Product States

This is the canonical treatment of product states: factorization tests, local statistics, dynamics, mixed-state distinctions, and multipartite scope. The composition-postulate prerequisite is Product States: Core First Encounter.

A pure state of a composite system is a product state when it assigns a pure state to each subsystem:

$$\lvert\Psi\rangle_{AB} = \lvert\psi\rangle_A \otimes \lvert\phi\rangle_B.$$

Product states describe independent pure preparations. They can contain local superpositions and other fully quantum behavior, but they contain no entanglement or correlation between $A$ and $B$.

Definition

Let

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

A nonzero vector $\lvert\Psi\rangle_{AB}$ is a product vector across the split $A\mid B$ if there exist nonzero vectors

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

such that

$$\lvert\Psi\rangle_{AB} = \lvert\psi\rangle_A \otimes \lvert\phi\rangle_B.$$

For a normalized joint state, the factors may be chosen normalized because

$$\lVert\Psi\rVert = \lVert\psi\rVert\,\lVert\phi\rVert.$$

The normalized factors are unique up to compensating phases. If

$$\lvert\Psi\rangle = \lvert\psi\rangle\otimes\lvert\phi\rangle = \lvert\psi'\rangle\otimes\lvert\phi'\rangle,$$

then, for normalized factors,

$$\begin{aligned} \lvert\psi'\rangle &= e^{i\alpha}\lvert\psi\rangle,\\ \lvert\phi'\rangle &= e^{-i\alpha}\lvert\phi\rangle \end{aligned}$$

for some real $\alpha$. An additional common phase changes only the representative of the physical ray.

A pure state that is not a product state across the chosen split is entangled.

Relative to a Subsystem Split

Productness is not a property of a vector without further structure. It is defined relative to a factorization

$$\mathcal H \simeq \mathcal H_A\otimes\mathcal H_B.$$

The same abstract Hilbert space can admit different identifications of subsystems. A vector may be product relative to one factorization and entangled relative to another.

Once the split is fixed, changing local bases does not change productness. If $U_A$ and $U_B$ are unitary, then

$$\begin{aligned} & (U_A\otimes U_B) ( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B )\\ &\qquad= U_A\lvert\psi\rangle_A \otimes U_B\lvert\phi\rangle_B, \end{aligned}$$

which is still a product state. Productness is therefore invariant under local unitary changes of basis.

A global unitary that does not factor as $U_A\otimes U_B$ can map product states to entangled states.

Coefficient Factorization

Choose product bases $\{\lvert i\rangle_A\}$ and $\{\lvert j\rangle_B\}$. A general pure state is

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

If the local factors are

$$\begin{aligned} \lvert\psi\rangle_A &= \sum_i a_i\lvert i\rangle_A,\\ \lvert\phi\rangle_B &= \sum_j b_j\lvert j\rangle_B, \end{aligned}$$

then bilinearity gives

$$\begin{aligned} \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B &= \sum_{i,j} a_i b_j \lvert i\rangle_A\otimes\lvert j\rangle_B. \end{aligned}$$

Thus a product state has coefficients

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

Organize the coefficients into the matrix

$$C=(c_{ij}).$$

The state is product exactly when the nonzero matrix $C$ has rank one. In that case,

$$C=\boldsymbol a\,\boldsymbol b^{\mathsf T},$$

where $\boldsymbol a$ and $\boldsymbol b$ collect the local amplitudes.

The matrix rank does not depend on local basis choices. Under local unitary changes,

$$C \longmapsto U_A C U_B^{\mathsf T},$$

and multiplication by invertible matrices preserves rank.

Equivalent Pure-State Criteria

For a normalized bipartite pure state, the following statements are equivalent:

1. $\lvert\Psi\rangle$ factors as $\lvert\psi\rangle_A\otimes\lvert\phi\rangle_B$.
2. Its coefficient matrix has rank one.
3. Its Schmidt rank is one.
4. Its reduced state $\rho_A$ has rank one.
5. Its reduced state $\rho_B$ has rank one.
6. Either reduced state is pure: $\operatorname{Tr}(\rho_A^2)=\operatorname{Tr}(\rho_B^2)=1$.
7. Every product-observable expectation factorizes.

