A colleague, Nicolas Broutin, who is not on MO, pointed me to a recent "preliminary report" of Gábor Pete, about joint work with Christophe Garban and Oded Schramm, on the scaling limit of the MST. In the references of that note I found a paper of Aizenmann et.al. which contains the following information about a suitably defined (subsequential) scaling limit of the MST (I won't get into the precise definition of the scaling limit as it's not particularly important for this answer):

The branches of all trees in $\mathcal{F}(\omega)$ are random curves $\mathcal{C}$ with Hausdorff dimensions bounded above and below: $d_{\mathrm{min}} \leq \mathrm{dim}~\mathcal{C} \leq d_{\mathrm{max}}$, for some non-random $1 < d_{\mathrm{min}} < d_{\mathrm{max}} < 2$.

A couple comments are in order.

First, the paper in fact shows that this result holds for any of three models: uniform spanning tree on the lattice (the limit is as the lattice spacing goes to zero); minimum spanning tree on the lattice with iid uniform bond weights (this is the one most naturally connected with percolation); and third, Euclidean MST on a Poisson point process with intensity $r$ (where the rescaled limit is as $r \to \infty$ appropriately). The last of these is the one most directly connected with my question.

Second, the reason for "all trees" is because the paper considers a process taking place in the whole plane. In this setting, the natural attempts to generalize the definition of "minimum spanning tree" can in principle end up yielding a forest. (I seem to recall that this doesn't occur in fact happen for the Euclidean MST on a Poisson point set, but I forget where I think I saw this -- maybe something by Aldous?).

Third, given the answer and comments of Bill Thurston, this suggests that the diameter should indeed scale like $n^{\alpha}$ for some $1/2 < \alpha < 1$. (Though, inspired by Peter Shor's comment, I will also mention that it doesn't a priori rule out scaling behaviour of the form $n^{\alpha}\log^{\beta} n$.)

A colleague, Nicolas Broutin, who is not on MO, pointed me to a recent "preliminary report" of Gábor Pete, about joint work with Christophe Garban and Oded Schramm, on the scaling limit of the MST. In the references of that note I found a paper of Aizenmann et.al. which contains the following information about a suitably defined (subsequential) scaling limit of the MST (I won't get into the precise definition of the scaling limit as it's not particularly important for this answer):

The branches of all trees in $\mathcal{F}(\omega)$ are random curves $\mathcal{C}$ with Hausdorff dimensions bounded above and below: $d_{\mathrm{min}} \leq \mathrm{dim}~\mathcal{C} \leq d_{\mathrm{max}}$, for some non-random $1 < d_{\mathrm{min}} < d_{\mathrm{max}} < 2$.

A couple comments are in order.

First, the paper in fact shows that this result holds for any of three models: uniform spanning tree on the lattice (the limit is as the lattice spacing goes to zero); minimum spanning tree on the lattice with iid uniform bond weights (this is the one most naturally connected with percolation); and third, Euclidean MST on a Poisson point process with intensity $r$ (where the rescaled limit is as $r \to \infty$ appropriately). The last of these is the one most directly connected with my question.

Second, the reason for "all trees" is because the scaling limit is in fact described in terms of a graded collection of finite trees spanning any finite collection of points in one-point compactification of $\mathbb{R}^2$, and satisfying certain consistency conditions. In the paper they show that the collection also turns out to "describe a single spanning tree" in the sense that for sets of points in $\mathbb{R}^2$ (i.e. not including the point at infinity), the tree in $\mathcal{F}(\omega)$ spanning them stays within some finite region in $\mathbb{R}^2$. (Edit: this explanation of "all trees" replaces an earlier false explanation which was based on an overhasty reading of the results.)

Third, given the answer and comments of Bill Thurston, this suggests that the diameter should indeed scale like $n^{\alpha}$ for some $1/2 < \alpha < 1$. (Though, inspired by Peter Shor's comment, I will also mention that it doesn't a priori rule out scaling behaviour of the form $n^{\alpha}\log^{\beta} n$.)

