4
$\begingroup$

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.)

$\endgroup$
  • 3
    $\begingroup$ You don't need to square the constraint function in the exponential, you can replace that term by $$i \mu \sum_i x_i$$ to obtain $Z(\mu)$ and then the constrained partition function can be expressed as a Fourier integral of $Z(\mu)$. $\endgroup$ – Count Iblis Dec 19 '16 at 1:04
  • $\begingroup$ Cross-post at Math.SE : Asymptotics for constrained partition function in a 1D lattice model. $\endgroup$ – user36790 Dec 19 '16 at 2:39
  • $\begingroup$ @MAFIA36790 Thanks for including the link; may I ask whether cross posts are frowned upon on this SE? (The vast majority of my SE experience is on Math.SE.) $\endgroup$ – Ian Dec 19 '16 at 2:42
  • $\begingroup$ Hi @Ian; I've included the link to the Maths post so that the community can have a easy notice on the post you mentioned at the end here; and regarding cross-posting, yes, it is generally frowned upon by the community and discouraged. Let the community decide its approach in this present context. $\endgroup$ – user36790 Dec 19 '16 at 2:45
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.