# Does H = XX+YY spontaneously break symmetry in 1D?

Hello,

I am working in 1D here. For the ferromagnetic Ising model ##H = -\sum_k X_k X_{k+1}## (or ##H = -YY##) we know that the ground state is gapped and has a twofold degeneracy due to SSB (spontaneous symmetry breaking) of the spin flip symmetry ##P = Z_1 Z_2 Z_3 \cdots##.

I am now interested in the Hamiltonian ##H = -XX - YY##. This is known to be gapless (as can be derived using a Jordan-Wigner transformation). However, is it known whether or not this displays spontaneous symmetry breaking? Note that it has a continuous symmetry ##SO(2) = U(1)##, and I am not asking whether it continuously breaks in 1D (as Coleman-Mermin-Wagner implies that does not happen) but rather whether there is still the discrete SSB similar to what happens for the above ##XX## Ising case. Moreover, how can one show it? (Analytically? Numerically?)

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

Physics Monkey
Homework Helper
Hey nonequilibrium,

This is the XY model. It is exactly solvable using free fermions and there is no symmetry breaking in the model. The fermion ground state is a Fermi gas which is translation invariant, time reversal invariant, and preserves spin rotation symmetry.

Note that, if I understood what you are suggesting correctly, it is not possible to "discretely" break a continuous symmetry. If you break a continuous symmetry you will get a Goldstone mode. You could imagine breaking some discrete symmetry, like time reversal or lattice translation, but that does not happen in this case.

Hope this helps.

DrDu