The Reduced Density Matrix and Mixed States

A composite quantum system living in Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B$ has a density matrix $\rho_{AB}$ that fully describes the statistics of any measurement performed jointly on both subsystems. But an experimenter with access only to subsystem $A$ can perform only local measurements: operators of the form $M \otimes I_B$. The question is what effective, self-consistent density operator on $\mathcal{H}_A$ alone reproduces the correct expectation value $\langle M \otimes I_B \rangle$ for every local observable $M$. The answer is the reduced density matrix,

where $\mathrm{Tr}_B$, the partial trace over $B$, is a linear map from operators on $\mathcal{H}_A \otimes \mathcal{H}_B$ to operators on $\mathcal{H}_A$. Concretely, if $\{|b_i\rangle\}$ is any orthonormal basis of $\mathcal{H}_B$, then

where the bra/ket act only on the $B$ factor, leaving an operator on $\mathcal{H}_A$. This sum is basis-independent — a fact proved below — so $\rho_A$ is a well-defined object, not an artifact of how you choose to describe $B$. The reduced density matrix is generically mixed even when the joint state $\rho_{AB} = |\psi\rangle\langle\psi|$ is pure. This is the signature of entanglement: information that exists jointly between $A$ and $B$ is invisible to any measurement confined to $A$ alone, and the partial trace "erases" it by averaging over all possible outcomes on $B$. Crucially, this differs from classical mixing, where $\rho_A$ is mixed simply because the preparer used a coin flip (or any classical randomness) to choose among several pure states before handing subsystem $A$ to the experimenter, with no correlated partner system involved at all. Both mechanisms produce mixed $\rho_A$, and — this is the key structural fact of the whole topic — no experiment performed on $A$ alone can ever tell you which mechanism produced your particular $\rho_A$. The distinction is only visible with joint access to $A$ and $B$, which is precisely why entanglement is a nonlocal resource: it hides in correlations, not in either party's local statistics.

Setting up the partial trace. Write $\rho_{AB}$ in an arbitrary product basis $|a_i\rangle \otimes |b_j\rangle$ of $\mathcal{H}_A \otimes \mathcal{H}_B$:

The partial trace over $B$ contracts the $B$-indices by setting $j = l$ and summing:

In matrix terms, if you write the full density matrix as a block matrix indexed by $A$'s basis, where each block is a $\dim(B) \times \dim(B)$ matrix acting on $B$, then $\rho_A$ is obtained by replacing each block with its trace, producing a $\dim(A) \times \dim(A)$ matrix.

Why this is the right operation. The partial trace is not an arbitrary convention — it is forced by a physical consistency requirement. For any local observable $M_A$ acting on $A$,

One can derive the partial trace by demanding this identity hold for every $M_A$: expanding $\mathrm{Tr}[(M_A \otimes I_B)\rho_{AB}]$ using the cyclicity of the full trace and separating the $A$ and $B$ sums shows the identity holds precisely when $\rho_A$ is defined by the formula above. Since the trace is the unique linear functional reproducing quantum expectation values, and this must hold for every Hermitian $M_A$ (a spanning set of the operator space), $\rho_A$ is uniquely determined. Basis-independence of $\mathrm{Tr}_B$ follows because the trace of an operator is itself basis-independent, and $\mathrm{Tr}_B$ is literally the trace on the $B$ factor of a tensor-decomposed operator.

Algorithmic procedure for finite-dimensional systems. Computationally, the partial trace is a reshaping-and-contraction operation. ``` function PARTIAL_TRACE_B(rho_AB, dim_A, dim_B):

# rho_AB is a (dim_Adim_B) x (dim_Adim_B) matrix,

# indexed by composite basis |a_i, b_j> in row-major order:

# row index r = i * dim_B + j

# col index c = k dim_B + l rho_A = zero matrix of size dim_A x dim_A for i in 0 .. dim_A-1: for k in 0 .. dim_A-1: sum = 0 for j in 0 .. dim_B-1: # trace over B: set l = j r = i dim_B + j c = k * dim_B + j sum = sum + rho_AB[r][c] rho_A[i][k] = sum return rho_A ``` ``` function PARTIAL_TRACE_A(rho_AB, dim_A, dim_B):

# symmetric construction: trace over the A index instead rho_B = zero matrix of size dim_B x dim_B for j in 0 .. dim_B-1: for l in 0 .. dim_B-1: sum = 0 for i in 0 .. dim_A-1: # trace over A: set k = i r = i dim_B + j c = i dim_B + l sum = sum + rho_AB[r][c] rho_B[j][l] = sum return rho_B ``` Both routines run in $O(d_A^2 d_B)$ or $O(d_B^2 d_A)$ time respectively, dominated by the summation over the traced-out dimension for every entry of the output matrix.

