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The $p$-adic Grand Canonical Ensemble

$p$-adic Electrostatics

See The Grand Canonical Ensemble for details about the general setup, but the basic idea distinguishing the canonical ensemble and the grand canonical ensemble is that in the latter we allow the number of particles to vary.

The grand canonical partition function can also be viewed as the generating function for the partition functions of the canonical generating function, and we write $$Z(t, \beta, \lambda) = \sum_{N=0}^N Z(N, \beta, \lambda) \frac{t^N}{N!}.$$ Note that the $t$ in $Z(t, \beta)$ distinguishes it from the canonical partition function $Z(N, \beta)$.

Theorem: $Z(N, \beta, \lambda) = Z(N, \beta, \lambda_1)^p.$


We use the theorem in The $p$-adic Canonical Ensemble, $$\frac{1}{N!} Z(N, \beta, \lambda) = \sum_{n_0 + \cdots + n_{p-1} = N} \prod_{j=0}^{p-1} \frac{1}{n_j!} Z(n_j, \beta, \lambda_1).$$ The sum here is over all vectors of non-negative integers $\mathbf n = (n_0, \ldots, n_{p-1})$ summing to $N$. It follows that $$Z(t, \beta, \lambda) = \sum_{N=0}^N t^N \sum_{n_0 + \cdots + n_{p-1} = N} \prod_{j=0}^{p-1} \frac{1}{n_j!} Z(n_j, \beta, \lambda_1).$$ We can slide the $t^N$ into the product by noting that $t^N = t^{n_0} \cdots t^{n_{p-1}}.$ We find, $$Z(t, \beta, \lambda) = \sum_{N=0}^N \sum_{n_0 + \cdots + n_{p-1} = N} \prod_{j=0}^{p-1} \frac{t^{n_j}}{n_j!} Z(n_j, \beta, \lambda_1).$$ We now are simply summing over all non-negative integers $n_0, n_1, \ldots, n_{p-1}$ as the sum over $N$ frees the restriction on the coordinates of $\mathbf n$. That is, $$Z(t, \beta, \lambda) = \sum_{n_0} \cdots \sum_{n_{p-1}} \prod_{j=0}^{p-1} \frac{t^{n_j}}{n_j!} Z(n_j, \beta, \lambda_1),$$ and the theorem follows.

This theorem has a simple physical explanation. The logarithm of the partition function is called the grand canonical potential and it represents the expected potential energy contained in the system. If we join two non-interacting systems then the grand canonical potential of the joined system is the sum of the grand canonical potentials of the constituent systems. From our discussion on the $p$-adic canonical ensemble, we know that particles in different cosets of $p\mathbb Z_p$ do not interact. In that ensemble the only dependence between the systems in each coset is the sum constraint $n_0 + \cdots + n_{p-1} = N$. When we eliminate that constraint the systems in each coset become independent and the grand canonical potential of the system consisting of all cosets is the sum of the grand canonical potentials for each coset. Exponentiating returns the identity in the theorem.

Lemma: $$Z(t, \beta, \lambda_1) = \sum_{N=0} \frac{t^N}{N!} p^{-\beta{N \choose 2} – N} Z(N, \beta, \lambda).$$


If $y \in p \mathbb Z_p$ then there exists $x \in \mathbb Z_p$ such that $y = p x$. Because $\lambda$ is a Haar measure, $d \lambda(p x) = |p| d\lambda(x) = p^{-1} d\lambda(x)$. Moreover, $d\lambda^N(p\mathbf x) = p^{-N} d\lambda^N(\mathbf x).$ By homogeneity, $$\prod_{m < n} |p x_n – p x_m|^{\beta} = p^{-\beta {N \choose 2}} \prod_{m < n} |x_n – x_m|^{\beta}.$$ Putting this together we see $$ $$

Corollary: $$\sum_N \frac{t^N}{N!} Z(N, \beta, \lambda) = \left( \sum_N \frac{t^N}{N!} p^{-\beta {N \choose 2} – N} Z(N, \beta, \lambda) \right)^p.$$


It is worth exploring a couple easy examples using physical intuition and mathematical reckoning. Physically, the parameter $\beta$ represents the inverse temperature of the system. When $\beta = \infty$, we are at absolute zero and we expect the system to stay in the state with lowest energy.

