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Quantum no-signalling says that, without communicating a measurement outcome, any local trace-preserving operation performed on subsystem $A$ leaves all outcome probabilities for measurements on subsystem $B$ unchanged; equivalently, it leaves the reduced density matrix ${\rho }_B$ unchanged.

The two Bell states are orthogonal. Applying the inverse Bell-state preparation circuit, namely a Controlled-NOT gate from $A$ to $B$ followed by a Hadamard gate on $A$, givesA computational-basis measurement therefore distinguishes them perfectly, soIf only $B$ is accessible, however,Every measurement on $B$ consequently has the same outcome distribution under both hypotheses. With equal prior probabilities no information is available, andthe error probability of a random guess.

No. Before any operation, Bob's reduced density matrix is $I/2$, independent of $|\alpha ⟩$. Any trace-preserving operation performed only by Alice leaves Bob's reduced state unchanged when her uncommunicated outcomes are averaged. This is the no-communication theorem, or quantum no-signalling.

If Bob received the unknown state, his reduced state would instead be the pure state $|\alpha ⟩⟨\alpha |$, which depends on Alice's input. That would let Alice signal by choosing different inputs, contradicting no-signalling. The two classical outcome bits are therefore essential.

The Schmidt decomposition states that every finite-dimensional bipartite pure state haswhere $s_j>0$, ${\sum }_j{s}_j^2=1$, and the two displayed families are orthonormal. The integer $r$ is the Schmidt rank.

To prove it, choose product bases and writeApply the singular value decomposition $C=US{V}^{†}$. Absorbing the columns of $U$ and the complex conjugates of the columns of $V$ into new orthonormal bases gives the stated sum, with the nonzero singular values as the Schmidt coefficients. Equivalently, $s_j^2$ are the common nonzero eigenvalues of the two reduced density matrices, so $r=\mathrm{rank}{\rho }_A=\mathrm{rank}{\rho }_B$.

Arrange the amplitudes of $|{\varphi }_1⟩$ as the matrixThe second state has $C_2=C_{1}X$, where $X$ swaps the $A_2$ basis states. If Bob receives $A_1$, his two reduced density matrices areand are identical. No measurement on $A_1$ contains any information about the shared state.

If Bob instead receives $A_2$,They differ because $a^2+c^2<b^2+d^2$. Bob can therefore distinguish them with better-than-random success. For equal priors,so the optimal success probability is $\frac{1}{2}(1+D)$. It is generally below one, so a single copy does not permit certain identification.

Without Alice's operation, Bob obtains outcome $b$ with probabilityIf Alice applies $U$ and then measures, but her outcome is not communicated, Bob's probability isbecause ${\sum }_a{\Pi }_a=I_A$. Equivalently, a trace-preserving local operation leaves Bob's reduced density matrix unchanged. This is the no-communication theorem: local operations cannot signal without communication of their outcomes.

The pure state $|\Psi {⟩}_{AB}$ is entangled when it cannot be written as a product state $|a{⟩}_A|b{⟩}_B$, equivalently when its Schmidt rank exceeds one. Its subsystem states are the reduced density matriceswhere the partial trace is characterized by $\mathrm{Tr}(M_A{\rho }_A)=⟨\Psi |M_A\otimes I_B|\Psi ⟩$ for every observable $M_A$.

Let Bob's measurement have Kraus operators $M_b$ satisfying ${\sum }_b{M}_b^{†}{M}_b=I_B$. If Alice does not learn Bob's random outcome, her state after the measurement isThis no-communication theorem means that Bob can change Alice's conditional state after she learns his outcome, but cannot change any local outcome distribution available to her alone. Standard nonrelativistic quantum mechanics is therefore operationally compatible with the prohibition of superluminal signalling in special relativity, despite its nonlocal conditional-state updates.

An exact nondisturbing state readout would destroy this protection if it reported the globally collapsed state on an absolute-time slice. For example, Alice and Bob may share Bell states. At a prearranged time Bob encodes a bit by measuring his qubit in either the computational or Hadamard basis. The usual projection postulate assigns Alice respectively one of $|0⟩,|1⟩$ or one of $|+⟩,|-⟩$. A device returning the exact pure state lets Alice identify which basis Bob chose without waiting for his outcome, producing a superluminal signal. Equivalently, Bob may choose whether to measure, and the device distinguishes the resulting proper pure state from the original improper maximally mixed local state.