The coefficient test is usually fastest for small pure states. Reduced-state purity is convenient when density operators are already in use. Schmidt rank is the most structural criterion. The proofs involving reduced states and Schmidt decomposition are developed in Reduced States and Schmidt Decomposition Overview.

These criteria apply to pure bipartite states. Mixed states require a different separability definition.

Two-Qubit Determinant Test

Write a general two-qubit pure state as

$$\begin{aligned} \lvert\Psi\rangle &= c_{00}\lvert00\rangle + c_{01}\lvert01\rangle\\ &\quad+ c_{10}\lvert10\rangle + c_{11}\lvert11\rangle. \end{aligned}$$

Its coefficient matrix is

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

A nonzero $2\times2$ matrix has rank one exactly when its determinant vanishes. Therefore

$$\begin{aligned} \lvert\Psi\rangle \text{ is product} &\quad\Longleftrightarrow\\ c_{00}c_{11}-c_{01}c_{10} &=0. \end{aligned}$$

For example, a candidate product

$$\left( a\lvert0\rangle+b\lvert1\rangle \right) \otimes \left( c\lvert0\rangle+d\lvert1\rangle \right)$$

expands to

$$\begin{aligned} ac\lvert00\rangle + ad\lvert01\rangle + bc\lvert10\rangle + bd\lvert11\rangle. \end{aligned}$$

The determinant condition follows:

$$(ac)(bd)-(ad)(bc)=0.$$

The test also reconstructs factors. If $c_{00}\neq0$ and the determinant vanishes, then

$$\begin{aligned} \lvert\Psi\rangle &= \left( c_{00}\lvert0\rangle+c_{10}\lvert1\rangle \right)\\ &\quad\otimes \left( \lvert0\rangle + \frac{c_{01}}{c_{00}}\lvert1\rangle \right). \end{aligned}$$

Overall normalization can then be redistributed between the two factors. If $c_{00}=0$, choose any nonzero row or column as the pivot.

Local States

For a normalized product pure state,

$$\rho_{AB} = \lvert\Psi\rangle\langle\Psi\rvert = \rho_A\otimes\rho_B,$$

where

$$\begin{aligned} \rho_A &= \lvert\psi\rangle\langle\psi\rvert,\\ \rho_B &= \lvert\phi\rangle\langle\phi\rvert. \end{aligned}$$

Taking either partial trace returns the other factor:

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

Each subsystem therefore has its own pure state vector. This is the defining contrast with an entangled pure state, whose reduced states are mixed.

The local vectors determine only the product ray after a phase convention is chosen. Observable predictions depend on $\rho_A$ and $\rho_B$, which are phase independent.

Factorized Local Statistics

Let $E_a$ be a measurement effect on $A$ and $F_b$ an effect on $B$. In a product state,

$$\begin{aligned} p(a,b) &= \operatorname{Tr}\!\left[ (\rho_A\otimes\rho_B) (E_a\otimes F_b) \right]\\ &= \operatorname{Tr}(\rho_AE_a) \operatorname{Tr}(\rho_BF_b)\\ &= p_A(a)p_B(b). \end{aligned}$$

This factorization holds for every pair of local measurements. More generally, for local observables $A$ and $B$,

$$\langle A\otimes B\rangle = \langle A\rangle_A \langle B\rangle_B.$$

Hence every connected local correlation vanishes:

$$\begin{aligned} \langle A\otimes B\rangle - \langle A\otimes I\rangle \langle I\otimes B\rangle &=0. \end{aligned}$$

One uncorrelated measurement setting does not prove that a state is product. The criterion requires factorization for a tomographically complete set, or equivalently for all local observables.

Local Superposition Is Not Entanglement

The state

$$\lvert0\rangle_A \otimes \frac{ \lvert0\rangle_B+\lvert1\rangle_B }{\sqrt2}$$

is a product state. Subsystem $B$ is in a coherent superposition, while subsystem $A$ is sharp in the computational basis.

