Tensor Products of Hilbert Spaces

Combining two quantum systems requires a space that includes both independent preparations and their coherent superpositions. The tensor product supplies this space: it preserves the linear structure of each subsystem while allowing joint states that do not factor. The introductory explanation gives a short first example.

For distinguishable quantum subsystems $A$ and $B$, the state space of the composite system is

$$\mathcal H_{AB} = \mathcal H_A\otimes\mathcal H_B.$$

The symbol $\otimes$ does more than place two state labels side by side. It defines a vector space in which independent subsystem states can be combined, superpositions remain linear in each factor, and general vectors need not factor into separate states of $A$ and $B$.

Helpful background. State vectors supply inner products and normalization; the tensor-product construction explains the underlying bilinear algebra.

Product Vectors

Given

$$\lvert\psi\rangle_A\in\mathcal H_A, \qquad \lvert\phi\rangle_B\in\mathcal H_B,$$

their tensor product is a vector

$$\lvert\psi\rangle_A \otimes \lvert\phi\rangle_B \in \mathcal H_A\otimes\mathcal H_B.$$

Such a vector is called a simple tensor or product vector. When both factors are nonzero, its ray represents a pure state that assigns one pure state to each subsystem. Unit factors give a unit joint vector; the zero vector represents no physical state.

The map from a pair of vectors to their tensor product is bilinear. In the first factor,

$$\left( \alpha\lvert\psi_1\rangle + \beta\lvert\psi_2\rangle \right) \otimes\lvert\phi\rangle = \alpha\lvert\psi_1\rangle\otimes\lvert\phi\rangle + \beta\lvert\psi_2\rangle\otimes\lvert\phi\rangle,$$

and similarly in the second factor:

$$\lvert\psi\rangle\otimes \left( \alpha\lvert\phi_1\rangle + \beta\lvert\phi_2\rangle \right) = \alpha\lvert\psi\rangle\otimes\lvert\phi_1\rangle + \beta\lvert\psi\rangle\otimes\lvert\phi_2\rangle.$$

Scalars can be moved between factors:

$$\begin{aligned} (c\lvert\psi\rangle)\otimes\lvert\phi\rangle &= \lvert\psi\rangle\otimes(c\lvert\phi\rangle)\\ &= c( \lvert\psi\rangle\otimes\lvert\phi\rangle ). \end{aligned}$$

Consequently, the factors of a product vector are not unique as vectors. For nonzero $c$,

$$(c\lvert\psi\rangle) \otimes \left(c^{-1}\lvert\phi\rangle\right) = \lvert\psi\rangle\otimes\lvert\phi\rangle.$$

Normalized product-state factors are unique only up to compensating phases.

Inner Products and Norms

The inner product of product vectors is defined by

$$\begin{aligned} \langle\psi\otimes\phi \mid \psi'\otimes\phi'\rangle &= \langle\psi\mid\psi'\rangle_A \langle\phi\mid\phi'\rangle_B. \end{aligned}$$

It follows that norms multiply:

$$\lVert \lvert\psi\rangle\otimes\lvert\phi\rangle \rVert = \lVert\psi\rVert\,\lVert\phi\rVert.$$

Thus normalized subsystem vectors produce a normalized composite vector. If either pair of local vectors is orthogonal, then the corresponding product vectors are orthogonal:

$$\begin{aligned} \langle\psi\mid\psi'\rangle_A=0 &\quad\Longrightarrow\\ \langle\psi\otimes\phi \mid \psi'\otimes\phi'\rangle &=0. \end{aligned}$$

The algebraic tensor product consists of finite sums of product vectors, subject to the bilinear relations above. On finite sums, the product-vector rule extends sesquilinearly:

$$\left\langle \sum_i \psi_i\otimes\phi_i \middle| \sum_j \psi'_j\otimes\phi'_j \right\rangle = \sum_{i,j} \langle\psi_i\mid\psi'_j\rangle_A \langle\phi_i\mid\phi'_j\rangle_B.$$

After identifying any zero-norm representatives, the Hilbert-space tensor product is the completion in the induced norm. Finite-dimensional tensor products are already complete; the completion step matters for infinite-dimensional systems.

