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  Quantum phase transitions in a finite lattice

+ 6 like - 0 dislike

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
asked Sep 21, 2018 in Theoretical Physics by Alex Buser (30 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

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 Greg Graviton (775 points) [ no revision ]

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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