Partial Trace and Reduced Density Matrices

Give two parties, Alice and Bob, a single joint quantum state $\rho_{AB}$ on the tensor-product space $\mathcal{H}_A\otimes\mathcal{H}_B$ , and then forbid Alice from ever touching Bob's half. Everything she can measure, every statistic she can gather, must be predicted by one compact object living on her space alone. That object is the reduced density matrix $\rho_A$ , and the operation that produces it from $\rho_{AB}$ is the partial trace $\mathrm{Tr}_B$ . It is the exact quantum analogue of computing a marginal distribution: where a classical joint law $p(a,b)$ collapses to $p(a)=\sum_b p(a,b)$ , the quantum joint state collapses to $\rho_A=\mathrm{Tr}_B(\rho_{AB})$ . The definition is not a matter of taste. It is forced by a single operational demand — for every observable $M_A$ on Alice's side, the expectation computed locally from $\rho_A$ must match the global one, $\mathrm{Tr}[M_A\rho_A]=\mathrm{Tr}[(M_A\otimes I_B)\rho_{AB}]$ . Exactly one map meets this for all $M_A$ ; it acts on product operators as $X_A\otimes Y_B\mapsto X_A\,\mathrm{Tr}(Y_B)$ , extended by linearity. The consequence that makes the partial trace indispensable is this: a globally pure entangled state has a mixed marginal. Trace out one qubit of a Bell pair and you are left with $I/2$ — maximally mixed, locally featureless as a coin flip, every trace of structure hidden in correlations neither party can see alone. This is the mathematical seed of decoherence (trace out the environment), of entanglement entropy (measure the mixedness), of purification (every mixed state is the shadow of a pure one), and of the no-signalling backbone of quantum theory (nothing Bob does can move $\rho_A$ ).

Key takeaways

The partial trace $\mathrm{Tr}_B$ is the unique linear map that sends a joint state $\rho_{AB}$ to the reduced state $\rho_A=\mathrm{Tr}_B(\rho_{AB})$ reproducing every local measurement statistic on subsystem $A$ .
Concretely $\rho_A=\sum_j (I_A\otimes\langle j|_B)\,\rho_{AB}\,(I_A\otimes|j\rangle_B)$ for any orthonormal basis $\{|j\rangle_B\}$ , and the answer does not depend on which basis you sum over.
$\rho_A$ is always a legitimate density operator — Hermitian, positive semidefinite, unit trace — even when $\rho_{AB}$ is pure; entanglement is precisely what makes $\rho_A$ mixed.
It is fixed by the operational identity $\mathrm{Tr}[(M_A\otimes I_B)\rho_{AB}]=\mathrm{Tr}[M_A\rho_A]$ for all $M_A$ ; this is the no-signalling property that makes the marginal physically meaningful.
For a pure bipartite state the Schmidt decomposition shows $\rho_A$ and $\rho_B$ share their nonzero spectrum, so their entropies and mixedness coincide and both measure the entanglement.
Key idea: The reduced density matrix $\rho_A=\mathrm{Tr}_B(\rho_{AB})$ is the unique operator that reproduces all local expectation values, $\mathrm{Tr}[M_A\rho_A]=\mathrm{Tr}[(M_A\otimes I_B)\rho_{AB}]$ for every $M_A$ ; the partial trace acts as $X_A\otimes Y_B\mapsto X_A\,\mathrm{Tr}(Y_B)$ and extends by linearity.

Where this lives today

The partial trace is the daily workhorse of quantum science. In quantum error correction and NISQ-era hardware, every decoherence model is a partial trace over an environment — the accessible qubit state is $\rho_S(t)=\mathrm{Tr}_E[U(t)(\rho_S\otimes\rho_E)U^\dagger(t)]$ , and the decay of its coherences is what codes are built to fight. In many-body physics, the scaling of $S(\rho_A)$ with subsystem size — an area law versus a volume law — diagnoses phases of matter, locates quantum critical points, and decides whether a tensor-network method can simulate a system at all; the same reduced-state spectrum feeds the density-matrix renormalization group. Most strikingly, the reduced state is now directly measurable: the 2019 Innsbruck trapped-ion experiments of Brydges and collaborators extracted the second Rényi entropy $S_2(\rho_A)=-\log_2\mathrm{Tr}(\rho_A^2)$ via randomized local measurements, turning "how mixed is Alice" into a laboratory number and watching entanglement grow under a quench. And in quantum gravity, the Ryu–Takayanagi formula equates a boundary region's entanglement entropy with the area of a bulk minimal surface — making the humble partial trace a probe of spacetime geometry, an open frontier where reduced density matrices meet holography.

