Bipartite Systems

A bipartite system is a composite system with two named subsystems, usually called $A$ and $B$. Its Hilbert space is

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

An amplitude now needs two labels: one for each part. Arranging those amplitudes in a matrix makes local probabilities and reduced states easy to calculate without constructing the larger joint density matrix.

Helpful background. Tensor products explains product-basis vectors; product states explains factorization. The matrix formulas below assume finite-dimensional distinguishable factors. The final position–spin example extends the notation to wavefunctions.

Subsystems A and B

The labels $A$ and $B$ identify the two tensor factors. They may denote two particles, two qubits, two modes, spin and position, a system and an ancilla, or a subsystem and an environment. The labels are not merely decorative: they determine which Hilbert space an operator acts on and how basis labels are ordered.

The standard convention in this chapter is

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

with the $A$ label written first and the $B$ label second. If a calculation uses the opposite order, it must say so before writing compact basis labels.

For more than two subsystems, the same idea generalizes, but bipartite systems are special because they have powerful tools such as the Schmidt decomposition.

For finite-dimensional factors, write

$$d_A=\dim\mathcal H_A, \qquad d_B=\dim\mathcal H_B.$$

Then

$$\dim\mathcal H_{AB}=d_A d_B.$$

The subsystem labels also determine dimensions. An operator on $A$ is a $d_A\times d_A$ matrix, whereas an operator on $AB$ is a $(d_A d_B)\times(d_A d_B)$ matrix.

Changing the order of tensor factors is not a typographical rearrangement. The swap map

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

is a unitary map from $\mathcal H_A\otimes\mathcal H_B$ to $\mathcal H_B\otimes\mathcal H_A$. Coordinates and matrix representations must be reordered with it. This relabeling does not require physically exchanging the subsystems. The coordinate convention is described in Tensor-Product Ordering.

Product Basis

Choose orthonormal bases

$$\bigl\{ \lvert i\rangle_A \bigr\}_{i=0}^{d_A-1}, \qquad \bigl\{ \lvert j\rangle_B \bigr\}_{j=0}^{d_B-1}.$$

Their tensor products

$$\lvert i,j\rangle \equiv \lvert i\rangle_A\otimes\lvert j\rangle_B$$

form an orthonormal basis of $\mathcal H_{AB}$. Indeed,

$$\langle i',j'\vert i,j\rangle = \delta_{i'i}\delta_{j'j},$$

and completeness reads

$$\sum_{i=0}^{d_A-1} \sum_{j=0}^{d_B-1} \lvert i,j\rangle\langle i,j\rvert = I_A\otimes I_B.$$

One often suppresses the comma and subsystem labels,

$$\lvert i,j\rangle = \lvert ij\rangle,$$

but the compact notation is meaningful only after the subsystem order has been fixed.

For two qubits, the computational product basis is

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

Here $\lvert01\rangle$ means

$$\lvert0\rangle_A\otimes\lvert1\rangle_B,$$

not a two-digit number and not a single qubit state.

For a qubit $A$ and a qutrit $B$, the product basis has six vectors:

$$\begin{gathered} \lvert0,0\rangle,\; \lvert0,1\rangle,\; \lvert0,2\rangle,\\ \lvert1,0\rangle,\; \lvert1,1\rangle,\; \lvert1,2\rangle. \end{gathered}$$

This example is a useful reminder that the two factors need not have equal dimensions.

Flattened coordinate order

With the $A$ index first and the $B$ index changing fastest, the pair $(i,j)$ is assigned the zero-based coordinate

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

For two qubits,

$$\begin{aligned} (0,0)&\longleftrightarrow 0, & (0,1)&\longleftrightarrow 1,\\ (1,0)&\longleftrightarrow 2, & (1,1)&\longleftrightarrow 3. \end{aligned}$$

Thus the ordered basis is

$$\bigl( \lvert00\rangle, \lvert01\rangle, \lvert10\rangle, \lvert11\rangle \bigr).$$

Other software or references may flatten indices differently. State the ordering before comparing component vectors or Kronecker-product matrices.

