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Give Alice the first qubit $A$ and Bob the second $B$. Compare the Bell state $|{\Phi }^{+}⟩$ with a preparation that tosses a fair classical bit and gives both parties its value. Their density operators, written in the computational basis ordered $00,01,10,11$, areThey have identical diagonal quantum state populations. The off-diagonal entries of the Bell state retain quantum coherence between $00$ and $11$; the classical mixture has none in this basis.

If both parties perform a Pauli measurement of $Z$, the Born rule gives $P(00)=P(11)=1/2$ and $P(01)=P(10)=0$ for both preparations. Perfect agreement here therefore cannot establish entanglement: an ordinary shared random bit already explains it.

Now both measure $X$, whose normalized eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Changing basis givesThe Bell state again has matching outcomes, each with probability $1/2$. For either product state $|00⟩$ or $|11⟩$, each of the four $X$ outcome pairs has probability $1/4$, so their classical mixture does too. Equivalently, the Pauli correlator $⟨X\otimes X⟩$ is one for the Bell state and zero for this mixture, while $⟨Z\otimes Z⟩$ is one for both.

Taking a partial trace gives ${\rho }_A={\rho }_B=I_2/2$ for both preparations. Thus every measurement confined to Alice's system has the same statistics in the two cases, and so does every measurement confined to Bob's. They reveal the difference by comparing records through classical communication. Conditional on Alice's result, Bob can predict a matching $X$ result for the Bell state; without that result, his outcomes remain evenly split. An uncommunicated local quantum channel, including a measurement with its outcome discarded, leaves his reduced density matrix unchanged. This is quantum no-signalling, even when conditional states change. ^[ref]

The Von Neumann entropy makes the joint-versus-local distinction numerical. With base-two logarithms, the Bell state has eigenvalues $1,0,0,0$, giving $S(AB)=0$, whereas the mixture has $1/2,1/2,0,0$, giving $S(AB)=1$. Both have $S(A)=S(B)=1$. Their quantum mutual information $I(A:B)=S(A)+S(B)-S(AB)$ is consequently two and one bits, respectively. The value two is not two readable classical bits in either local qubit: the matching $Z$ records have ordinary mutual information one bit in both cases. Random records can come from a globally pure state. ^[ref]

These two matching-axis measurements distinguish the specified preparations; they alone are not a test of a Bell inequality. A local hidden-variable theory could assign shared random answers separately to these two axes. Bell theorem concerns incompatible correlations across a suitable larger collection of measurement settings. Bell's 1964 paper established the obstruction for local hidden-variable descriptions of the predicted correlations. ^[ref]

A quantum system with two specified subsystems has a tensor product Hilbert space. A joint density operator determines subsystem $A$'s measurement statistics through its partial trace ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$. A joint pure state need not give a pure reduced state: the Bell state $(|00⟩+|11⟩)/\sqrt{2}$ gives ${\rho }_A=I/2$.

With normalized Choi state $C_{\Lambda }$ and the output factor first,Expand $C_{\Lambda }=d^{-1}{\sum }_{i,j}\Lambda (|i⟩⟨j|)\otimes |i⟩⟨j|$. The partial trace multiplies each coefficient by $\mathrm{Tr}(|i⟩⟨j|X^T)=X_{ij}$, reconstructing $\Lambda (X)$ by linearity.

For a bipartite input density operator ${\rho }_{AC}^x$ and a POVM $\{E_y\}$ on $A$, let $q(y|x)=\mathrm{Tr}[(E_y\otimes I){\rho }^x]$. For $q(y|x)>0$, the conditioned input on $C$ isIts positivity follows from the positive sandwich before the partial trace. A quantum channel on $C$ acts on these conditional density operators even if the original input is an entangled state.

For every quantum channel $T$,For a partial trace, this says that discarding a subsystem cannot make two density operators more distinguishable in quantum relative entropy.

The $n$-qubit Dicke state with $m$ excitations is the normalized equal superpositionwhere $|x|$ is the Hamming weight in the computational basis. The binomial coefficient counts the mutually orthogonal summands, proving normalization. Every permutation of the qubits leaves the vector unchanged. Splitting the sum according to the first bit and using $\left(\genfrac{}{}{0ex}{}{n-1}{m}\right)/\left(\genfrac{}{}{0ex}{}{n}{m}\right)=(n-m)/n$ givesTerms with zero coefficient are omitted at the endpoints. This recursion is useful for partial traces and for distributing a logical qubit among many symmetric physical qubits.

Physical Review Letters 71, 666–669. Equations (1)–(5) explicitly solve the two-coordinate-coupled-oscillator ground state, partial trace and effective thermal oscillator spectrum. The effective frequency need not equal either bare subsystem frequency.

Suppose two density operators on $R\otimes E$ have the same reduced density matrix ${\sigma }_R$. Every measurement confined to $R$ then has the same outcome probabilities for the two preparations: a POVM effect $F_R$ has probability $\mathrm{Tr}(F_R{\sigma }_R)$ in either case. This remains true when $R$ contains several laboratories brought together for a joint measurement.