The physical picture

Alice holds qubit $A$ , Bob holds qubit $B$ , and they share a joint state $\rho_{AB}$ . Alice will never touch Bob's qubit, yet she must predict any experiment on her own — and a single operator on her two-dimensional space does the whole job. The partial trace is the recipe that averages out everything Bob could ever do (every measurement, unitary, or choice not to report) and leaves exactly the marginal Alice experiences. The first figure shows the flow: a correlated pair goes in, Bob's half is discarded, Alice's reduced state comes out carrying all of her local physics.

Figure

Quantum marginalization: the joint state ρ_AB on H_A ⊗ H_B is fed through the partial trace Tr_B, which discards Bob's subsystem and returns Alice's reduced state ρ_A. The output is fixed by the operational demand Tr(M_A ρ_A) = Tr((M_A ⊗ I_B) ρ_AB) for every local observable M_A — the exact quantum analogue of the classical marginal p(a) = Σ_b p(a,b)

Tracing out $B$ in a fixed basis means: for each diagonal label $|j\rangle\langle j|_B$ of Bob's register, keep the $A$ -operator content in that block and add the blocks up. The off-diagonal $B$ -coherences $|j\rangle\langle k|_B$ with $j\ne k$ contribute nothing, because $\langle k|j\rangle=0$ — those coherences are the correlations that vanish from any local description, which is why entanglement looks to Alice like ordinary noise. The second figure carries this to numbers: the Bell state as a $4\times4$ matrix, partitioned into $2\times2$ blocks, with the partial trace reading off the trace of each diagonal block to give $\rho_A=I/2$ and a full bit of entropy.

Figure

The partial trace as a block trace, worked in numbers: the Bell state Φ⁺ written as a 4×4 density matrix (prefactor ½) splits into four 2×2 B-blocks. Each entry of ρ_A is the trace of the corresponding block, so summing the two green diagonal blocks gives ρ_A = ½I. Its equal eigenvalues ½, ½ give entanglement entropy S(ρ_A) = 1 bit — a pure global state with a maximally mixed marginal

Because a Bell state reduces to $\rho_A=I/2$ , a maximally entangled pair is locally as blank as physics allows: all of its structure lives in correlations invisible to either party. That is the precise content of the slogan "entanglement is information stored non-locally," and it is why measuring one half of an EPR pair cannot send a message to the other half.

The core equations

The whole subject rests on one definition and its corollaries. The cleanest working form of the partial trace is the basis-sum: sandwich the joint state between the bras and kets of a $B$ -basis and add up the results, contracting only the $B$ factor.

\rho_A \;=\; \mathrm{Tr}_B(\rho_{AB}) \;=\; \sum_{j} \big(I_A\otimes\langle j|_B\big)\,\rho_{AB}\,\big(I_A\otimes|j\rangle_B\big) \qquad (1)

where: $\rho_{AB}$ — density operator on $\mathcal{H}_A\otimes\mathcal{H}_B$ ; $\{|j\rangle_B\}$ — any orthonormal basis of $\mathcal{H}_B$ ; $I_A$ — identity on $\mathcal{H}_A$ ; $\rho_A$ — reduced operator on $\mathcal{H}_A$ . Reading: summing $\langle j|_B\,\cdots\,|j\rangle_B$ over a full $B$ -basis is the trace over $B$ . What fixes $\mathrm{Tr}_B$ on every operator is its action on product operators — factor the $A$ -part through and reduce the $B$ -part to its scalar trace; linearity then extends it to all $\rho_{AB}$ .

\mathrm{Tr}_B(X_A\otimes Y_B)=X_A\,\mathrm{Tr}(Y_B) \qquad (2)

where: $X_A$ — operator on $\mathcal{H}_A$ ; $Y_B$ — operator on $\mathcal{H}_B$ ; $\mathrm{Tr}(Y_B)$ — its scalar trace. Reading: the partial trace "eats" the $B$ factor and leaves $A$ untouched. The reason to single out this map is operational: it alone preserves every local expectation value, inserting $I_B$ as "do nothing on Bob's side."

\mathrm{Tr}\big[(M_A\otimes I_B)\,\rho_{AB}\big]=\mathrm{Tr}\big[M_A\,\rho_A\big]\quad\forall\,M_A \qquad (3)

where: $M_A$ — any observable on $A$ ; $I_B$ — identity on $B$ . Reading: this is the defining property — no other operator on $\mathcal{H}_A$ satisfies it (proved below). The output is always a bona fide state — the partial trace is completely positive and trace-preserving.

