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Under the intended assumption that one use of the quantum channel transmits one qubit without noise, Alice can communicate two classical bits with certainty if she and Bob already share a Bell pair. She uses superdense coding. The prior distribution of the Bell pair must be a resource available before the permitted channel use; distributing it through that same channel would consume another use.

Take the shared Bell state to be $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, with Alice holding the first qubit. For a two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies the local unitary operation $U_{ab}=X^a{Z}^b$ to her qubit. The four encoded pure states are, up to irrelevant global phases,They are orthogonal: for any matrix $V$ on the first qubit, $⟨{\Phi }^{+}|(V\otimes I)|{\Phi }^{+}⟩=\mathrm{Tr}V/2$, and the four Pauli operators are orthogonal in the trace inner product. Explicitly,Alice sends her qubit through the quantum channel once. Bob now holds both qubits and performs a Bell-basis measurement. Its four orthogonal projectors identify $(a,b)$ with certainty, givingBefore the transmission, Bob's reduced density matrix is $I/2$ for every message, so shared entanglement alone does not convey the message.

Without shared entanglement, a noiseless single-qubit transmission cannot perfectly distinguish four messages: perfect discrimination requires orthogonal state supports, and its two-dimensional Hilbert space has room for at most two nonzero mutually orthogonal supports. The Holevo bound also limits its accessible classical information to one bit. For the usual protocol starting with a shared two-qubit pure state and using local unitary encodings, exact two-bit transmission requires a maximally mixed Bob marginal. Indeed all encoded joint states are pure and have the same ${\rho }_B$, while Subadditivity of Von Neumann entropy gives their average entropy at most $1+S({\rho }_B)$. Four equally likely perfectly distinguishable messages have average Von Neumann entropy two, forcing $S({\rho }_B)=1$ and hence a maximally entangled state.

The specification “quantum channel” alone does not determine its dimension or noise, so the one-qubit, noiseless interpretation is needed for this particular conclusion. A noiseless four-dimensional channel could send four orthogonal states without any shared entanglement; a completely depolarizing qubit channel cannot transmit the message even with a shared Bell pair, since its output is independent of Alice's encoding. Shared entanglement is therefore the extra resource for the intended single-qubit channel, not a guarantee for every unspecified quantum channel.

Write the unknown pure photon polarization qubit asThe polarizing beam splitter sends the $H$ component down the lower path and the $V$ component up the upper path. After rotating the upper photon polarization to $H$, both paths have the same internal quantum state and can interfere at BS2. Absorb fixed apparatus phases into a calibrated zero for the tunable quantum phase. With the preceding reflection convention, the quantum state before BS2 is $(\alpha |\ell ⟩+i\beta e^{i\theta }|u⟩)|H⟩$, so the two output amplitudes areConsequentlyRepeated counts while scanning $\theta$ determine the complex quantum coherence $\gamma$. The interference visibility is $2|\gamma |$, and the fringe position determines the relative quantum phase $\arg \beta -\arg \alpha$ whenever both components are nonzero.

A fringe alone is not enough to determine an arbitrary photon polarization quantum state: it fixes $|\alpha \beta |$ but not which of $|\alpha {|}^2$ and $|\beta {|}^2$ is larger. Indeed, $|H⟩$ and $|V⟩$ both give the flat fringe $P_1=1/2$. For a general pure quantum state, the two possible population differences areResolve this ambiguity by temporarily detecting the two paths before BS2. Their relative counts give $h=|\alpha {|}^2$ and $1-h=|\beta {|}^2$ using the same polarizing beam splitter and detectors. Taking $\alpha =\sqrt{h}$ as a global-phase choice, the reconstructed quantum state for $h>0$ isIf $h=0$, it is $|V⟩$ up to a global phase. This completes interferometric polarization tomography. The counts are obtained from many identically prepared photons; they do not identify an arbitrary quantum state from one photon.

Use a concatenated quantum error-correcting code: each of the five outer-code qubits is encoded in its own five-qubit inner block. Independent physical errors and ideal inner correction make each decoded block act as one outer qubit with error probability $q=f(p)$. Errors in distinct inner blocks remain independent in this model. Outer recovery succeeds if at most one of the five decoded blocks is bad, givingThus the two-level logical failure probability isThis improves the leading error order from $p$ for a bare qubit to $p^2$ for one level and $p^4$ for two. At least two physical errors must occur in each of two different inner blocks to defeat the prescribed two-stage decoder. Many error patterns with more than one error in the full 25-qubit register are therefore corrected: one error in each of all five blocks is removed at the inner level, and a single failed inner block is corrected at the outer level. Concentrating several physical errors in one block can be easier for this decoder than placing two errors in each of two blocks.