Product Bases

Let

$$\left\{ \lvert i\rangle_A \right\}_{i=1}^{d_A}$$

and

$$\left\{ \lvert j\rangle_B \right\}_{j=1}^{d_B}$$

be orthonormal bases. Pairing every basis vector from $A$ with every basis vector from $B$ gives the product basis

$$\left\{ \lvert i\rangle_A\otimes\lvert j\rangle_B \right\}_{i,j}.$$

Its orthonormality follows immediately:

$$\begin{aligned} \langle i,j\mid k,\ell\rangle &= \langle i\mid k\rangle_A \langle j\mid\ell\rangle_B\\ &= \delta_{ik}\delta_{j\ell}. \end{aligned}$$

For finite-dimensional spaces,

$$\dim(\mathcal H_A\otimes\mathcal H_B) = d_A d_B.$$

Every vector in the composite space has an expansion

$$\lvert\Psi\rangle = \sum_{i=1}^{d_A} \sum_{j=1}^{d_B} c_{ij}\lvert i\rangle_A\otimes\lvert j\rangle_B.$$

The basis vectors are product vectors. A general superposition of them need not itself be a product vector.

Compact Ket Notation

Subsystem labels and tensor symbols are often compressed:

$$\begin{aligned} \lvert i\rangle_A\otimes\lvert j\rangle_B &= \lvert i\rangle_A\lvert j\rangle_B\\ &= \lvert i,j\rangle_{AB}\\ &= \lvert ij\rangle. \end{aligned}$$

The last form is safe only when the subsystem order and label boundaries are clear. For multi-digit labels, $\lvert i,j\rangle$ avoids ambiguity.

Bras follow the same convention:

$$\langle i,j\rvert = \langle i\rvert_A\otimes\langle j\rvert_B.$$

For product bras and kets,

$$\begin{aligned} (\langle a\rvert\otimes\langle b\rvert) (\lvert\psi\rangle\otimes\lvert\phi\rangle) &= \langle a\mid\psi\rangle \langle b\mid\phi\rangle. \end{aligned}$$

Default Ordering Convention

We write the subsystem labels in the same order as the tensor factors. If the space is written

$$\mathcal H_A\otimes\mathcal H_B,$$

then the first ket label belongs to $A$ and the second to $B$:

$$\lvert ij\rangle \equiv \lvert i\rangle_A\otimes\lvert j\rangle_B.$$

For two qubits, the default computational-basis order is

$$\lvert00\rangle,\quad \lvert01\rangle,\quad \lvert10\rangle,\quad \lvert11\rangle.$$

The left label changes more slowly than the right label. With zero-based indices, the pair $(i,j)$ is assigned the coordinate

$$k=i\,d_B+j.$$

Basis ordering determines the coordinates of vectors and matrices. Different software libraries may use little-endian, register-specific, or reversed conventions. Translation between them is a permutation of coordinates, not a physical operation, but formulas from different conventions cannot be mixed without that permutation.

For a worked permutation between conventions, see Tensor Product Ordering.

Kronecker Coordinates

In chosen bases, the tensor product of coordinate columns is their Kronecker product. For two qubit vectors,

$$\lvert\psi\rangle = \begin{pmatrix} \alpha\\ \beta \end{pmatrix}, \qquad \lvert\phi\rangle = \begin{pmatrix} \gamma\\ \delta \end{pmatrix},$$

the default ordering gives

$$\lvert\psi\rangle\otimes\lvert\phi\rangle = \begin{pmatrix} \alpha\gamma\\ \alpha\delta\\ \beta\gamma\\ \beta\delta \end{pmatrix}.$$

Expanding the same result in kets,

$$\lvert\psi\rangle\otimes\lvert\phi\rangle = \alpha\gamma\lvert00\rangle + \alpha\delta\lvert01\rangle + \beta\gamma\lvert10\rangle + \beta\delta\lvert11\rangle.$$

This coordinate rule is a convenient factorization test. A two-qubit coefficient array

$$\begin{pmatrix} c_{00}\\ c_{01}\\ c_{10}\\ c_{11} \end{pmatrix}$$

is a nonzero product vector exactly when the coefficient matrix

$$C= \begin{pmatrix} c_{00}&c_{01}\\ c_{10}&c_{11} \end{pmatrix}$$

has rank one. Product States develops this factorization test.

Tensor Products of Operators

Here take $A,B$ to be bounded operators (in particular, finite matrices). Then $A\otimes B$ is defined first on product vectors by

$$(A\otimes B) ( \lvert\psi\rangle\otimes\lvert\phi\rangle ) = (A\lvert\psi\rangle) \otimes (B\lvert\phi\rangle),$$

and then extended linearly and continuously to the completed tensor product. For unbounded operators, this action initially requires factors in the relevant domains; domains and closures must be specified before using operator identities.

