Tensor-Product Rotation Representations

Angular momentum addition begins with a tensor product. If one subsystem carries angular momentum $j_1$ and another carries angular momentum $j_2$, the combined Hilbert space is

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

A single physical rotation acts on both factors at once. The resulting tensor-product representation is usually reducible, so the same Hilbert space can be reorganized into total-angular-momentum sectors:

$$\mathcal H_{j_1}\otimes\mathcal H_{j_2} \cong \bigoplus_{J=|j_1-j_2|}^{j_1+j_2} \mathcal H_J.$$

This is the structural statement behind singlets, triplets, Clebsch–Gordan coefficients, spin–orbit coupling, hyperfine structure, and many-body spin decompositions.

Required background. Tensor products combine the two state spaces. The general representation construction explains how an operator acts on each factor.

Tensor Products Are the Composite Space

For fixed angular momenta, the dimensions are

$$\dim\mathcal H_{j_1}=2j_1+1, \qquad \dim\mathcal H_{j_2}=2j_2+1.$$

The composite dimension is the product:

$$\dim(\mathcal H_{j_1}\otimes\mathcal H_{j_2}) = (2j_1+1)(2j_2+1).$$

The tensor-product space is the place where product states, superpositions, entangled states, and total-angular-momentum eigenstates all live. The decomposition into total-$J$ sectors is a way to reorganize this same space by rotational symmetry; it does not replace the tensor-product structure.

Diagonal Rotation Action

Let $U_1(R)$ be the rotation operator on $\mathcal H_{j_1}$ and $U_2(R)$ the rotation operator on $\mathcal H_{j_2}$. A physical rotation $R$ of the whole composite system acts as

$$U(R) = U_1(R)\otimes U_2(R).$$

On a product state,

$$U(R) ( \lvert\psi_1\rangle\otimes\lvert\psi_2\rangle ) = U_1(R)\lvert\psi_1\rangle \otimes U_2(R)\lvert\psi_2\rangle.$$

For half-integer spins, $R$ here includes a consistent choice of lift to $SU(2)$; the two factors use the same rotation path. The same rotation appears in both factors. This is different from applying arbitrary independent local unitaries to the two subsystems. Angular momentum addition concerns the diagonal rotational action.

Total Generators

Differentiating the rotation action gives the total angular momentum operators. Component by component,

$$J_i = J_{1i}\otimes I_2 + I_1\otimes J_{2i}.$$

The identity factors are part of the definition: $J_{1i}$ acts only on the first factor, and $J_{2i}$ acts only on the second factor. Since operators on different factors commute,

$$[J_{1i}\otimes I_2,\ I_1\otimes J_{2j}] = 0.$$

Using the angular momentum algebra on each factor, one finds

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

Thus the total generators again form an angular momentum algebra. The operator-level consequences of this definition are treated in Total Angular Momentum.

Decomposition into Total J

Take each factor to be one irreducible spin representation. If both spins are nonzero, their tensor product is reducible; if one spin is zero, it is equivalent to the other factor. For $SU(2)$ angular momentum,

$$\mathcal H_{j_1}\otimes\mathcal H_{j_2} \cong \mathcal H_{j_1+j_2} \oplus \mathcal H_{j_1+j_2-1} \oplus \cdots \oplus \mathcal H_{|j_1-j_2|}.$$

Equivalently,

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

Each allowed $J$ appears once for two angular momenta. Dimension counting checks the decomposition:

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

The right-hand side is a direct sum of irreducible rotational sectors. It is not a tensor product of smaller physical subsystems.

The First Example

For two spin-$1/2$ systems,

$$\mathcal H = \mathbb C^2\otimes\mathbb C^2, \qquad \dim\mathcal H=4.$$

The angular-momentum decomposition is

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

In representation-space language,

$$\mathcal H_{1/2}\otimes\mathcal H_{1/2} \cong \mathcal H_1\oplus\mathcal H_0.$$

The dimensions match:

$$2\cdot2=3+1.$$

The three-dimensional spin-$1$ sector is the triplet; the one-dimensional spin-$0$ sector is the singlet. Their explicit states are developed in Two Spin-1/2 Particles and Singlet and Triplet States.

Product States and Coupled Sectors

An uncoupled product basis uses separate labels:

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

A coupled basis uses total labels:

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

Both are bases of the same tensor-product Hilbert space. The coupled basis states are often superpositions of product-basis states, and they may be entangled when the tensor factors are physical subsystems.

