Discussion
Addition of angular momentum states \(\vec{j}_{1} = 1/2\) and \(\vec{j}_{2} = 1/2\) will result in 4 states, of which number of linearly dependent states with magnetic quantum number m = 0 is :
Answer: C
The keyed count is 2.
Coupling two spin-1/2 angular momenta gives total j = 1 (a triplet, m = 1, 0, -1) and j = 0 (a singlet, m = 0), four states in all. Two of these have m = 0: the triplet state \(|1, 0\rangle\) = \((|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle)/\sqrt{2}\) and the singlet \(|0, 0\rangle\) = \((|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)/\sqrt{2}\).
Both are built as linear combinations of the same two product states \(|\uparrow\downarrow\rangle\) and \(|\downarrow\uparrow\rangle\), which is the sense of the question; as states they are orthogonal to each other. The m = 1 and m = -1 states are each a single product state, so the number of m = 0 states is 2, not zero, 1 or 4.
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