A causal alternative is a quantum state readout device located at a spacetime point $x$ that reports the local quantum state under objective collapse: take reduced states on spacelike hypersurfaces through $x$ and let those hypersurfaces approach the past light cone of $x$. The result includes localized collapses in $J^{-}(x)$ and excludes spacelike-separated collapses. This assumes a fixed Minkowski spacetime, localized collapse events, ordinary local unitary dynamics between them, and outputs that may control only operations in their causal future. The postulate is logically consistent because it adds a classical record of this local state without changing the state or any standard measurement probability. It also cannot support an indirect signalling algorithm. Inductively through the algorithm's events, every readout at $x$ depends only on operations, collapses, and earlier readouts in $J^{-}(x)$; any operation selected from that output lies in the readout's future light cone. Composing readouts, unitary evolutions, and measurements therefore never carries a controllable dependence outside a future light cone, so relativistic causality is preserved.

Expanding either Bell state and using $⟨0|1{⟩}_B=0$ removes the cross terms:Thus the two expectation values agree. Equivalently, both states have the same reduced density matrix $I/2$ on qubit $A$; the positivity of $Q$ makes it a possible measurement effect but is not needed for the algebraic equality.

For $|J,J⟩$, tracing out subsystem 2 gives the reduced density matrixwhich is pure, so its entanglement entropy is zero.

For $|J,J-1⟩$, the two subsystem-2 factors in part (iii) are orthogonal. The partial trace therefore removes the cross terms and givesIts nonzero eigenvalues are $j_1/J$ and $j_2/J$, so the Von Neumann entropy isWhen $j_1=j_2$, both eigenvalues are $1/2$ and $S=\log 2$, the maximal entropy for a rank-two reduced state.

Let a projective measurement have mutually orthogonal projections $P_i$ satisfying ${\sum }_i{P}_i=I$. On a density operator $\rho$, the Born rule and the Lüders rule giveconditional on outcome $i$. On the first part of a bipartite quantum system, replace $P_i$ by $P_i\otimes I_2$.

The subsystems are isolated when there is no interaction term coupling them. Their Hamiltonian operator has the formso their subsequent unitary time evolution factorizes as $U_1\otimes U_2$.

The reduced density matrix of subsystem 1 iswhere the partial trace is characterized byfor every local observable $M_1$. Thus ${\rho }_1$ contains exactly the statistics accessible by measurements on subsystem 1.

Suppose a projective measurement $\{Q_j\}$ is performed on subsystem 2 and its outcome is not communicated. The resulting nonselective state isFor every $M_1$, the cyclic property of the trace and ${\sum }_j{Q}_j^2=I_2$ giveHence ${\rho }_1^{\prime }={\rho }_1$. No local projective measurement on subsystem 1 can reveal whether the remote unreported measurement occurred. This is quantum no-signalling, which prevents a choice made at a spacelike separation from transmitting information faster than light and makes the measurement formalism compatible with relativistic causality.

The proposed nondisturbing device would violate no information without disturbance. In an ordinary quantum instrument, let $M_{i\alpha }$ be the Kraus operators associated with classical output $i$. If every pure state $|\psi ⟩$ remains unchanged even conditional on the displayed outcome, every nonzero $M_{i\alpha }|\psi ⟩$ must be parallel to $|\psi ⟩$. A linear operator for which every vector is an eigenvector is a scalar multiple of the identity, so $M_{i\alpha }=c_{i\alpha }I$. Its output probabilityis independent of the state. It cannot equal $⟨\psi |P_i|\psi ⟩$ for arbitrary projectors. Equivalently, repeated nondisturbing samples would permit quantum state tomography of one specimen and then quantum cloning, contradicting ordinary quantum theory.

Such devices would not necessarily enable superluminal signalling. One consistent operational extension could make every sequence of outputs depend only on the local reduced density matrix and local settings. Since an unreported remote measurement leaves that matrix unchanged, all local device statistics would remain unchanged too. Other extensions could add nonlocal outcome-dependent rules and permit signalling, but that behavior is additional to the device specification rather than forced by it.

