Coupled and Uncoupled Bases

Composite angular momentum systems have two natural bases. The uncoupled basis labels each subsystem separately:

$$|j_1,m_1\rangle|j_2,m_2\rangle.$$

The coupled basis labels total angular momentum:

$$|j_1,j_2;J,M\rangle.$$

Both bases span the same tensor-product Hilbert space for fixed $j_1$ and $j_2$. The difference is which commuting observables are diagonal.

Tensor-Product Setting

Let the first subsystem carry angular momentum $\mathbf J_1$ and the second carry $\mathbf J_2$. The composite Hilbert space is

$$\mathcal H = \mathcal H_{j_1}\otimes\mathcal H_{j_2},$$

with dimension

$$\dim\mathcal H = (2j_1+1)(2j_2+1).$$

Operators for the two subsystems act on different tensor factors. The total angular momentum is

$$\mathbf J=\mathbf J_1+\mathbf J_2,$$

with

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

The total components again satisfy the angular momentum algebra:

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

The purpose of angular momentum addition is to understand how the tensor product decomposes into irreducible total-$J$ multiplets.

The Uncoupled Basis

The uncoupled basis diagonalizes

$$J_1^2,\quad J_{1z},\quad J_2^2,\quad J_{2z}.$$

Its basis states obey

$$J_1^2|j_1,m_1\rangle|j_2,m_2\rangle = \hbar^2j_1(j_1+1) |j_1,m_1\rangle|j_2,m_2\rangle,$$

and

$$J_{1z}|j_1,m_1\rangle|j_2,m_2\rangle = \hbar m_1 |j_1,m_1\rangle|j_2,m_2\rangle,$$

with analogous equations for $J_2^2$ and $J_{2z}$.

This basis is natural when the Hamiltonian treats the two angular momenta separately, for example in a field that couples independently to $J_{1z}$ and $J_{2z}$.

The Coupled Basis

The coupled basis diagonalizes

$$J_1^2,\quad J_2^2,\quad J^2,\quad J_z.$$

Its basis states obey

$$J^2|j_1,j_2;J,M\rangle = \hbar^2J(J+1)|j_1,j_2;J,M\rangle,$$

and

$$J_z|j_1,j_2;J,M\rangle = \hbar M|j_1,j_2;J,M\rangle.$$

The labels $j_1$ and $j_2$ are usually kept in the ket because the same total $J$ value can arise from different constituent angular momenta in larger problems.

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

$$J = |j_1-j_2|, |j_1-j_2|+1, \ldots, j_1+j_2.$$

For each $J$,

$$M=-J,-J+1,\ldots,J.$$

The dimension check is

$$(2j_1+1)(2j_2+1) = \sum_{J=|j_1-j_2|}^{j_1+j_2}(2J+1).$$

Change of Basis

The coupled and uncoupled bases are related by Clebsch–Gordan coefficients:

$$|j_1,j_2;J,M\rangle = \sum_{m_1,m_2} \langle j_1,m_1;j_2,m_2|J,M\rangle |j_1,m_1\rangle|j_2,m_2\rangle.$$

The inverse expansion is

$$|j_1,m_1\rangle|j_2,m_2\rangle = \sum_{J,M} \langle J,M|j_1,m_1;j_2,m_2\rangle |j_1,j_2;J,M\rangle.$$

The transformation is unitary. It is a change of basis, not a physical approximation.

Because

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

the coefficient vanishes unless

$$M=m_1+m_2.$$

The detailed coefficient conventions, signs, and table-reading rules are in Clebsch–Gordan Coefficients.

For three or more angular momenta, different orders of pairwise coupling are related by Recoupling and Wigner Symbols. Which order is physically useful is discussed in Angular Momentum Coupling Schemes.

Which Basis Is Natural

The preferred basis depends on the observables or Hamiltonian.

The uncoupled basis is natural for questions such as:

• What are the two individual $z$ components?
• What happens when each spin couples to a different local field?
• What is the tensor-product state before an interaction is turned on?

The coupled basis is natural for questions such as:

• What is the total angular momentum?
• Which irreducible rotational multiplet is the state in?
• How does an interaction depending on $\mathbf J_1\cdot\mathbf J_2$ split the space?