\rho_A\succeq 0,\qquad \rho_A=\rho_A^\dagger,\qquad \mathrm{Tr}(\rho_A)=1 \qquad (4)

where: $\succeq 0$ — positive semidefinite; $\rho_A^\dagger$ — Hermitian conjugate; unit trace inherited from $\mathrm{Tr}(\rho_{AB})=1$ . Reading: $\rho_A$ is a legitimate state even when $\rho_{AB}$ is pure and $\rho_A$ comes out mixed. For a pure bipartite state the Schmidt decomposition diagonalizes $|\psi\rangle$ , and tracing out either side leaves the coefficients as the marginal's eigenvalues.

\rho_A=\mathrm{Tr}_B\big(|\psi\rangle\langle\psi|\big)=\sum_i\lambda_i\,|i\rangle\langle i|_A,\qquad |\psi\rangle=\sum_i\sqrt{\lambda_i}\,|i\rangle_A|i\rangle_B \qquad (5)

where: $\lambda_i$ — squared Schmidt coefficients, $\lambda_i\ge0$ , $\sum_i\lambda_i=1$ ; $|i\rangle_A,|i\rangle_B$ — Schmidt bases. Reading: the rank of $\rho_A$ (the Schmidt number) counts the entanglement of $|\psi\rangle$ . Since $\rho_A$ and $\rho_B$ share these nonzero eigenvalues, their von Neumann entropies are equal — even for unequal dimensions — making the marginal's entropy the entanglement measure for pure states.

S(\rho_A)=-\mathrm{Tr}(\rho_A\log_2\rho_A)=-\sum_i\lambda_i\log_2\lambda_i=S(\rho_B) \qquad (6)

where: $S(\rho_A)$ — von Neumann entropy in bits; equality with $S(\rho_B)$ holds for any pure $|\psi\rangle_{AB}$ . Reading: entanglement entropy is symmetric — "how mixed Alice looks" equals "how mixed Bob looks." For a single qubit the reduced state is three real numbers, the local Pauli expectations, which are its Bloch-vector components — the practical form when $A$ is one qubit of a register.

\rho_A=\tfrac{1}{2}\big(I + \langle X\rangle\,X + \langle Y\rangle\,Y + \langle Z\rangle\,Z\big) \qquad (7)

where: $\langle P\rangle=\mathrm{Tr}[(P\otimes I_B)\rho_{AB}]$ — local Pauli expectation; $X,Y,Z$ — Pauli operators on $A$ . Reading: the three local Pauli averages are single-qubit tomography — reconstructing $\rho_A$ exactly and nothing more.

Derivation AWhy the reduced density matrix is unique and basis-independent

The partial trace is defined by a formula that names an orthonormal basis of $B$ , which raises two worries. Does the answer secretly depend on the basis we picked, and is the operator we get even the "right" marginal? We settle both at once: first the operational identity of equation (3) pins down $\rho_A$ uniquely, so the marginal is defined by its job rather than any formula; then the basis-sum recipe of equation (1) is shown to do that job, and to give the same operator no matter which basis of $B$ we sum over.

Step 1. Demand that some operator on $\mathcal{H}_A$ satisfy $\mathrm{Tr}[(M_A\otimes I_B)\rho_{AB}]=\mathrm{Tr}[M_A\,\rho_A]$ for every Hermitian $M_A$ . Suppose two operators $\rho_A$ and $\sigma_A$ both satisfy it. Subtracting, $\mathrm{Tr}[M_A(\rho_A-\sigma_A)]=0$ for all $M_A$ . Since the Hilbert–Schmidt pairing $\langle M,X\rangle=\mathrm{Tr}[M^\dagger X]$ is non-degenerate, the only operator orthogonal to every $M_A$ is zero, so $\rho_A=\sigma_A$ . Why: this defines "the marginal" by what it must do — reproduce every local expectation — so any correct recipe is forced to land on one and the same operator.

Step 2. Check that the basis-sum recipe delivers exactly this operator. Take $\rho_A=\sum_j (I_A\otimes\langle j|)\rho_{AB}(I_A\otimes|j\rangle)$ and compute $\mathrm{Tr}[M_A\rho_A]=\sum_j\mathrm{Tr}_A\!\big[M_A(I_A\otimes\langle j|)\rho_{AB}(I_A\otimes|j\rangle)\big]=\sum_j\langle j|\,\mathrm{Tr}_A[(M_A\otimes I_B)\rho_{AB}]\,|j\rangle$ , and summing the diagonal over the full $B$ -basis is the $B$ -trace, giving $\mathrm{Tr}[(M_A\otimes I_B)\rho_{AB}]$ . Why: summing $(I_A\otimes\langle j|)\,\cdots\,(I_A\otimes|j\rangle)$ over a complete basis is literally the trace over $B$ , so the recipe computes precisely the number the operational identity requires.

