Spin & Angular Momentum Addition

Spin is an intrinsic form of angular momentum carried by elementary particles. Unlike orbital angular momentum, spin is not associated with spatial motion. It is a purely quantum mechanical property.

The spin operators $\hat{S}_x$, $\hat{S}_y$, $\hat{S}_z$ satisfy the same algebra as orbital angular momentum: $[\hat{S}_i, \hat{S}_j] = i\hbar \epsilon_{ijk} \hat{S}_k$. The total spin operator is $\hat{S}^2 = \hat{S}_x^2 + \hat{S}_y^2 + \hat{S}_z^2$, with eigenvalues $\hbar^2 s(s+1)$.

For a given spin quantum number $s$, the magnetic quantum number $m_s$ can take $2s+1$ values: $m_s = -s, -s+1, \ldots, s$. The raising and lowering operators $\hat{S}_\pm = \hat{S}_x \pm i \hat{S}_y$ act as:

The most important case is spin-$1/2$ (electrons, protons, neutrons), where $s = 1/2$ and $m_s = \pm 1/2$. The spin operators are represented by the Pauli matrices divided by $2$.

For spin-$1/2$, the basis states are $|\uparrow\rangle = |1/2, 1/2\rangle$ and $|\downarrow\rangle = |1/2, -1/2\rangle$. In matrix form, we write:

The spin operators are:

where $\sigma_x$, $\sigma_y$, $\sigma_z$ are the Pauli matrices. These matrices satisfy $\sigma_i \sigma_j = \delta_{ij} I + i\epsilon_{ijk} \sigma_k$, which implies $[\sigma_i, \sigma_j] = 2i\epsilon_{ijk} \sigma_k$.

Any spin-$1/2$ state can be written as a linear combination $|\chi\rangle = a |\uparrow\rangle + b |\downarrow\rangle$, with $|a|^2 + |b|^2 = 1$. The state can also be represented as a point on the Bloch sphere: $|\chi\rangle = \cos(\theta/2)|\uparrow\rangle + e^{i\phi} \sin(\theta/2)|\downarrow\rangle$.

When two angular momenta $\hat{J}_1$ and $\hat{J}_2$ combine, the total angular momentum $\hat{J} = \hat{J}_1 + \hat{J}_2$ can take values from $|j_1 - j_2|$ to $j_1 + j_2$ in integer steps. For each total $j$, the magnetic quantum number $m_j$ ranges from $-j$ to $j$.

The tensor product basis $|j_1, m_1\rangle \otimes |j_2, m_2\rangle$ and the total angular momentum basis $|j, m_j\rangle$ are related by Clebsch-Gordan coefficients:

The Clebsch-Gordan coefficients are determined by the requirement that $\hat{J}^2$ and $\hat{J}_z$ are diagonal in the total angular momentum basis, combined with the phase convention that the coefficient for the highest weight state ($m_1 = j_1$, $m_2 = j_2$) is positive.

The most important case is adding two spin-$1/2$ particles. The total spin can be $j = 1$ (triplet, symmetric) or $j = 0$ (singlet, antisymmetric). The states are:

The total spin states have definite symmetry under exchange of the two particles. The triplet states ($j=1$) are symmetric: $\hat{P}_{12} |1,m\rangle = |1,m\rangle$. The singlet state ($j=0$) is antisymmetric: $\hat{P}_{12} |0,0\rangle = -|0,0\rangle$.

For identical particles, the total wave function must be symmetric under exchange for bosons (integer spin) and antisymmetric for fermions (half-integer spin). The spin symmetry is correlated with the spatial symmetry:

If the spatial part is symmetric, the spin part must be symmetric for bosons and antisymmetric for fermions. If the spatial part is antisymmetric, the spin part must be antisymmetric for bosons and symmetric for fermions.

For two electrons (fermions) in a symmetric spatial state, the spin state must be the antisymmetric singlet. For two electrons in an antisymmetric spatial state, the spin state can be any of the three symmetric triplet states.

An electron has orbital angular momentum $l = 1$ and spin $s = 1/2$. Find the possible values of total $j$ and construct the $|j, m_j\rangle$ states.

Step 1 — Possible $j$ values. $j = |l - s|$ to $l + s = 1/2, 3/2$. So $j = 1/2$ and $j = 3/2$.

Step 2 — $j = 3/2$ states (4 states, $m_j = 3/2, 1/2, -1/2, -3/2$). The highest weight state is $|3/2, 3/2\rangle = |1,1\rangle|1/2,1/2\rangle = |1,1;\uparrow\rangle$.

Step 3 — Apply $\hat{J}_- = \hat{L}_- + \hat{S}_-$ to get lower states. $\hat{J}_- |3/2, 3/2\rangle = \hbar\sqrt{3} |3/2, 1/2\rangle = \hat{L}_- |1,1;\uparrow\rangle + \hat{S}_- |1,1;\uparrow\rangle = \hbar\sqrt{2} |1,0;\uparrow\rangle + \hbar |1,1;\downarrow\rangle$. Thus $|3/2, 1/2\rangle = \sqrt{2/3}|1,0;\uparrow\rangle + \sqrt{1/3}|1,1;\downarrow\rangle$.

Step 4 — $j = 1/2$ states (2 states, $m_j = 1/2, -1/2$). $|1/2, 1/2\rangle$ must be orthogonal to $|3/2, 1/2\rangle$ and have the same $m_j$. So $|1/2, 1/2\rangle = \sqrt{1/3}|1,0;\uparrow\rangle - \sqrt{2/3}|1,1;\downarrow\rangle$. This can be verified to be an eigenstate of $\hat{J}^2$ with eigenvalue $(3/4)\hbar^2$.