Coefficient Matrix

In a chosen product basis, a finite-dimensional bipartite pure state has the unique expansion

$$\lvert\Psi\rangle = \sum_{i=0}^{d_A-1} \sum_{j=0}^{d_B-1} C_{ij}\, \lvert i\rangle_A\otimes\lvert j\rangle_B.$$

The amplitudes form a $d_A\times d_B$ coefficient matrix $C$: the $A$ index labels rows and the $B$ index labels columns. In the flattened order above, the coordinate column of $\lvert\Psi\rangle$ is

$$\begin{pmatrix} C_{00}\\ C_{01}\\ \vdots\\ C_{0,d_B-1}\\ C_{10}\\ \vdots\\ C_{d_A-1,d_B-1} \end{pmatrix}.$$

Normalization is the Frobenius-norm condition

$$\langle\Psi\vert\Psi\rangle = \sum_{i,j}\lvert C_{ij}\rvert^2 = \operatorname{Tr}(C^\dagger C) =1.$$

If another state $\lvert\Phi\rangle$ has coefficient matrix $D$, then

$$\langle\Phi\vert\Psi\rangle = \sum_{i,j}D_{ij}^*C_{ij} = \operatorname{Tr}(D^\dagger C).$$

For two qubits,

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

and the coefficient matrix is

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

The matrix depends on the chosen local bases, but its rank is invariant under invertible changes of local basis. This is why matrix rank is useful for diagnosing product versus entangled pure states.

The coefficient matrix also packages partial amplitudes. Projecting the $B$ factor onto $\lvert j\rangle_B$ selects column $j$:

$$\bigl( I_A\otimes{}_B\langle j\rvert \bigr) \lvert\Psi\rangle = \sum_i C_{ij}\lvert i\rangle_A.$$

Projecting $A$ onto $\lvert i\rangle_A$ selects row $i$:

$$\bigl( {}_A\langle i\rvert\otimes I_B \bigr) \lvert\Psi\rangle = \sum_j C_{ij}\lvert j\rangle_B.$$

The matrix $C$ is a reshaping of a state vector. It is not the joint density matrix, which is

$$\rho_{AB} = \lvert\Psi\rangle\langle\Psi\rvert$$

and has dimension $d_A d_B$ by $d_A d_B$.

Product Versus Entangled States

A product pure state has the form

$$\lvert\Psi\rangle = \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B.$$

If

$$\lvert\psi\rangle_A = \sum_i a_i\lvert i\rangle_A, \qquad \lvert\phi\rangle_B = \sum_j b_j\lvert j\rangle_B,$$

then

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

So a product pure state has a rank-one coefficient matrix. Conversely, a nonzero finite-dimensional bipartite pure state with rank-one coefficient matrix is product.

For two qubits, this becomes the determinant test:

$$\det C = c_{00}c_{11}-c_{01}c_{10}.$$

The state is product exactly when

$$c_{00}c_{11}=c_{01}c_{10}.$$

If the coefficient matrix has rank greater than one, the state is entangled across the chosen $A$-versus-$B$ split. The Schmidt Decomposition Overview is the basis-independent way to put this statement into canonical form.

Operators on Two Subsystems

An operator that acts only on $A$ is embedded into the joint space as

$$X_A\longmapsto X_A\otimes I_B.$$

Similarly, an operator local to $B$ is

$$Y_B\longmapsto I_A\otimes Y_B.$$

Their action on a product vector is

$$\bigl( X_A\otimes Y_B \bigr) \bigl( \lvert\psi\rangle_A \otimes \lvert\phi\rangle_B \bigr) = \bigl( X_A\lvert\psi\rangle_A \bigr) \otimes \bigl( Y_B\lvert\phi\rangle_B \bigr).$$

Matrix elements factor in a product basis:

$$\langle i',j'\rvert \bigl( X_A\otimes Y_B \bigr) \lvert i,j\rangle = \langle i'\rvert X_A\lvert i\rangle \times \langle j'\rvert Y_B\lvert j\rangle.$$

Operators local to different factors commute:

$$\left[ X_A\otimes I_B,\, I_A\otimes Y_B \right] =0.$$

This algebraic commutativity does not say that all joint states are uncorrelated. It says that the two local operations can be applied in either order.

If $\lvert\Psi\rangle$ has coefficient matrix $C$, then

$$\bigl( X_A\otimes Y_B \bigr) \lvert\Psi\rangle \quad\longleftrightarrow\quad C' = X_A C Y_B^{\mathsf T}.$$

The transpose appears because the $B$ index labels the columns of $C$. Consequently,

$$\langle\Psi\rvert X_A\otimes Y_B \lvert\Psi\rangle = \operatorname{Tr} \left( C^\dagger X_A C Y_B^{\mathsf T} \right).$$

A general joint operator need not factor as one tensor product. In finite dimensions it can be expanded as a sum,

$$O_{AB} = \sum_\mu X_A^{(\mu)} \otimes Y_B^{(\mu)}.$$

Terms that cannot be reduced to an $A$-only contribution plus a $B$-only contribution describe genuinely joint couplings or observables. The detailed operator calculus belongs to Operators on Composite Systems.