Step 3. Now change the $B$ -basis from $\{|j\rangle\}$ to $\{|j'\rangle\}=\{U|j\rangle\}$ with $U$ unitary. Expanding $\langle j'|=\sum_k U_{kj}^*\langle k|$ and $|j'\rangle=\sum_l U_{lj}|l\rangle$ , the new sum becomes $\sum_{j,k,l}U_{kj}^*U_{lj}\,(I\otimes\langle k|)\rho_{AB}(I\otimes|l\rangle)$ . The coefficient collapses because $\sum_j U_{lj}U_{kj}^*=(UU^\dagger)_{lk}=\delta_{lk}$ . Why: the two change-of-basis unitaries meet as $UU^\dagger=I$ and cancel — the arbitrary choice of basis leaves no fingerprint.

Step 4. With $\delta_{lk}$ enforced, the double sum reduces to $\sum_k (I\otimes\langle k|)\rho_{AB}(I\otimes|k\rangle)$ , the original expression. So $\mathrm{Tr}_B$ is basis-independent, and by Step 1 it is the unique operator meeting the operational demand. Why: uniqueness plus basis-independence together certify that " $\rho_A$ " is a well-defined physical object, not an artifact of bookkeeping. *Why basis-independence holds in practice: recomputing the Bell marginal in the Hadamard basis of $B$ returns the same $I/2$ — a concrete check that the arbitrary choice of $B$ -basis leaves no fingerprint on $\rho_A$ .*

Go deeper · Derivation B · 1 minThe reduced state of a general two-qubit pure state and its concurrence

Derivation BThe reduced state of a general two-qubit pure state and its concurrence

We compute the marginal of the most general two-qubit pure state and watch entanglement fall out as a determinant. Write $|\psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$ with $|a|^2+|b|^2+|c|^2+|d|^2=1$ , trace out $B$ , and read off $\rho_A$ with its purity and determinant. The payoff: one scalar built from the amplitudes — the concurrence $\mathcal{C}=2|ad-bc|$ — controls how mixed the marginal is, so a local quantity (mixedness of $\rho_A$ ) and a global one (entanglement of $|\psi\rangle$ ) are the same number in disguise.

Step 2. Tracing out $B$ leaves $(\rho_A)_{mn}=\langle m|_A\,\rho_A\,|n\rangle_A=\langle u_n|u_m\rangle_B$ , i.e. the reduced-matrix entries are the $B$ -inner products of the conditional vectors. Evaluating: $\langle u_0|u_0\rangle=|a|^2+|b|^2$ , $\langle u_1|u_1\rangle=|c|^2+|d|^2$ , and the off-diagonal $\langle u_1|u_0\rangle=ac^*+bd^*$ . Why: the overlap measures how distinguishable Bob's two conditional states are — vanishing overlap means Alice's outcome carries a distinguishable flag on Bob, the mark of strong correlation.

Step 3. Assemble $\rho_A=\begin{pmatrix}|a|^2+|b|^2 & ac^*+bd^*\\ a^*c+b^*d & |c|^2+|d|^2\end{pmatrix}$ , which is manifestly Hermitian with unit trace. Its determinant simplifies by the Lagrange identity to $\det\rho_A=|ad-bc|^2$ . Why: the determinant of a unit-trace $2\times2$ state controls its mixedness, and $|ad-bc|^2$ is zero exactly when $|\psi\rangle$ factorizes — the algebraic signature of "no entanglement."

Step 4. Define the concurrence $\mathcal{C}=2|ad-bc|$ , so $\det\rho_A=\mathcal{C}^2/4$ , and compute the purity $\mathrm{Tr}(\rho_A^2)=(\mathrm{Tr}\rho_A)^2-2\det\rho_A=1-2|ad-bc|^2=1-\tfrac12\mathcal{C}^2$ . As $\mathcal{C}$ runs from $0$ (product) to $1$ (Bell), the purity falls from $1$ to $\tfrac12$ and $\rho_A$ slides from pure to maximally mixed. Why: the master link — local mixedness $1-\mathrm{Tr}(\rho_A^2)$ and global entanglement $\mathcal{C}$ are readings of the same $|ad-bc|$ , so Alice's purity measures her entanglement with Bob. *Two boundary checks: setting $a=d=1/\sqrt2,\,b=c=0$ (the Bell state) recovers $\rho_A=I/2$ , $\mathcal{C}=1$ , purity $1/2$ ; setting $a=1,\,b=c=d=0$ (a product state) gives $\rho_A=|0\rangle\langle0|$ , $\mathcal{C}=0$ , purity $1$ — the two extremes of the entanglement range.*

Go deeper · Derivation C · 1 minNo local operation on B can change Alice's marginal

Derivation CNo local operation on B can change Alice's marginal

Finally we prove the property that elevates the partial trace from a bookkeeping device to a law of nature: whatever Bob does to his subsystem — a unitary, a measurement he never reports, any physical process — Alice's reduced state does not budge. Let Bob apply an arbitrary quantum channel $\mathcal{E}_B$ , a completely positive trace-preserving map. Pushing it through the partial trace, we show it vanishes from Alice's marginal, so quantum mechanics is non-signalling: correlations show only when the parties compare notes, never through local statistics alone.

