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 !

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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.

This post imported from StackExchange Physics at 2014-04-01 16:37 (UCT), posted by SE-user user346