AbstractWe consider the case that four spin-3 atoms are confined in an optical trap. The temperature is so low that the spatial degrees of freedom have been frozen. Exact numerical and analytical solutions for the spin-states have been both obtained. Two kinds of phase-diagrams for the ground states (g.s.) have been plotted. In general, the eigen-states with the total-spin S (a good quantum number) can be expanded in terms of a few basis-states $$f_{S,i}$$
f
S
,
i
. Let $$P_{f_{S,i}}^{\lambda }$$
P
f
S
,
i
λ
be the probability of a pair of spins coupled to $$\lambda =0, 2, 4$$
λ
=
0
,
2
,
4
, and 6 in the $$f_{S,i}$$
f
S
,
i
state. Obviously, when the strength $$g_{\lambda }$$
g
λ
of the $$\lambda $$
λ
-channel is more negative, the basis-state with the largest $$P_{f_{S,i}}^{\lambda }$$
P
f
S
,
i
λ
would be more preferred by the g.s.. When two strengths are more negative, the two basis-states with the two largest probabilities would be more important components. Thus, based on the probabilities, the spin-structures (described via the basis-states) can be understood. Furthermore, all the details in the phase-diagrams, say, the critical points of transition, can also be explained. Note that, for $$f_{S,i}$$
f
S
,
i
, $$P_{f_{S,i}}^{\lambda }$$
P
f
S
,
i
λ
is completely determined by symmetry. Thus, symmetry plays a very important role in determining the spin-structure of the g.s..