Step 1. Write Bob's channel in Kraus form, $(I_A\otimes\mathcal{E}_B)(\rho_{AB})=\sum_k(I_A\otimes E_k)\,\rho_{AB}\,(I_A\otimes E_k^\dagger)$ , where the Kraus operators obey $\sum_k E_k^\dagger E_k=I_B$ (trace preservation). Why: every physical operation Bob can perform has such a representation, so a generic $\{E_k\}$ covers all of them at once.

Step 2. Trace out $B$ using the basis-sum definition and combine the $B$ -operators inside the sum. The key move is cyclicity within the $B$ -factor: $\mathrm{Tr}_B[(I\otimes E_k)\,X\,(I\otimes E_k^\dagger)]=\mathrm{Tr}_B[(I\otimes E_k^\dagger E_k)\,X]$ . Why: cyclicity lets Bob's operators reunite as $E_k^\dagger E_k$ , the combination trace preservation constrains.

Step 3. Sum over the Kraus index and apply $\sum_k E_k^\dagger E_k=I_B$ : the reduced state becomes $\mathrm{Tr}_B[(I_A\otimes\sum_k E_k^\dagger E_k)\rho_{AB}]=\mathrm{Tr}_B[(I_A\otimes I_B)\rho_{AB}]=\mathrm{Tr}_B(\rho_{AB})=\rho_A$ . Bob's channel has left no trace on Alice's side. Why: trace preservation — Bob's operation conserving total probability — is the very condition that makes his action invisible to Alice.

Step 4. Read off the consequence. Even an unrecorded projective measurement on $B$ is such a channel, so Alice cannot tell whether Bob measured, what he found, or whether he did anything at all. Only classical communication of his outcome lets her update to a conditional state. Why: the no-signalling theorem in one line — entanglement carries correlation, not communication, so it cannot signal faster than light. *The unitary special case is a clean check: $\mathcal{E}_B(\cdot)=U_B(\cdot)U_B^\dagger$ gives $\mathrm{Tr}_B[U_B\rho_{AB}U_B^\dagger]=\mathrm{Tr}_B(\rho_{AB})$ directly by cyclicity of the trace, confirming that even a unitary rotation by Bob leaves Alice's marginal unchanged.*

Limiting cases

The reduced state behaves in sharply characteristic ways at the corners of the parameter space, and these are the fastest sanity checks on any calculation.

Case	Result	Meaning
Product state	$\rho_{AB}=\rho_A\otimes\rho_B\Rightarrow\mathrm{Tr}_B(\rho_{AB})=\rho_A$	For an uncorrelated pair the partial trace returns the genuine local state; nothing about $A$ is lost.
Maximally entangled pure state	$\|\Phi^+\rangle=\tfrac{1}{\sqrt2}(\|00\rangle+\|11\rangle)\Rightarrow\rho_A=\tfrac{I}{2}$	A pure global state gives the maximally mixed marginal: maximal entanglement, zero local information.
Schmidt form	$\|\psi\rangle=\sum_i\sqrt{\lambda_i}\|i\rangle_A\|i\rangle_B\Rightarrow\rho_A=\sum_i\lambda_i\|i\rangle\langle i\|_A$	$\rho_A$ and $\rho_B$ share the spectrum $\{\lambda_i\}$ ; equal entropies $S(\rho_A)=S(\rho_B)$ quantify entanglement.
Iterated trace	$\mathrm{Tr}_A(\mathrm{Tr}_B(\rho_{AB}))=\mathrm{Tr}(\rho_{AB})=1$	Tracing out every subsystem in turn recovers the ordinary trace, confirming normalization is consistent.
Classically correlated	$\tfrac12(\|00\rangle\langle00\|+\|11\rangle\langle11\|)\Rightarrow\rho_A=\tfrac{I}{2}$	A separable mixture can give the same $I/2$ as a Bell state — the marginal alone cannot see entanglement.

