Why is the ground state of the ferromagnetic tetrahedron threefold degenerate?

Question

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.

Answer 1

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 !

                        Cheers,

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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.

Answer 2

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...