Because <normal ordering> removes contractions within either exponential, <Wick theorem> sums all cross-contractions into
$$
\begin{aligned}
G_{k,-k}
&=\exp\left[(ik)(-ik)
\langle X(z,\bar z)X(w,\bar w)\rangle\right]\\
&=\exp\left[-\frac{k^2}{2}\log|z-w|^2\right]
=|z-w|^{-k^2}.
\end{aligned}
$$
Under $z\mapsto\lambda z$, this scales as $\lambda^{-k^2}$. Each operator has scaling dimension $\Delta=h+\widetilde h=k^2/2$, so a two-point function of two such primaries must scale as $\lambda^{-2\Delta}=\lambda^{-k^2}$. The two calculations agree.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 306 / 3 / c / ii / Solution

Expand $\delta=a\delta_1+a^2\delta_2+a^3\delta_3+\cdots$. For a Gaussian linear density field, odd linear correlators vanish and <Wick theorem> reduces the fourth-order terms to products of the linear <power spectrum>. Hence
$$
P_{\rm 1-loop}(k)
=a^2P(k)+a^4[P_{22}(k)+2P_{13}(k)].
$$
The two contractions of the two quadratic fields give
$$
\boxed{P_{22}(k)=2\int\frac{d^3q}{(2\pi)^3}
F_2(\mathbf q,\mathbf k-\mathbf q)^2
P(q)P(|\mathbf k-\mathbf q|)}.
$$
The three choices for which argument of $F_3$ carries the external momentum give
$$
\boxed{P_{13}(k)=3P(k)\int\frac{d^3q}{(2\pi)^3}
F_3(\mathbf k,\mathbf q,-\mathbf q)P(q)}.
$$
Thus the contribution conventionally called $P_{13}+P_{31}$ is $2P_{13}=6P(k)\int F_3P$. These are the two one-loop diagrams of the <one-loop matter power spectrum>.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 312 / 2 / ii / Solution

{c}
{wiki=Wick's_theorem}

Wick theorem expresses a product of free fields as its normal-ordered product plus all possible contractions.


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