Addition of Angular Momenta

In the previous posts, I treated the electron spin by itself. A hydrogen atom also contains a proton with spin $1/2$, so the next question is unavoidable: what states are available when two angular momenta are combined?

For a magnetic moment $\boldsymbol{\mu}=\gamma\mathbf{S}$, the gyromagnetic ratio $\gamma$ depends on the particle. The proton magnetic moment is much smaller than the electron magnetic moment, which is why proton-spin effects can be negligible in some approximations. That does not mean the proton has no spin. When spin-spin coupling matters, both spins must be included.

This post develops the addition of two spin-$1/2$ angular momenta. To keep the notation general, write

$$ \mathbf{J}=\mathbf{J}_1+\mathbf{J}_2, $$

with $j_1=j_2=1/2$. For the electron-proton spin pair in hydrogen, write $\mathbf{K}=\mathbf{I}+\mathbf{S}$, where $\mathbf{I}$ is the proton spin and $\mathbf{S}$ is the electron spin. In standard atomic notation the hyperfine total is $\mathbf{F}=\mathbf{I}+\mathbf{J}_{\mathrm e}$; for an $S$-state with zero orbital angular momentum, $\mathbf{J}_{\mathrm e}=\mathbf{S}$. The addition mathematics is the same.

One spin-½ system

For either constituent,

$$ \begin{aligned} \left|\uparrow\right\rangle &=\left|\frac12,\frac12\right\rangle =\begin{pmatrix}1\\0\end{pmatrix},\\[4pt] \left|\downarrow\right\rangle &=\left|\frac12,-\frac12\right\rangle =\begin{pmatrix}0\\1\end{pmatrix}. \end{aligned} $$

The arrows refer to eigenstates of the relevant $z$-component, not to tiny classical arrows literally pointing in space.

For a spin-½ particle,

$$ J_i^2\left|\frac12,m_i\right\rangle =\frac34\hbar^2\left|\frac12,m_i\right\rangle, \qquad J_{iz}\left|\frac12,m_i\right\rangle =m_i\hbar\left|\frac12,m_i\right\rangle. $$

In the $\{|\uparrow\rangle,|\downarrow\rangle\}$ basis, the component operators are

$$ J_x=\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad J_y=\frac{\hbar}{2}\begin{pmatrix}0&-i\\i&0\end{pmatrix},\qquad J_z=\frac{\hbar}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$

Their action on the basis states is

$$ \begin{array}{lll} J_x|\uparrow\rangle=\dfrac{\hbar}{2}|\downarrow\rangle, & J_y|\uparrow\rangle=\dfrac{i\hbar}{2}|\downarrow\rangle, & J_z|\uparrow\rangle=\dfrac{\hbar}{2}|\uparrow\rangle,\\[6pt] J_x|\downarrow\rangle=\dfrac{\hbar}{2}|\uparrow\rangle, & J_y|\downarrow\rangle=-\dfrac{i\hbar}{2}|\uparrow\rangle, & J_z|\downarrow\rangle=-\dfrac{\hbar}{2}|\downarrow\rangle. \end{array} $$

These relations are worth keeping nearby because they make the two-spin calculation almost mechanical.

The uncoupled basis

The tensor-product space has four natural basis states:

$$ \begin{aligned} |\uparrow\uparrow\rangle &=\left|\frac12,\frac12\right\rangle_1 \left|\frac12,\frac12\right\rangle_2,\\ |\uparrow\downarrow\rangle &=\left|\frac12,\frac12\right\rangle_1 \left|\frac12,-\frac12\right\rangle_2,\\ |\downarrow\uparrow\rangle &=\left|\frac12,-\frac12\right\rangle_1 \left|\frac12,\frac12\right\rangle_2,\\ |\downarrow\downarrow\rangle &=\left|\frac12,-\frac12\right\rangle_1 \left|\frac12,-\frac12\right\rangle_2. \end{aligned} $$

This is the uncoupled basis because each ket specifies $m_1$ and $m_2$ separately. It diagonalizes $J_{1z}$, $J_{2z}$, and therefore

$$ J_z=J_{1z}+J_{2z}. $$

However, the two mixed states $|\uparrow\downarrow\rangle$ and $|\downarrow\uparrow\rangle$ are not eigenstates of the total-$J^2$ operator.

Constructing the total-angular-momentum operator

Squaring $\mathbf{J}=\mathbf{J}_1+\mathbf{J}_2$ gives

$$ \begin{aligned} J^2 &=(\mathbf{J}_1+\mathbf{J}_2)^2\\ &=J_1^2+J_2^2+2\mathbf{J}_1\!\cdot\!\mathbf{J}_2\\ &=J_1^2+J_2^2 +2\left(J_{1x}J_{2x}+J_{1y}J_{2y}+J_{1z}J_{2z}\right). \end{aligned} $$

It is often cleaner to introduce $J_{i\pm}=J_{ix}\pm iJ_{iy}$. Then

$$ J^2=J_1^2+J_2^2+2J_{1z}J_{2z}+J_{1+}J_{2-}+J_{1-}J_{2+}. $$

Applying this operator to the uncoupled basis gives

$$ \begin{aligned} J^2|\uparrow\uparrow\rangle &=2\hbar^2|\uparrow\uparrow\rangle,\\ J^2|\uparrow\downarrow\rangle &=\hbar^2\left(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\right),\\ J^2|\downarrow\uparrow\rangle &=\hbar^2\left(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\right),\\ J^2|\downarrow\downarrow\rangle &=2\hbar^2|\downarrow\downarrow\rangle. \end{aligned} $$

