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Mermin-Wagner theorem and 2D magnetism

  1. Jul 31, 2013 #1
    I am a little confused about the Mermin-Wagner theorem, and the assumptions it makes, and I would like to better understand the 'exceptions' to the rule that occur in practical (non-ideal) scenarios.

    As is my understanding,
    The Mermin-Wagner theorem states that there cannot be spontaneous symmetry breaking in 2d (or less) systems at any finite temperature T > 0. Essentially what this means is that you cannot have, for example, ferromagetic order in 2D systems since the continuous rotational (and translational?) symmetry of the spins are broken by the formation of ordered domains where one direction is preferred over another. The reason for this is that, in such low-dimensionality systems, the energy cost to excite a defect/disordered in the lattice is infinitesimal small and thus infinitely many excited modes are created that destroy the order.

    But there are exceptions to the rule, so to speak. Since this theorem only applies to continuous symmetries, and only says that the ordered states are forbidden in 2 or fewer dimensions. In real life, though, it is is very difficult to achieve such ideal circumstances. 'Accidentally' symmetry breaking might occur to do a small biasing fields, the presence of a substrate that supports the 2d material, or the fact that a thin film still has a small, but finite, thickness and is therefor not really two dimensional.


    My question is about what constitutes enough of a symmetry breaking for the Mermin-Wagner theorem to not apply. Does the non-continuous symmetry from the lattice break the symmetry? Spins located at discretely spaced lattice points are not continuously translationally symmetric. And the lattice type should also break the symmetry as well, whether it is a square, triangular, honeycomb, … lattice, would break rotational symmetry. Is this enough? I think the answer is “no,” (explained below) but I am not exactly sure why.

    The Ising model is often used as an example of an 'exception' to the Mermin-Wagner theorem in that it has a discrete symmetry, and therefore has a 2d ferromagnetic state, compared to the 2d Heisenberg/XY model, where does not exhibit as ferromagnetic state in 2 dimensions. Both of these models involve spins being located at discrete lattice points, but while the Ising model constrains the spins to point only “up” or “down”, the Heisenberg model allows the spins to point any direction.
     
  2. jcsd
  3. Aug 2, 2013 #2
    Any help?
     
  4. Aug 2, 2013 #3

    king vitamin

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    It sounds like you're confusing the symmetry of the lattice to the internal spin symmetry of the interactions. The continuous symmetry in the Heisenberg model is the rotation of the spin at every single site by the same angle simultaneously. Obviously, since the model is basically defined to be O(3) symmetric with respect to short-range spin interactions, this is a continuous symmetry of the Hamiltonian. It has nothing to do with translational or point-group symmetries, which will generally be discrete. Looking at it this way, we see that the Ising model does not possess a continuous symmetry since the spins cannot be rotated, only flipped by a sign.
     
  5. Aug 3, 2013 #4
    By breaking the continuous O(3) rotational symmetry of the Heisenberg model, and forcing one of two discrete states in the Ising model, allows for a ferromagnetic state. Does it need to be such an extreme case of symmetry breaking, going from continuous to only two allowed states? What is I was to break the symmetry by allowing (for example) 360 possible spin orientations...one for each degree of rotation, but no allowed states between those, would that be enough? Is there a way to know what the minimum amount of symmetry breaking is required?
     
  6. Aug 3, 2013 #5

    king vitamin

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    I'm a little bit confused as to what you're asking. By saying that a continuous symmetry is broken, we mean that the ground state of the system is not invariant under the symmetry transformation. However, because the Hamiltonian is invariant under the symmetry, there must be a degeneracy of ground states which are mapped to each other by the symmetry transformation. (Often in the literature, one specifies "assume the ground state is unique" to specify the assumption that SSB does not occur).

    For definiteness, consider an O(3) symmetric spin model at T=0 with a ferromagnetic ground state. All of the spins are pointing in the same direction; take this direction to be the z-coordinate (this defines our coordinates). The hamiltonian is a function of the each spin, [itex]\mathcal{H} = \mathcal{H}(\vec{S}_1,\vec{S}_2,...,\vec{S}_N)[/itex]. By assumption, the state where [itex]\vec{S}_i = S\vec{e}_z[/itex] minimizes the hamiltonian, but we also assumed [itex]\mathcal{H}(R \vec{S}_i) = \mathcal{H}(\vec{S}_i)[/itex] where [itex]R \in O(3)[/itex]. Therefore, by rotating our spins by an arbitrary angle, ANY state where the spin of each site points in the same direction is a ground state since it also minimizes the hamiltonian. So we are not only allowing 2 spin states, or 360 spin states, but a degenerate manifold of spin ground states that can break the symmetry. We simply defined our coordinates in accordance with the decision that the system made (in practice one usually introduces a field H in the z-direction to break the symmetry in a preferred direction, and then take the limit H -> 0).
     
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