For independent copies, the accessible states are both ${\sigma }_R^{\otimes n}$. No collective measurement on those copies can distinguish the original preparations. Adaptive local processing also cannot help, since applying the same quantum operation to equal inputs gives equal outputs. Access to an additional environment measurement record is a change of resources, and can reveal differences hidden by the partial trace.

Averaging conjugation by all Heisenberg-Weyl operators completely depolarizes a system:Summing over phases $Z^m$ kills off-diagonal matrix elements, and averaging shifts $X^k$ makes all diagonal elements equal. Acting on subsystem $B$ of a bipartite operator gives $({\mathrm{Tr}}_{B}A)\otimes I_B/d$. This identity turns joint convexity of quantum relative entropy into monotonicity under partial trace.

The reduced density matrix obtained by taking the partial trace of a larger entangled state. It predicts the local measurement statistics, but its mixed form does not imply that the total state has collapsed into one of the displayed components. Different ensembles can represent the same reduced density matrix.

A theoretical physicist working on quantum information theory.^[ref]

Homepage: www.preskill.caltech.edu/

Under a product unitary operator $U_A\otimes U_B$, the reduced density matrix of $A$ becomes $U_A{\rho }_A{U}_A^{†}$. The unitary on $B$ cancels under the partial trace. Unitary conjugation preserves eigenvalues, so the squared Schmidt coefficients of a bipartite pure state are unchanged. Conversely, two normalized bipartite pure states with the same squared Schmidt coefficients, including multiplicities and zero padding, have Schmidt decompositions carried into each other by local unitaries mapping their Schmidt bases. Thus reduced-state spectra completely classify bipartite pure states up to local unitary operations.

If $\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1\to 0$, thenIn finite dimension this follows from multipartite superadditivity of quantum relative entropy, additivity of quantum relative entropy, contraction of trace distance under partial trace, and uniform continuity on the compact state space.

The partial trace over $B$ is the linear operation characterized byApplied to a density operator, it produces a reduced density matrix.

The vectors need not form orthonormal bases. With the inner product linear in its second entry, expand the rank-one operator $|\Psi ⟩⟨\Psi |$ and take the trace on the second factor. This gives the displayed reduced positive operator. For every observable $A$ on the first factor,Positivity also follows directly from $⟨v,S_{1}v⟩=\Vert {\sum }_j⟨v,{\xi }_j⟩{\eta }_j{\Vert }^2$. Thus the physical density operator is $S_1/\Vert \Psi {\Vert }^2$ when the nonzero source vector is not normalized. The Gram matrix of the second-factor vectors supplies the cross coefficients; replacing those coefficients by Kronecker deltas without checking orthonormality changes the subsystem state. This formula is useful when an efficient product expansion uses nonorthogonal vectors, even though the partial trace itself is basis-independent.

The partial trace of a positive operator is positive by this sum of nonnegative expectations in an orthonormal basis of the traced factor. Its trace equals the original trace, using the product orthonormal basis. Basis independence follows from $\mathrm{Tr}[({\mathrm{Tr}}_B\rho )A]=\mathrm{Tr}[\rho (A\otimes I)]$, which uniquely determines the reduced operator. Thus reductions of density matrices remain density matrices. For a positive trace-class operator the same argument uses convergent nonnegative basis sums.

Charlie cannot send a bit to Alice with the stated preparation, fixed interaction and final local measurements. Write $V={\mathrm{CNOT}}_{B\to A}$ and $W={\mathrm{CNOT}}_{B\to C}$, so the complete gate is $U=WV$ with the rightmost gate acting first. Alice has finished interacting before the gate involving Charlie acts.

For arbitrary product input density matrices, put ${\tau }_{AB}=V({\rho }_A\otimes {\rho }_B)V^{†}$. The final reduced density matrix of Alice isThe equality follows because a unitary operator on the traced-out systems leaves the partial trace unchanged. The right-hand side is independent of Charlie's preparation. Therefore every outcome probability of every local measurement by Alice is independent of Charlie's message. Quantum no-signalling applies after Alice's last interaction, even though phase kickback can allow communication between the two qubits of an individual CNOT gate.

No test on Alice and Bob's systems gives any information about which global state was prepared, even with arbitrarily many independent copies. In the pure GHZ density matrix, the off-diagonal terms contain $|0⟩⟨1{|}_C$ or its adjoint. Their partial traces vanish. Hence both preparations have the same reduced density matrixFor any POVM effect $E$ acting jointly on $AB$, the outcome probability is $\mathrm{Tr}(E{\sigma }_{AB})$ in either case. Equal reduced states give identical local statistics, where local here means local to the entire accessible laboratory $AB$, not separate operations on $A$ and $B$.^[ref]

For $n$ independent copies, taking the partial trace of each inaccessible $C$ system yields ${\sigma }_{AB}^{\otimes n}$ under either hypothesis. Even an arbitrary collective quantum measurement on all $2n$ accessible qubits therefore has identical outcome probabilities. Adaptive local processing cannot change equality of the accessible density matrices. With equal prior probabilities the optimal guessing probability remains $1/2$; with other priors, no observation updates their ratio.

Expand the density matrix of the Bell state and take the partial trace over Alice's qubit:The orthonormality of the computational basis removes the off-diagonal terms. Thus Bob's reduced density matrix is the maximally mixed state