The Reduced Density Matrix and Partial Trace

Two parties share a quantum state. What is the honest description of one party's half, on its own? The answer is the reduced density matrix, obtained by the partial trace — summing the joint density operator over an orthonormal basis of the party you are ignoring. This one operation is the hinge of the entire entanglement chapter: it is how a pure whole comes to have mixed parts, how "looking locally" is defined, and why no message can be sent by entanglement alone. The article owns four results. First, the partial trace $\rho_A = \mathrm{Tr}_B\rho_{AB}$ is basis-independent — the answer does not depend on which basis of $B$ you sum over, because a change of that basis is a unitary on $B$ and the trace is blind to it. Second, the local-observable theorem: $\rho_A$ is the unique operator reproducing $\langle\hat A\otimes\mathbb 1\rangle = \mathrm{Tr}(\rho_A\hat A)$ for every local observable $\hat A$ — it is defined precisely to be everything a local experimenter can ever measure, no more, no less. Third, and the chapter's structural hinge: for a pure joint state, $\rho_A$ is mixed if and only if the state is entangled. Product states leave both parts pure; any entanglement forces local mixedness, quantified by the purity drop. Half of a Bell pair is maximally mixed — an experimenter holding it sees pure noise, though the global state is perfectly definite. Fourth, the no-signalling theorem: any operation Bob performs on $B$ alone — a unitary, a measurement whose result he keeps — leaves $\rho_A$ completely unchanged. Alice cannot tell what Bob did, so entanglement carries correlation but not causation, and quantum mechanics respects relativity's speed limit. The decoherence mechanism (tracing out an environment) is the same partial trace read as noise, developed in Quantum Decoherence and the Density Matrix by name.

Key takeaways

Partial trace: $\rho_A = \mathrm{Tr}_B\rho_{AB}$ , $\langle i|\rho_A|j\rangle_A = \sum_k\langle i,k|\rho_{AB}|j,k\rangle$ — the local state of $A$ .
Basis-independence (owned): the result is independent of the basis of $B$ traced over (a change of that basis is a unitary on $B$ , invisible to $\rho_A$ ).
Local-observable theorem (owned): $\rho_A$ is the unique operator with $\langle\hat A\otimes\mathbb 1\rangle = \mathrm{Tr}(\rho_A\hat A)$ for all local $\hat A$ — it is the local physics.
Entanglement ⇒ local mixedness (owned): for pure $|\Psi\rangle$ , $\rho_A$ mixed $\iff$ entangled. Product $\Rightarrow$ $\rho_A$ pure; entangled $\Rightarrow$ $\mathrm{Tr}\,\rho_A^2 < 1$ . Half a Bell pair: $\rho_A = \mathbb 1/2$ , $S_A = \ln 2$ .
No-signalling (owned): any local operation on $B$ leaves $\rho_A$ unchanged ( $\mathrm{Tr}_B(U_B\rho U_B^\dagger) = \mathrm{Tr}_B\rho$ ); entanglement transmits correlation, not causation.
Worked: Bell $\to\rho_A = \mathbb 1/2$ , $S_A = \ln 2$ ; $\sqrt{0.8}\,|00\rangle + \sqrt{0.2}\,|11\rangle\to\rho_A = \mathrm{diag}(0.8,0.2)$ , purity $0.68$ , and $\sigma_x$ on $B$ leaves it fixed.
Key idea: the partial trace is the definition of "looking locally"; it turns entanglement into local mixedness and enforces no-signalling.

The core insight

The reduced density matrix $\rho_A = \mathrm{Tr}_B\rho_{AB}$ is the unique operator giving every local expectation value. It is basis-independent, it is mixed exactly when the joint pure state is entangled, and it is untouched by anything the other party does — so it is at once the definition of a subsystem, the source of local mixedness, and the guarantee that entanglement cannot signal.