Common trapThe partial trace is not "deleting the B rows and columns"

It is tempting to compute $\rho_A$ by striking out the rows and columns of $\rho_{AB}$ that carry $B$ -labels. That is wrong, and not by a little. The correct operation sums the diagonal blocks of $\rho_{AB}$ over a $B$ -basis; the cross terms cancel only through basis orthogonality $\langle k|j\rangle=\delta_{kj}$ , not because we discard them by hand. Deleting entries is not even a positive map, so it can return negative eigenvalues — not a state at all. A subtler trap: $\rho_A$ genuinely loses information. Many distinct global states, entangled and separable, share the same marginal, so $\rho_A$ alone can never tell you whether $A$ is entangled with $B$ — a maximally mixed $\rho_A=I/2$ fits a Bell pair and a coin-flip mixture, and only the global purity resolves which.

Worked examples

1 · Reduced state, purity, and entanglement of a partially entangled state (fully worked)

Alice and Bob share the normalized state $|\psi\rangle=\tfrac{1}{\sqrt3}\big(|00\rangle+|01\rangle+|11\rangle\big)$ . Find Alice's reduced density matrix, its eigenvalues, its purity, its von Neumann entropy, and the concurrence — and check the numbers hang together.

Step 1. Read off the amplitudes in the convention $|\psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$ : here $a=b=d=\tfrac{1}{\sqrt3}$ and $c=0$ . Normalization checks: $|a|^2+|b|^2+|c|^2+|d|^2=\tfrac13+\tfrac13+0+\tfrac13=1$ . Good.

Step 2. Apply the general result from the derivation above. The diagonal entries are $|a|^2+|b|^2=\tfrac23$ and $|c|^2+|d|^2=\tfrac13$ ; the off-diagonal is $ac^*+bd^*=0+\tfrac{1}{\sqrt3}\cdot\tfrac{1}{\sqrt3}=\tfrac13$ . So $\rho_A=\begin{pmatrix}2/3 & 1/3\\ 1/3 & 1/3\end{pmatrix}$ , with $\mathrm{Tr}\,\rho_A=1$ as required. The off-diagonal coherence 1/3 is nonzero, so Alice's state is not simply diagonal.

Step 3. Diagonalize. For a $2\times2$ state the eigenvalues are $\lambda_\pm=\tfrac12\big(1\pm\sqrt{1-4\det\rho_A}\big)$ with $\det\rho_A=\tfrac23\cdot\tfrac13-(\tfrac13)^2=\tfrac29-\tfrac19=\tfrac19$ . Then $\sqrt{1-4/9}=\sqrt{5}/3=0.7454$ , giving $\lambda_+=0.8727$ and $\lambda_-=0.1273$ . The marginal is mixed — its eigenvalues are 0.873 and 0.127, not 1 and 0.

Step 4. Purity: $\mathrm{Tr}(\rho_A^2)=1-2\det\rho_A=1-\tfrac29=\tfrac79\approx0.778$ . Cross-check via the eigenvalues: $0.8727^2+0.1273^2=0.762+0.016=0.778$ . Purity = 7/9 ≈ 0.778 (between the pure value 1 and the maximally mixed value 1/2).

Step 5. Concurrence: $\mathcal{C}=2|ad-bc|=2\big|\tfrac{1}{\sqrt3}\cdot\tfrac{1}{\sqrt3}-0\big|=\tfrac23\approx0.667$ . Consistency with the purity formula: $1-\tfrac12\mathcal{C}^2=1-\tfrac12\cdot\tfrac49=1-\tfrac29=\tfrac79$ — matches Step 4 exactly.

Step 6. Von Neumann entropy: $S(\rho_A)=-0.8727\log_2 0.8727-0.1273\log_2 0.1273=0.171+0.379=\mathbf{0.550\ bits}$ . Between $0$ (product) and $1$ (Bell), so $|\psi\rangle$ is partially entangled. By equation (6) Bob's marginal has the identical entropy $0.550$ bits, even though we only ever touched Alice. Answer: ρ_A has eigenvalues 0.873 and 0.127, purity 7/9, concurrence 2/3, and entanglement entropy S = 0.550 bits.

2 · The block-trace mechanic, and why the marginal cannot see entanglement

Do the Bell partial trace as an explicit block operation on the $4\times4$ matrix, then compare it to a classically correlated mixture that yields the same marginal.

Step 1. In the basis order $|00\rangle,|01\rangle,|10\rangle,|11\rangle$ , the Bell state is $|\Phi^+\rangle\langle\Phi^+|=\tfrac12\big(|00\rangle+|11\rangle\big)\big(\langle00|+\langle11|\big)$ , i.e. the $4\times4$ matrix with $\tfrac12$ in the four corners and zeros elsewhere. Partition it into four $2\times2$ blocks $G_{mn}$ indexed by Alice's qubit $m,n\in\{0,1\}$ , so $A$ is the outer index and each block acts on Bob's qubit.