Reduced States from the Coefficient Matrix

For a pure bipartite state,

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

the reduced density matrices can be read directly from $C$. Their components are

$$\begin{aligned} (\rho_A)_{ik} &= \sum_j C_{ij}C_{kj}^*,\\ (\rho_B)_{j\ell} &= \sum_i C_{ij}C_{i\ell}^*. \end{aligned}$$

With the convention that $A$ labels rows and $B$ labels columns,

$$\rho_A=CC^\dagger, \qquad \rho_B=C^{\mathsf T}C^*.$$

The transpose in the second formula is convention-dependent. It is the correct form for the coefficient matrix and row-column assignment used on this page. Writing the component formulas first is the safest way to translate between conventions.

Both reduced states are positive, have unit trace, and share the same nonzero eigenvalues. These eigenvalues are the squared singular values of $C$, or equivalently the squared Schmidt coefficients.

For a product state $C=\mathbf a\mathbf b^{\mathsf T}$, both reduced states have rank one. For an entangled pure state, their rank exceeds one. The Reduced States page explains their physical interpretation; Partial Trace: First Encounter derives the trace operation itself.

Local Measurements

A measurement performed only on subsystem $A$ is represented on the composite system by adding the identity on $B$. If $\{E_a\}$ is a POVM on $A$, then

$$E_a\ge0, \qquad \sum_a E_a=I_A,$$

and the corresponding joint-space effects are

$$E_a\otimes I_B.$$

For a joint density operator $\rho_{AB}$, the probability of outcome $a$ is

$$p(a) = \operatorname{Tr}_{AB} \left[ \rho_{AB} \bigl( E_a\otimes I_B \bigr) \right].$$

If $A$ and $B$ are measured with local POVMs $\{E_a\}$ and $\{F_b\}$, with no intervening interaction or outcome-dependent change of settings, their joint probability is

$$p(a,b) = \operatorname{Tr}_{AB} \left[ \rho_{AB} \bigl( E_a\otimes F_b \bigr) \right].$$

The marginal probability for $A$ follows by summing over the complete set of $B$ outcomes:

$$\begin{aligned} \sum_b p(a,b) &= \operatorname{Tr}_{AB} \left[ \rho_{AB} \left( E_a\otimes\sum_b F_b \right) \right]\\ &= \operatorname{Tr}_{AB} \left[ \rho_{AB} \bigl( E_a\otimes I_B \bigr) \right]\\ &= p(a). \end{aligned}$$

The local statistics of $A$ can therefore be computed from its reduced state

$$\rho_A=\operatorname{Tr}_B\rho_{AB}$$

using

$$p(a) = \operatorname{Tr}_A(\rho_A E_a).$$

Conditional outcomes and states

When $p_B(b)>0$, the conditional probability is

$$\begin{aligned} p(a\mid b) &= \frac{p(a,b)}{p_B(b)},\\ p_B(b) &= \sum_a p(a,b). \end{aligned}$$

For a projective measurement $\{Q_b\}$ on $B$, the unnormalized conditional state of $A$ is

$$\widetilde{\rho}_{A|b} = \operatorname{Tr}_B \left[ \bigl( I_A\otimes Q_b \bigr) \rho_{AB} \bigl( I_A\otimes Q_b \bigr) \right].$$

Its trace is the outcome probability,

$$p_B(b) = \operatorname{Tr}_A \widetilde{\rho}_{A|b},$$

and the normalized conditional state is

$$\rho_{A|b} = \frac{ \widetilde{\rho}_{A|b} }{ p_B(b) }.$$

The state of the measured subsystem $B$, or of the full pair after measurement, requires an instrument. In contrast, the conditional state of the untouched $A$ depends only on the effects and the initial joint state: for an effect $F_b$, replace each $Q_b$ in the displayed partial trace by $\sqrt{F_b}$. All local instruments with that effect give the same reduced conditional state of $A$; see the reference-system calculation.

Marginal and conditional statistics answer different questions. Entanglement can make $\rho_{A|b}$ depend strongly on $b$, while the unconditioned state $\rho_A$ remains independent of which complete local measurement is performed on $B$. That is the operational core of no-signaling.

Examples

Product two-qubit state

The two-qubit product state

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

has coefficient matrix

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

Its determinant vanishes:

$$(\alpha\gamma)(\beta\delta) - (\alpha\delta)(\beta\gamma) = 0.$$

The Bell state

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

has coefficient matrix

$$C_{\Phi^+} = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$

Its determinant is $1/2$, so the state is entangled.

