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Let $\Psi \ne 0$ and take the reduced positive operator ${\rho }_1={\mathrm{Tr}}_2|\Psi ⟩⟨\Psi |$. Choose orthonormal eigenvectors ${\xi }_j$ for its positive eigenvalues ${\lambda }_j$. DefineThe defining identity of the partial trace gives $⟨v_i,v_j⟩={\lambda }_j{\delta }_{ij}$, so the ${\eta }_j$ are orthonormal. Expansion in a full eigenbasis of ${\rho }_1$ yields $\Psi ={\sum }_j{\xi }_j\otimes v_j$; zero-eigenvalue terms have zero squared norm and contribute nothing. ThusThe zero vector has the empty decomposition. Unit normalization would give ${\sum }_j{c}_j^2=1$, but no normalization is needed for the theorem itself.

For the given vector, the coefficient matrix is $M=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right)$. Its second reduced operator is $M^{\ast }M=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, whose eigenvalues are $3,1$ and eigenvectors ${\eta }_{+}=(E_1+E_2)/\sqrt{2}$, ${\eta }_{-}=(E_1-E_2)/\sqrt{2}$. The corresponding normalized first-factor vectors areTheir norms are one and their inner product is zero. HenceExpanding the two terms returns the original three contributions. The vector has squared norm $4$; if a normalized state is wanted, divide by two and the Schmidt coefficients become $\sqrt{3}/2$ and $1/2$.

Alice's outcome is random. Conditional on knowing it, Bob may assign a different quantum state, but without that classical record he sees the same density matrix. For the displayed pure two-qubit state, his reduced density matrix before Alice's measurement isAfter Alice's computational-basis measurement with the outcome unannounced, the weighted mixture $p_0{\rho }_{B|0}+p_1{\rho }_{B|1}$ is exactly this same matrix. More generally, for any local trace-preserving operation with measurement operators $K_a$,The equality follows from the cyclic property of the partial trace for operators acting on Alice's subsystem. Bob's every local outcome probability therefore remains unchanged. This is the no-communication theorem: $\boxed{\text{entanglement alone cannot send information faster than light}}$. Conditional state changes become usable correlations only when the classical outcomes are compared.

An apparent collapse is the transition from predicting several possible recorded outcomes to describing a particular recorded result. Whether that transition is a physical discontinuity, a conditional state update or an effective description depends on what the wave function is taken to represent. The practical success of collapse-based calculation does not by itself settle that interpretive question.

The difficulty is exposed by unitary time evolution. An ideal apparatus that reliably measures a basis hasLinearity then makes a superposed input evolve asThis is an entangled state, not automatically one pointer packet or the other. A separate projection rule would instead select one recorded outcome, with Born rule weights, and assign its corresponding conditional state.

Quantum decoherence explains much of the practical appearance of a classical record. Interaction with an environment correlates the packets with nearly orthogonal environment states. The subsystem's reduced density matrix then has very small interference terms between the pointer alternatives. Local experiments on that subsystem approximately agree with a diagonal mixture. But this is an improper mixture: the total state can remain a pure superposition. Taking a partial trace has not selected one term of the total state. Suppression of observable interference and the occurrence of a unique absolute outcome are distinct claims.

There is a measurement problem if one simultaneously insists that the wave function is the complete physical description, that its evolution is universally unitary, and that a measurement has exactly one absolute outcome of the ordinary recorded kind. The ideal model shows why simply applying the same unitary law to a larger apparatus does not make that combination work. Nor does calling collapse a change in knowledge solve the problem for someone who has already taken the wave function to be a complete ontic state.

Different interpretations revise different parts of this package. Bohmian mechanics adds an actual configuration and obtains effective subsystem collapse while the total wave evolves unitarily. The many-worlds interpretation retains the total superposition and treats each record as definite relative to its decohering branch. A stochastic collapse theory instead modifies the physical dynamics. Everett's original proposal explicitly separates his relative-state account from an additional collapse postulate.^[ref]

Thus collapse is not problematic merely because conditional probabilities change when a record is learned. The problem is the status of that record and of the dynamical rule used to describe it. A defensible interpretation must explain when its updating rule is applicable, what makes a record definite in its own ontology, and why the ordinary measurement statistics follow. No appeal to human consciousness is required by the displayed unitary measurement model.

The stated annihilation conditions can be solved algebraically, even though (b) shows they are not ground-state conditions for the full printed Hamiltonian. Invert the canonical transformation:For $|{\psi }_{\lambda }⟩=\mathcal{N}{e}^{\lambda a^{†}{b}^{†}}|0,0⟩$, the commutators giveThus both $A$ and $B$ annihilate it precisely whenThe normalized two-mode squeezed vacuum isThe factorial from the exponential cancels those generated by the two creation operators. Normalization follows from a geometric series; the problem does not require its value, but it makes the subsequent partial trace explicit.

For comparison, the actual ground wave function of the printed stable Hamiltonian isIt follows directly from the two normal-mode Gaussian vacua. The proposed $|{\psi }_{\lambda }⟩$ is instead the exact ground state of the distinct pair-coupling Hamiltonian $\hslash \omega (a^{†}a+b^{†}b+1)+g(ab+a^{†}{b}^{†})$ under its stability condition. Omitting the remaining terms of $kxy$ silently would change the problem.