Where this lives today

The partial trace is the single most-used operation in modern quantum information. Decoherence theory is the partial trace read as noise: trace out the environment a system got entangled with, and the reduced state's off-diagonals decay — the mechanism by which classical behavior emerges, owned by Quantum Decoherence and the Density Matrix by name. Entanglement entropy — $S(\rho_A)$ of a reduced state — has become a central diagnostic across physics: it detects quantum phase transitions in condensed matter, obeys "area laws" that make tensor-network simulation possible, and, through the Ryu–Takayanagi formula, equals a geometric area in holographic quantum gravity, tying spacetime itself to the partial trace. No-signalling is a foundational constraint on any physical theory: device-independent cryptography and the study of "post-quantum" correlations (PR boxes) are organized around exactly which correlations are achievable subject to no-signalling, the theorem this article derives. From cryptographic security proofs to the emergence of spacetime, the humble sum-over- $B$ is doing the work.

Figure

No-signalling: whatever Bob does to B, Alice's ρ_A never moves

The physical picture

Entanglement forced the density operator on us: the previous chapters showed that half of a Bell pair has no wavefunction of its own. The partial trace is the tool that computes what it does have. Think of it operationally: an experimenter holds qubit $A$ , has no access to $B$ , and asks what her measurements will show. Any measurement she can do is a local observable $\hat A\otimes\mathbb 1$ — it acts on $A$ and ignores $B$ . The partial trace is engineered so that the single operator $\rho_A$ reproduces the statistics of every such measurement. It is the maximal information she can extract, packaged as a density operator, and the local-observable theorem says it is unique: two different global states with the same $\rho_A$ are locally indistinguishable, forever. The striking consequence is that a globally definite state can look locally random. The Bell state $|\Phi^+\rangle$ is a single, pure, fully-specified vector — there is nothing unknown about the pair — yet each half, computed by partial trace, is $\mathbb 1/2$ , maximally mixed, the state of an unpolarized beam. The information that makes the pair definite lives entirely in the correlations between the halves, invisible to either one alone. This is the deepest source of the density operator, and it is what the whole chapter is about: mixedness of a part measuring entanglement of the whole.

Figure

The partial trace: entanglement of the whole becomes mixedness of the part

The partial trace and its basis-independence

The composite Hilbert space is $\mathcal H_A\otimes\mathcal H_B$ , and a joint density operator $\rho_{AB}$ lives on it. The partial trace over $B$ produces an operator on $\mathcal H_A$ alone.

Derivation AThe partial trace, and why the basis of B does not matter

We define $\rho_A = \mathrm{Tr}_B\rho_{AB}$ by its matrix elements and prove the result is independent of which orthonormal basis of $B$ is summed over — a prerequisite for $\rho_A$ to mean anything. The tool is that any two orthonormal bases of $\mathcal H_B$ are related by a unitary, together with the cyclicity of the (full) trace.

Step 1. Define it. Fixing orthonormal bases $\{|i\rangle_A\}$ and $\{|k\rangle_B\}$ , the reduced operator has matrix elements

\langle i|\rho_A|j\rangle = \sum_k\langle i,k|\rho_{AB}|j,k\rangle, \qquad \rho_A = \sum_k\langle k|_B\,\rho_{AB}\,|k\rangle_B \qquad (1)

where: the sum "closes the $B$ -legs" — you sandwich $\rho_{AB}$ between $\langle k|_B$ and $|k\rangle_B$ and add over $k$ , leaving free indices only on $A$ . It is a partial version of the ordinary trace: trace only the $B$ factor, keep the $A$ factor as operator-valued.

Step 2. Change the basis of $B$ . Let $\{|k'\rangle_B\}$ be another orthonormal basis, $|k'\rangle_B = \sum_k U_{kk'}|k\rangle_B$ for a unitary $U$ . Substituting into (1),

\sum_{k'}\langle k'|_B\rho_{AB}|k'\rangle_B = \sum_{k',k,k''}U^*_{kk'}U_{k''k'}\langle k|_B\rho_{AB}|k''\rangle_B = \sum_{k,k''}\delta_{kk''}\langle k|_B\rho_{AB}|k''\rangle_B \qquad (2)

using $\sum_{k'}U^*_{kk'}U_{k''k'} = \delta_{kk''}$ (unitarity of $U$ ). The right side is exactly the original sum. Why this had to work: the partial trace is contraction of the $B$ -index, and contraction of a repeated index is a scalar operation invisible to any relabelling of that index — the unitary rows sum to a Kronecker delta, collapsing the double sum.

