Addition of Orbital and Spin Angular Momentum

A particle with spatial motion and spin carries two angular momenta: orbital angular momentum $\mathbf L$ and spin angular momentum $\mathbf S$. The generator of simultaneous rotations of the full one-particle state is

$$\mathbf J=\mathbf L+\mathbf S.$$

This page explains the one-particle coupled basis. For an electron or any spin-$1/2$ particle in an orbital state with angular momentum $\ell$, the total angular momentum labels are

$$j=\ell+\frac12 \qquad \text{or} \qquad j=\ell-\frac12,$$

with the lower value absent when $\ell=0$.

Required background. Orbital angular momentum acts on spatial wavefunctions; spin acts on an internal factor. Their coupled basis will let us calculate both projection probabilities and spin–orbit shifts.

One-Particle Hilbert Space

In the simplest nonrelativistic model, a spin-$s$ particle in three dimensions has Hilbert space

$$\mathcal H = L^2(\mathbb R^3)\otimes\mathcal H_s.$$

Orbital angular momentum acts on the spatial factor. Spin acts on the internal factor. Therefore

$$[L_i,S_j]=0.$$

Because both $\mathbf L$ and $\mathbf S$ satisfy angular momentum commutation relations and commute with each other, their sum

$$\mathbf J=\mathbf L+\mathbf S$$

also satisfies

$$[J_i,J_j] = i\hbar\sum_k\epsilon_{ijk}J_k.$$

The total operator $\mathbf J$ is the generator of rotations of the complete spinor wavefunction: its spatial dependence and its spin index rotate together.

Uncoupled and Coupled Labels

The uncoupled one-particle basis labels orbital and spin projections separately:

$$\lvert n,\ell,m_\ell\rangle \lvert s,m_s\rangle.$$

Here $n$ represents radial or additional labels; it is not part of the angular-momentum addition itself. This basis diagonalizes

$$L^2,\quad L_z,\quad S^2,\quad S_z.$$

The coupled basis instead labels total angular momentum:

$$\lvert n,\ell,s;j,m_j\rangle.$$

It diagonalizes

$$L^2,\quad S^2,\quad J^2,\quad J_z.$$

The change of basis is

$$\lvert n,\ell,s;j,m_j\rangle = \sum_{m_\ell,m_s} \langle \ell,m_\ell;s,m_s\vert j,m_j\rangle \lvert n,\ell,m_\ell\rangle \lvert s,m_s\rangle.$$

The coefficient vanishes unless

$$m_j=m_\ell+m_s.$$

Spin-One-Half Case

For a spin-$1/2$ particle, $s=1/2$. The allowed total angular momenta are

$$j = \left|\ell-\frac12\right|, \ldots, \ell+\frac12.$$

Thus for $\ell>0$,

$$j=\ell-\frac12,\quad \ell+\frac12.$$

For $\ell=0$, only

$$j=\frac12$$

occurs. Dimension counting checks the result. For $\ell>0$,

$$(2\ell+1)(2s+1) = (2\ell+1)2 = 4\ell+2,$$

while the coupled multiplets have dimensions

$$\left[2\left(\ell+\frac12\right)+1\right] + \left[2\left(\ell-\frac12\right)+1\right] = (2\ell+2)+2\ell = 4\ell+2.$$

For $\ell=0$, the product dimension is $1\cdot2=2$, matching the single $j=1/2$ multiplet.

Spectroscopic Examples

For an $s$ orbital, $\ell=0$, so the only total angular momentum is

$$s_{1/2}:\qquad j=\frac12.$$

For a $p$ orbital, $\ell=1$, so

$$p_{1/2} \quad \text{and} \quad p_{3/2}$$

are possible. Their multiplet dimensions are

$$2j+1=2 \qquad \text{and} \qquad 2j+1=4.$$

Together they contain the six spin-orbital states from the uncoupled $p$-orbital product space:

$$(2\ell+1)(2s+1)=3\cdot2=6.$$

For a $d$ orbital, $\ell=2$, the allowed total angular momenta are $j=3/2$ and $j=5/2$, with dimensions $4$ and $6$.

Spinor Spherical Harmonics

In a central potential, spatial wavefunctions separate into radial and angular parts. With spin included, the angular-spin part of the coupled basis is a spinor spherical harmonic:

$$\Omega_{\ell j m_j}(\theta,\phi) = \sum_{m_\ell,m_s} \langle \ell,m_\ell;\tfrac12,m_s\vert j,m_j\rangle Y_\ell^{m_\ell}(\theta,\phi) \chi_{m_s}.$$

Here $\chi_{m_s}$ is a two-component spin basis vector. A corresponding separated state has the schematic form

$$\Psi_{n\ell j m_j}(r,\theta,\phi) = R_{n\ell j}(r) \Omega_{\ell j m_j}(\theta,\phi).$$

The radial function can depend on $j$ if a central spin-dependent interaction is present; it is independent of $j$ for a spin-independent central Hamiltonian. This notation packages the Clebsch–Gordan expansion into a spinor-valued angular function. It is the natural nonrelativistic basis for central problems with spin, and it is also the angular language that reappears in relativistic central-potential problems.

