Past exam of the quantum physics course of the University of Oxford / 2016 / 6

For two spin one-half particles, $s_1^2=s_2^2=3{\hslash }^2/4$. Squaring $\mathbf{S}={\mathbf{s}}_1+{\mathbf{s}}_2$ givesTake $a,b$ real so that $f$ is Hermitian. Since $[S^2,S_z]=0$, the displayed polynomial implies $[f,S^2]=[f,S_z]=0$. There is a common orthonormal eigenbasis, so all three observables can be simultaneously certain in a common eigenstate. Commutation does not mean that every quantum superposition has definite values of all three.

Suppress the fixed individual quantum numbers $s_1=s_2=1/2$. In the coupled basis, ordered on both rows and columns as $|1,1⟩,|1,0⟩,|1,-1⟩,|0,0⟩$, the total-spin eigenvalues giveHere the first three entries form the triplet state sector and the last is the singlet state. For the product basis, useThe first term has diagonal entries ${\hslash }^2(1,-1,-1,1)/4$. Each nonzero exchange term flips $|\uparrow \downarrow ⟩$ to $|\downarrow \uparrow ⟩$ or back, with coefficient ${\hslash }^2/2$. With rows and columns ordered as $|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩$,The change of basis is made explicit by $|1,0⟩=(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$ and $|0,0⟩=(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$, with $|1,\pm 1⟩=|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩$. The resulting eigenvalues reproduce the diagonal coupled matrix.