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For a Bell pair $|{\varphi }_{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, Alice's basic measurement in the entrywise complex conjugate of any desired Bob orthonormal basis prepares each Bob pure state with probability $1/2$. Indeed, if $|u⟩={\sum }_{j=0}^1{u}_j|j⟩$, then $⟨\overline{u}|j⟩=u_j$, which proves the displayed contraction identity. The undisclosed-outcome density matrix stays $I/2$; quantum steering changes the conditional quantum state ensemble without violating the no-communication theorem.

For two systems of equal dimension $d\ge 2$, no joint pure state guarantees equal outcome labels whenever both systems undergo the same arbitrary basic measurement. Equal outcomes in the computational basis would force $|\Psi ⟩={\sum }_j{c}_j|jj⟩$. On any two coordinates j,k, changing both bases to the real sum/difference pair makes the different-outcome amplitude $(c_j-c_k)/2$, so all coefficients must agree. Changing both to the imaginary-phase pair makes a different-outcome amplitude $(c_j+c_k)/2$, so those coefficients must also be negatives. They must all vanish, contradicting normalization. The one-dimensional case is an exception.

The requirement of all complex bases matters. A maximally entangled vector $d^{-1/2}{\sum }_j|jj⟩$ does produce matching labels if one party uses a basis and the other its coordinatewise complex conjugate basis. Conjugate-state steering from a maximally entangled pair gives this conjugation explicitly. Identical real bases are a special case, not evidence for identical arbitrary complex bases.

Suppose one equally likely action returns a known normalized pure state $|s⟩$ unchanged, while another returns the maximally mixed state $I/d$. A basic measurement in any orthonormal basis containing $|s⟩$ distinguishes them with success $1-1/(2d)$: guess the pure-state action on its basis label and the reset action on the remaining $d-1$ labels. The first action succeeds with probability one and the second with probability $(d-1)/d$.

The Holevo–Helstrom theorem also proves optimality for these returned states. The Hermitian difference $|s⟩⟨s|-I/d$ has eigenvalues $1-1/d$ and $-1/d$ with multiplicity $d-1$, hence trace norm $2(1-1/d)$. Its optimal equal-prior discrimination success is $1/2+\Vert {\rho }_0-{\rho }_1{\Vert }_1/4$. Random computational-basis permutations fix the uniform superposition, so they supply an example of the invariant action. This compares the returned states in the stated unassisted protocol; it makes no assertion about a more general channel-discrimination problem with an entangled ancillary probe.^[ref]

A basic measurement is specified by an orthonormal basis $\{|b_j⟩{\}}_{j=0}^{d-1}$, with outcome $j$ associated with the rank-one orthogonal projection $P_j=|b_j⟩⟨b_j|$. For a normalized pure state $|\psi ⟩$, the Born rule givesThe probabilities sum to one because ${\sum }_j{P}_j=I$. Conditional on an outcome with $p(j)>0$, the post-measurement state iswhich represents $|b_j⟩$ up to a global phase. An outcome of zero probability has no conditional state to normalize. For a density matrix $\rho$, the corresponding expressions are $p(j)=\mathrm{Tr}(P_j\rho )$ and $P_j\rho P_j/p(j)$. If the outcome is ignored, the nonselective projective measurement instead produces ${\sum }_j{P}_j\rho P_j$, generally a mixed state.^[ref]

Let $p=|\alpha {|}^2$ and $1-p=|\beta {|}^2$. The first basic measurement has outcome probabilities $p$ and $1-p$, obtained by conjugating the amplitudes in the bras. Conditional on its first outcome, the next computational-basis measurement returns 0 with probability $p$; conditional on its second outcome, that probability is $1-p$. The law of total probability therefore givesConsequently the minimum is $1/2$, attained exactly when $|\alpha |=|\beta |=1/\sqrt{2}$. The relative complex phases do not affect this result. The alternatives after the first measurement form a classical mixture; their amplitudes must not be added coherently as if no measurement occurred.

Suppose $|W⟩=(U_A\otimes U_B\otimes U_C)|GHZ⟩$. Measure C in the orthonormal basis $\{U_C|0⟩,U_C|1⟩\}$. Its two AB conditional vectors would be proportional torespectively. Both are product states, because measuring the original GHZ state's C factor in the computational basis selects $|00⟩$ or $|11⟩$. This contradicts part (i), which applies to every basic measurement basis of the W state's C factor. Therefore no local unitary transformation maps GHZ to W.

An additional invariant makes the distinction visible without measurement records. Across the partition AB versus C, the GHZ reduced density matrix on C has eigenvalues $1/2,1/2$, whereas the W state's are $2/3,1/3$. By local-unitary invariance of reduced-state spectra, these spectra cannot be changed by local unitaries. This supplies the same impossibility and emphasizes that the proposed transformation is a product of unitaries, not arbitrary postselected local processing.

Identification is stronger than quantum state exclusion. Here an outcome only needs to rule out its associated preparation. Use the following orthonormal basis for a basic measurementIts defining overlaps satisfy $⟨b_m|{\varphi }_m⟩=0$ for $m=0,1,2$. By the Born rule, observing $b_m$ makes preparation ${\varphi }_m$ impossible. Measure in this basis and send its outcome label $m$; the prize is guaranteed.

For completeness, the conditional probability matrix, with rows $m$ and columns the prepared-state label $j$, isEvery column sums to one and the diagonal vanishes. Each result leaves two possible preparations, so this projective state exclusion does not contradict the impossibility of perfect identification. The family consists of antidistinguishable quantum states.

