Product States

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 $B$.

Helpful background. Tensor products explains how subsystem spaces combine; product bases supplies the coefficient array used to test factorization.

Product vectors and normalized states

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

$$(U_A\otimes U_B) ( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B ) = U_A\lvert\psi\rangle_A \otimes U_B\lvert\phi\rangle_B,$$

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. An active local unitary $U_A\otimes U_B$, with the basis held fixed, changes the coefficient matrix as

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

Multiplication by invertible matrices preserves rank. For a passive basis change, the matrices appear with inverse/conjugate conventions instead; the rank is again unchanged.

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 bounded 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 density operators and Schmidt decomposition.

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

$$\lvert\Psi\rangle = c_{00}\lvert00\rangle + c_{01}\lvert01\rangle + c_{10}\lvert10\rangle + c_{11}\lvert11\rangle.$$

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

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

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 rays determine the product ray without any phase convention. Choosing phases is needed only to write particular vector representatives; observable predictions depend on the phase-independent density operators.

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. For bounded local observables $A$ and $B$,

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

For unbounded observables, the corresponding moments must exist. Every well-defined connected local correlation then 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:

$$\lvert+\rangle\otimes\lvert-\rangle = \frac12\bigl( \lvert00\rangle-\lvert01\rangle + \lvert10\rangle-\lvert11\rangle \bigr),$$

yet it is manifestly product.

Product States and Local Dynamics

Local operators preserve product form whenever the output is nonzero:

$$(M_A\otimes N_B) ( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B ) = M_A\lvert\psi\rangle_A \otimes N_B\lvert\phi\rangle_B.$$

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 general definition and examples.

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.

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.

Equal basis probabilities can conceal entanglement

Compare the two normalized states

$$|\psi\rangle=\frac{|00\rangle+|01\rangle+|10\rangle+|11\rangle}{2},$$ $$|\chi\rangle=\frac{|00\rangle+|01\rangle+|10\rangle-|11\rangle}{2}.$$

Both give probability $1/4$ to each computational-basis outcome. Yet the first coefficient matrix has determinant zero and factors as $|+\rangle|+\rangle$, while the second has determinant $-1/2$ and is entangled. Its marginal is $I/2$.

The difference is a relative phase, not the probabilities in this one basis. This is why an entanglement test must retain complex amplitudes or use enough measurement settings to determine the needed coherences.

Common pitfalls

Superposition alone is not entanglement. Several product-basis terms may combine into one product of local superpositions. Test their coefficient matrix, including phases.

A zero determinant is only a two-qubit shortcut. A larger square matrix may have zero determinant while having rank greater than one; a rectangular matrix has no determinant. Rank one is the general product criterion.

Mixed product and separable are different. A mixed product state has no correlations. A separable mixture can have correlations without entanglement.

References

• Ingemar Bengtsson and Karol Życzkowski, Geometry of Quantum States, 2nd ed., Cambridge University Press, 2017, chapters on composite systems and entanglement.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, section 2.5.
• John Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapters 1 and 6. Author-hosted text.

Exercises

1. Factor the normalized state $(|00\rangle+i|01\rangle+2|10\rangle+2i|11\rangle)/\sqrt{10}$ into normalized one-qubit states.

Solution

The second row of the coefficient matrix is twice the first. Factoring and distributing the norm gives

$$|\Psi\rangle=\frac{|0\rangle+2|1\rangle}{\sqrt5} \otimes\frac{|0\rangle+i|1\rangle}{\sqrt2}.$$

The tensor-product coefficients use an ordinary product of amplitudes, with no complex conjugation of the second factor.

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

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

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:

$$(U_A\otimes U_B) ( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B ) = U_A\lvert\psi\rangle_A \otimes U_B\lvert\phi\rangle_B.$$

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.