Step 3. Read the guarantee. $\rho_A$ is a well-defined operator on $\mathcal H_A$ , independent of the arbitrary choice of $B$ -basis. It inherits the credentials of a density operator — Hermitian (from $\rho_{AB}$ ), unit trace ( $\mathrm{Tr}\,\rho_A = \mathrm{Tr}\,\rho_{AB} = 1$ ), positive (a partial trace of a positive operator is positive) — so it is itself a legal state, exactly as a subsystem should be. Verified numerically: for a random pure two-qubit state, the partial trace over $B$ computed in the computational basis equals the one computed after a Hadamard (and after a random unitary) on $B$ , to machine precision.

The local-observable theorem and no-signalling

Two theorems make $\rho_A$ physically inevitable rather than merely convenient.

Go deeper · Derivation B · 2 minρ_A is exactly the local physics — and Bob cannot change it

Derivation Bρ_A is exactly the local physics — and Bob cannot change it

We prove that $\rho_A$ is the unique operator reproducing every local expectation value, then that no operation on $B$ alters it — the no-signalling theorem. The first is a defining property; the second is its immediate consequence and the operational heart of the whole chapter.

Step 1. Local means factor-times-identity. Any observable an $A$ -only experimenter can measure is $\hat A\otimes\mathbb 1_B$ : it acts on $A$ , ignores $B$ . Its expectation in the joint state is

\langle\hat A\otimes\mathbb 1\rangle = \mathrm{Tr}\bigl[\rho_{AB}\,(\hat A\otimes\mathbb 1)\bigr] = \mathrm{Tr}_A\bigl[(\mathrm{Tr}_B\rho_{AB})\,\hat A\bigr] = \mathrm{Tr}(\rho_A\hat A) \qquad (3)

Why the partial trace appears: the $\mathbb 1_B$ inside the full trace sums $\rho_{AB}$ over $B$ before the $A$ -trace acts — that inner sum is precisely $\rho_A$ . So $\rho_A$ is built to make (3) hold. Uniqueness follows because the set of all $\hat A$ spans the local operators: any two operators giving the same $\mathrm{Tr}(\cdot\,\hat A)$ for every $\hat A$ are equal. $\rho_A$ is the local physics, entirely.

Step 2. Bob acts on $B$ . Suppose Bob applies a unitary $U_B = \mathbb 1_A\otimes V_B$ , so $\rho_{AB}\to U_B\rho_{AB}U_B^\dagger$ . Then

\rho_A' = \mathrm{Tr}_B\bigl[(\mathbb 1\otimes V)\rho_{AB}(\mathbb 1\otimes V^\dagger)\bigr] = \mathrm{Tr}_B\bigl[\rho_{AB}\,(\mathbb 1\otimes V^\dagger V)\bigr] = \mathrm{Tr}_B\,\rho_{AB} = \rho_A

using cyclicity within the $B$ -trace and $V^\dagger V = \mathbb 1$ . Why it vanishes: Bob's unitary hides under the $B$ -trace, where cyclicity lets $V^\dagger V$ annihilate — his rotation of $B$ never reaches $A$ . Alice's local state is unchanged.

Step 3. Bob measures $B$ and keeps the result. If Bob performs a projective measurement $\{\hat\Pi_m\}$ on $B$ and does not tell Alice, her state is the average $\sum_m(\mathbb 1\otimes\hat\Pi_m)\rho_{AB}(\mathbb 1\otimes\hat\Pi_m)$ reduced on $A$ ; because $\sum_m\hat\Pi_m = \mathbb 1$ , the partial trace again returns $\rho_A$ unchanged (the linearity-no-go of Measurement and Wavefunction Collapse recalled by name — an unread measurement cannot change local statistics). No-signalling: nothing Bob does to $B$ — unitary or measurement — is visible in $\rho_A$ , so entanglement cannot carry a message.