Step 2. The reduced-matrix rule from equation (1) is exactly "trace each block": $(\rho_A)_{mn}=\mathrm{Tr}(G_{mn})$ . Here $G_{00}=\tfrac12\begin{pmatrix}1&0\\0&0\end{pmatrix}$ has trace $\tfrac12$ ; $G_{11}=\tfrac12\begin{pmatrix}0&0\\0&1\end{pmatrix}$ has trace $\tfrac12$ ; and the off-diagonal blocks $G_{01},G_{10}=\tfrac12\begin{pmatrix}0&1\\0&0\end{pmatrix},\tfrac12\begin{pmatrix}0&0\\1&0\end{pmatrix}$ each have trace $0$ . So $\rho_A=\begin{pmatrix}1/2&0\\0&1/2\end{pmatrix}=I/2$ — a pure global state with a maximally mixed marginal, entropy exactly 1 bit.

Step 3. Now take the separable mixture $\rho_{AB}=\tfrac12\big(|00\rangle\langle00|+|11\rangle\langle11|\big)$ — a classically correlated coin: both qubits are $0$ or both are $1$ , chosen by a hidden fair coin, with no coherence. Trace out $B$ the same way: the diagonal blocks are $\tfrac12|0\rangle\langle0|$ and $\tfrac12|1\rangle\langle1|$ , giving again $\rho_A=I/2$ .

Step 4. Draw the moral. The Bell state (pure, maximally entangled) and the coin mixture (mixed, separable) are utterly different global states, yet hand Alice the identical marginal $I/2$ . No local measurement can distinguish "entangled with Bob" from "classically correlated with Bob" from "just noisy" — the entanglement lives in the global coherences the partial trace discards. This is the constructive twin of the no-signalling result: not only Bob's actions but the very nature of their correlation is invisible to Alice alone.

Physics of the portrait

The reduced density matrix and the partial trace are the load-bearing timbers of quantum information, and they were laid down at the theory's founding. The portrait belongs to the mathematician who gave quantum mechanics its operator-and-trace language — and whose name is on the entropy that measures exactly the mixedness this article computes.

Origins · 1 minJohn von Neumann and how this was discovered

John von Neumann

1903–1957 · density operators, the trace formalism, and the entropy of quantum states

John von Neumann built the Hilbert-space scaffolding on which all of this stands. In 1927 — the same year Lev Landau independently reached the density matrix while describing a subsystem coupled to its surroundings, the reduced-state idea in embryo — von Neumann introduced the density operator to place statistical mixtures and quantum states on the same footing, and defined the entropy $S(\rho)=-\mathrm{Tr}(\rho\log\rho)$ that bears his name and that equation (6) evaluates for a marginal. His 1932 treatise Mathematische Grundlagen der Quantenmechanik fixed the trace, the projection postulate, and the operator algebra as the permanent grammar of the subject; the partial trace is the natural offspring of that grammar — the trace performed on a single tensor factor. His reach was vast even by his century's standards, from the minimax theorem to the stored-program computer, but for quantum information the density-operator formalism is the gift that keeps giving: purification, entanglement entropy, quantum channels, and the whole theory of open systems are written in the language he set down. Every time a physicist traces out an environment to watch coherence decay, they are using von Neumann's machinery to compute a von Neumann entropy.

History

1927

— Landau and von Neumann independently introduce the density matrix, Landau to describe a subsystem of a larger quantum system

1932

— von Neumann's Mathematische Grundlagen formalizes the trace, density operators, and the entropy S = −Tr(ρ log ρ)

1935

— Schrödinger names entanglement (Verschränkung) and notes that the reduced state of one half of a pure pair is generically mixed

1970

— Araki and Lieb prove the entropy triangle inequality relating S(A), S(B), and S(AB) for bipartite systems

1998

— Wootters derives the concurrence, a closed-form two-qubit entanglement measure read directly from the reduced state

Going further

Texts. Nielsen & Chuang, Quantum Computation and Quantum Information, §2.4.3 (density operators and the partial trace) · Preskill, Lecture Notes on Quantum Computation, Ch. 2–3 · Watrous, The Theory of Quantum Information, Ch. 1–2 · Wilde, Quantum Information Theory, Ch. 4–5 · Kitaev, Shen & Vyalyi, Classical and Quantum Computation, Ch. 11 · Mermin, Quantum Computer Science, Ch. 1–2. The key anchor numbers are: the Bell marginal $I/2$ with entropy $1$ bit; the general two-qubit relations $\det\rho_A=|ad-bc|^2$ , $\mathcal{C}=2|ad-bc|$ , purity $1-\tfrac12\mathcal{C}^2$ ; the worked example $\tfrac{1}{\sqrt3}(|00\rangle+|01\rangle+|11\rangle)$ giving eigenvalues $0.873,0.127$ , purity $7/9$ , $\mathcal{C}=2/3$ , $S=0.550$ bits; and the no-signalling identity following from $\sum_k E_k^\dagger E_k=I_B$ .

