Subtract the zero-temperature term by using $\coth(x/2)=1+2/(e^x-1)$. In three dimensions the thermal part is
$$
\langle\theta^2\rangle_T=
\frac1{\chi v}\int\frac{d^3k}{(2\pi)^3}
\frac{1}{k}\frac1{e^{vk/T}-1}
=\frac{T^2}{12\chi v^3},
$$
where the last integral uses the <Bose-Einstein distribution>. Gaussian phase fluctuations reduce the <order parameter> by $\langle e^{-i\theta}\rangle=e^{-\langle\theta^2\rangle/2}$. Taking symmetry restoration to occur when the thermal variance is of order one gives
$$
T_*\sim\sqrt{12\chi v^3}.
$$
The effective action in part c has $\chi=1/(2\lambda)$, and hence $T_*\sim\sqrt{6v^3/\lambda}$. This is an order-of-magnitude estimate because near restoration the phase-only <effective field theory> omits large amplitude fluctuations and sensitivity to its <ultraviolet cutoff>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 337 / 1 / e / Solution

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An ultraviolet cutoff removes modes above a chosen energy or momentum scale. An <effective field theory> encodes the influence of those omitted short-distance modes in its couplings.

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 Incoming links: Ultraviolet cutoff