# Quantum phase transitions in a finite lattice

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Sachdev begins his book on Quantum Phase Transitions by asserting that, for a system on a finite lattice, the ground state energy of a Hamiltonian H(g) (where g is some coupling) is a smooth, analytic function of g. This paper: https://arxiv.org/pdf/1603.03943.pdf seems to suggest an exception to this rule in quantum optics, and mentions a similar paper dealing with a QPT in the finite Rabi model.

Why must a finite lattice model have a ground state energy which is an analytic function of the coupling? Is Sachdev making some implicit assumptions that are violated in these studies? How does one show in general that a system possessing a level crossing in the thermodynamic limit opens up a spectral gap when taken to a finite lattice?

This post imported from StackExchange Physics at 2018-09-23 14:29 (UTC), posted by SE-user Alex Buser

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This is a very general argument: On a finite lattice with finitely many degrees of freedom, the Hamiltonian becomes a finite-dimensional matrix. The spectrum of a finite matrix is always a set of isolated points, and we can apply Cauchy's formula to the resolvent to obtain the projector onto the lowest eigenspace. If the Hamiltonian $H(g)$ is analytic in $g$, then the resolvent and hence the projector are also analytic in $g$.

So, strictly speaking, non-analyticity of the ground state can only be seen in the limit of infinitely many degrees of freedom. But it is certainly possible for a sequence of analytic functions (one for each system size) to converge pointwise to a function that is no longer analytic (the infinite system limit). (Example: $\lim_{n\to\infty} x^n$ on the interval $[0,1]$).

answered Sep 24, 2018 by (775 points)

It's worth mentioning that the examples considered by Hwang et al. always involve harmonic oscillators, i.e. infinite dimensional systems, for which the Hamiltonian cannot be written as a finite-dimensional matrix even if the number of lattice sites is just one (see e.g. https://arxiv.org/abs/1503.03090). The dimension of the Hilbert space is more important than the physical volume.

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