For example, the two-spin singlet

$$\lvert0,0\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \right)$$

belongs to a one-dimensional irreducible rotational sector, but it is still a vector in the original two-spin tensor-product space.

Why This Matters for Hamiltonians

A Hamiltonian that respects joint rotations is naturally organized by total angular momentum. For example, if

$$H = A\,\mathbf J_1\cdot\mathbf J_2,$$

then

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

So the Hamiltonian is diagonal in sectors of fixed $J$, because $J^2,J_1^2,J_2^2$ are all diagonal there. This is the algebra behind singlet–triplet splittings, spin–orbit coupling, hyperfine coupling, and many effective exchange Hamiltonians.

If a Hamiltonian instead treats the two factors separately, an uncoupled basis may be more convenient. The symmetry question tells you which basis is natural.

Multiplicity for Three Spins

A two-factor product contains each allowed $J$ once. With three spin-$1/2$ factors,

$$(\tfrac12\otimes\tfrac12)\otimes\tfrac12 =(1\oplus0)\otimes\tfrac12 =\tfrac32\oplus\tfrac12\oplus\tfrac12.$$

The dimensions are $8=4+2+2$. The two $J=1/2$ multiplets are independent copies, distinguished by whether the first pair had $j_{12}=1$ or $0$. A rotationally invariant Hamiltonian may mix these copies while still preserving total $J$. Recoupling relates this choice of pair to others.

Common Mistakes

• Treating $j_1\otimes j_2=J$ as ordinary multiplication rather than representation decomposition.
• Forgetting identity operators in $J_{1i}\otimes I+I\otimes J_{2i}$.
• Confusing the tensor-product space with the direct-sum decomposition into irreducible sectors.
• Thinking that a singlet is outside the two-particle Hilbert space because it is a one-dimensional rotational sector.
• Assuming the coupled basis is always better. The natural basis depends on the Hamiltonian and the observables.
• Forgetting that tensor-product basis states can be reorganized without changing the underlying physical Hilbert space.

Exercises

1. Decompose $j_1=1$ and $j_2=1/2$ into total-$J$ sectors and check dimensions.

Solution

The allowed total angular momenta are

$$J = |1-1/2|,\ldots,1+1/2 = \frac12,\frac32.$$

Thus

$$1\otimes\frac12 = \frac32\oplus\frac12.$$

The dimensions are

$$(2\cdot1+1)(2\cdot\frac12+1) = 3\cdot2 = 6,$$

and

$$(2\cdot\frac32+1) + (2\cdot\frac12+1) = 4+2 = 6.$$

2. Show that the total generators satisfy the angular momentum algebra.

Solution

Use

$$J_i = J_{1i}\otimes I + I\otimes J_{2i}.$$

Operators on different factors commute. Therefore

$$\begin{aligned} [J_i,J_j] &= [J_{1i},J_{1j}]\otimes I + I\otimes[J_{2i},J_{2j}] \\ &= i\hbar\sum_k\epsilon_{ijk} \left( J_{1k}\otimes I + I\otimes J_{2k} \right) \\ &= i\hbar\sum_k\epsilon_{ijk}J_k. \end{aligned}$$

3. Why is $\frac12\otimes\frac12=1\oplus0$ not ordinary arithmetic?

Solution

The symbols label irreducible angular-momentum representations, not numbers being multiplied. The statement means that the tensor product of two spin-$1/2$ representation spaces decomposes into a spin-$1$ irreducible sector and a spin-$0$ irreducible sector:

$$\mathcal H_{1/2}\otimes\mathcal H_{1/2} \cong \mathcal H_1\oplus\mathcal H_0.$$

The dimensions are $2\cdot2=3+1$, which is the ordinary numerical check.

4. A state lies in the spin-$0$ singlet sector of two spins. Does that mean the tensor-product structure has disappeared?

Solution

No. The singlet sector is a one-dimensional irreducible subspace inside the original tensor-product Hilbert space. The state is still a vector in $\mathcal H_{1/2}\otimes\mathcal H_{1/2}$. The direct-sum decomposition organizes the tensor-product space by total angular momentum; it does not remove the tensor-product origin of the composite system.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• B. C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, 2nd ed., Springer, 2015.
• 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.