The basic algebraic rules are

$$\begin{aligned} (A\otimes B)(C\otimes D) &= AC\otimes BD,\\ (A\otimes B)^\dagger &= A^\dagger\otimes B^\dagger. \end{aligned}$$

For boundedly invertible factors,

$$(A\otimes B)^{-1} = A^{-1}\otimes B^{-1}.$$

For finite-dimensional operators,

$$\operatorname{Tr}(A\otimes B) = \operatorname{Tr}(A)\operatorname{Tr}(B).$$

The matrix of $A\otimes B$ is the Kronecker product of the matrices. If

$$A= \begin{pmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{pmatrix},$$

then

$$A\otimes B = \begin{pmatrix} a_{11}B&a_{12}B\\ a_{21}B&a_{22}B \end{pmatrix}.$$

The blocks and their ordering follow the same product-basis convention used for state vectors.

Local Operators

An operator acting only on $A$ is represented on the whole space as

$$A\otimes I_B.$$

It acts as

$$(A\otimes I_B) ( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B ) = (A\lvert\psi\rangle_A) \otimes\lvert\phi\rangle_B.$$

Likewise, a $B$-local operator is $I_A\otimes B$. Operators on distinct factors commute:

$$[ A\otimes I_B,\, I_A\otimes B ] =0.$$

For two qubits,

$$\begin{aligned} (X\otimes I)\lvert00\rangle &= \lvert10\rangle,\\ (I\otimes X)\lvert00\rangle &= \lvert01\rangle. \end{aligned}$$

This pair is a quick ordering check: the left operator changes the first ket label, and the right operator changes the second.

Product operators probe joint properties. For example,

$$Z\otimes Z$$

has eigenvalue $+1$ when the two computational-basis labels agree and $-1$ when they differ.

Density Operators and Partial Traces

Independent subsystem density operators combine as

$$\rho_{AB} = \rho_A\otimes\rho_B.$$

Their trace is normalized because

$$\begin{aligned} \operatorname{Tr}(\rho_A\otimes\rho_B) &= \operatorname{Tr}(\rho_A) \operatorname{Tr}(\rho_B)\\ &=1. \end{aligned}$$

For trace-class factors $A,B$ (automatically true for finite matrices), the partial trace obeys the useful product rule

$$\operatorname{Tr}_B(A\otimes B) = A\,\operatorname{Tr}(B).$$

By linearity, this determines the partial trace of any finite sum of product operators. Reduced Density Operators explains its state interpretation; Partial Trace works through the index calculation.

Reversing and Regrouping Factors

The spaces $\mathcal H_A\otimes\mathcal H_B$ and $\mathcal H_B\otimes\mathcal H_A$ are naturally isomorphic, but their vectors are not literally identified without a map. The swap unitary is defined by

$$S_{AB} ( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B ) = \lvert\phi\rangle_B\otimes\lvert\psi\rangle_A.$$

Writing

$$\lvert\psi\rangle_A\otimes\lvert\phi\rangle_B \neq \lvert\phi\rangle_B\otimes\lvert\psi\rangle_A$$

without a swap map is generally incorrect: the two sides live in differently ordered tensor products.

For three factors, there is a canonical associating isomorphism

$$\begin{aligned} (\mathcal H_A\otimes\mathcal H_B) \otimes\mathcal H_C &\simeq \mathcal H_A\otimes (\mathcal H_B\otimes\mathcal H_C). \end{aligned}$$

Parentheses are therefore often suppressed:

$$\mathcal H_A\otimes\mathcal H_B\otimes\mathcal H_C.$$

The factor order is still retained. For $N$ qubits, the dimension is

$$\dim\left[ (\mathbb C^2)^{\otimes N} \right] = 2^N.$$

This exponential growth is one of the central practical facts of many-body quantum mechanics and quantum information.

Product Versus Nonproduct Vectors

The tensor-product space is spanned by product vectors, but not every vector is one product vector. For example,

$$\lvert\Phi^+\rangle = \frac{ \lvert00\rangle+\lvert11\rangle }{\sqrt2}$$

is a sum of product-basis vectors but cannot itself be written as

$$\lvert\psi\rangle_A\otimes\lvert\phi\rangle_B.$$

This distinction is analogous to matrices: every matrix is a sum of rank-one matrices, but not every matrix has rank one. For bipartite pure states, product vectors correspond to rank-one coefficient matrices; higher rank signals entanglement.

Entangled States develops the physical distinction; Schmidt Decomposition gives a systematic factorization test.

