Two spins $\tfrac12$ make totals 0 and 1: three states with $S=1$, the triplet, and one with $S=0$, the singlet. The triplet is $\ket{\uparrow\uparrow}$, $(\ket{\uparrow\downarrow}+\ket{\downarrow\uparrow})/\sqrt{2}$ and $\ket{\downarrow\downarrow}$, all unchanged when the two spins are swapped. The singlet, $(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow})/\sqrt{2}$, changes sign under the swap. An interaction of the form $A\,\hat{\mathbf{S}}_{1}\cdot\hat{\mathbf{S}}_{2}/\hbar^{2}$ raises the triplet by $A/4$ and lowers the singlet by $3A/4$. In the ground state of hydrogen the electron’s and the proton’s spins interact this way; the gap between singlet and triplet, 5.9 μeV, gives the 21 cm radio line by which astronomers map the hydrogen of the galaxy.

Two identical bells ringing together can ring in step or exactly out of step. Swapping the bells leaves in-step ringing as it was and flips the out-of-step pattern: the triplet is in step, the singlet out of step.

Four states of two spins $\tfrac12$ under $A\,\hat{\mathbf{S}}_{1}\cdot\hat{\mathbf{S}}_{2}/\hbar^{2}$ with : the three triplet states rise by and the singlet falls by . Weighted by the number of states, the centre stays where it was.

Writing $\ket{\uparrow\downarrow}$ for $\ket{\uparrow}\ket{\downarrow}$, the coupled states of two spins $\tfrac12$ are

• $\hat{\mathbf{S}}_{1}\cdot\hat{\mathbf{S}}_{2}=\tfrac12(\hat{S}^{2}-\hat{S}_{1}^{2}-\hat{S}_{2}^{2})=\tfrac{\hbar^{2}}{2}\bigl[S(S+1)-\tfrac32\bigr]$: on the triplet, on the singlet.

The top state $\ket{\uparrow\uparrow}$ is alone at $M=1$, so it is $\ket{1,1}$. Lowering it with the total $\hat{S}_{-}$ gives , and lowering again gives . At there are two product states; the combination orthogonal to is the singlet.

The only product state with $M=1$.

$\hat{S}_{-}\ket{S,M}=\hbar\sqrt{S(S+1)-M(M-1)}\,\ket{S,M-1}$ with .

Each part lowers its own spin: $\hat{S}_{-}\ket{\uparrow}=\hbar\ket{\downarrow}$.

Equate the two and divide by $\hbar\sqrt2$.

The combination at $M=0$ orthogonal to $\ket{1,0}$, with the sign of $\ket{\uparrow\downarrow}$ chosen positive.

The dot product of the two spins is fixed by the total: half the difference between the total squared and the two parts squared.

The only combination of two spins $\tfrac12$ with total zero is the difference of the two mixed products, divided by $\sqrt{2}$.

Swapping the two spins turns $\ket{\uparrow\downarrow}$ into $\ket{\downarrow\uparrow}$. The triplet states are unchanged by the swap; the singlet changes sign.

With a coupling $A\,\hat{\mathbf{S}}_{1}\cdot\hat{\mathbf{S}}_{2}/\hbar^{2}$, the three triplet states share one energy and the singlet has another.

Great work! You now know how to:

• build the triplet and the singlet by lowering the top state
• find their energies under a coupling of the two spins
• tell them apart by their symmetry under a swap