I'm preparing a presentation on Spin-Ice, but something's been bugging me for a while. On the Wikipedia page for Geometrical Frustration, it says the following about easy spins on a tetrahedron with ferromagnetic interactions:

There are three different equivalent arrangements with two spins out and two in, so the ground state is three-fold degenerate.

I just don't understand why.

  • We're considering easy spins with ferromagnetic interactions, so any ground state needs 2 spins pointing towards the center (in) and 2 spins pointing away from the center (out).
  • Given a tetrahedron, I can think of 6 different ways to distribute 2 spins in and 2 spins out on the vertexes. $(iioo,ioio,oiio,iooi,oioi,ooii)$
  • You can rotate any of the above states to reproduce another, which might mean they are not different.

But then, why does wiki state there are three different arrengements?

Note Each of the above configurations has a total magnetic moment in a different direction. We can place our $(x,y,z)$ axis so that each one of the first three cases has a total moment in the positive direction of one of the axis, and each of the last three has a total moment in the negative direction of one of the axis.

  • $\begingroup$ I have to say I am also puzzled. Wikipedia article probably implicitly allows some symmetries and not others to identify some (but not all) of the six possibilities. $\endgroup$
    – Marek
    Dec 15, 2010 at 15:03
  • $\begingroup$ Just a random thought: if you add up the spin vectors, you'll obtain some total magnetization vector. There are in total three possibilities for its axis (if you forget orientation). So you factor the original 6 element set by the inversion of $i \leftrightarrow o$. Perhaps this is what wikipedia is talking about. $\endgroup$
    – Marek
    Dec 15, 2010 at 15:08
  • $\begingroup$ @Marek: Perhaps. It's strange that it would consider states reachable by rotation as different, but states reachable by i-o inversion as equivalent. $\endgroup$
    – Malabarba
    Dec 15, 2010 at 16:16
  • $\begingroup$ i-o inversion is not equivalent if they assume an external magnetic field to be present. Is that the case? $\endgroup$ Dec 15, 2010 at 16:28
  • $\begingroup$ @Raskolnikov: I believe not. Judging from the Wikipedia page, it seems they are considering a single tetrahedral cell, with only internal energies. But if i-o inversion was an equivalency, why wouldn't rotations be as well (thus making all 6 states equivalent). $\endgroup$
    – Malabarba
    Dec 15, 2010 at 19:32

2 Answers 2


What you have to take into account here is the discrete rotational symmetries of the tetrahedron. For instance let us write the state of the tetrahedron as $\mid i_1, i_2, i_3, i_4\rangle $ where $i_k$ is the spin on the $k^{\textrm{th}}$ vertex. The state in the figure you show above can then be written as $\mid o,i,i,o \rangle$ (with $o$ and $i$ meaning "outward pointing" and "inward pointing" respectively).

In the absence of any anisotropies which break the rotational symmetry, the state $\mid i_3, i_1, i_2, i_4\rangle $ can be obtained from the state $\mid i_1, i_2, i_3, i_4\rangle $ by rotating the tetradhedron by $2\pi/3$ around the axis passing through the 4th vertex ($v_4$) and the center of triangle $\Delta_{123}$, i.e.:

$$ \mid i,o,i,o \rangle = \hat R_4(2\pi/3) \mid o,i,i,o \rangle $$

where $\hat R_i (\theta)$ is the operator for rotations by $\theta$ around the $i^\textrm{th}$ axis.

alternatively you can also obtain $\mid i,o,i,o \rangle$ by performing a reflection across the axis passing through $v_3$ and bisecting the edge ($e_{12}$) between $v_1$ and $v_2$:

$$ \mid i,o,i,o \rangle = \hat S_{123} \mid o,i,i,o \rangle $$

where $\hat S_{ijk} $ is the generator of reflections through the axis passing through $v_k$ and bisecting the edge ($e_{ij}$). Similarly we have:

$$ \mid i,i,o,o \rangle = \hat R_4(4\pi/3) \mid o,i,i,o \rangle $$

Thus, w.r.t these discrete symmetries the six-states you mention are not independent. We must take suitable linear combinations of these states to obtain a set of independent basis vectors which are invariant under the action of these symmetries. When you do this correctly the six states will reduce to three states:

$$ \mid \Psi_4 \rangle = \frac{1}{\sqrt{3}}\left(\mid v_1, v_2, v_3, v_4\rangle + \mid v_3, v_1, v_2, v_4\rangle + \mid v_2, v_3, v_1, v_4\rangle \right) $$

and likewise for $ \mid \Psi_3 \rangle $ and $ \mid \Psi_2 \rangle $. There are only three such states, and not four (we have four triangles), because the fourth state (in this case $\mid \Psi_1 \rangle$ ) can be written as a linear sum of the other three !