Example: Position and Spin

A nonrelativistic spin-$1/2$ particle has a spatial Hilbert space and a spin Hilbert space:

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

Equivalently, a state may be represented as a two-component spinor wavefunction:

$$\Psi(\mathbf x) = \begin{pmatrix} \psi_\uparrow(\mathbf x)\\ \psi_\downarrow(\mathbf x) \end{pmatrix}.$$

The inner product is

$$\langle\Phi\vert\Psi\rangle = \int_{\mathbb R^3} d^3x\, \bigl[ \phi_\uparrow(\mathbf x)^*\psi_\uparrow(\mathbf x) +\phi_\downarrow(\mathbf x)^*\psi_\downarrow(\mathbf x) \bigr].$$

This example shows that tensor factors need not correspond to separate particles. They may represent different degrees of freedom of one particle.

Example: Two Distinguishable Particles on a Line

For one spinless particle on a line, the Hilbert space is $L^2(\mathbb R)$. For two distinguishable spinless particles on a line,

$$\mathcal H_{12} = L^2(\mathbb R)\otimes L^2(\mathbb R) \cong L^2(\mathbb R^2).$$

A wavefunction can be written as

$$\Psi(x_1,x_2).$$

A product state has the special form

$$\Psi(x_1,x_2) = \psi(x_1)\phi(x_2).$$

More general wavefunctions are not products. Interactions, boundary conditions, and preparation procedures can produce states whose spatial degrees of freedom are entangled.

For identical particles, one must further restrict to symmetric or antisymmetric subspaces. That symmetrization postulate is separate from the bare tensor-product construction for distinguishable systems.

Relationship to Probability Spaces

Classical probability for two random variables uses a joint distribution $p_{ij}$. Independent distributions factor:

$$p_{ij}=p_i q_j.$$

Quantum theory assigns amplitudes before probabilities. In a product basis,

$$\lvert\Psi\rangle = \sum_{ij} c_{ij} \lvert i\rangle_A\lvert j\rangle_B.$$

Born probabilities are $|c_{ij}|^2$, but the amplitudes $c_{ij}$ also contain phase information. A product state has coefficients of the form

$$c_{ij}=a_i b_j.$$

An entangled pure state cannot be written that way. This is why entanglement is not merely “correlation with complex numbers.” It is a statement about factorization in Hilbert space before measurement probabilities are extracted.

Common pitfalls

A product basis does not imply a product state. Every finite-dimensional joint state expands in products of basis vectors. Factorization requires the entire coefficient matrix to have rank one, not merely a convenient choice of basis labels.

A change of ordering changes coordinates. A column or matrix built in AB order cannot be reused unchanged in BA order. Transform both state and operators with the same permutation.

Independent preparation is an assumption. Local devices can share a random preparation label and produce a correlated separable mixture. A product density operator describes preparations with no such shared correlations; see Separable Mixed States.

References

• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013, chapter 19, “Systems and Subsystems, Multiple Particles.” Publisher.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, section 2.1.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapter 1. Author manuscript.

Exercises

1. Expand

$$\left( \alpha\lvert0\rangle+\beta\lvert1\rangle \right) \otimes \left( \gamma\lvert0\rangle+\delta\lvert1\rangle \right)$$

in the default two-qubit basis.

Solution

Apply bilinearity in both factors:

$$\left( \alpha\lvert0\rangle+\beta\lvert1\rangle \right) \otimes \left( \gamma\lvert0\rangle+\delta\lvert1\rangle \right) = \alpha\gamma\lvert00\rangle + \alpha\delta\lvert01\rangle + \beta\gamma\lvert10\rangle + \beta\delta\lvert11\rangle.$$

The coordinate column is

$$\begin{pmatrix} \alpha\gamma\\ \alpha\delta\\ \beta\gamma\\ \beta\delta \end{pmatrix}.$$

2. In the default basis order, write the coordinate vector of

$$\frac{\lvert0\rangle+i\lvert1\rangle}{\sqrt2} \otimes \lvert1\rangle.$$

Solution

Expand the product:

$$\begin{aligned} \frac{\lvert0\rangle+i\lvert1\rangle}{\sqrt2} \otimes \lvert1\rangle &= \frac{1}{\sqrt2}\lvert01\rangle + \frac{i}{\sqrt2}\lvert11\rangle. \end{aligned}$$

In the order $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$, the coordinate vector is

$$\frac{1}{\sqrt2} \begin{pmatrix} 0\\ 1\\ 0\\ i \end{pmatrix}.$$

3. Compute $X\otimes Z$ in the default two-qubit basis and state its action on $\lvert01\rangle$.

Solution

Since

$$X= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}, \qquad Z= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix},$$

the block rule gives

$$X\otimes Z = \begin{pmatrix} 0&0&1&0\\ 0&0&0&-1\\ 1&0&0&0\\ 0&-1&0&0 \end{pmatrix}.$$

