Self Avoiding Walk Pair Correlation

Question

Let $\gamma(i)$ be a self avoiding walk (SAW) on a 2D lattice $L$ (a square lattice for example) starting at a predefined origin ( $\gamma(0)=(0,0)$ ) and having length $n:=\ell(\gamma)$. Furthermore, suppose $\gamma$ ends at a point $u$, so that $\gamma(n)=u$. Let $N(u)$ be the set of neighboring vertices of $u$ (that are connected to $u$). If I were to further condition the SAW by clamping the second-to-last-step, i.e. $\gamma(n-1)=v$, what can be said about the probability that the walk visited any of the other neighbors of $u$. Naturally, the length of the SAW and the distance of $u$ from the origin will be interlocked in a tug of war. I am primarily interested in seeing when the probability of visiting adjacent neighbors tends to zero.

Answer 1

If the walks are sampled uniformly at random, then the probability of another neighbour of $u$ being visited is related to the "atmosphere" of a walk, and the connective constant $\mu$. The mean number of additional neighbours approaches $3-\mu \approx 0.361841469\cdots$ for SAWs on $Z^2$, while the probability of there being at least one additional neighbour approaches $1-0.71114 = 0.28886$ (from eq. 12, Owczarek and Prellberg, ref. below).

A recent article gave probabilities for having 0, 1, 2, and 3 unoccupied neighbours, A L Owczarek and T Prellberg, "Scaling of the atmosphere of self-avoiding walks", J. Phys. A: Math. Theor. 41 (2008) 375004 (6pp).

The most accurate estimate of $\mu$ for $Z^2$ comes from self-avoiding polygon enumerations by Iwan Jensen, "A parallel algorithm for the enumeration of self-avoiding polygons on the square lattice", J. Phys. A: Math. Gen. 36 (2003) 5731–5745, also on the arXiv: http://arXiv.org/abs/cond-mat/0301468v1

If you're wondering about the scaling behaviour as $|u|$ is varied, then I think that given $x = |u|/\langle R_e^2 \rangle^{1/2}$, the scaling function $P(x)$ for the probability of more than one neighbour being occupied must go to zero as $x \rightarrow \infty$, but I don't have any intuition yet as to how it will approach zero (I think this is a nice question).