The Hamiltonian for the spin 1/2 ferromagnetic Heisenberg spin chain is $H=-J\sum_i \vec \sigma_i \cdot \vec\sigma_{i+1}$ with $J>0$ and $\vec\sigma_i$ the Pauli matrices acting on ith lattice site.

The state with spin up at every lattice site is an eigenstate. This can be seen since clearly it is an eigenstate of the $-J \sigma^z_i \sigma^z_{i+1}$ term and the $-J(\sigma^x_i \sigma^x_{i+1}+\sigma^y_i \sigma^y_{i+1})$ terms cancel acting on it. And of course there is nothing special about the z-axis. The state with all spin up belongs to the irreducible representation with net spin $N/2$ where $N$ is the number of lattice sites. Any state in this space should have the same energy by rotational symmetry.

Now my question is what happens in the thermodynamic limit at zero temperature? I thought that the Mermin-Wagner-Coleman theorem precludes spontaneous symmetry breaking even at zero temperature for a quantum system with one spatial dimension. Canonical examples include the 1+1 dimensional sigma models on a sphere which are closely related to the classical Heisenberg model in 2D.

But this model seems to have obvious spontaneous symmetry breaking for any finite $N$. Is there a loophole here or is there something technical going on in taking the thermodynamic limit that restores symmetry?

  • 2
    $\begingroup$ First of all there is a semantic issue: depending on whether you ask Anderson or Peierls, the Heisenberg ferromagnet does (not) have spontaneous symmetry breaking. But anyway, the usual Mermin-Wagner result does apply to classical ferromagnets in 2D at finite temperature. Usually we invoke the quantum-classical correspondence to relate this to zero-temperature 1D quantum systems. However, the 1D FM quantum Heisenberg chain is special cause the ground state has zero entanglement. So I'd guess that tis classical analogue is in fact at zero temperature, and hence Mermin-Wagner does not apply. $\endgroup$ May 8, 2018 at 12:55
  • 2
    $\begingroup$ Note that it is different to the 1D AFM quantum Heisenberg chain: that has a lot of entanglement in the ground state, hence it would correspond to a 2D classical system at finite-temperature, such that Mermin-Wagner applies and hence indeed the 1D system has no ordering. The FM is thus very special in that there is a complete absence of entanglement. $\endgroup$ May 8, 2018 at 12:57


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.