The Quantum de Finetti theorem says that the fixed-$k$ reduced density matrix of an exchangeable $N$-particle state approachesas $N\to \infty$. Finite versions bound the trace norm error by a constant of order $d_{\mathrm{loc}}^{2}k/N$. On a bipartite lattice, the corresponding two-sublattice form is a mixtureThis is the mean-field ansatz from the quantum de Finetti theorem.

The bond energy is a linear functional of ${\rho }_{AB}$. A convex combination cannot have energy below its lowest product component, so it remains only to minimizeThe Cauchy-Schwarz inequality gives ${\mathbf{r}}_A\cdot {\mathbf{r}}_B\ge -|{\mathbf{r}}_A||{\mathbf{r}}_B|\ge -1$, with equality for pure antiparallel vectors. This reproduces the Néel state and $-J/4$ per bond found in part (a).

Yes. The limiting state saturates the de Finetti mean-field lower bound: the minimizing product state belongs to the allowed de Finetti mixture, and the finite-de-Finetti error tends to zero as $z\to \infty$. At finite $z$ the theorem gives only an approximation; entanglement and correlated fluctuations can lower the energy below the product-state value by corrections that vanish in the infinite-coordination limit.

For a bipartite density operator ${\rho }_{SA}$, the reduced density matrix of $S$ isIt is characterized byfor every observable $M_S$ on $S$. Equivalently, in any orthonormal basis $\{|a_k⟩\}$ of $A$, the partial trace isand the result is independent of that basis.

For positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-particle Schrodinger equation isNeglecting packet spreading and writing $|a{⟩}_i$ for ${\psi }_{ai}$, branchwise evolution givesup to local kinetic phases. This branch-dependent Newtonian gravitational potential energy produces gravitationally induced entanglement.

Under the stated distance approximation, only the three branches $a=b$ acquire an appreciable common phaseThe coefficient matrix iswhere $J$ is the $3\times 3$ all-ones matrix. The reduced density matrix isIt equals $I/3$ when $\cos \varphi =-1/2$, so the first maximally entangled state occurs at $\varphi =2\pi /3$. Therefore

The state does not remain entangled for every $t>0$. Whenever $\varphi =2\pi k$, all branch phases again agree and the state returns to its initial product state. The revival period is

For a subsystem $A$ of a multipartite system with density operator ${\rho }_{A{E}_1\cdots E_n}$, its reduced density matrix is the unique operatorsuch that $\mathrm{Tr}(M_A{\rho }_A)=\mathrm{Tr}[(M_A\otimes I_E){\rho }_{AE}]$ for every observable $M_A$ on $A$. In an orthonormal basis $\{|e_j⟩\}$ of the other subsystems, the partial trace iswhich is independent of the chosen basis.

A bipartite density operator is a product state when it factors as ${\rho }_A\otimes {\rho }_B$. A pure product state factors as $|{\psi }_A⟩\otimes |{\psi }_B⟩$; its reduced density matrices have rank one and zero entanglement entropy. Mixed product states can have nonzero marginal Von Neumann entropy.

Quantum decoherence is the suppression of observable phase coherence when a system becomes entangled with unobserved environmental degrees of freedom. In a preferred basis, tracing out the environment damps off-diagonal entries of the system's reduced density matrix while leaving the corresponding populations unchanged.

The Schmidt coefficients of a normalized bipartite pure state are the nonnegative numbers $s_j$ in its Schmidt decomposition; they satisfy ${\sum }_j{s}_j^2=1$ and their squares are the nonzero eigenvalues of either reduced density matrix.

The Schmidt rank is the number of nonzero terms in a Schmidt decomposition, equivalently the rank of either reduced density matrix of a bipartite pure state.

The binary Lüders rule measurement of singlet versus triplet preserves the entire triplet subspace, but its nonselective channel permits signalling. Starting with $|00⟩$, Bob retains $|0⟩$. If Alice instead applies a local Pauli X gate, the input is $|10⟩$ and the channel outputs the equal mixture of $|{\Psi }^{+}⟩$ and $|{\Psi }^{-}⟩$; Bob has reduced density matrix $I/2$. Thus no spacelike local implementation can realize this channel. A Bell-state nondemolition measurement has a finer triplet readout and a different nonselective channel, so this obstruction does not exclude it.