At order

g_{0}^{2}

, use the two-point sunset diagram: two quartic vertices are joined by three internal propagators, with one external line attached to each vertex. Unlike the one-vertex tadpole diagram, its self-energy

Σ (p)

depends nontrivially on the external momentum

p

p^{2}

in the expansion of

Σ (p)

changes the kinetic term, so normalizing that term requires wave-function renormalization and gives a nonzero field anomalous dimension.

Solution

No. With only

λ_{0} ϕ^{6}

ϕ \mapsto - ϕ

. Integrating out fast modes and rescaling preserve that symmetry, whereas

ϕ^{5}

is odd. Therefore no five-point vertex and no correction to

γ (ζ)

can be generated at any order in

λ_{0}

In the displayed connected Feynman diagram, each sextic vertex carries two slow external legs and the four remaining legs at each vertex are paired across the vertices. The second cumulant expansion has a factor

- 1/2

, the slow legs can be selected in

(2 6)^{2}

ways, and the four cross-contractions can be paired in

4!

ways. The coefficient is therefore

- \frac{1}{2} (2 6)^{2} 4! = - 2700.

S_{ζ} = {q : Λ/ ζ < ∣ q ∣ < Λ}

, the local zero-external-momentum contribution is

Δ g (ζ) = - 2700 ζ^{4 - d} λ_{0}^{2} \int_{S_{ζ}} \frac{d ^{d} q _{1} d ^{d} q _{2} d ^{d} q _{3}}{( 2 π ) ^{3 d}} G_{0} (q_{1}) G_{0} (q_{2}) G_{0} (q_{3}) G_{0} (q_{1} + q_{2} + q_{3}),

with the integral restricted further so that

q_{1} + q_{2} + q_{3} \in S_{ζ}

. The three independent internal momenta agree with the diagram's loop order

L = 4 - 2 + 1 = 3

v

Classify the connected contractions by the numbers

(r, 4 - r)

of external slow legs on the two sextic vertices and by the number

c

of fast propagators joining them. Besides the displayed

(2, 2; c = 4)

graph, the distinct topologies are:

$(2, 2; c = 2)$ : two joining lines and one tadpole on each vertex.
$(1, 3; c = 3)$ : three joining lines and one tadpole on the one-external-leg vertex.
$(0, 4; c = 2)$ : two joining lines and two tadpoles on the vertex with no external legs.
$(1, 3; c = 1)$ : one joining line, two tadpoles on the one-external-leg vertex, and one tadpole on the three-external-leg vertex.

Exchanging the two vertices gives no new topology. All four are connected Feynman diagrams selected by the logarithm in the cumulant expansion. With an ideal sharp momentum shell and a projection at exactly zero external momentum, the single joining line in the last topology cannot carry shell momentum, so that topology gives zero to the local quartic coupling; it is still the remaining formal connected contraction.