To understand this question, one can make use of the fact that that QM is equivalent to 1D statistical mechanics. The relation is, that the trace over the quantum mechanical time evolution operator taken at imaginary time corresponds to the (canonical) partition function of an 1D statistical mechanics system at finite temperature

\(Z(\beta) = \text{trace} \{ U(i\beta)\} = \sum\limits_{\alpha}\langle \alpha | e^{-\frac{H}{k_BT}}|\alpha\rangle = \sum\limits_{\alpha}e^{-\beta\epsilon_{\alpha}}\)

In statistical mechanics, a phase transition corresponds to a qualitative change in the behavior of the many particle system. The partition sum diverges and the correlation length becomes infinite.

For the example of an 1D Gaussian model

\(Z = \text{Tr}(q)\prod\limits_{j=1}^N\exp(w(q_j,q_{j+1}))\)

with

\(w(q,q') = -\frac{1}{4} (q^2 + q'^2) + Kqq'\)

it can be shown by applying the transfer matrix method that the correlation length is given by

\(\xi = \frac{a}{\epsilon_1 -\epsilon_0}\)

where $a$ corresponds to the lattice spacing and $\epsilon_0$ and $\epsilon_1$ are the energy of the ground and first excited state of the corresponding quantum system respectively.

From this one can see, that the correlation length becomes infinite as the ground state gets degenerated.

\((\epsilon_1 - \epsilon_0) \rightarrow 0\)

Physically, this means in this example that a symmetry of the quantum system gets restored.