Its reduced states follow immediately from the coefficient matrix:

$$\rho_A = C_{\Phi^+}C_{\Phi^+}^\dagger = \frac12 I_A,$$

and likewise $\rho_B=I_B/2$.

Qubit–qutrit system

Let $\mathcal H_A=\mathbb C^2$ and $\mathcal H_B=\mathbb C^3$, and consider

$$\lvert\Psi\rangle = \frac{ \lvert0,0\rangle +\lvert0,2\rangle +\lvert1,1\rangle }{\sqrt3}.$$

Its $2\times3$ coefficient matrix is

$$C = \frac{1}{\sqrt3} \begin{pmatrix} 1&0&1\\ 0&1&0 \end{pmatrix}.$$

The two rows are linearly independent, so the state is entangled. Measuring $A$ in its computational basis gives

$$p_A(0)=\frac23, \qquad p_A(1)=\frac13.$$

Conditioned on outcome $0$, the qutrit is in

$$\lvert\phi_{B|0}\rangle = \frac{ \lvert0\rangle_B+\lvert2\rangle_B }{\sqrt2}.$$

Conditioned on outcome $1$, it is in $\lvert1\rangle_B$. The reduced states are

$$\rho_A = \begin{pmatrix} 2/3&0\\ 0&1/3 \end{pmatrix},$$

and

$$\rho_B = \frac13 \begin{pmatrix} 1&0&1\\ 0&1&0\\ 1&0&1 \end{pmatrix}.$$

Their nonzero eigenvalues are $2/3$ and $1/3$, as required.

Position and spin

A bipartite split need not describe two particles. For one spin-$1/2$ particle,

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

separates position and spin. A general state can be written formally as

$$\begin{aligned} \lvert\Psi\rangle &= \int d^3x\, \lvert\mathbf x\rangle \otimes \lvert\chi(\mathbf x)\rangle,\\ \lvert\chi(\mathbf x)\rangle &= \psi_\uparrow(\mathbf x)\lvert\uparrow\rangle + \psi_\downarrow(\mathbf x)\lvert\downarrow\rangle. \end{aligned}$$

Normalization requires

$$\int d^3x\, \left( \lvert\psi_\uparrow(\mathbf x)\rvert^2 + \lvert\psi_\downarrow(\mathbf x)\rvert^2 \right) =1.$$

The probability of spin up, without resolving position, is

$$p(\uparrow) = \int d^3x\, \lvert\psi_\uparrow(\mathbf x)\rvert^2.$$

This is the continuous-index analogue of summing over the unobserved position index (the first tensor factor in this example). The state is product across position and spin only when both component wavefunctions are proportional to one common spatial wavefunction.

Common Pitfalls

Confusing $C$ with the density matrix. The coefficient matrix has size $d_A\times d_B$ and describes a pure-state vector. The joint density matrix has size $d_Ad_B\times d_Ad_B$ and also describes mixed states.

Losing the complex conjugation in a reduced state. In this convention, $\rho_B=C^{\mathsf T}C^*$, not $C^\dagger C$. They have the same eigenvalues but generally different entries. Exercise 3 exposes the difference using a complex product state.

Using one determinant for a rectangular matrix. The two-qubit determinant test is a shortcut for a $2\times2$ coefficient matrix. For larger factors, use rank or the singular values.

Confusing a marginal with a selected subensemble. Summing all remote outcomes recovers the original local state. Selecting one outcome can change the conditional local state, but requires access to that record.

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.

Exercises

1. In the $A$-first convention, what does $\lvert10\rangle$ mean?

Solution

It means

$$\lvert10\rangle = \lvert1\rangle_A\otimes\lvert0\rangle_B.$$

The first label belongs to subsystem $A$, and the second belongs to subsystem $B$.

2. For

$$\lvert\Psi\rangle = \frac{\lvert00\rangle+\lvert01\rangle}{\sqrt2},$$

write the coefficient matrix and decide whether the state is product.

Solution

The coefficient matrix is

$$C = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}.$$

It has rank one, so the state is product. Explicitly,

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

3. For $|\Psi\rangle=|0\rangle_A(|0\rangle_B+i|1\rangle_B)/\sqrt2$, find $C$, $\rho_A$, and $\rho_B$. What incorrect state results if $C^\dagger C$ is used for $\rho_B$?