A colleague, Nicolas Broutin, who is not on MO, pointed me to a recent "preliminary report" of Gábor Pete, about joint work with Christophe Garban and Oded Schramm, on the scaling limit of the MST. In the references of that note I found a paper of Aizenmann et.al. which contains the following information about a suitably defined (subsequential) scaling limit of the MST (I won't get into the precise definition of the scaling limit as it's not particularly important for this answer):

The branches of all trees in $\mathcal{F}(\omega)$ are random curves $\mathcal{C}$ with Hausdorff dimensions bounded above and below: $d_{\mathrm{min}} \leq \mathrm{dim}~\mathcal{C} \leq d_{\mathrm{max}}$, for some non-random $1 < d_{\mathrm{min}} < d_{\mathrm{max}} < 2$.

A couple comments are in order.

First, the paper in fact shows that this result holds for any of three models: uniform spanning tree on the lattice (the limit is as the lattice spacing goes to zero); minimum spanning tree on the lattice with iid uniform bond weights (this is the one most naturally connected with percolation); and third, Euclidean MST on a Poisson point process with intensity $r$ (where the rescaled limit is as $r \to \infty$ appropriately). The last of these is the one most directly connected with my question.

Second, the reason for "all trees" is because the paper considers a process taking place in the whole plane. In this setting, the natural attempts to generalize the definition of "minimum spanning tree" can in principle end up yielding a forest. (I seem to recall that this doesn't occur in fact happen for the Euclidean MST on a Poisson point set, but I forget where I think I saw this -- maybe something by Aldous?).

Third, given the answer and comments of Bill Thurston, this suggests that the diameter should indeed scale like $n^{\alpha}$ for some $1/2 < \alpha < 1$. (Though, inspired by Peter Shor's comment, I will also mention that it doesn't a priori rule out scaling behaviour of the form $n^{\alpha}\log^{\beta} n$.

A colleague, Nicolas Broutin, who is not on MO, pointed me to a recent "preliminary report" of Gábor Pete, about joint work with Christophe Garban and Oded Schramm, on the scaling limit of the MST. In the references of that note I found a paper of Aizenmann et.al. which contains the following information about a suitably defined (subsequential) scaling limit of the MST (I won't get into the precise definition of the scaling limit as it's not particularly important for this answer):

The branches of all trees in $\mathcal{F}(\omega)$ are random curves $\mathcal{C}$ with Hausdorff dimensions bounded above and below: $d_{\mathrm{min}} \leq \mathrm{dim}~\mathcal{C} \leq d_{\mathrm{max}}$, for some non-random $1 < d_{\mathrm{min}} < d_{\mathrm{max}} < 2$.

A couple comments are in order.

First, the paper in fact shows that this result holds for any of three models: uniform spanning tree on the lattice (the limit is as the lattice spacing goes to zero); minimum spanning tree on the lattice with iid uniform bond weights (this is the one most naturally connected with percolation); and third, Euclidean MST on a Poisson point process with intensity $r$ (where the rescaled limit is as $r \to \infty$ appropriately). The last of these is the one most directly connected with my question.

Second, the reason for "all trees" is because the paper considers a process taking place in the whole plane. In this setting, the natural attempts to generalize the definition of "minimum spanning tree" can in principle end up yielding a forest. (I seem to recall that this doesn't occur in fact happen for the Euclidean MST on a Poisson point set, but I forget where I think I saw this -- maybe something by Aldous?).

Third, given the answer and comments of Bill Thurston, this suggests that the diameter should indeed scale like $n^{\alpha}$ for some $1/2 < \alpha < 1$. (Though, inspired by Peter Shor's comment, I will also mention that it doesn't a priori rule out scaling behaviour of the form $n^{\alpha}\log^{\beta} n$.)