Step 4. Bank the operational meaning. Alice and Bob can share entanglement and generate correlations that violate every classical bound, yet neither can use those correlations to signal, because each sees only their reduced state, which the partner cannot budge. Correlation without causation is the precise content — the resolution of the "spooky action" worry, developed further in The EPR Paradox and Local Realism by name. Verified numerically: for $\sqrt{0.8}\,|00\rangle + \sqrt{0.2}\,|11\rangle$ , applying $\sigma_x$ on $B$ leaves $\rho_A = \mathrm{diag}(0.8,0.2)$ exactly; a full unread measurement on $B$ of a Bell state leaves $\rho_A = \mathbb 1/2$ .

The hinge: entanglement equals local mixedness

For a pure joint state the reduced state's purity is a perfect entanglement detector. The Schmidt decomposition (developed in The Schmidt Decomposition by name) writes any pure $|\Psi\rangle = \sum_k\sqrt{\lambda_k}|k\rangle_A|k\rangle_B$ , and the partial trace then gives $\rho_A = \sum_k\lambda_k|k\rangle\langle k|_A$ — its eigenvalues are the Schmidt weights. So $\rho_A$ is pure ( $\mathrm{Tr}\,\rho_A^2 = 1$ ) exactly when one $\lambda_k = 1$ , i.e. $|\Psi\rangle = |k\rangle_A|k\rangle_B$ is a product; and $\rho_A$ is mixed exactly when two or more $\lambda_k$ are nonzero, i.e. $|\Psi\rangle$ is entangled. The clean one-parameter family makes it quantitative: for $|\Psi(\theta)\rangle = \cos\theta|00\rangle + \sin\theta|11\rangle$ , $\rho_A = \mathrm{diag}(\cos^2\theta, \sin^2\theta)$ and $\mathrm{Tr}\,\rho_A^2 = 1 - \tfrac12\sin^2 2\theta$ — purity $1$ at the product endpoints $\theta = 0,\pi/2$ , dipping to $\tfrac12$ at the maximally entangled Bell point $\theta = \pi/4$ . The dip is the entanglement, and this " $E = S(\rho_A)$ " reading is the seed of the measure theory in Entanglement Entropy and Measures of Entanglement by name.

Limiting cases

joint state	$\rho_A$	reading
product $\|\alpha\rangle_A\|\beta\rangle_B$	pure $\|\alpha\rangle\langle\alpha\|$	no entanglement, no local mixing
Bell $\|\Phi^+\rangle$	$\mathbb 1/2$ , maximally mixed	maximal entanglement, $S_A = \ln 2$
$\cos\theta\|00\rangle + \sin\theta\|11\rangle$	$\mathrm{diag}(\cos^2\theta,\sin^2\theta)$	tunable entanglement, purity $1 - \tfrac12\sin^2 2\theta$
separable mixed $\sum_i p_i\rho_A^i\otimes\rho_B^i$	$\sum_i p_i\rho_A^i$	classical correlation only (no signalling either)
trace out an environment	decohered $\rho_A$	off-diagonals suppressed $=$ decoherence (owned downstream)
Bob applies any $U_B$ or measures $B$	$\rho_A$ unchanged	no-signalling

Common trapThe reduced state is not "the state the system is really in" — and it hides the correlations

Two traps. First, $\rho_A = \mathbb 1/2$ for half a Bell pair does not mean qubit $A$ is "secretly $\uparrow$ or $\downarrow$ with a coin flip." It means $A$ 's local statistics match such a coin flip; the actual state has no local description at all, and the missing information lives in correlations with $B$ that no local measurement reads. Treating $\rho_A$ as a hidden classical mixture leads straight to the local-hidden-variable fallacy that Bell's theorem refutes (developed in Bell's Theorem and the CHSH Inequality by name). Second, the partial trace discards the correlations irreversibly: from $\rho_A$ and $\rho_B$ alone you cannot reconstruct $\rho_{AB}$ (the product $\rho_A\otimes\rho_B$ is generally a different, less-correlated state). So $\rho_A$ answers "what does $A$ see locally?" and nothing else — it is silent about, and cannot recover, the entanglement it was computed from. This one-way loss of correlations, when $B$ is an environment, is exactly decoherence.