No. Perfect discrimination of pure states requires orthogonality, whereasTo see the obstruction directly for a basic measurement, let $\{|b_j⟩\}$ be its orthonormal basis. Certain identification would require disjoint sets of possible outcome labels for two different preparations. For every $j$, at least one of $⟨b_j|{\varphi }_0⟩$ and $⟨b_j|{\varphi }_1⟩$ would therefore vanish. Completeness would then givecontradicting the computed overlap. Failure to distinguish just this pair already rules out certain identification of the three-state family.

A basic measurement of a finite-dimensional quantum system is specified by an orthonormal basis $\{|b_j⟩{\}}_j$ of its Hilbert space. For a normalized input pure state $|\chi ⟩$, the Born rule givesThe post-measurement state is therefore $|b_j⟩$ up to a global phase. An outcome of zero probability has no conditional post-measurement state. For a general input density matrix $\rho$, the same basic measurement has $p_j=⟨b_j|\rho |b_j⟩$ and conditional density matrix $|b_j⟩⟨b_j|$ whenever $p_j>0$.

If Alice's basic measurement returns a normalized vector $|u{⟩}_A$, form Bob's unnormalized conditional vectorThe Born rule and normalization giveThe joint post-measurement state is $|u{⟩}_A\otimes |{\chi }_u{⟩}_B$. For $|u{⟩}_A=\alpha |0{⟩}_A+\beta |1{⟩}_A$ with real coefficients and ${\alpha }^2+{\beta }^2=1$, contraction of the Bell state givesThus Bob's normalized conditional state is $\alpha |0{⟩}_B+\beta |1{⟩}_B$. For complex coefficients the contraction instead contains their complex conjugates; this is the same conjugation underlying conjugate-basis steering of a Bell pair.

Use the entanglement-assisted promised equality protocol. Alice applies the phase oracle $O_x|i⟩=(-1{)}^{x_i}|i⟩$ on $A$, and Bob applies $O_y|i⟩=(-1{)}^{y_i}|i⟩$ on $B$. These are unitary operators because their diagonal entries have modulus one. After Alice sends $A$ to Bob, the joint pure state isLet $|\Omega ⟩=n^{-1/2}{\sum }_i|ii⟩$ be the original maximally entangled state. If $d=d_H(x,y)$ is the Hamming distance, thenBob performs the projective measurement with projectors $P=|\Omega ⟩⟨\Omega |$ and $I-P$. Under the promise,He declares equal strings for the first outcome and distance $n/2$ for the other. This is exact because the promised outputs occupy orthogonal subspaces. A basic measurement also suffices: extend $|\Omega ⟩$ to an orthonormal basis and group all its other outcomes together. Alice transmits one $n$-dimensional system, realizable using $\lceil {\log }_{2}n\rceil$ qubits; $n$ need not be a power of two.

A basic measurement is specified by an orthonormal basis $\{|b_j⟩{\}}_{j=0}^{d-1}$. The Born rule for the density matrix $\rho$ gives the outcome probabilitiesThese probabilities are nonnegative by positive semidefiniteness and sum to one by the resolution of the identity and unit trace. In the computational basis, the answers are simply the diagonal entries ${\rho }_{jj}$.^[ref]

Use deterministic two-qubit entanglement dilution, retaining the specified complex probability amplitudes. Alice implements a quantum instrument on $A$ with Kraus operatorsThe Kraus representation completeness condition holds even for complex $\alpha ,\beta$:To realize this using the permitted quantum gates and a basic measurement, Alice initializes an ancilla qubit $C$ to $|0⟩$. Conditional on her computational-basis state being $j$, she applies to $C$ the unitary matrixEach has orthonormal columns. Therefore $W=|0⟩⟨0{|}_A\otimes V_0+|1⟩⟨1{|}_A\otimes V_1$ is a unitary gate, withA computational-basis measurement of $C$ implements the desired quantum instrument. Its two unnormalized branches on $AB$ areEach has probability $1/2$. Alice sends her outcome $m$ as one classical bit. For $m=0$, both parties keep their qubits unchanged. For $m=1$, both apply a Pauli X gate, givingThus both outcomes produce the exact target using one classical bit and local operations. The construction includes $\alpha =0$ or $\beta =0$ and arbitrary relative quantum phases; it requires no quantum channel between the parties.

For $d$ labelled pure states in dimension $d$, an orthonormal basis $\{|b_j⟩{\}}_{j=0}^{d-1}$ supplies error-free projective state exclusion when $⟨b_j|{\psi }_j⟩=0$ for every $j$. The Born rule gives zero probability to reporting $j$ on input ${\psi }_j$; orthonormal-basis completeness ensures that an outcome is always obtained. Conversely, zero error for this labelled basic measurement forces precisely those zero overlaps. The matrix of transition probabilities is therefore a matrix of column probability distributions with zero diagonal.

For the Bell state $|{\varphi }_{+}⟩$, let both zero-input measurements use the computational basis, and let the one-input bases be rotated by $\theta$ for Alice and $-\theta$ for Bob in the real qubit plane. In the CHSH game, the four conditional winning probabilities are $1,{\cos }^2\theta ,{\cos }^2\theta ,{\sin }^2(2\theta )$. They follow by contraction of the Bell state with vectors from the real orthonormal bases used for the basic measurements. Their average is the displayed expression. On $0\le \theta \le \pi /2$, this beats $3/4$ exactly when $0<\theta <\pi /4$ and reaches $13/16$ at $\theta =\pi /6$, since $t-t^2=1/4-(t-1/2{)}^2$.