Expanding the tensor product gives

$$\frac{ \lvert00\rangle+\lvert01\rangle }{\sqrt2}.$$

The presence of several product-basis terms is not an entanglement test. What matters is whether their coefficients factor into one set of amplitudes for $A$ and one for $B$.

Similarly,

$$\lvert+\rangle_A\otimes\lvert-\rangle_B$$

contains four computational-basis terms:

$$\begin{aligned} \lvert+\rangle\otimes\lvert-\rangle &= \frac12\bigl( \lvert00\rangle-\lvert01\rangle\\ &\quad+ \lvert10\rangle-\lvert11\rangle \bigr), \end{aligned}$$

yet it is manifestly product.

Product States and Local Dynamics

Local operators preserve product form whenever the output is nonzero:

$$\begin{aligned} & (M_A\otimes N_B) ( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B )\\ &\qquad= M_A\lvert\psi\rangle_A \otimes N_B\lvert\phi\rangle_B. \end{aligned}$$

In particular, local unitary evolution

$$U_A(t)\otimes U_B(t)$$

cannot create entanglement from a product state. A coupling or other nonproduct operation is required to turn an initially product pure state into an entangled one.

The converse is also useful: local unitaries cannot remove entanglement from an entangled pure state. They change local bases while preserving Schmidt coefficients.

Product States Versus Product Measurements

A product state is a property of a preparation:

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

A product measurement is a property of a measurement effect:

$$E_a\otimes F_b.$$

These notions are independent:

• product measurements can be performed on entangled states;
• product states can be measured by global, nonproduct measurements;
• product measurements on product states give factorized probabilities;
• product measurements on nonproduct states may or may not reveal correlations in a chosen setting.

The apparatus structure does not determine whether the input state is product.

Pure Product, Mixed Product, and Separable

For density operators, a product state has the form

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

The factors may themselves be mixed. This is stronger than separability.

A separable state may be a correlated mixture of products:

$$\rho_{AB} = \sum_k p_k\, \rho_A^{(k)} \otimes \rho_B^{(k)}.$$

For example,

$$\rho_{\mathrm{cc}} = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert$$

is separable but not product. Its computational-basis outcomes are classically correlated.

Thus the pure-state dichotomy

$$\text{product} \quad\text{or}\quad \text{entangled}$$

does not extend by replacing kets with arbitrary density operators. Mixed-state entanglement is defined by the absence of every separable decomposition. Continue to Separable Mixed States for the canonical treatment.

Multipartite Product Structure

For $N$ labeled subsystems, a fully product pure state is

$$\lvert\Psi\rangle = \bigotimes_{r=1}^{N} \lvert\psi_r\rangle.$$

Multipartite states can have intermediate factorization structure. For example,

$$\lvert0\rangle_A \otimes \frac{ \lvert00\rangle_{BC}+\lvert11\rangle_{BC} }{\sqrt2}$$

is product across $A\mid BC$ but is not fully product because $B$ and $C$ are entangled.

Therefore a multipartite claim should specify the partition: fully product, product across a chosen bipartition, or entangled across that bipartition.

Practical Factorization Workflow

1. State the subsystem split, such as $A\mid B$.
2. Put coefficients in a consistently ordered product basis.
3. Build the coefficient matrix $C=(c_{ij})$.
4. Check whether $C$ has rank one.
5. For two qubits, use $\det C=0$.
6. Reconstruct local factors from a nonzero row or column.
7. Normalize the factors and track compensating phases.
8. If density matrices are available, verify that either reduced state is pure.

Product States Under Interactions

An interaction term can turn a product state into an entangled state. For example,

$$H_{\text{int}} = J\,\sigma_z\otimes\sigma_z$$

produces phases that depend on joint computational-basis labels. Acting for a suitable time on a product superposition, such a coupling can produce a nonproduct state.