Solution

The matrices are

$$C=\frac1{\sqrt2}\begin{pmatrix}1&i\\0&0\end{pmatrix},\qquad \rho_A=CC^\dagger=\begin{pmatrix}1&0\\0&0\end{pmatrix},$$

and

$$\rho_B=C^{\mathsf T}C^* =\frac12\begin{pmatrix}1&-i\\i&1\end{pmatrix} =|+y\rangle\langle+y|.$$

Using $C^\dagger C$ gives the complex-conjugate matrix, the projector onto $|-y\rangle$, so it predicts the opposite certain outcome for a $Y$ measurement. Both matrices are positive with eigenvalues $1,0$; checking eigenvalues alone would miss the conjugation error.

4. Let a local measurement on $A$ have projector $P_a$. What full-system projector represents the same outcome on $\mathcal H_A\otimes\mathcal H_B$?

Solution

The full-system projector is

$$P_a\otimes I_B.$$

The identity acts on subsystem $B$, indicating that the measurement is local to $A$.

5. A qubit $A$ and qutrit $B$ use the flattened index $k=i\,d_B+j$. Which coordinate corresponds to $\lvert1,2\rangle$? Find the ordered coordinate vector of the state

$$\lvert\Psi\rangle = \frac{ \lvert0,1\rangle+\lvert1,2\rangle }{\sqrt2}.$$

Solution

Here $d_B=3$, so

$$k=1\cdot3+2=5.$$

In the ordered basis

$$\bigl( \lvert0,0\rangle, \lvert0,1\rangle, \lvert0,2\rangle, \lvert1,0\rangle, \lvert1,1\rangle, \lvert1,2\rangle \bigr),$$

the coordinate vector is

$$\lvert\Psi\rangle \longleftrightarrow \frac{1}{\sqrt2} \begin{pmatrix} 0\\ 1\\ 0\\ 0\\ 0\\ 1 \end{pmatrix}.$$

6. For two qubits, evaluate

$$\bigl( \sigma_x\otimes I \bigr) \lvert01\rangle, \qquad \bigl( I\otimes\sigma_x \bigr) \lvert01\rangle.$$

Then verify on this basis vector that the two local operators commute.

Solution

The first operator flips subsystem $A$, while the second flips subsystem $B$:

$$\begin{aligned} \bigl( \sigma_x\otimes I \bigr) \lvert01\rangle &= \lvert11\rangle,\\ \bigl( I\otimes\sigma_x \bigr) \lvert01\rangle &= \lvert00\rangle. \end{aligned}$$

Applying both in either order gives

$$\begin{aligned} \bigl( \sigma_x\otimes I \bigr) \bigl( I\otimes\sigma_x \bigr) \lvert01\rangle &= \lvert10\rangle,\\ \bigl( I\otimes\sigma_x \bigr) \bigl( \sigma_x\otimes I \bigr) \lvert01\rangle &= \lvert10\rangle. \end{aligned}$$

Thus their commutator annihilates this vector, as expected from the general identity.

7. For the qubit–qutrit state

$$\lvert\Omega\rangle = \frac{ \lvert0,2\rangle+\lvert1,0\rangle }{\sqrt2},$$

find $C$, $\rho_A$, and $\rho_B$. What are the local computational-basis probabilities?

Solution

The coefficient matrix is

$$C = \frac{1}{\sqrt2} \begin{pmatrix} 0&0&1\\ 1&0&0 \end{pmatrix}.$$

Therefore

$$\rho_A = CC^\dagger = \frac12 \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix},$$

while

$$\rho_B = C^{\mathsf T}C^* = \frac12 \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&1 \end{pmatrix}.$$

Hence

$$p_A(0)=p_A(1)=\frac12,$$

and

$$p_B(0)=p_B(2)=\frac12, \qquad p_B(1)=0.$$

8. The two qubits are prepared in

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

Subsystem $B$ is measured in the $x$ basis $\{\lvert+\rangle,\lvert-\rangle\}$. Find the conditional state of $A$ for each outcome and compare their average with $\rho_A$.

Solution

In the $x$ basis,

$$\lvert\Phi^+\rangle = \frac{ \lvert+\rangle_A\lvert+\rangle_B + \lvert-\rangle_A\lvert-\rangle_B }{\sqrt2}.$$

Each outcome on $B$ has probability $1/2$. The conditional states are

$$\rho_{A|+} = \lvert+\rangle\langle+\rvert, \qquad \rho_{A|-} = \lvert-\rangle\langle-\rvert.$$

If the outcome is not retained, their probability-weighted average is

$$\begin{aligned} \rho_A &= \frac12 \lvert+\rangle\langle+\rvert + \frac12 \lvert-\rangle\langle-\rvert\\ &= \frac12 I. \end{aligned}$$

The conditional state depends on the remote outcome, while the unconditioned marginal remains maximally mixed.