Allowed Total-J Values

The range of allowed totals

When two angular momenta $j_1$ and $j_2$ combine, the total quantum number $j$ does not take a single value. It ranges over

stepping by one. This is the triangle rule: $j$, $j_1$, and $j_2$ must be able to form the sides of a (possibly degenerate) triangle. The maximum, $j_1 + j_2$, is the stretched case of maximally aligned momenta; the minimum, $|j_1 - j_2|$, is the maximally anti-aligned case.

Why the maximum is $j_1 + j_2$

The largest possible $z$-projection is $m_{\max} = j_1 + j_2$, achieved by the single uncoupled state $|j_1, j_1; j_2, j_2\rangle$. A state of total $j$ has projections up to $m = j$, so the largest total present is exactly $j = j_1 + j_2$. Lowering this stretched state generates the full $j_1 + j_2$ multiplet. The remaining states reorganize into successively smaller $j$, each new family appearing because the multiplicity of a given $m$ in the uncoupled basis exceeds the number already accounted for.

Dimension counting as a check

The reorganization cannot create or destroy states, so the multiplet sizes must add up to the dimension of the combined space:

This identity is a reliable sanity check on any addition problem. For two spin-$\tfrac12$, the right side is $2\cdot2 = 4$, and the left side is $(2\cdot1+1) + (2\cdot0+1) = 3 + 1 = 4$ — the triplet plus the singlet, as we found directly.

Worked example: $j_1 = 1,\ j_2 = \tfrac12$

This is the coupling of an orbital $p$-state to a spin-$\tfrac12$ electron. The allowed totals are

The multiplets have sizes $2\cdot\tfrac12+1 = 2$ and $2\cdot\tfrac32+1 = 4$, summing to $6$, which matches $(2\cdot1+1)(2\cdot\tfrac12+1) = 3\cdot2 = 6$.

Try it

This is a numerical exercise — your code should return a number. For $j_1 = 1$, $j_2 = \tfrac12$, loop over the allowed $j$ values, sum the multiplet sizes $2j+1$, and return the total. Confirm it equals the uncoupled-space dimension.

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