The key identity is

$$\mathbf J_1\cdot\mathbf J_2 = \frac12 \left( J^2-J_1^2-J_2^2 \right).$$

Therefore an interaction proportional to $\mathbf J_1\cdot\mathbf J_2$ is diagonal in the coupled basis, because $J^2$, $J_1^2$, and $J_2^2$ are all diagonal there.

Two Spin-One-Half Example

For two spin-$1/2$ systems,

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

The uncoupled basis is

$$|\uparrow\uparrow\rangle,\quad |\uparrow\downarrow\rangle,\quad |\downarrow\uparrow\rangle,\quad |\downarrow\downarrow\rangle.$$

The coupled basis is a triplet plus a singlet:

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

The full derivation starts in Two Spin-1/2 Particles, and the rotational scalar/vector structure is developed in Singlet and Triplet States.

Orbital Plus Spin Example

For one electron in an orbital state with angular momentum $\ell$ and spin $s=1/2$, the uncoupled basis is

$$|\ell,m_\ell\rangle|s,m_s\rangle.$$

The coupled basis is

$$|\ell,s;j,m\rangle,$$

where

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

when $\ell>0$. For $\ell=0$, only $j=1/2$ occurs.

This basis becomes natural when the Hamiltonian contains a spin–orbit term proportional to

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

The angular momentum algebra makes the structure transparent even before one studies the detailed atomic fine-structure Hamiltonian. The dedicated angular-momentum treatment is Spin–Orbit Coupling.

Common Mistakes

• Thinking coupled states live in a different Hilbert space from uncoupled states. They are different bases of the same space.
• Forgetting that $M=m_1+m_2$ for every nonzero Clebsch–Gordan coefficient.
• Treating $J$ as if it were always $j_1+j_2$; lower total values also occur.
• Dropping the constituent labels $j_1,j_2$ when they are needed to distinguish sectors.
• Assuming the coupled basis is always better. It is better for rotationally invariant interactions, not for every Hamiltonian.
• Confusing a product state with a state of definite total angular momentum.

Cross-Links

• Tensor Product Representations in Angular Momentum
• Total Angular Momentum
• Two Spin-1/2 Particles
• Singlet and Triplet States
• Clebsch–Gordan Coefficients
• Angular Momentum Coupling Schemes
• Spin–Orbit Coupling
• Angular Momentum Algebra
• Eigenvalues of J² and Jz
• Spin-1/2 Hilbert Space
• Tensor Products
• Complete Sets of Commuting Observables
• Tensor Product Representations
• Clebsch–Gordan Coefficients as Representation Coefficients

References

• 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.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.

Exercises

1. For $j_1=1$ and $j_2=1$, list the allowed total $J$ values and check the dimension count.

Solution

The allowed values are

$$J=|1-1|,\ldots,1+1,$$

so

$$J=0,1,2.$$

The uncoupled space has dimension

$$(2j_1+1)(2j_2+1)=3\cdot3=9.$$

The coupled multiplet dimensions add to

$$(2\cdot0+1)+(2\cdot1+1)+(2\cdot2+1) = 1+3+5 = 9.$$

2. Why must $\langle j_1,m_1;j_2,m_2|J,M\rangle$ vanish unless $M=m_1+m_2$?

Solution

The product state is an eigenstate of

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

with eigenvalue

$$\hbar(m_1+m_2).$$

The coupled state $|j_1,j_2;J,M\rangle$ is an eigenstate of $J_z$ with eigenvalue $\hbar M$. States with different $J_z$ eigenvalues are orthogonal, so the overlap can be nonzero only when

$$M=m_1+m_2.$$

3. Show why $\mathbf J_1\cdot\mathbf J_2$ is diagonal in the coupled basis.

Solution

Use

$$J^2 = (\mathbf J_1+\mathbf J_2)^2 = J_1^2+J_2^2+2\mathbf J_1\cdot\mathbf J_2.$$

Therefore

$$\mathbf J_1\cdot\mathbf J_2 = \frac12 \left( J^2-J_1^2-J_2^2 \right).$$

The coupled basis diagonalizes $J^2$, $J_1^2$, and $J_2^2$, so it also diagonalizes this scalar product.

4. For $\ell=1$ and $s=1/2$, list the allowed total $j$ values and their multiplet dimensions.

Solution

The allowed total values are

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

so

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

Their dimensions are

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

The total dimension is $4+2=6$, matching

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