Worked examples

Example 1 — the Bell state, member by member (confidence builder). Take $|\Phi^+\rangle = (|00\rangle + |11\rangle)/\sqrt2$ and compute the local state of qubit $A$ .

Step 1. Build the joint operator. $\rho_{AB} = |\Phi^+\rangle\langle\Phi^+| = \tfrac12(|00\rangle + |11\rangle)(\langle00| + \langle11|)$ — a pure state, $\mathrm{Tr}\,\rho_{AB}^2 = 1$ , containing the coherences $|00\rangle\langle11|$ and $|11\rangle\langle00|$ that link the branches.

Step 2. Trace out $B$ . Summing over $|0\rangle_B,|1\rangle_B$ : the diagonal blocks each drop half a projector onto $A$ , and the coherence blocks need $\langle0|1\rangle_B = 0$ and die. The result is $\rho_A = \tfrac12|0\rangle\langle0| + \tfrac12|1\rangle\langle1| = \boxed{\mathbb 1/2}$ (machine-verified).

Step 3. Read it. $\rho_A$ is maximally mixed: purity $\tfrac12$ , $\langle\sigma_{\hat n}^A\rangle = 0$ on every axis, and its entropy $S_A = -\tfrac12\ln\tfrac12 - \tfrac12\ln\tfrac12 = \boxed{\ln 2}$ (one bit). Alice holding half a Bell pair sees pure noise — no local measurement reveals anything — even though the pair is a single definite pure state. The full bit of information sits in the $A$ – $B$ correlation, and its magnitude $\ln 2$ is exactly the entanglement.

Example 2 — a partially entangled state and no-signalling (exam level). Take $|\Psi\rangle = \sqrt{0.8}\,|00\rangle + \sqrt{0.2}\,|11\rangle$ . Find $\rho_A$ , its purity and $\langle\sigma_z^A\rangle$ , then show a $\sigma_x$ on $B$ changes nothing for $A$ .

Step 1. Reduce. Tracing out $B$ (the state is already in Schmidt form): $\rho_A = 0.8\,|0\rangle\langle0| + 0.2\,|1\rangle\langle1| = \mathrm{diag}(0.8, 0.2)$ (machine-verified). Diagonal — the Schmidt basis diagonalizes it.

Step 2. Read the numbers. Purity $\mathrm{Tr}\,\rho_A^2 = 0.64 + 0.04 = \boxed{0.68}$ — partially mixed, so the state is partially entangled ( $0.68$ sits between the Bell dip $0.5$ and the product ceiling $1$ ). $\langle\sigma_z^A\rangle = 0.8 - 0.2 = \boxed{0.6}$ — a net $z$ -polarization, because the state is not maximally entangled.

Step 3. Test no-signalling. Let Bob apply $\sigma_x$ to $B$ , sending $|\Psi\rangle\to\sqrt{0.8}\,|01\rangle + \sqrt{0.2}\,|10\rangle$ , a completely different global state. Recompute $\rho_A = \mathrm{Tr}_B$ of the new state: still $\mathrm{diag}(0.8, 0.2)$ , identical (machine-verified). Bob's flip is invisible to Alice: her purity, her polarization, every local statistic is unchanged. Correlation with $B$ has been rearranged, but Alice's local physics — all $\rho_A$ knows — cannot register it. Entanglement carries no message.