Parity

Spin is an internal axial degree of freedom and does not change the spatial parity of the orbital wavefunction. Taking the fixed intrinsic parity of the particle as $+1$, a state with orbital label $\ell$ has parity

$$P=(-1)^\ell.$$

For a species assigned intrinsic parity $\eta_P$, multiply this orbital factor by $\eta_P$. The total $j$ label does not determine parity by itself. For example, $j=1/2$ can arise from $\ell=0$ or $\ell=1$ when spin $1/2$ is present, and those two possibilities have opposite parity.

This is why one often needs both $j$ and an orbital or parity label.

Spin–Orbit Identity

The same basis diagonalizes the scalar product $\mathbf L\cdot\mathbf S$ because

$$J^2 = L^2+S^2+2\mathbf L\cdot\mathbf S.$$

Thus

$$\mathbf L\cdot\mathbf S = \frac12 \left( J^2-L^2-S^2 \right).$$

On a coupled state,

$$\mathbf L\cdot\mathbf S \longrightarrow \frac{\hbar^2}{2} \left[ j(j+1)-\ell(\ell+1)-s(s+1) \right].$$

This identity is the algebra behind fine-structure splittings when a spin–orbit Hamiltonian is present. The energy-shift application is treated in Spin–Orbit Coupling.

Relativistic Bridge

In the Dirac equation with a central potential, total angular momentum remains central. The angular spinors are closely related to the spinor spherical harmonics above, but relativistic wavefunctions have upper and lower components with different radial functions and orbital labels $\ell$ and $2j-\ell$. Those angular factors have opposite orbital parity, while the parity operator on a Dirac spinor also acts with $\gamma^0$; the full state can still have definite parity.

The nonrelativistic statement to remember is:

$$\mathbf J=\mathbf L+\mathbf S$$

is the total rotation generator. Relativistic theory preserves the importance of total angular momentum while changing the representation carried by the wavefunction. For formula-level reference, see Dirac Equation and Dirac Hamiltonian.

Common Mistakes

• Treating $j=\ell+1/2$ as the only possibility. The $j=\ell-1/2$ branch exists when $\ell>0$.
• Forgetting that the $\ell=0$ case has only $j=1/2$.
• Confusing $m_j$ with $m_\ell$ or $m_s$. The rule is $m_j=m_\ell+m_s$ for nonzero coefficients.
• Assuming that $j$ determines parity. The orbital label $\ell$ determines parity.
• Thinking spinor spherical harmonics are new dynamics. They are angular-spin basis functions.
• Applying spin–orbit splitting formulas without checking whether the Hamiltonian actually contains a scalar $\mathbf L\cdot\mathbf S$ term.

Exercises

1. List the total $j$ values and dimensions for a spin-$1/2$ particle with $\ell=2$.

Solution

For $\ell=2$ and $s=1/2$,

$$j=\ell-\frac12,\ \ell+\frac12,$$

so

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

The dimensions are

$$2\cdot\frac32+1=4, \qquad 2\cdot\frac52+1=6.$$

Together they give $4+6=10$, matching

$$(2\ell+1)(2s+1)=5\cdot2=10.$$

2. In the stated Clebsch–Gordan convention, the $\ell=1,j=1/2,m_j=1/2$ spinor harmonic is

$$\Omega_{1,1/2,1/2} =\begin{pmatrix}-\sqrt{1/3}\,Y_1^0\\\sqrt{2/3}\,Y_1^1\end{pmatrix}.$$

Check its normalization and find the probabilities for spin up and down along $z$ after spatial direction is unobserved.

Solution

Each harmonic has unit norm, and the spin components are orthogonal. Thus $\int\Omega^\dagger\Omega\,d\Omega=1/3+2/3=1$. The spin probabilities are $P(\uparrow)=1/3$ and $P(\downarrow)=2/3$. In each term, $m_\ell+m_s=1/2$, so the total projection remains definite even though the spin projection is not.

3. Compute $\mathbf L\cdot\mathbf S$ for $\ell=1$, $s=1/2$, and the two allowed values of $j$.

Solution

Use

$$\mathbf L\cdot\mathbf S = \frac{\hbar^2}{2} \left[ j(j+1)-\ell(\ell+1)-s(s+1) \right].$$

For $\ell=1$ and $s=1/2$,

$$\ell(\ell+1)=2, \qquad s(s+1)=\frac34.$$

For $j=3/2$,

$$j(j+1)=\frac32\cdot\frac52=\frac{15}{4},$$

so

$$\mathbf L\cdot\mathbf S = \frac{\hbar^2}{2} \left( \frac{15}{4}-2-\frac34 \right) = \frac{\hbar^2}{2}.$$

For $j=1/2$,

$$j(j+1)=\frac12\cdot\frac32=\frac34,$$

so

$$\mathbf L\cdot\mathbf S = \frac{\hbar^2}{2} \left( \frac34-2-\frac34 \right) = -\hbar^2.$$

4. Can two states with the same $j=1/2$ have opposite parity?

Solution

Yes. For spin $1/2$, $j=1/2$ can arise from $\ell=0$ or $\ell=1$. The parity is determined by $\ell$:

$$P=(-1)^\ell.$$

The $\ell=0$ state is even, while the $\ell=1$ state is odd. Therefore $j$ alone does not determine parity.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• W. Greiner, Relativistic Quantum Mechanics: Wave Equations, 3rd ed., Springer, 2000.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon Press, 1977.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.