Proposition: $$Z(t, \infty, \lambda) = \left(1 + \frac{t}{q}\right)^p.$$

Proof. The integrand is non-zero only when there is at most one particle in each coset, and in this situation the integrand is identically one. Thus, $$Z(t, \infty, \lambda) = \sum_{N=0}^p Z(N, \infty, \lambda) {p \choose N} \left(\frac{t}{p}\right)^N.$$ The sum is over the number of particles is restricted to $N=0, \ldots, p$ to ensure there are at most one particle in each coset. The ${p \choose N}$ term arises when we choose which cosets out of $p$ we distribute our $N$ particles into. This collection of occupied states has volume $p^{-N}$. Finally, it is trivial to verify that $Z(N, \infty, \lambda)=1$ so we have $$Z(t, \infty, \lambda) = \sum_{N=0}^p {p \choose N} \left(\frac{t}{p}\right)^N = \left(1 + \frac{t}{p}\right)^p.$$

We may also use our theorem to evaluate $Z(t, \infty, \lambda) = Z(t, \infty, \lambda_1)^p$. It is easy to see that $Z(N, \infty, \lambda_1) \neq 0$ only when $N = 0$ or $N=1$, otherwise there are two particles in $p \mathbb Z_p$ and the distance between those points is less than 1. A quick calculation reveals $$Z(t, \infty, \lambda_1) = Z(0, \infty, \lambda_1) + t Z(1, \infty, \lambda_1) = 1 + \frac{t}{p}.$$

In the other direction, we can look at $\beta = 0$ which represents infinite temperature. In this situation we expect that the thermal energy far outweighs the interaction energy between particles, and thus their positions should be independent. The following proposition puts this observation into mathematical form.

Proposition: $Z(t, 0, \lambda) = e^t.$

Proof. When $\beta = 0$ the integrands for each of the $Z(N, \beta, \lambda)$ are equal to 1. And hence, because the integral is with respect to the probability measure $\lambda$, we get $$Z(t, 0, \lambda) = \sum_{N=0}^N \frac{t^N}{N!} = e^t.$$

The theorem is trivial to verify in this case as $e^t = (e^{t/p})^p.$

Some Basic Probabilities

If we want an explicit description of the event “all particles in $p \mathbb Z_p$” then we may describe it as the disjoint union $$\Omega(p \mathbb Z_p) := \bigsqcup_N (p \mathbb Z_p)^N,$$ and thus $$\mathbb P( \Omega(p \mathbb Z_p) ) = \frac{Z(t, \beta, \lambda_1)}{Z(t, \beta, \lambda)}.$$

More generally, given any Borel set $B$ we define $\Omega(B) = \bigsqcup_N B^N$ to be the event “all particles are in $B$”. Then the probability $\Omega(B)$ can be computed as $$\mathbb P(\Omega(B)) = \frac{Z(t, \beta, \boldsymbol 1_B \lambda)}{Z(t, \beta, \lambda)}.$$ When $B$ is a ball we can explicitly evaluate this probability using the related probability from the canonical ensemble.

Proposition: Suppose $B$ is a ball of radius $p^{-r}$ in $\mathbb Z_p$. Then $$\mathbb P(\Omega(B)) =\frac{1}{Z(t, \beta, \lambda)} \sum_{N=0}^{\infty} \frac{t^N}{N!} p^{-r{N \choose 2} – Nr} Z(N, \beta, \lambda).$$

This leads to the amusing result for gap probabilities. The gap probability for a ball $B$ is the probability that it contains no particles, $\mathbb P\{N_B = 0 \}$. The $p$-power in the theorem came from our distribution of particles into all $p$ cosets of $p \mathbb Z_p$. If we distribute the particles into all but one coset, then a $p-1$ power would arise instead.

Proposition: Let $B$ be a coset of $p \mathbb Z_p$ then $$\mathbb P\{N_B = 0 \} = \frac{1}{Z(t, \beta, \lambda_1)}.$$

Suppose $A_N \in \mathcal B(\mathbb Z_p^N)$, and define $A = A_1 \sqcup A_2 \sqcup \cdots.$ $A$ is a generic event of interest in the grand canonical ensemble.

Lemma: $$\mathbb P(A) = \frac{1}{Z(t, \beta, \lambda)} \sum_N \frac{t^N}{N!} Z(N, \beta, \boldsymbol 1_{A_N} \lambda^N).$$

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