The bookkeeping that became a subsystem

Origins · 1 minreduced density matrix and how this was discovered

reduced density matrix

1927–1935 · Göttingen, Oxford

The reduced density matrix has three founding moments, each approaching it from a different need. Lev Landau reached it in 1927 by asking what describes a subsystem coupled to radiation — the partial-trace problem in disguise — and Paul Dirac's 1930 Principles of Quantum Mechanics gave the density matrix its systematic operator calculus, the language in which the trace bookkeeping became routine. But the reduced state's meaning was crystallized by Erwin Schrödinger in 1935, in the same burst of papers that coined "entanglement." Schrödinger saw that for an entangled pair the whole can be maximally known while the parts cannot — "the best possible knowledge of a whole does not include the best possible knowledge of its parts" — which is exactly the statement that the reduced state $\rho_A$ is mixed though $\rho_{AB}$ is pure. He even discovered "steering": that a distant party's measurement choice fixes which ensemble realizes the local reduced state, without changing the reduced state itself — the seed of the no-signalling theorem this article proves. (The naming of "entanglement" and the EPR provocation that prompted it are owned by Quantum Entanglement and Non-Separability and The EPR Paradox and Local Realism by name.) The partial trace, a piece of linear-algebra bookkeeping, turned out to be the precise definition of what it means to be a part of an entangled world.

History

1927

— Landau describes a coupled subsystem by a density matrix (the reduced-state problem)

1930

— Dirac's Principles gives the density matrix its operator calculus

1935

— Schrödinger: the whole can be known while the parts cannot; "steering" of the reduced state

1970

s — decoherence theory reads the partial trace over an environment as the classical limit

2006

— Ryu–Takayanagi tie the entropy of a reduced region to a geometric area in holography

Going further

References. Nielsen & Chuang, §2.4.3 (partial trace), §2.4.3–2.5 · Sakurai & Napolitano, §3.4, §3.9 · Cohen-Tannoudji, Diu & Laloë, Complement $E_{III}$ · Preskill, Lecture Notes, Ch. 2 · Peres, Quantum Theory: Concepts and Methods, Ch. 5 · Ryu & Takayanagi, Phys. Rev. Lett. 96, 181602 (2006) (holographic entanglement entropy).

Provenance. All results are standard. The density-operator credentials and the trace formula $\langle\hat A\rangle = \mathrm{Tr}(\rho\hat A)$ are owned by The Density Operator and Mixed States, recalled by name; the linearity-no-go of unread measurement by Measurement and Wavefunction Collapse. Owned and derived here: (1) the partial trace (1) and its basis-independence, machine-verified by tracing over the computational, Hadamard, and a random basis of $B$ with identical results; (2) the local-observable theorem (3), $\langle\hat A\otimes\mathbb 1\rangle = \mathrm{Tr}(\rho_A\hat A)$ with uniqueness, checked on $\hat A = 0.3\sigma_x + 0.5\sigma_z$ ; (3) the no-signalling theorem, with $\rho_A$ unchanged under a $B$ -unitary and under an unread $B$ -measurement (checked on $\sqrt{0.8}\,|00\rangle + \sqrt{0.2}\,|11\rangle$ and $|\Phi^+\rangle$ ); and (4) the entanglement ⇒ local mixedness hinge for pure states, $\rho_A = \mathrm{diag}(\cos^2\theta,\sin^2\theta)$ , purity $1 - \tfrac12\sin^2 2\theta$ (checked over $\theta\in[0,\pi/2]$ ). Worked numbers recomputed: Bell $\to\rho_A = \mathbb 1/2$ , $S_A = \ln 2 = 0.6931$ ; $\sqrt{0.8}\,|00\rangle + \sqrt{0.2}\,|11\rangle\to\rho_A = \mathrm{diag}(0.8,0.2)$ , purity $0.68$ , $\langle\sigma_z^A\rangle = 0.6$ , invariant under $\sigma_x$ on $B$ . The Schmidt decomposition is recalled by name from The Schmidt Decomposition; the entropy measure of entanglement from Entanglement Entropy and Measures of Entanglement; the decoherence reading from Quantum Decoherence and the Density Matrix; the environment-purification (Stinespring) from Generalized Measurements and POVMs; the hidden-variable warning from Bell's Theorem and the CHSH Inequality — all by name.

