Tensor Product Representations

A tensor product representation is the natural representation carried by a tensor-product vector space when symmetries act on the factors. It is the representation-theory mechanism behind combining subsystems, adding angular momenta, decomposing singlet and triplet sectors, and organizing coupled versus uncoupled bases.

The linear-algebra construction is Tensor Products. This page explains what happens when the tensor-product factors already carry group or Lie-algebra representations.

Why Quantum Mechanics Needs It

Composite quantum systems have tensor-product Hilbert spaces. If each subsystem carries a symmetry representation, the composite system carries a representation too.

For rotations, this is the origin of angular momentum addition. Two spins may be described first by separate labels $j_1,m_1$ and $j_2,m_2$, but the same tensor-product space can be decomposed into total-angular-momentum sectors labeled by $J,M$.

The key structural idea is:

$$\text{tensor product of spaces} \quad\Longrightarrow\quad \text{representation that may decompose as a direct sum}.$$

So tensor products and direct sums both appear: tensor products combine subsystems or factors, while direct sums describe the irreducible sectors inside the combined representation.

Product Group Action

Let

$$\rho:G\to \operatorname{GL}(V), \qquad \sigma:H\to \operatorname{GL}(W)$$

be representations. Their external tensor product is a representation of the product group $G\times H$ on $V\otimes W$:

$$(\rho\boxtimes\sigma)(g,h) = \rho(g)\otimes\sigma(h).$$

On a product vector,

$$(\rho\boxtimes\sigma)(g,h)(v\otimes w) = \rho(g)v\otimes\sigma(h)w.$$

The group law works because operator tensor products multiply factor by factor:

$$\begin{aligned} &(\rho(g_1)\otimes\sigma(h_1)) (\rho(g_2)\otimes\sigma(h_2))\\ &\qquad = \rho(g_1g_2)\otimes\sigma(h_1h_2). \end{aligned}$$

This is the right structure when two independent symmetry groups act on two factors.

Diagonal Action of One Group

Often the same group acts on both factors. If

$$\rho_1:G\to \operatorname{GL}(V_1), \qquad \rho_2:G\to \operatorname{GL}(V_2),$$

then the tensor product representation of $G$ on $V_1\otimes V_2$ is

$$(\rho_1\otimes\rho_2)(g) = \rho_1(g)\otimes\rho_2(g).$$

This is the diagonal action: one group element $g$ is applied to both factors at once.

For quantum rotations, this is the physically relevant action when a single spatial rotation acts on every subsystem. If $U_1(R)$ and $U_2(R)$ are the rotation representations on two factors, the composite rotation is

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

This formula should not be confused with an arbitrary local operation $U_1\otimes U_2$ where the two unitary operators can be chosen independently. A symmetry rotation uses the same rotation parameter on both factors.

Lie Algebra Generator Rule

For a Lie group representation, differentiating the diagonal tensor product gives the Lie-algebra action.

Let $X\in\mathfrak g$. If $d\rho_1(X)$ acts on $V_1$ and $d\rho_2(X)$ acts on $V_2$, then on $V_1\otimes V_2$:

$$d(\rho_1\otimes\rho_2)(X) = d\rho_1(X)\otimes I + I\otimes d\rho_2(X).$$

This is the algebraic reason total generators are sums. For two angular momenta,

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

meaning componentwise

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

The identity factors are part of the statement. They specify which tensor factor each operator acts on.

Decomposition into Irreducibles

A tensor product of irreducible representations is usually not irreducible. For compact groups and finite-dimensional unitary representations, it decomposes as a direct sum of irreducible pieces:

$$V_a\otimes V_b \cong \bigoplus_c N_{ab}^{\ \ c}V_c.$$

Here $V_a,V_b,V_c$ denote irreducible representation spaces, and the nonnegative integers $N_{ab}^{\ \ c}$ are multiplicities. They count how many times each irreducible representation appears.

This is representation decomposition, not factorization of states. The left side is a tensor-product space. The right side is the same vector space reorganized as a direct sum of symmetry sectors.

For angular momentum, the relevant compact group is $SU(2)$, and the decomposition is especially simple:

$$V_{j_1}\otimes V_{j_2} \cong \bigoplus_{J=\lvert j_1-j_2\rvert}^{j_1+j_2} V_J.$$

Each allowed $J$ appears once. Dimension counting checks the formula:

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

The change-of-basis coefficients between the uncoupled product basis and the coupled direct-sum basis are Clebsch–Gordan Coefficients. The angular-momentum derivation is developed in the Symmetry, Angular Momentum, and Spin treatment.

Uncoupled and Coupled Bases

For two angular momenta, the uncoupled basis is

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

It diagonalizes $J_1^2,J_{1z},J_2^2,J_{2z}$.

The coupled basis is

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

It diagonalizes $J_1^2,J_2^2,J^2,J_z$, where

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

Both bases describe the same tensor-product space. The difference is which commuting observables have been diagonalized. The Clebsch–Gordan coefficients are the entries of the unitary change-of-basis matrix between them.

