## Long-range order in a scalar field at thermal equilibrium

This calculation works through an example of a system in which long-range order (in a sense made precise below) appears in spatial dimensions $d\geq 3$ but not in dimensions $d=1,2$:

Suppose that a real scalar field $u$ is defined on a $d$-dimensional cube $C$ of side length $L$, with boundary condition $u\vert_{\partial C}\equiv 0$ and an upper bound on the frequency components of the field, as would be the case if we were trying to approximate a scalar field defined on a lattice. Let the system be governed by a Hamiltonian given by

\begin{aligned} H(u)=\frac{c}{2}\int_C\nabla u\cdot\nabla u\hspace{2pt}d^dr. \end{aligned}

At nonzero temperatures, for points $x$ and $y$ close to the center of the cube, the thermal average of the squared field variation has behavior varying with dimension:

\begin{aligned} \left\langle(u(x)-u(y))^2\right\rangle_\beta&\sim\left\vert x-y\right\vert\hspace{20pt}&d=1\\ &\leq \text{const}&d\geq 3 \end{aligned}

so that for dimensions greater than two, the field is roughly constant no matter how far away you look, whereas for dimension one, the thermal fluctuations destroy this long-range order. The $d=2$ case apparently gives logarithmic dependence on separation (see e.g. these lecture notes), but I don’t show it here.

We’ll demonstrate that this behavior exists in a few steps:

1. Mode decomposition of Hamiltonian

We can start by decomposing the scalar field into Fourier modes:

$u(x)=\sum_n\hat{u}_{n}\phi_n(x)\hspace{20pt}\phi_n(x)=\prod_{i=1}^{d}\sin\left(\frac{n_i\pi x_i}{L}\right)$

where $n$ is a $d$-dimensional vector of integers ranging from 1 to some cutoff $N$. In this basis, evaluating the Hamiltonian is simple:

\begin{aligned} H(u)&=-\frac{c}{2}\int_Cu\nabla^2ud^dx=\frac{c\pi^2}{2L^2}\sum_{nm}n^2\hat{u}_n\hat{u}_m\int_C\phi_n\phi_md^dx=\frac{c\pi^2}{2L^2}\sum_{nm}n^2\hat{u}_n\hat{u}_m\left(\frac{L}{2}\right)^d\delta_{nm}\\ &=2^{1-d}c\pi^2L^{d-2}\sum_{n}n^2\hat{u}_n^2. \end{aligned}

The first equality resulted from the given Hamiltonian by integration by parts. The modes are non-interacting, which simplifies the following calculation.

2. Calculation of frequency-space correlation function

Consider the following thermal correlation function:

$\left\langle\hat{u}_{m_1}\hat{u}_{m_1}\right\rangle_\beta=\frac{\left(\prod_n\int_{-\infty}^{\infty}d\hat{u}_n\right)\hat{u}_{m_1}\hat{u}_{m_1}e^{-\beta H(u)}}{\left(\prod_n\int_{-\infty}^{\infty}d\hat{u}_n\right)e^{-\beta H(u)}}.$

By the symmetry of the integrand, it’s clear that the correlation function vanishes unless $m_1=m_2$. In this case:

$\left\langle\hat{u}_{m}^2\right\rangle_\beta=\frac{\int_{-\infty}^{\infty}d\hat{u}_m\hat{u}_m^2e^{-c\pi^2L^{d-2}\beta m^2\hat{u}_m^2/2^{d+1}}}{\int_{-\infty}^{\infty}d\hat{u}_me^{-c\pi^2L^{d-2}\beta m^2\hat{u}_m^2/2^{d+1}}}=\frac{1}{2}\left(c\pi^2L^{d-2}\beta m^2/2^{d+1}\right)^{-1}=\left(\frac{2^d}{c\pi^2}\right)\beta^{-1}m^{-2}L^{2-d},$

so that overall we get

$\left\langle\hat{u}_{m}\hat{u}_{n}\right\rangle_\beta=\delta_{mn}\left(\frac{2^d}{c\pi^2}\right)\beta^{-1}n^{-2}L^{2-d}$.

Since the factorization of the partition function is a direct consequence of the absence of mode-coupling terms in the Hamiltonian, we just had to do a single-mode calculation and then write this answer down.

3. Calculation of average squared disorder

Expanding the squared difference between field values at points $x$ and $y$ in terms of the Fourier modes, we get

\begin{aligned} (u(x)-u(y))^2&=\left(\sum_n\hat{u}_n\left(\phi_n(x)-\phi_n(y)\right)\right)^2\\ &=\sum_{nm}\hat{u}_n\hat{u}_m\left(\phi_n(x)-\phi_n(y)\right)\left(\phi_m(x)-\phi_m(y)\right). \end{aligned}

Taking the expectation using the frequency-space correlation function:

\begin{aligned} \left\langle(u(x)-u(y))^2\right\rangle_\beta&=\sum_{nm}\left(\phi_n(x)-\phi_n(y)\right)\left(\phi_m(x)-\phi_m(y)\right)\delta_{mn}\left(\frac{2^d}{c\pi^2}\right)\beta^{-1}n^{-2}L^{2-d}\\ &=\left(\frac{2^d}{c\pi^2}\right)\beta^{-1}L^{2-d}\sum_{n}\frac{\left(\phi_n(x)-\phi_n(y)\right)^2}{n^2} \end{aligned}

Note that up to this point, everything has been exact (for the continuum theory defined in the problem statement, which itself could be used as an approximation of a lattice system).

4. Approximations for different values of $d$

For $d=1$:
\begin{aligned} \left\langle(u(x)-u(y))^2\right\rangle_\beta&=\left(\frac{2^d}{c\pi^2}\right)\beta^{-1}L^{2-d}\sum_{n=1}^N\frac{\left(\sin\left(\frac{n\pi x}{L}\right)-\sin\left(\frac{n\pi y}{L}\right)\right)^2}{n^2} \end{aligned}

For very large $L$ and $x$, $y$ close to $L/2$ on the scale of $L$, the summand varies slowly in $n$, so we can approximate the sum by an integral. Since for sufficiently large $N$, which assuming a fixed frequency cutoff is achieved by taking $L$ large enough, the integrand is heavily suppressed, we can extend the upper limit to infinity without too much error and evaluate the resulting integral (I did it in Mathematica):

\begin{aligned} \left\langle(u(x)-u(y))^2\right\rangle_\beta&\approx\left(\frac{2^d}{c\pi^2}\right)\beta^{-1}L^{2-d}\int_0^\infty\frac{\left(\sin\left(\frac{n\pi x}{L}\right)-\sin\left(\frac{n\pi y}{L}\right)\right)^2}{n^2}dn\\ &=\left(\frac{2}{L}\right)^{d-1}c^{-1}\beta^{-1}\left\vert x-y\right\vert \end{aligned}

For $d\geq 3$ the cutoff is significant, as the integrand grows with increasing $n$ (due to the factor of $n^{d-1}$ in the measure). The largest the numerator of the integrand can be is four, since it’s the square of a difference between products of sines. Fixing the integrand to its maximum value and integrating up to the cutoff frequency gives an upper bound independent of $x$ and $y$.