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

Problem:

(1) (a) Show that in ϕ^{4} theory the anomalous dimension of ϕ vanishes in the one loop approximation and that the one-loop beta function is therefore given by a one-loop diagram that we have computed before. The one-loop beta function is in particular β(λ) = Aλ^{2} where λ is the ϕ^{4} coupling, and A is the coefficient of the one-loop divergence.

(b) Convince yourself that the sign of the logarithmic divergence in this one-loop diagram (whose significance we by now appreciate) can be determined without any computation; it is determined by the fact that in free scalar field theory the two point function ⟨ϕ^{2}(x)ϕ^{2}(0)⟩ is positive. (This is actually a point that has been made previously.)

Solution:

(a) The anomalous dimension of ϕ gets contributions from 2-point function diagrams. All such (connected) diagrams have naive quadratic divergences. Let p be the momentum flowing in (and by conservation flowing out) and m the mass. Then with a cutoff Λ the diagram diverges as

f(\frac{p}{m})Λ^{2} + g(\frac{p}{m}) log (\frac{Λ}{m})p^{2} + h(\frac{p}{m}) log (\frac{Λ}{m})m^{2}

for some functions f, g, h. This follows by dimensional analysis. (In fact, since the second derivative of a diagram with respect to p is finite, we see that f, g, h are simple polynomials.) Now the only 1-loop diagram is shown, and the loop integral is independent

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of p. For this diagram, then, g ≡ 0 and there is no renormalization to the coefficient of |dϕ|^{2} and so no contribution to the anomalous dimension of ϕ.

The diagram which contributes to the 1-loop β-function is the integral

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D̂(p_{1}, p_{2}, p_{3}, p_{4}) = λ^{2} ∫ \frac{d^{4}q}{(2π)^{4}} \frac{1}{(q^{2}+m^{2})((q−p_{1}−p_{2})^{2}+m^{2})} · δ(p_{1} + p_{2} + p_{3} + p_{4})

which has a leading divergence Aλ^{2} log \frac{Λ}{m} if an ultraviolet cutoff Λ is imposed. (Here Γ̂ is a distribution supported on the subspace (p_{1} + p_{2} + p_{3} + p_{4} = 0.)

(b) We want to understand a priori that A > 0. For this it is best to Fourier transform to position space. This is straightforward and leads to

D(x_{1}, x_{2}, x_{3}, x_{4}) = G(x − y)^{2}δ(x_{1} − x)δ(x_{2} − x)δ(x_{3} − y)δ(x_{4} − y),

where G is the propagator. (So D is supported on the subspace x_{1} = x_{2}, x_{3} = x_{4} and its value there is the square of the Green’s function.) In general, the position space prescription for computing a one particle irreducible Feynman diagram as part of the effective action is the same as usual — we attach propagators to the legs and integrate over the positions in the internal vertices — except that we replace the propagators on the external legs by δ-functions. For convenience we take y = 0 and write D(x) = G(x)^{2}.

Now introduce the composite operator 𝒪(x) = ϕ^{2}(x). Then it is immediate that the two point function ⟨𝒪(x)𝒪(0)⟩_{free} in the free theory is computed by the diagram shown.

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We evaluate this diagram as G(x)^{2}, so it is precisely equal to D(x). It is manifest that D(x) ≥ 0, of course, since it is written as the square of a real function. But we would like to invoke instead reflection positivity since the argument is useful in more general situations.

Reflection positivity asserts that if 𝒪_{Ω} is a (complex) operator in a Euclidean field theory, supported in some set Ω lying on one side of a hyperplane, and if Ω̄ denotes the reflection of Ω in the hyperplane, then

⟨𝒪_{Ω}\overline{𝒪Ω̄}⟩ ≥ 0.

If 𝒪_{Ω} takes values in a vector space V, then the assertion is that ⟨𝒪_{Ω}\overline{𝒪Ω̄}⟩ is a nonnegative hermitian form on V^{∗}.

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 Ω Ω̄  π X_{+} X_{−}

We can understand this heuristically by writing a path integral over the half-space. Namely, suppose for simplicity that we are dealing with a complex scalar field ϕ with action S(ϕ). Then for ϕ_{0} a function on π we set

Z_{X_{+}}(ϕ_{0}) = ∫_{\begin{matrix}ϕ: X_{+}→ℂ \\ ϕ _{π}=ϕ_{0} \end{matrix}} 𝒪_{Ω}(ϕ)e^{−S_{X_{+}} (ϕ)}Dϕ.

This should be viewed as a vector in some Hilbert space of functions of ϕ_{0}. Now X_{−} ≅ X̄_{+}, where X̄_{+} is X_{+} with the reversed orientation. The Euclidean action has the property that SȲ = \overline{S_{Y}} for any spacetime Y: orientation reversal conjugates the action. It follows that

\overline{Z_{X_{+}}(ϕ_{0})} = ∫_{\begin{matrix}ϕ: X_{+}→ℂ \\ ϕ _{π}=ϕ_{0} \end{matrix}} \overline{𝒪_{Ω}(ϕ)}e^{−\overline{S_{X_{+}} (ϕ)}}Dϕ

= ∫_{\begin{matrix}ϕ: X_{−}→ℂ \\ ϕ _{π}=ϕ_{0} \end{matrix}} \overline{𝒪_{Ω}(ϕ)}e^{−S_{X_{−}} (ϕ)}Dϕ

The norm square of Z_{X_{+}}(ϕ) in ℋ is computed by integrating over ϕ_{0}:

∥Z_{X_{+}}(ϕ_{0})∥_{ℋ}^{2} = ∫_{ϕ_{0}: π→ℂ} Z_{X_{+}}(ϕ_{0})\overline{Z_{X_{+}}(ϕ_{0})}Dϕ_{0}

= ∫_{ϕ:X→ℂ} 𝒪_{Ω}(ϕ)\overline{𝒪Ω̄(ϕ)}e^{−S_{X}(ϕ)}Dϕ

= ⟨𝒪_{Ω}\overline{𝒪Ω̄}⟩ ,

where the measure is normalized so that the partition function is one. The left hand side is obviously positive, hence so is the right.

In our example 𝒪_{(x)} = ϕ^{2}(x) is real, and to show that D(x) = ⟨𝒪(x)𝒪(0)⟩_{free} ≥ 0 we use reflection positivity for the perpendicular bisector of the line segment connecting 0 and x. Now D(x) diverges as x → 0, and since D(x) ≥ 0 the coefficient of the divergence is positive. This implies that the divergence in D with a momentum cutoff Λ has a positive coefficient.