Using the operator action directly,

$$\begin{aligned} (X\otimes Z)\lvert01\rangle &= X\lvert0\rangle \otimes Z\lvert1\rangle\\ &= -\lvert11\rangle. \end{aligned}$$

4. For bounded $A$ and $B$, prove that $A\otimes I_B$ and $I_A\otimes B$ commute.

Solution

Use the multiplication rule:

$$\begin{aligned} (A\otimes I_B)(I_A\otimes B) &= A\otimes B,\\ (I_A\otimes B)(A\otimes I_B) &= A\otimes B. \end{aligned}$$

The two products are equal, so

$$[ A\otimes I_B,\, I_A\otimes B ] =0.$$

5. Let $\lvert a\rangle,\lvert a'\rangle\in\mathcal H_A$ and $\lvert b\rangle,\lvert b'\rangle\in\mathcal H_B$. If $\langle a\mid a'\rangle=0$, show that the corresponding product vectors are orthogonal regardless of $\langle b\mid b'\rangle$.

Solution

The product inner rule gives

$$\begin{aligned} \langle a\otimes b \mid a'\otimes b'\rangle &= \langle a\mid a'\rangle \langle b\mid b'\rangle\\ &= 0. \end{aligned}$$

One zero local overlap is enough to make the joint overlap vanish.

6. Determine whether

$$\lvert\Psi\rangle = \frac12\lvert00\rangle + \frac12\lvert01\rangle + \frac12\lvert10\rangle - \frac12\lvert11\rangle$$

is a product vector.

Solution

The coefficient matrix is

$$C= \frac12 \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}.$$

Its determinant is

$$\det C = -\frac12,$$

which is nonzero. Therefore $C$ has rank two, so the state is not a product vector. It is entangled.

7. Let $S_{AB}$ be the swap unitary and let $A,B$ be bounded operators. Show that

$$S_{AB}(A\otimes B)S_{AB}^\dagger = B\otimes A$$

by checking its action on a product vector.

Solution

Take $\lvert\phi\rangle_B\otimes\lvert\psi\rangle_A$ in the reversed-order space. Since $S_{AB}^\dagger$ swaps it into $A\otimes B$ order,

$$\begin{aligned} S_{AB}(A\otimes B)S_{AB}^\dagger \left( \lvert\phi\rangle_B\otimes\lvert\psi\rangle_A \right) &= S_{AB}(A\otimes B) \left( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B \right) \\ &\quad= S_{AB} \left( A\lvert\psi\rangle_A \otimes B\lvert\phi\rangle_B \right) \\ &\quad= B\lvert\phi\rangle_B \otimes A\lvert\psi\rangle_A. \end{aligned}$$

This is exactly the action of $B\otimes A$ in the reversed order.

8. Three subsystems have dimensions $2$, $3$, and $4$. Find the composite dimension and explain why regrouping the tensor products does not change it.

Solution

The dimension is

$$2\cdot3\cdot4=24.$$

Regrouping gives either

$$(2\cdot3)\cdot4=24$$

or

$$2\cdot(3\cdot4)=24.$$

The canonical associating isomorphism pairs

$$\left( \lvert i\rangle_A\otimes\lvert j\rangle_B \right) \otimes\lvert k\rangle_C$$

with

$$\lvert i\rangle_A \otimes \left( \lvert j\rangle_B\otimes\lvert k\rangle_C \right).$$

It changes parentheses, not factor order or physical content.

9. Decide whether the two-qubit state

$$\frac{1}{2} \bigl( \lvert00\rangle+\lvert01\rangle +\lvert10\rangle+\lvert11\rangle \bigr)$$

is a product state.

Solution

Yes. It factors as

$$\frac{1}{\sqrt2} \bigl( \lvert0\rangle+\lvert1\rangle \bigr) \otimes \frac{1}{\sqrt2} \bigl( \lvert0\rangle+\lvert1\rangle \bigr).$$

10. Why is $L^2(\mathbb R)\otimes L^2(\mathbb R)$ naturally identified with a space of functions of two variables?

Solution

Product wavefunctions have the form $\psi(x_1)\phi(x_2)$, which is a function of the pair $(x_1,x_2)$. Linear combinations of such products give more general square-integrable functions of two variables, and the Hilbert-space completion yields $L^2(\mathbb R^2)$.