Worked Example: Two Spin-1/2 Spaces

Let $V_{1/2}\cong\mathbb C^2$ be the spin-$1/2$ representation space. The tensor product has dimension four:

$$\dim(V_{1/2}\otimes V_{1/2}) = 2\cdot2 = 4.$$

The $SU(2)$ decomposition is

$$V_{1/2}\otimes V_{1/2} \cong V_1\oplus V_0.$$

The dimensions match:

$$4 = 3+1.$$

The three-dimensional sector $V_1$ is the spin-$1$ triplet. The one-dimensional sector $V_0$ is the spin-$0$ singlet. A representative singlet vector is

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

The full coupled basis and its physical interpretation are developed in Two Spin-1/2 Particles.

Tensor Products Versus Entanglement

Tensor product representations organize the space on which a symmetry acts. They do not say that every state is a product state.

The tensor-product space $V_1\otimes V_2$ contains product vectors such as $v_1\otimes v_2$, but it also contains linear combinations that cannot be factored. In quantum mechanics, those are entangled states when the tensor factors are physical subsystems.

The direct-sum decomposition into irreducible sectors is a different structure. For example, the singlet/triplet decomposition reorganizes the two-spin tensor-product space by total spin. The singlet state is still a vector in the original tensor-product Hilbert space.

Common Mistakes

• Confusing the product group action $G\times H$ with the diagonal action of one group $G$ on both factors.
• Forgetting identity factors in total generators such as $J_{1i}\otimes I+I\otimes J_{2i}$.
• Treating $j_1\otimes j_2=J_1\oplus J_2\oplus\cdots$ as arithmetic rather than representation decomposition.
• Assuming a tensor product of irreducible representations is automatically irreducible.
• Confusing a direct-sum decomposition into symmetry sectors with a tensor-product subsystem split.
• Treating Clebsch–Gordan coefficients as dynamics instead of change-of-basis coefficients.
• Forgetting that phase conventions affect the signs of coupled basis vectors.

Cross-Links

• Tensor Products
• Direct Sums
• Representations
• Unitary Representations
• Lie Algebras
• SU(2)
• Angular Momentum Algebra
• Ladder Operators as Lie Algebra Tools
• Clebsch–Gordan Coefficients
• Wigner 3j, 6j, and 9j Symbols
• Two Spin-1/2 Particles
• Clebsch–Gordan Coefficients: Angular Momentum Derivation
• Tensor Products of Hilbert Spaces
• Direct Sums versus Tensor Products

References

• W. Fulton and J. Harris, Representation Theory: A First Course, Springer, 1991.
• B. C. Hall, Lie Groups, Lie Algebras, and Representations, 2nd ed., Springer, 2015.
• J. F. Cornwell, Group Theory in Physics, Vol. 1, Academic Press, 1984.
• 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.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. Verify that the external tensor product is a representation of $G\times H$.

Solution

Compute

$$\begin{aligned} &(\rho\boxtimes\sigma)(g_1,h_1) (\rho\boxtimes\sigma)(g_2,h_2)\\ &\qquad = (\rho(g_1)\otimes\sigma(h_1)) (\rho(g_2)\otimes\sigma(h_2))\\ &\qquad = \rho(g_1g_2)\otimes\sigma(h_1h_2)\\ &\qquad = (\rho\boxtimes\sigma)(g_1g_2,h_1h_2). \end{aligned}$$

The identity element maps to $I_V\otimes I_W$, so the representation axioms hold.

2. Derive the Lie-algebra generator rule for a diagonal tensor product.

Solution

For a one-parameter subgroup $g(t)=\exp(tX)$,

$$(\rho_1\otimes\rho_2)(g(t)) = \rho_1(g(t))\otimes\rho_2(g(t)).$$

Differentiate at $t=0$ and use the product rule:

$$\left.\frac{d}{dt}\right|_{t=0} \rho_1(g(t))\otimes\rho_2(g(t)) = d\rho_1(X)\otimes I + I\otimes d\rho_2(X).$$

3. Check the dimension identity for $j_1=1$ and $j_2=1/2$.

Solution

The tensor-product dimension is

$$(2j_1+1)(2j_2+1) = 3\cdot2 = 6.$$

The allowed total angular momenta are

$$J=\frac12,\frac32.$$

The sum of sector dimensions is

$$(2\cdot\tfrac12+1) + (2\cdot\tfrac32+1) = 2+4 = 6.$$

4. Explain why $V_{1/2}\otimes V_{1/2}\cong V_1\oplus V_0$ is not an equation of ordinary numbers.

Solution

The symbols $V_{1/2}$, $V_1$, and $V_0$ denote representation spaces, not numbers. The left side is the four-dimensional tensor-product representation of two spin-$1/2$ spaces. The right side is the same representation decomposed into a three-dimensional spin-$1$ irreducible sector and a one-dimensional spin-$0$ irreducible sector.

5. In the diagonal rotation action on two spin spaces, why is the total generator $J_i=J_{1i}\otimes I+I\otimes J_{2i}$ rather than $J_{1i}\otimes J_{2i}$?

Solution

A small rotation acts as

$$U_1(t)\otimes U_2(t) = \left(I-\frac{i}{\hbar}tJ_{1i}+O(t^2)\right) \otimes \left(I-\frac{i}{\hbar}tJ_{2i}+O(t^2)\right).$$

Keeping only first-order terms gives

$$I\otimes I - \frac{i}{\hbar}t \left( J_{1i}\otimes I+I\otimes J_{2i} \right) + O(t^2).$$

The product $J_{1i}\otimes J_{2i}$ appears only at order $t^2$, not as the infinitesimal generator.