Builds on: Density Operators · Dirac Bra-Ket Notation · The Born Rule · Expectation Values and Variance · Inner Product, Norm, and Normalization. Leads to: Quantum Channels and CPTP Maps · Entanglement Entropy and Schmidt Decomposition · Purification and the Stinespring Dilation · Decoherence and Open Quantum Systems.

Ask the tutor: Ask me to trace out a subsystem of any state you name — a Bell pair, a GHZ or W state, a noisy qubit coupled to an environment — and I will compute the reduced density matrix, its entropy, and what it does and does not reveal about the entanglement.

What is the difference between the full trace and the partial trace?

The full trace $\mathrm{Tr}$ contracts all indices and returns a scalar. The partial trace $\mathrm{Tr}_B$ contracts only the indices of subsystem $B$ and returns an operator on the remaining space $\mathcal{H}_A$ . Where the full trace over $\rho_{AB}$ gives $1$ (total probability), the partial trace over $B$ gives $\rho_A$ , a full-fledged density operator you can still measure, evolve, or trace again. Iterating, $\mathrm{Tr}_A(\mathrm{Tr}_B(\rho_{AB}))$ recovers the full trace, which is why normalization is preserved at every stage.

Why does a pure entangled state give a mixed reduced state?

Because the purity of the whole says nothing about the purity of a part. When $|\psi\rangle_{AB}$ is entangled, no local property of $A$ is definite on its own — the information lives in correlations between $A$ and $B$ . Tracing out $B$ discards that correlated information, and the leftover uncertainty about $A$ shows up as mixedness. The Schmidt form makes it exact: $\rho_A=\sum_i\lambda_i|i\rangle\langle i|_A$ is pure (rank one) only when a single $\lambda_i=1$ , i.e. only when $|\psi\rangle$ is a product; any spread in the $\lambda_i$ makes $\rho_A$ mixed.

Does the answer depend on which basis of B I trace over?

No — this is a theorem, not a convention. Switch Bob's basis from $\{|j\rangle\}$ to any rotated basis $\{U|j\rangle\}$ with $U$ unitary, and the change-of-basis factors appear as $\sum_j U_{lj}U_{kj}^*=(UU^\dagger)_{lk}=\delta_{lk}$ and cancel exactly, leaving the same $\rho_A$ . It must be so: the basis describing Bob's discarded qubit is an arbitrary coordinate choice, and no prediction about Alice can depend on it. Verify it on the Bell state — trace out $B$ in the computational or the Hadamard basis and you get $I/2$ either way.

How is the partial trace related to quantum channels?

It is the simplest non-trivial quantum channel — the one for discarding or losing a subsystem — and it is completely positive and trace-preserving (CPTP), so it maps valid states to valid states. More than an analogy: the Stinespring dilation theorem says every quantum channel is "attach an ancilla, apply a joint unitary, then partial-trace the ancilla away." Decoherence is exactly this — a system evolves unitarily with its environment, and tracing the environment out yields the noisy channel the system experiences.

If Alice's marginal can't reveal entanglement, how do we detect it?

By looking at more than one marginal, or at the global state. Local tomography on $A$ gives you $\rho_A$ and nothing else — and a maximally mixed $\rho_A=I/2$ is consistent with a Bell pair, a classical coin-flip mixture, or plain local noise. To certify entanglement you need joint information: measure correlated observables on both qubits (a Bell-inequality test, or full two-qubit state tomography), or compare $S(\rho_A)$ against the global entropy $S(\rho_{AB})$ . For a pure global state the rule is clean — any $S(\rho_A)>0$ means entanglement — but for a mixed global state you need entanglement witnesses or measures like the concurrence, which depend on the whole $\rho_{AB}$ , not on any single marginal.

Is tracing out a subsystem the same as measuring it and forgetting the result?

Almost, and the difference is instructive. If Bob measures his qubit and discards the outcome, Alice's state becomes $\sum_k p_k\,\rho_A^{(k)}$ , a mixture over his possible results, and by the no-signalling derivation this equals $\rho_A=\mathrm{Tr}_B(\rho_{AB})$ exactly. So an unrecorded measurement gives the same marginal as the partial trace. The partial trace is basis-independent while a measurement picks a basis; they still agree because averaging over any complete measurement, with outcomes forgotten, reproduces the basis-independent trace. Only when Bob records and communicates his outcome can Alice condition on it and collapse to a specific $\rho_A^{(k)}$ — that extra classical information, not the measurement itself, is what changes her state.