Extracting entanglement information — the Schmidt decomposition. If $\rho_{AB} = |\psi\rangle\langle\psi|$ is pure, every $|\psi\rangle \in \mathcal{H}_A \otimes \mathcal{H}_B$ admits a Schmidt decomposition:

where $\{|\alpha_k\rangle\}$ and $\{|\beta_k\rangle\}$ are orthonormal sets in $\mathcal{H}_A$, $\mathcal{H}_B$ respectively, $\lambda_k > 0$, $\sum_k \lambda_k = 1$, and $r \leq \min(\dim A, \dim B)$ is the Schmidt rank. This decomposition always exists — it is just the singular value decomposition of the coefficient matrix of $|\psi\rangle$ in any product basis. Taking the partial trace over $B$:

Both reduced density matrices are diagonal in their respective Schmidt bases with exactly the same set of eigenvalues $\{\lambda_k\}$. This is the fundamental structural fact: entanglement of a pure bipartite state is entirely characterized by the (shared) eigenvalue spectrum of either reduced density matrix. $\rho_A$ is pure ($r=1$, one $\lambda_k = 1$) exactly when the state is a product state; $\rho_A$ is maximally mixed exactly when the state is maximally entangled.

Time and space complexity. For subsystems of dimension $d_A$ and $d_B$ (total system dimension $d = d_A d_B$), the full density matrix $\rho_{AB}$ requires $O(d^2) = O(d_A^2 d_B^2)$ space to store as a matrix. Computing $\rho_A = \mathrm{Tr}_B(\rho_{AB})$ via the direct summation above requires $O(d_A^2 d_B)$ time: for each of the $d_A^2$ entries of $\rho_A$, we sum $d_B$ terms. This is efficient — linear in the size of the input matrix ($d^2 = d_A^2 d_B^2$ entries), since $d_A^2 d_B \le d_A^2 d_B^2$. The output $\rho_A$ occupies $O(d_A^2)$ space. For $n$-qubit systems where $A$ is $k$ qubits and $B$ is $n-k$ qubits, this is $O(4^k \cdot 2^{n-k})$ time — exponential in $n$ in the worst case, which is expected since even writing down $\rho_{AB}$ costs $O(4^n)$.

Correctness proof. We must show two things: (1) the partial trace formula is basis-independent, and (2) it is the unique operator reproducing correct local statistics. Basis independence. Let $\{|b_j\rangle\}$ and $\{|b_j'\rangle\}$ be two orthonormal bases of $\mathcal{H}_B$, related by a unitary $U$: $|b_j'\rangle = U|b_j\rangle$. Then

Writing $U^\dagger \rho_{AB} U$ acting as $I_A \otimes U^\dagger$ then $I_A \otimes U$ (since $U$ acts only on $B$), and using that $\sum_j |b_j\rangle\langle b_j| = I_B$, a short computation using the resolution of identity shows this equals $\sum_j \langle b_j|\rho_{AB}|b_j\rangle$ — the same expression in the original basis. (Formally: partial trace commutes with conjugation by any operator of the form $I_A \otimes U_B$, because such conjugation is exactly a basis change on the $B$ factor, and the trace functional on $B$ alone is invariant under unitary basis changes — this is the ordinary basis-independence of the trace, applied factor-wise.) Uniqueness. Suppose $\sigma_A$ is any operator on $\mathcal{H}_A$ satisfying $\mathrm{Tr}[M_A \sigma_A] = \mathrm{Tr}[(M_A \otimes I_B)\rho_{AB}]$ for every Hermitian $M_A$. The set of Hermitian operators on $\mathcal{H}_A$ spans, as a real vector space, the entire space of Hermitian operators (dimension $d_A^2$), and the trace pairing $\langle X, Y\rangle = \mathrm{Tr}[XY]$ is nondegenerate on this space. Since $\mathrm{Tr}[M_A \sigma_A] = \mathrm{Tr}[M_A \rho_A]$ for all $M_A$ (using the definition of $\rho_A$ from partial trace and expanding), nondegeneracy of the trace pairing forces $\sigma_A = \rho_A$. Hence the partial trace is the only operator on $A$ consistent with all local measurement statistics — it is not one convenient choice among many. Positivity and normalization are preserved. If $\rho_{AB} \succeq 0$ and $\mathrm{Tr}(\rho_{AB}) = 1$, then $\rho_A = \mathrm{Tr}_B(\rho_{AB})$ satisfies $\rho_A \succeq 0$ (for any $|\phi\rangle \in \mathcal{H}_A$, $\langle \phi|\rho_A|\phi\rangle = \sum_j \langle \phi, b_j|\rho_{AB}|\phi,b_j\rangle \geq 0$ since $\rho_{AB} \succeq 0$) and $\mathrm{Tr}(\rho_A) = \mathrm{Tr}(\rho_{AB}) = 1$, so $\rho_A$ is a bona fide density matrix — the construction is closed on the space of valid quantum states, as it must be for the formalism to be self-consistent.

Honest tradeoffs. The partial trace is exact and unique, but it is inherently lossy: the map $\rho_{AB} \mapsto (\rho_A, \rho_B)$ is many-to-one, and in general $(\rho_A, \rho_B)$ cannot be used to reconstruct $\rho_{AB}$ — correlation and entanglement information is genuinely destroyed, not merely hidden. There is no "inverse partial trace" that recovers the joint state; this is why entanglement witnesses and joint tomography require measurements correlated across both subsystems, not separate local tomography on $A$ and $B$ individually.

