />

The interaction and source terms in the path integral do not have a factor<br />

∆ coming out naturally, but we may simply define the continuum limits by<br />

redefining the objects in the action :<br />

λ 4 → ∆λ 4 , J n → ∆J(x) , (2.30)<br />

so that the continuum limit of the full action, including this time also the sources,<br />

becomes 13<br />

∫ [ 1<br />

S[ϕ, J] =<br />

2 m2 ϕ(x) 2 + 1 2 ϕ′ (x) 2 + λ ]<br />

4<br />

4! ϕ(x)4 − J(x)ϕ(x) dx . (2.31)<br />

Note the notation with square brackets: the action is now no longer a number<br />

depending on (a countably infinite set of) numbers, but rather on the functions<br />

ϕ(x) and J(x) ; this is called a functional.<br />

2.3.3 The continuum limit of the classical equation<br />

For the discrete action, there is an obvious classical equation :<br />

∂<br />

∂ϕ n<br />

S({ϕ}) = 0 ∀n , (2.32)<br />

where, again, the source terms have been subsumed into the action. For the ϕ 4<br />

model of Eq.(2.1), the classical equation is therefore<br />

µϕ n − γ(ϕ n+1 + ϕ n−1 ) + ∆λ 4<br />

ϕ 3 n = ∆J n (2.33)<br />

3!<br />

for all n, and the extra factor ∆ in the coupling constant and the sources have<br />

been taken into account. The Weyl prescription leads us to write<br />

µϕ n − γ(ϕ n+1 + ϕ n−1 ) ≈ m 2 ∆ϕ(x) − ∆ϕ ′′ (x) , (2.34)<br />

so that the continuum limit of the classical field equation takes the form<br />

m 2 ϕ(x) − ϕ ′′ (x) + λ 4<br />

3! ϕ(x)3 = J(x) . (2.35)<br />

This is precisely the Euler-Lagrange equation, that can also be obtained immediately<br />

from the continuum form of the action by taking functional derivatives.<br />

To see this, let us assume that the action of a theory can be written as<br />

∫ ( )<br />

S[ϕ] = F ϕ(x); ϕ ′ (x) dx . (2.36)<br />

13 Strictly speaking, the Weyl ordering requires the replacement of J n not by ∆J(x) but by<br />

∆J(x)+∆ 2 J ′ (x)/2. The additional term, however, vanishes in the continuum limit as ∆ → 0,<br />

as do the higher powers of ∆ involved in the ϕ n 4 term.

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