Edit: Following a suggestion by @bruce, just want to clarify that each $ \mid \Psi_i \rangle $ is invariant only under the action of the permutation group on the triangle dual (opposite) to the vertex $v_i$. This is a subgroup of the full symmetry group of the tetrahedron.

  • $\begingroup$ @space_cadet: Is the same true for classical rotation? I can really take a tetrahedron and rotate it to all six states listed by OP. So the degeneracy should be one now. $\endgroup$
    – unsym
    Dec 16, 2010 at 1:45
  • $\begingroup$ @hwlau can you write down some linear combination of the six states which is invariant under the discrete symmetries of the tetrahedron? that is only way you'll get one non-degenerate state. $\endgroup$
    – user346
    Dec 16, 2010 at 2:05
  • $\begingroup$ So, what you're saying is that two states can only be considered different if they cannot be obtained from one another through symmetry operations? I think your last paragraph is just a little misleading. It sounds like each of the $|\Psi_i\rangle$ states are invariant under any of the simmetries, but I think you mean each of them is invariant under the $R_i$ rotations. Also, it sounds like they are only linearly independent, while it's also important the reader realizes they are also independent w.r.t. rotations. $\endgroup$
    – Malabarba
    Dec 16, 2010 at 4:01
  • $\begingroup$ Also I think it's "plane" instead of "axis" on the sentence reflection across the axis. Other than that, this answer really hit the spot. Thanks =) $\endgroup$
    – Malabarba
    Dec 16, 2010 at 4:03
  • 1
    $\begingroup$ That is the traditional recipe - for eg. we identify elementary particles with the irreps of a group. You cannot use a group operation to go from one irrep to another, right? And yeah each $|\Psi_i\rangle$ is invariant only under the permutation group of the triangle formed by the remaining three vertices. I guess I should edit the answer a little. Thanks ! $\endgroup$
    – user346
    Dec 16, 2010 at 4:06

It also bugs me that the description of the triangular lattice in the Geometric frustration

The third spin cannot simultaneously minimize its interactions with both of the other two. Thus the ground state is twofold degenerate.

Within all eight state, except all up and all down, the ground state is should be six folds degenerate, instead of two folds. I guess something wrong here.

In general, it is uncommon to consider the rotational symmetry of a lattice because the counting cannot be scaled for large $N$. So I guess they may consider the following symmetry:

$$ s_i \rightarrow -s_i $$

If you consider the Hamiltonian of a antiferromagnetic Ising model or Heisenberg model: $$H=J\sum_{i,j}\mathbf{s}_{i}\cdot\mathbf{s}_{j}$$ This Hamiltonian is invariant under the above symmetry, so every state $\{s_i\}$ has the corresponding state $\{-s_i\}$. This symmetry should correspond to the $i\leftrightarrow o$ inversion in your counting.

Here is a reason why the above symmetry is considered. Considering the magnetization of a system: $$M=\sum_i s_i$$ However, because of the symmetry above, the ensemble average of this value is always 0 for all spin model with no external magnetic field. A remedy to this situation is to add a small external magnetic field to break the symmetry. Another one might be considering only half of the ensemble under the symmetry...

  • $\begingroup$ "In general, it is uncommon to consider the rotational symmetry of a lattice because the counting cannot be scaled for large N." - but that is exactly what will give you the correct degeneracy. For the triangular lattice, as you say, we have six states. But three of those are equivalent under permutations of the vertices and likewise for the other three. So you have take linear combinations and you get two states in the end. $\endgroup$
    – user346
    Dec 16, 2010 at 1:17
  • $\begingroup$ @space_cadet: I think that there should be three states, not two states written in the wiki. $\endgroup$
    – unsym
    Dec 16, 2010 at 1:43
  • $\begingroup$ that's fine, but you have to give some reason for you believe that. $\endgroup$
    – user346
    Dec 16, 2010 at 1:58
  • $\begingroup$ In the end, it's a matter of what constitutes different states. The wiki is talking about a single cell, so it (implicitly) defines different states as being those that can't be obtained from one another from symmetry operations (in the case of the triangle, they are just the permutations). Like you said, there are 6 ground states. 3 of them have a total spin of +1, and you can change between them by permuting a pair of vertices (so they are not different). The other three have a total spin of -1, and the same is true for them. $\endgroup$
    – Malabarba
    Dec 16, 2010 at 4:12
  • $\begingroup$ These two states that are left are considered different, because one has positive spin and the other has negative. $s\rightarrow-s$ is indeed a symmetry of the Hamiltonian, but that only implies there is degeneracy in energy, it doesn't mean those 2 states are equal. They are different because they have different total spins, and the switch $s\rightarrow -s$ has nothing to do with triangle simmetries, it simply transforms a state into another different state (with same energy). $\endgroup$
    – Malabarba
    Dec 16, 2010 at 4:18

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