Assume the composite pure state has unit norm. Put its amplitudes into the matrixThe partial trace contracts the same $B$ index between ket and bra:ThusIt is Hermitian, positive semidefinite and has trace one. Its determinant is particularly useful:This follows from multiplicativity of the determinant, or by expanding the two-by-two expression. If the supplied amplitudes were not normalized, the physical reduced state would be $C{C}^{†}/\mathrm{tr}(C{C}^{†})$ instead.

Fix the mode ordering with the $b^{†}$ operator before the $c^{†}$ operator, as in the given state. In the sector with one fermion in each subsystem, use the ordered basisThe state vector is $v=(1/2,1/\sqrt{2},0,1/2{)}^T$, whose squared norm is one. Thus the global density matrix is the rank-one projectorIt has trace one and satisfies ${\rho }_{BC}^2={\rho }_{BC}$, so the global state is pure.

Take the partial trace over $B$. The conditional vectors in $C$ are $v_1=(1/2,1/\sqrt{2}{)}^T$ for $B_1$ and $v_2=(0,1/2{)}^T$ for $B_2$. Therefore the reduced density matrix isIts determinant is $1/16$, giving eigenvaluesBoth are positive; in particular, $\mathrm{Tr}{\rho }_C^2=7/8<1$, so the reduced state is mixed despite global purity. The entanglement entropy is obtained by diagonalizing this reduced matrix and evaluatingFor a globally pure bipartite state this equals the entropy of either reduced subsystem. No numerical evaluation is needed.

Taking the partial trace over the left atom in the singlet state gives the right-hand reduced density matrixThus every measurement on the right stream alone has the same statistics as B or C. The joint pure state is compatible with locally mixed states; the correlations reside in the full density operator.

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.

For common classical weights, the matrix logarithm on each positive-weight block is $\log (p_i{\rho }_i)=(\log p_i)I+\log {\rho }_i$ on its support. The two logarithms of $p_i$ cancel in the quantum relative entropy, giving the displayed identity. Zero-weight blocks are omitted, including if their unweighted relative entropy is infinite. If a positive-weight block violates support inclusion, both sides are infinite. Applying the data-processing inequality for quantum relative entropy to the partial trace over the flag gives joint convexity of quantum relative entropy.

The nonzero eigenvalues, including multiplicities, of either reduced density operator of a bipartite pure state. A Schmidt decomposition $|\psi ⟩={\sum }_k\sqrt{p_k}|a_k{b}_k⟩$ gives ${\rho }_A={\sum }_k{p}_k|a_k⟩⟨a_k|$ and the corresponding expression for ${\rho }_B$ by partial trace. Thus unequal marginal spectra rule out a pure joint state. The amplitudes $\sqrt{p_k}$ are often called Schmidt coefficients; the spectrum here uses the probabilities $p_k$.

Tracing one oscillator out of a symmetric coordinate-coupled Gaussian ground state gives a thermal density operator for an effective oscillator of frequency $\nu$. Its geometric ratio is $q=[(\sqrt{{\omega }_{+}}-\sqrt{{\omega }_{-}})/(\sqrt{{\omega }_{+}}+\sqrt{{\omega }_{-}}){]}^2$. It is generally squeezed relative to the original bare-frequency number basis. The covariance ratio fixes the effective frequency rather than permitting an arbitrary thermal-basis identification.

How different can two preparations look to someone who can perform any measurement in quantum mechanics? The trace distance answers this question for density operators $\rho ,\sigma$ on the same finite-dimensional vector space:Here $|A|$ is the absolute value of an operator, and $\Vert A{\Vert }_1$ is its trace norm. Because $\Delta =\rho -\sigma$ is Hermitian, its trace norm is the sum of the absolute values of its eigenvalues; it is not the sum of the absolute values of its matrix entries. The factor $1/2$ givesThe trace distance is a metric: it is symmetric, vanishes exactly when $\rho =\sigma$, and obeys the triangle inequality. Identical preparations have trace distance zero; preparations with orthogonal occupied eigenspaces have trace distance one and can be identified perfectly in a single measurement in quantum mechanics.

Operationally, the trace distance is the largest total variation distance between the outcome probability distributions obtainable by a POVM. Equivalently, the variational characterization of trace distance says it is the largest possible probability gap for one event in a binary measurement in quantum mechanics. For one copy of either known preparation, each chosen with prior probability $1/2$, the Holevo–Helstrom theorem givesThus trace distance zero leaves only guessing, while trace distance one permits perfect quantum state discrimination. The optimum ranges over measurements; different measurement bases can expose different amounts of distinguishability. ^[ref]

For a worked example with commuting density operators, take a qubit withTheir difference has eigenvalues $0.5,-0.5$, so $D=(|0.5|+|-0.5|)/2=0.5$. Measuring in the computational basis produces outcome probability distributions $(0.8,0.2)$ and $(0.3,0.7)$, whose total variation distance is also $0.5$. Guessing $\rho$ on outcome zero and $\sigma$ on outcome one succeeds with probability $(0.8+0.7)/2=0.75$. A projective measurement in the Hadamard basis instead produces $(0.5,0.5)$ for both preparations and gives no discrimination advantage.

The trace-distance contraction under quantum channels ensures that a common quantum channel, including the partial trace, cannot increase the trace distance. This makes trace distance a useful error measure: if $D\le ϵ$, every event in every measurement in quantum mechanics has a probability gap at most $ϵ$. Losing access to a subsystem can hide distinguishability, but cannot create more of it.