# The result

As a mathematician I am currently struggling to understand Tóth's Phase transition in an interacting Bose system. An application of the theory of Ventsel' and Freidlin. The free energy (for bosons) is defined as $$\beta f_{0,B}(\rho,\beta):=-\lim_{N,V\to\infty\atop N/V\to\rho}\frac{1}{V}\log Z(\beta,N,V),$$ with $$Z(\beta,N,V)$$ being the trace on the symmetric subspace of the Hamiltonian describing $$N$$ non-interacting bosons on the complete graph $$\{1,2,\dots,V\}$$. (I.e. the $$N$$-fold tensorised Laplacian $$-\Delta$$ defined by $$-\Delta \phi(x):=\frac{1}{V}\sum_{y=1}^V(\phi(x)-\phi(y))$$.)

Now it is pointed out that one gets "for Bose-Einstein statistics" that $$\beta f_{0,B}(\rho,\beta)=-(\rho+1)\log(\rho+1)+\rho\log\rho+\beta\rho\quad\text{if } \rho<\rho_c:=\frac{1}{e^\beta-1}$$ and $$\beta f_{0,B}(\rho,\beta)=-(\rho_c+1)\log(\rho_c+1)+\rho_c\log\rho_c+\beta\rho_c\quad\text{if }\rho\geq \rho_c.$$

# My attempt

I must have some misunderstanding of the physics part though because that's not what I get at all:

Following the proof of the Feynman-Kac formula (see e.g. here), setting the interaction potential to 0, I get that \begin{align*}Tr_{sym}e^{-\beta H_{V,N}} &= \frac{1}{N!}\sum_{\sigma\in S_N}\sum_{x_1,\dots,x_N\in V}\mathbb P(X_1(\beta)=\sigma(x_1),\dots,X_N(\beta)=\sigma(x_N)|X_1(0)=x_1,\dots,X_N(0)=x_N)\\ &={V\choose N}\mathbb E_\sigma\mathbb P(X_1(\beta)=\sigma(1),\dots,X_N(\beta)=\sigma(N)|X_1(0)=1,\dots,X_N(0)=N), \end{align*} where $$X_i$$ are independent continuous time simple random walks on the complete graph $$\{1,\dots,V\}$$ (with rate 1) and $$\mathbb E_\sigma$$ is the expectation w.r.t. the uniform measure on random permutations $$\sigma\in S_N$$.

It's not hard to carry out explicit calculations, but intuitively it is already clear this will not give the desired result because of the following: While $$V\choose N$$ (in the $$N/V\to\rho$$ limit) grows like some $$O(1)^V$$, the second term grows like $$O\left(\frac{1}{V}\right)^{N-O(1)}$$ (note that the walks are independent so the probability for each fixed $$\sigma$$ factorises, $$\sigma$$ has only finitely many fixed points a.s. and once a walk took a step -- this happens with positive probability -- the probability of reaching the desired point is basically $$1/V$$). Hence taking logarithm and dividing by $$V$$, this term would trivially go to $$-\infty$$.

I didn't manage to find a derivation of this result online, but I'd also be happy with a link. Of course I'd be extra grateful if somebody could also point out my mistake.