Not every interaction entangles every product state. If the initial product state is an eigenstate of the interaction, or if the coupling acts trivially on the occupied subspace, the state may remain product. Entanglement generation is a dynamical question, not a property of the Hamiltonian name alone.

Examples Across Degrees of Freedom

Product structure is not restricted to qubit labels. For position and spin,

$$\Psi(\mathbf x) = \psi(\mathbf x)\chi$$

is product across the position–spin split when the spinor $\chi$ is independent of $\mathbf x$. A position-dependent spinor generally does not factor this way.

For two distinguishable spinless particles on a line,

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

is product across the particle split. These examples emphasize that factorization is always relative to a specified tensor-product decomposition, not to a preferred notation or basis.

Non-Examples

The Bell state

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr)$$

is not a product state. Its coefficient matrix has rank two.

The state

$$\frac{1}{\sqrt2} \bigl( \lvert01\rangle+\lvert10\rangle \bigr)$$

is also not a product state. It is a coherent superposition of two product basis states, and the coefficient array cannot be factored as $a_i b_j$.

Common Mistakes

• Thinking that every superposition in a product basis is entangled.
• Thinking that product means classical or non-quantum.
• Forgetting to specify the subsystem split.
• Assuming productness depends on the chosen local basis.
• Confusing a product state with a product basis, operator, or measurement.
• Using one uncorrelated measurement setting as proof that a state is product.
• Applying the two-qubit determinant test to a coefficient array that is not $2\times2$.
• Forgetting that rank one, not zero determinant alone, is the general-dimensional criterion.
• Calling every separable mixed state a product state.
• Assuming local unitaries can create or destroy pure-state entanglement.
• Assigning separate pure state vectors to the parts of an entangled pure state.

Cross-Links

• Composite Systems
• Tensor Products
• Bipartite Systems
• Entangled States
• Reduced States
• Schmidt Decomposition Overview
• Subsystems and Local Observables
• Local Unitary Equivalence

References

• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, ch. 10.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, ch. 3.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer Academic, 1995, chs. 3 and 5.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, sec. 2.5.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, ch. 1.
• I. Bengtsson and K. Życzkowski, Geometry of Quantum States, 2nd ed., Cambridge University Press, 2017, chs. 9 and 15.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press (1958).

Exercises

1. Factor

$$\frac{ \lvert00\rangle+\lvert01\rangle }{\sqrt2}$$

into one-qubit states.

Solution

Both terms have the first qubit in $\lvert0\rangle$, so

$$\frac{ \lvert00\rangle+\lvert01\rangle }{\sqrt2} = \lvert0\rangle \otimes \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}.$$

2. Use the determinant test to classify

$$\lvert\Psi\rangle = \frac{ \lvert00\rangle+\lvert01\rangle +\lvert10\rangle+\lvert11\rangle }{2}.$$

Solution

The coefficient matrix is

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

Its determinant vanishes, so it has rank one. Explicitly,

$$\lvert\Psi\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2} \otimes \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}.$$

3. Determine whether

$$\lvert\chi\rangle = \frac12 \left( \lvert00\rangle-\lvert01\rangle +\lvert10\rangle-\lvert11\rangle \right)$$

is product, and find its local factors.

Solution

The coefficient matrix is

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

Its rows are identical, so it has rank one. Factoring gives

$$\begin{aligned} \lvert\chi\rangle &= \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2} \otimes \frac{ \lvert0\rangle-\lvert1\rangle }{\sqrt2} \\ &= \lvert+\rangle\otimes\lvert-\rangle. \end{aligned}$$

4. Let $\rho_{AB}=\rho_A\otimes\rho_B$. Prove that local measurement effects $E_a$ and $F_b$ have factorized joint probabilities.

Solution

The joint probability is

$$\begin{aligned} p(a,b) &= \operatorname{Tr}\!\left[ (\rho_A\otimes\rho_B) (E_a\otimes F_b) \right]\\ &= \operatorname{Tr}\!\left[ (\rho_AE_a)\otimes(\rho_BF_b) \right]\\ &= \operatorname{Tr}(\rho_AE_a) \operatorname{Tr}(\rho_BF_b)\\ &= p_A(a)p_B(b). \end{aligned}$$

The third line uses the trace rule for tensor-product operators.