Thus, in the ordered basis

$$ \mathcal B_{\mathrm u} =\{|\uparrow\uparrow\rangle, |\uparrow\downarrow\rangle, |\downarrow\uparrow\rangle, |\downarrow\downarrow\rangle\}, $$

the matrix is

$$ [J^2]_{\mathcal B_{\mathrm u}} =\hbar^2 \begin{pmatrix} 2&0&0&0\\ 0&1&1&0\\ 0&1&1&0\\ 0&0&0&2 \end{pmatrix}. $$

The middle $2\times2$ block,

$$ \hbar^2\begin{pmatrix}1&1\\1&1\end{pmatrix}, $$

has eigenvalues $2\hbar^2$ and $0$, with normalized eigenvectors proportional to $(1,1)^T$ and $(1,-1)^T$. Therefore the correct $m=0$ eigenstates are

$$ \frac{1}{\sqrt2}\left(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\right) \quad\text{and}\quad \frac{1}{\sqrt2}\left(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\right). $$

This change of basis is the small block-diagonalization problem hiding inside the calculation.

The coupled basis: triplet and singlet

For a simultaneous eigenstate $|j,m\rangle$,

$$ J^2|j,m\rangle=j(j+1)\hbar^2|j,m\rangle, \qquad J_z|j,m\rangle=m\hbar|j,m\rangle. $$

The four two-spin states become

$$ \begin{aligned} |1,1\rangle &=|\uparrow\uparrow\rangle,\\ |1,0\rangle &=\frac{1}{\sqrt2}\left(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\right),\\ |1,-1\rangle &=|\downarrow\downarrow\rangle,\\ |0,0\rangle &=\frac{1}{\sqrt2}\left(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\right). \end{aligned} $$

The three $j=1$ states form the triplet, and the single $j=0$ state is the singlet. In the coupled basis

$$ \mathcal B_{\mathrm c} =\{|1,1\rangle,|1,0\rangle,|1,-1\rangle,|0,0\rangle\}, $$

both $J^2$ and $J_z$ are diagonal:

$$ [J^2]_{\mathcal B_{\mathrm c}} =\hbar^2\operatorname{diag}(2,2,2,0), \qquad [J_z]_{\mathcal B_{\mathrm c}} =\hbar\operatorname{diag}(1,0,-1,0). $$

This is the central result: the four-dimensional product space decomposes as

$$ \frac12\otimes\frac12=1\oplus0. $$

The first three states are symmetric under interchange of the two spin labels, while the singlet is antisymmetric. For an electron and a proton, the particles are distinguishable, so this symmetry is a useful property of the spin wavefunction rather than an identical-particle exchange constraint.

A picture of the change in viewpoint

When we discussed one electron, the state label referred to that single particle.

Now the state label describes the proton-electron pair as one composite system.

The uncoupled labels $(m_1,m_2)$ and the coupled labels $(j,m)$ describe the same four-dimensional state space. They are simply two different orthonormal bases.

Ladder operators and the relative signs

The total ladder operators obey

$$ J_\pm|j,m\rangle =\hbar\sqrt{(j\mp m)(j\pm m+1)}\,|j,m\pm1\rangle. $$

Choose the standard Condon-Shortley phase convention and start from the highest-weight state

$$ |1,1\rangle=|\uparrow\uparrow\rangle. $$

Applying $J_-=J_{1-}+J_{2-}$ gives

$$ \sqrt2\hbar|1,0\rangle =\hbar\left(|\downarrow\uparrow\rangle+|\uparrow\downarrow\rangle\right), $$

so

$$ |1,0\rangle =\frac{|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle}{\sqrt2}. $$

The normalized $m=0$ state orthogonal to it is

$$ |0,0\rangle =\frac{|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle}{\sqrt2}. $$

The relative plus and minus signs distinguish the triplet and singlet. Multiplying an entire ket by one overall phase would not change its physics, but changing only one term’s relative sign would produce a different state. This is why a stated phase convention matters when tables are compared.

Clebsch-Gordan coefficients

In general, a coupled state is expanded in the uncoupled basis as

$$ |j,m\rangle =\sum_{m_1,m_2} \langle j_1m_1;j_2m_2|jm\rangle |j_1,m_1\rangle|j_2,m_2\rangle, $$

where $m=m_1+m_2$. The numbers

$$ \langle j_1m_1;j_2m_2|jm\rangle $$

are the Clebsch-Gordan coefficients. The factors $1/\sqrt2$ above are the simplest nontrivial examples.

For arbitrary angular momenta $j_1$ and $j_2$, the allowed total values are

$$ j=|j_1-j_2|,\ |j_1-j_2|+1,\ldots,j_1+j_2, $$

and for each $j$,

$$ m=-j,-j+1,\ldots,j. $$

As a check, combining $j_1=3/2$ and $j_2=2$ gives

$$ j=\frac72,\frac52,\frac32,\frac12. $$

The dimensions also match:

$$ (2j_1+1)(2j_2+1)=4\times5=20 $$

and

$$ 8+6+4+2=20. $$

That dimension check is a quick way to catch a missing multiplet.

What to remember

For two spin-½ angular momenta:

$$ \boxed{ \begin{aligned} |1,1\rangle&=|\uparrow\uparrow\rangle,\\ |1,0\rangle&=\frac{|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle}{\sqrt2},\\ |1,-1\rangle&=|\downarrow\downarrow\rangle,\\ |0,0\rangle&=\frac{|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle}{\sqrt2}. \end{aligned}} $$

The product basis is convenient when the individual $z$-components matter. The coupled basis is convenient when $J^2$, rotational symmetry, or an interaction proportional to $\mathbf{J}_1\cdot\mathbf{J}_2$ matters. Clebsch-Gordan coefficients are the change-of-basis coefficients connecting those two descriptions.