9.3: Two indistinguishable particles with spin 1/2

If we have two identical fermions of spin 1/2, confined in the same region, what is the appropriate wavefunction? In the scattering case we could measure spins far from the interaction, and if we knew that the total spins is conserved, spins can be associated with each particle. In the bound state we cannot tell which particle we are measuring, so the ket must contain both spin and spatial wavefunctions of both particles.

Assuming the spins do not interact, we can separate the two-particle spin wavefunction into $\sigma (1, 2) = \sigma_1\sigma_2$. We also know the appropriate one particle basis states $\uparrow_1, \downarrow_1, \uparrow_2, \downarrow_2$, where $\uparrow_1$ represents “particle 1” in spinor state $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$. The combinations for indistinguishable particles are then:

$$\uparrow_1\uparrow_2, \quad \downarrow_1\downarrow_2, \quad (\uparrow_1\downarrow_2 + \downarrow_1\uparrow_2)/ \sqrt{2}, \quad (\uparrow_1\downarrow_2 − \downarrow_1\uparrow_2)/ \sqrt{2} \nonumber$$

Operating on these with $\hat{P}_{12}$ yields eigenvalues 1, 1, 1 and -1 respectively. ${\bf S}^2 = S(S + 1)$ yields 2, 2, 2 and 0, $S_z$ yields 1,-1,0 and 0. Thus the demands of indistinguishability couples the spins of two identical particles into a triplet (S=1) and a singlet (S=0). The spin-1 vector has three possible $M_s$ component values - hence the triplet.