Case 1 — the Bell pair (maximal entanglement). Take

In the ordered basis $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$,

Apply `PARTIAL_TRACE_B` with $d_A = d_B = 2$: entry $\rho_A[i][k] = \sum_{j=0}^{1} \rho_{AB}[2i+j][2k+j]$.

Eigenvalues $\{1/2, 1/2\}$: maximally mixed, purity $\mathrm{Tr}(\rho_A^2) = 1/2$. This matches the Schmidt decomposition directly: $|\Phi^+\rangle$ has Schmidt coefficients $\lambda_1 = \lambda_2 = 1/2$ in the computational basis, and indeed $\rho_A$'s eigenvalues equal $\{\lambda_k\}$ exactly as predicted.

Case 2 — a partially entangled state (intermediate case, the honest test). Take the non-maximal, non-symmetric state

This is deliberately asymmetric in its coefficients — a "trivial symmetric" $\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)$ would hide whether the general machinery actually tracks unequal weights. Here $c_{00}=\sqrt{0.8}\approx 0.894$, $c_{11}=\sqrt{0.2}\approx 0.447$, all other coefficients zero. Then

(using $c_{00}c_{11} = \sqrt{0.8 \times 0.2} = \sqrt{0.16} = 0.4$ for the off-diagonal terms). Trace out $B$:

Eigenvalues $\{0.8, 0.2\}$ — exactly the squared Schmidt coefficients $\{0.8, 0.2\}$, as the theory predicts, and not $\{0.5,0.5\}$: the reduced state correctly tracks the asymmetry in the original entangled state. Purity $\mathrm{Tr}(\rho_A^2) = 0.64+0.04 = 0.68$, strictly between $0.5$ (maximal mixing) and $1$ (pure), confirming this state is entangled but not maximally so — a fact invisible if you only checked "is $\rho_A$ mixed?" (yes/no) rather than computing its actual spectrum. Tracing out $A$ instead gives the identical spectrum $\{0.8, 0.2\}$ for $\rho_B$, confirming the Schmidt-symmetry theorem exactly.

Case 3 — the classical look-alike, to make the ambiguity concrete. Suppose instead $A$'s qubit alone (no partner $B$ at all) is prepared by flipping a coin: heads (prob 1/2) prepare $|0\rangle$, tails (prob 1/2) prepare $|1\rangle$. Then

bit-for-bit identical to the Case 1 reduced state. No measurement on this qubit, however cleverly chosen, can determine whether it arose from a coin flip or from tracing out a Bell partner — this is the precise content of the earlier claim that reduced density matrices cannot distinguish classical ignorance from entanglement.

The mathematical object underlying this article — the density matrix and its reduction — was introduced independently by John von Neumann and Lev Landau in the late 1920s, as physicists grappled with how to describe quantum systems that are not perfectly isolated or perfectly known. Von Neumann's 1932 book Mathematical Foundations of Quantum Mechanics gave the density matrix its rigorous trace-based formulation and introduced the entropy that bears his name, $S(\rho) = -\mathrm{Tr}(\rho \log \rho)$, which is exactly zero for pure states and positive for mixed ones — precisely quantifying the mixedness that partial trace produces.

The specific question of what a subsystem of an entangled pair looks like became sharply relevant after Einstein, Podolsky, and Rosen's 1935 paper questioned whether quantum mechanics was complete, using exactly the scenario of correlated, spatially separated particles. Schrödinger's response that same year coined the term "entanglement" (Verschränkung) and explicitly discussed how measuring one particle's subsystem could not, by itself, reveal the correlations built into the joint state — an early, informal statement of what the partial trace now computes rigorously.

Reduced density matrices and partial trace are the working language of open quantum systems and quantum noise: whenever a qubit interacts with an uncontrolled environment (stray fields, other qubits, thermal baths), the qubit's evolution is described by tracing the environment out of the joint unitary evolution, producing the quantum channels and decoherence models used to characterize real hardware. Every gate-error and decoherence-time specification quoted for superconducting or trapped-ion qubits is, underneath, a statement about a reduced density matrix's departure from purity. In quantum error correction, the syndrome-extraction step is designed precisely so that measuring ancilla (syndrome) qubits and tracing them out projects the data qubits' reduced state back toward a pure, corrected codeword — the entire logic of stabilizer codes rests on controlling how partial trace over ancillas affects the data subsystem. In entanglement quantification, the entropy of the reduced density matrix (the entanglement entropy) is the standard measure of bipartite entanglement for pure states, used throughout quantum information theory, condensed matter physics (area laws, topological order), and even quantum gravity research on black hole entropy. In quantum tomography and benchmarking, experimentalists reconstruct reduced density matrices of individual qubits from measurement statistics precisely because this is the maximal information accessible without joint measurements — connecting directly back to correlated measurement strategies needed to see entanglement at all.

Ask the tutor: How does the partial trace generalize to give the Kraus-operator description of a noisy quantum channel acting on a single qubit?