Builds on: The Density Operator and Mixed States · Pure versus Mixed States and the Purity Criterion · Properties and Time Evolution of the Density Matrix · Completeness Relation and Resolution of Identity

Leads to: The Schmidt Decomposition · Entanglement Entropy and Measures of Entanglement · Quantum Entanglement and Non-Separability · The EPR Paradox and Local Realism

Is the partial trace a physical operation, or just a mathematical convenience?

It is the mathematical image of a very physical situation: having access to $A$ but not $B$ . Any experiment confined to $A$ measures $\hat A\otimes\mathbb 1$ , and (3) shows all such statistics are governed by $\rho_A = \mathrm{Tr}_B\rho_{AB}$ . So the partial trace is not an approximation or a bookkeeping trick — it is the exact and complete description of the local physics, forced by the local-observable theorem. It is "physical" in the strongest sense: it is what a local observer's world actually is.

If ρ_A doesn't change when Bob measures B, how do the famous EPR correlations arise?

The correlations are in the joint statistics, not the local ones. Bob's measurement does update his knowledge of what Alice will find — conditioned on his outcome, Alice's state is "steered" to a definite one — but the unconditioned $\rho_A$ , averaged over Bob's outcomes (which Alice doesn't know), is unchanged. Alice sees no local effect; only when the two compare notes do the correlations appear. This is exactly the steering-versus-signalling distinction: Bob steers Alice's conditional state but cannot move her reduced state, so no message is sent. The full story is owned by The EPR Paradox and Local Realism by name.

Why does the coherence between branches die under the partial trace?

Because the coherences of an entangled state connect terms in which the unobserved party occupies orthogonal states. In $|\Phi^+\rangle$ the coherence $|00\rangle\langle11|$ links a branch with $B = |0\rangle$ to one with $B = |1\rangle$ ; the partial trace sandwiches this between $\langle k|_B$ and $|k\rangle_B$ and sums, which requires $\langle 0|1\rangle_B$ — zero. So any coherence certifiable only by touching both sides is invisible to a one-sided trace. This is the microscopic mechanism of decoherence: an environment plays the role of $B$ , and every coherence that entangled with a distinguishable environment state is erased locally.

Can two different global states give the same ρ_A?

Constantly — that is the whole point. $|\Phi^+\rangle$ and the classically-correlated mixture $\tfrac12|00\rangle\langle00| + \tfrac12|11\rangle\langle11|$ both give $\rho_A = \mathbb 1/2$ , yet one is maximally entangled and the other separable. The reduced state cannot tell them apart, because it captures only local statistics, and they agree locally. Distinguishing them requires joint measurements on both parties (Bell tests do exactly this). So $\rho_A$ is a projection that deliberately forgets the correlations — which is why it enforces no-signalling but cannot, by itself, detect entanglement in a mixed global state.

How is tracing out a subsystem the same thing as decoherence?

Decoherence is the partial trace applied to an environment. A system $S$ interacts with an environment $E$ and becomes entangled, $|\psi\rangle_S\otimes|E_0\rangle\to\sum_n c_n|n\rangle_S|E_n\rangle$ . An observer with access only to $S$ computes $\rho_S = \mathrm{Tr}_E$ , and the off-diagonal element $\rho_{S,nm}\propto\langle E_m|E_n\rangle$ is suppressed by the overlap of the environment states — which becomes tiny fast for a macroscopic environment. The system's coherences "die" not because they were destroyed but because they leaked into inaccessible $S$ – $E$ correlations. The mechanism, rates, and pointer-basis selection are owned by Quantum Decoherence and the Density Matrix by name.

Does the partial trace lose information, and where does it go?

Locally, yes — $\rho_A$ carries strictly less information than $\rho_{AB}$ , and you cannot invert the map. The lost information is the $A$ – $B$ correlations, which are not destroyed but relocated: they are perfectly present in the joint state and recoverable by anyone with access to both parties. This is the difference between "forgotten" and "erased." For a closed universe nothing is ever truly lost (unitary evolution is reversible); a subsystem's apparent information loss is always correlations with the rest, waiting in the global state. That principle — information hides in correlations, never vanishes — underlies everything from decoherence to the black-hole information puzzle.