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Suppose that I have three non-interacting spin-1/2 particles such that I can represent the combined system in a basis of

\begin{align} D^{(1/2)}_1 \otimes D^{(1/2)}_2 \otimes D^{(1/2)}_3 & =\left(D^{(1)}_{12} \oplus D^{(0)}_{12}\right)\otimes D^{(1/2)}_3 \\ & = \left(D^{(1)}_{12} \otimes D^{(1/2)}_3\right) \oplus \left(D^{(0)}_{12} \otimes D^{(1/2)}_3\right) \\ & =D^{(3/2)}_{123} \oplus D^{(1/2)}_{123} \oplus D^{(1/2)}_{123}. \end{align}

Given a particular Hamiltonian, how does one then calculate the energy eigenvalues using this group theory representation of the system's basis?

Also, how does one distinguish between recurring terms in the group theory representation? For example, there are two $$D^{(1/2)}_{123}$$ terms above. What does this mean physically?

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  • $\begingroup$ It's not pretty clear what the $D_i^{(1/2)}$ are; are they the Hilbert spaces where the particles live? However, whenever the Hamiltonian is the direct sum of single non-interacting Hamiltonians the corresponding eigenvalues are always the sum of the single Hamiltonian eigenvalues. $\endgroup$ – gented Dec 11 '15 at 8:24
  • $\begingroup$ Yes they are the Hilbert spaces where the particles live. How does this representation via the direct sums relate to the Hamiltonian? $\endgroup$ – Loonuh Dec 11 '15 at 8:27
  • $\begingroup$ The Hamiltonian must be given, it is not something that one can derive (in fact different Hamiltonians may correspond to the same representations). $\endgroup$ – gented Dec 11 '15 at 9:59
  • $\begingroup$ Closely related: Adding 3 electron spins. It turns out it is much harder to find bases for the $D^{(1/2)}_{123}$ factors which fully respect the electron exchange symmetry than one would think. $\endgroup$ – Emilio Pisanty Jan 4 '16 at 15:52
  • $\begingroup$ Racah coeffs. $\endgroup$ – Cosmas Zachos Oct 29 '16 at 0:55
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If the particles are not interacting, using the decoupled basis where states are direct products $$\vert \textstyle\frac{1}{2} m_1\rangle \vert \textstyle\frac{1}{2} m_2\rangle \vert \textstyle\frac{1}{2} m_3\rangle $$ or the coupled basis $$ \vert \textstyle\frac{1}{2}M;(j_1j_2)j_{12}j_3\rangle\, ,\qquad \vert \textstyle\frac{3}{2}M\rangle $$ with $j_{12}=0,1$, does not make a difference.

Once cannot distinguish the two $j=\textstyle\frac{1}{2}$ using angular momentum arguments alone, i.e. by rotating the states or using $\hat L_\pm$ and $\hat L_z$.

Of course it is possible to make the copies apart using operations that are not related to rotations, and in this specific case these operations are those of the permutation group. In particular, the states with $j_{12}=0$ will be antisymmetric under permutation of the first and second particles, while the states with $j_{12}=1$ will be symmetric under such permutations. For other permutations the states will transform one into a linear combination of the two.

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