Witten’s Problems, Set Six — N^{o}. 1 (solution written by D. Freed)

Problem:

(1) In free scalar field theory in four dimensions, with mass m, calculate the operator product expansion ϕ^{2}(x)ϕ^{2}(x′)|_{x→x′} up to and including all operators of dimension four or less. (This means, relative to what we did in class, working out contributions of some operators with derivatives. This should not be a long exercise.)

Solution:

We represent ϕ^{2}(x) by ⋀^{x_{⊗}} and then there are 3 diagrams in ϕ^{2}(x)ϕ^{2}(x′):

∧^{⊗^{x}}  ∧^{⊗^{x′}}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}_{.}^{⊗}^{⊗}^{x}  ^{x′}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{⊗}  ^{⊗}^{x}  ^{x′}_{I  II  III}

The first diagram is the ordinary Taylor series, which to operators of dimension four is simply

I = ϕ(x)^{4} + ⋯

The second diagram occurs 4 times and is

II = 4G(x − x′)ϕ(x)ϕ(x′)
= 4G(x − x′)[ϕ(x)^{2} + (x′ − x)^{μ}ϕ(x)∂_{μ}ϕ(x) + \frac{1}{2}(x′ − x)^{μ}(x′ − x)^{ν}ϕ(x)∂_{μ}∂_{ν}ϕ(x)
+ ⋯].

The third diagram occurs twice and is simply

III = 2G(x − x′)^{2}.

Since G(x − x′) ∼ \frac{1}{|x−x′|^{2}} as x′ → x, we have the operator product expansion up to operators of dimension four:

ϕ^{2}(x)ϕ^{2}(x′) ∼_{x′→x} \frac{2}{|x−x′|^{4}} + \frac{4ϕ(x)^{2}}{|x−x′|^{2}} + 4\frac{(x−x′)^{μ}}{|x−x′|^{2}} ϕ(x)∂_{μ}ϕ(x)
+ \{ϕ(x)^{4} + 4\frac{(x−x′)^{μ}(x−x′)^{ν}}{|x−x′|^{2}} ϕ(x)∂_{μ}∂_{ν}ϕ(x)\}
+ ⋯