5. Show that the Bell state

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

is not product by computing the reduced state of subsystem $A$.

Solution

The joint density operator is

$$\begin{aligned} \rho_{AB} &= \frac12\bigl( \lvert00\rangle\langle00\rvert + \lvert00\rangle\langle11\rvert\\ &\quad+ \lvert11\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \bigr). \end{aligned}$$

Tracing out $B$ removes the cross terms because $\langle0\mid1\rangle=0$:

$$\rho_A = \frac12\lvert0\rangle\langle0\rvert + \frac12\lvert1\rangle\langle1\rvert = \frac{I}{2}.$$

Its purity is

$$\operatorname{Tr}(\rho_A^2)=\frac12<1.$$

A bipartite pure state is product only if its reduced states are pure, so the Bell state is entangled.

6. Explain why

$$\rho_{\mathrm{cc}} = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert$$

is separable but not product.

Solution

It is separable because the displayed expression is a convex mixture of the two product states $\lvert00\rangle$ and $\lvert11\rangle$.

Its reduced states are both $I/2$. If it were the product of those marginals, then

$$\rho_A\otimes\rho_B = \frac{I}{2}\otimes\frac{I}{2} = \frac{I_4}{4},$$

which assigns probability $1/4$ to all four computational-basis outcomes. By contrast, $\rho_{\mathrm{cc}}$ assigns probability $1/2$ to $00$ and $11$ and zero to $01$ and $10$. It is correlated and therefore not product.

7. Prove that a local unitary $U_A\otimes U_B$ maps every product pure state to another product pure state. Does the converse evolution create entanglement?

Solution

Apply the operator to the product:

$$\begin{aligned} & (U_A\otimes U_B) ( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B )\\ &\qquad= U_A\lvert\psi\rangle_A \otimes U_B\lvert\phi\rangle_B. \end{aligned}$$

The result is explicitly product. The inverse is also local:

$$(U_A\otimes U_B)^\dagger = U_A^\dagger\otimes U_B^\dagger.$$

Therefore local unitary evolution neither creates entanglement from a product state nor removes entanglement from an entangled pure state.

8. Consider

$$\lvert\Omega\rangle = \lvert0\rangle_A \otimes \frac{ \lvert00\rangle_{BC}+\lvert11\rangle_{BC} }{\sqrt2}.$$

Classify its product structure across $A\mid BC$ and across $B\mid AC$.

Solution

Across $A\mid BC$, the state is explicitly product:

$$\lvert\Omega\rangle = \lvert0\rangle_A \otimes \lvert\Phi^+\rangle_{BC}.$$

Across $B\mid AC$, write

$$\lvert\Omega\rangle = \frac{ \lvert0\rangle_B\lvert00\rangle_{AC} + \lvert1\rangle_B\lvert01\rangle_{AC} }{\sqrt2}.$$

The two $AC$ states are orthogonal, so this expression has Schmidt rank two across $B\mid AC$. It is entangled across that split.

The state is therefore product across one bipartition without being fully product.

Additional exercises retained from the earlier canonical treatment

1. Show that

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

is a product state.

Solution

It factors as

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

2. For a product state $\rho_{AB}=\rho_A\otimes\rho_B$, compute $\operatorname{Tr}_B\rho_{AB}$.

Solution

Using trace factorization,

$$\operatorname{Tr}_B(\rho_A\otimes\rho_B) = \rho_A\,\operatorname{Tr}_B(\rho_B).$$

Since $\rho_B$ is normalized, $\operatorname{Tr}_B(\rho_B)=1$, so the result is $\rho_A$.

3. Is the state

$$\frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert10\rangle \bigr)$$

product or entangled?

Solution

It is product:

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

It is a superposition in the product basis, but it factors into one-qubit states.