Asymptotics of a 1D lattice model's partition function

Question

So the quantity I would like to understand is:

$$\sum_{x \in \{ 0,1,\dots,m \}^n : \sum_{i=1}^n x_i=m} \exp \left ( -\beta \sum_{i=1}^n |x_{i+1}-x_i| \right )$$

where $m$ is a positive integer, $\beta=\frac{1}{k_B T}>0$, and we define $x_{n+1}=x_1$ (periodic boundary conditions). This is the partition function of a certain lattice model under a total mass constraint. The case I'm interested in is a thermodynamic limit: $m,n \to \infty$, but in particular with $m \gg n$ (high density).

If I drop the mass constraint, i.e. if I consider

$$\sum_{x \in \{ 0,1,\dots,m \}^n} \exp \left ( -\beta \sum_{i=1}^n |x_{i+1}-x_i| \right )$$

then I can actually compute the leading order asymptotic for the partition function when $m \to \infty$. Similar to the 1D Ising model, this can be done through the transfer matrix method: the partition function in this case is the trace of $Y^n$ where $Y$ is a $(m+1) \times (m+1)$ matrix with $y_{ij}=\exp(-\beta |i-j|)$. There is theory out there to compute the asymptotics of the trace for a nice family of matrices like this.

I have thought about computing the constrained partition function by introducing an "artificial field" with an intensity parameter $\mu$, and then sending $\mu \to \infty$. For example, I could look at:

$$\sum_{x \in \{ 0,1,\dots,m \}^n} \exp \left ( -\beta \sum_{i=1}^n |x_{i+1}-x_i| -\mu \left ( m - \sum_{i=1}^n x_i \right )^2 \right ).$$

The problem is that I don't see how to write this in the form $\operatorname{Tr}(Y^n)$ for some $Y$, because now all of the $x_i$ interact with each other (as can be seen by expanding the square).

If this is too hard, I would also be very interested to see useful approximations, for example a mean field approximation to my perturbed energy.

(I apologize if this question is too mathematical in character; I have actually asked the same question in different language on MSE already.)

Answer 1

I write some details following CountIblis's suggestion from the comments. Fix $m,n$ and let $N=mn+1$. We compute the partition function, call it $Z(m')$, for $m'=0,1,\dots,m$ in the following way. First compute the sums

$$\hat{Z}(k)=\sum_{x \in \{ 0,1,\dots,m \}^n} \exp \left ( -\beta \sum_{j=1}^n |x_{j+1}-x_j| - \frac{2 \pi i k}{N} \sum_{j=1}^n x_j \right )$$

where $k=0,1,\dots,N-1$. In physical language this amounts to artificially introducing an imaginary chemical potential. Mathematically it is just a rearrangement of

$$\sum_{m'=0}^{N-1} e^{-\frac{2 \pi i k m'}{N}} \sum_{x \in \{ 0,1,\dots,m \}^n : \sum_{j=1}^n x_j=m'} \exp \left ( -\beta \sum_{j=1}^n |x_{j+1}-x_j| \right )$$

which is basically the discrete Fourier transform of $Z(m')$. This is not quite right; it is the discrete Fourier transform of a function which agrees with $Z(m')$ between $0$ and $m$ but has a larger domain. Still, this will be enough to compute these values of $Z(m')$, which is what we wanted to do anyway.

These can be computed using the transfer matrix technique as discussed in the question. A symmetric way to do that is to take $y_{jk}=\exp \left ( -\beta |j-k| - \frac{2 \pi i(j+k) \ell}{2N} \right )$ for $j,k=0,1,\dots,m$ and $\ell=0,1,\dots,N-1$. Then $Z(m')$ is the inverse discrete Fourier transform of $\hat{Z}(\ell)$. Numerical tests on small examples show that this works and is much faster than direct summation, and there is some hope to be able to analytically estimate the result.