As stated in the title, why does a Heisenberg magnet break the $O(3)$ symmetry while degrees of freedom of the underlying spins are $SU(2)$?

  • 6
    $\begingroup$ While someone familiar with the Heisenberg model will almost surely know this, it is generally better to make the question self-contained. What is "the $\mathrm{O}(3)$ symmetry, and what is the supposed breaking mechanism? There's no need to have questions be one-liners, and a bit of context makes a question also accessible to people who might not off-hand remember how exactly the Heisenberg model works. $\endgroup$ – ACuriousMind Feb 28 '16 at 11:13

The Heisenberg model: $$H = \sum_{i \ne j} \mathbf{S}_i\cdot \mathbf{S}_j$$ has an $O(3)$ symmetry group rather than $SU(2)$ even though it is expressed in terms of spin $\frac{1}{2}$ operators $$\mathbf{S}_i = [\frac{\sigma^x_i}{2}, \frac{\sigma^y_i}{2}, \frac{\sigma^z_i}{2}]$$ It is because the Pauli-matrices transform according to the adjoint representation of $SU(2)$, and the adjoint representation of $SU(2)$ is not faithful (rather it is a faithful representation of an $SO(3)$ subgroup) because the element $$g_1 = \begin{pmatrix} -1&0\\ 0& -1 \end{pmatrix}$$ is represented by $1$ on the Pauli matrices, since they transform according $$g_1 \circ \sigma^a_i = g_1^{-1}\sigma^a_i g_1 = \sigma^a_i $$ Thus the largest $SU(2)$ symmetry subgroup of the Heisenberg model is $SO(3)$. However, the Heisenberg model is also symmetric under the parity operator: $$P\circ \sigma^a_i = -\sigma^a_i $$ This operator lies outside $SU(2)$ (it has a determinant of $-1$). It combines with $SO(3)$ to form an $O(3)$ group which is the full symmetry group of the Heisenberg model.

  • $\begingroup$ Then does the spin nematic break the $SU(2)$ or $O(3)$ symmetry? $\endgroup$ – R.Wigner Feb 29 '16 at 22:29
  • 2
    $\begingroup$ @hongchaniyi Spin nematic order breaks O(3) symmetry too. You just need to remember, a spin system does not have SU(2) symmetry in any case. You can talk about SU(2) only for electron systems where the symmetry is fractionalized in some sense. $\endgroup$ – Everett You Feb 29 '16 at 23:22
  • 1
    $\begingroup$ @hongchaniyi Indeed, the point symmetry group of a nematic state is $D_{\infty h}$ , a subgroup of $O(3)$. $\endgroup$ – David Bar Moshe Mar 2 '16 at 9:57
  • $\begingroup$ Wouldn't it be more standard to call your operator the time-reversal rather than parity operator, since it flips vectors in spin rather than in real space? (I know this can be a subtle issue, since high-energy and condensed-matter physicists sometimes use different definitions for the time-reversal operator.) $\endgroup$ – tparker Jun 4 '18 at 3:45
  • 1
    $\begingroup$ @Quantum spaghettification, yes the group acts on the generators of its Lie algebra (in any representation) by means of the adjoint representation, because the Lie algebra as a vector space forms a basis of the adjoint representation. The infinitesimal version of this action is just the commutation relations. $\endgroup$ – David Bar Moshe Jun 4 '18 at 14:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.