More generally, $\ell$ levels use $5^{\ell }$ physical qubits with recursion $p_{\ell +1}=f(p_{\ell })$. For $0<p<p_{\ast }$, one has $f(f(p))<f(p)<p$. Repeated concatenation decreases to zero: its limit must be a fixed point of the continuous function $f$, and below $p_{\ast }$ the only fixed point is zero. A simple quantitative bound is $f(x)\le 10x^2$, obtained by the union bound over the ten pairs of possible error positions. Induction giveswhich proves rapid suppression when $p<1/10$; the exact threshold is the larger value $p_{\ast }$. Above $p_{\ast }$ this recursion amplifies errors instead.

Below threshold, concatenation suppresses logical errors while allowing an entire bad inner block to be corrected at the next level. The cost is additional physical qubits, gates, syndrome extraction and recovery. The calculated threshold concerns the independent transmission-noise model with perfect encoding and recovery; including noisy operations or correlated channel errors requires a different error analysis.

Assume independent physical-qubit errors with common probability $p$, ideal encoding and recovery, and the specified rule that every block with two or more errors fails. The number of physical errors has a binomial distribution with parameters $5,p$. Correct recovery occurs for zero or one error, soThe corresponding logical failure probability isAn unencoded qubit succeeds with probability $1-p$. Thus encoding improves this probability precisely when $f(p)<p$. Factoring givesOn $0<p<1$, the sign is the sign of $g(p)=4p^3-11p^2+9p-1$. Its derivative is $12p^2-22p+9$: it is positive until $(11-\sqrt{13})/12$ and negative afterwards on $[0,1]$. Since $g(0)=-1$, $g(1)=1$ and its interior maximum is positive, there is exactly one zero in $(0,1)$. The five-qubit concatenation threshold is thereforeAt $p=p_{\ast }$ the two success probabilities agree; above it this decoder makes the error rate worse. At $p=0$ both schemes succeed perfectly, and at $p=1$ both fail under the stated error model. The small-$p$ behavior $f(p)=10p^2+O(p^3)$ exhibits cancellation of every single-qubit-error contribution.

The Bloch vector representation of a qubit density matrix is $\rho =\frac{1}{2}(I+r_{x}X+r_{y}Y+r_{z}Z)$, where $r_j=\mathrm{Tr}(\rho {\sigma }_j)$ and $|\mathbf{r}|\le 1$. Pure states have $|\mathbf{r}|=1$, while the origin represents the maximally mixed state $I/2$, not a pure state.

For the two preparations, direct evaluation of the Pauli matrices givesTheir dot product is $-1/2$, so their Bloch vectors meet at angle $\boxed{2\pi /3}$, or $120$ degrees. This is the angle between Bloch vectors; the angle determined by the modulus of the Hilbert-space overlap is different.

Completeness of the POVM requires $E_3=I-E_1-E_2$. In the computational basis, subtraction givesAn overall phase of ${\varphi }_3$ is immaterial. The matrix has eigenvalues $2/3$ and zero, so it is positive; together with $E_1,E_2$ it is a valid trine qubit POVM.

The Born rule gives $\mathrm{Pr}(j\mid {\rho }_i)=\mathrm{Tr}(E_j{\rho }_i)=\frac{2}{3}|⟨{\varphi }_j|{\psi }_i⟩{|}^2$. The conditional probabilities areOutcome one identifies ${\rho }_2$, since ${\varphi }_1$ is orthogonal to ${\psi }_1$; outcome two identifies ${\rho }_1$, since ${\varphi }_2$ is orthogonal to ${\psi }_2$. Outcome three is inconclusive. If Alice's prior probability for ${\rho }_1$ is $q$, the unconditional outcome probabilities are $(1-q)/2$, $q/2$, and $1/2$ respectively. Because outcome three has the same likelihood for both preparations, it leaves the prior odds unchanged. Bob's conclusive identification succeeds with probability one half and is never wrong; he does not identify the state on the inconclusive trials. This is unambiguous quantum state discrimination.

A linear map $\Phi$ is a completely positive map if ${\mathrm{id}}_R\otimes \Phi$ sends every positive operator to a positive operator for every finite-dimensional auxiliary system $R$. Positivity of $\Phi$ alone tests only inputs without an auxiliary system and is weaker.

For the Kraus representation, let $\Omega \ge 0$ be any operator on the auxiliary system and the input system. ThenFor any vector $v$, its expectation is ${\sum }_k⟨(I_R\otimes A_k^{†})v,\Omega (I_R\otimes A_k^{†})v⟩\ge 0$. Every amplification is therefore positive, proving the Kraus-form map is completely positive. No normalization condition on the $A_k$ is needed for this positivity proof. To make it a deterministic quantum channel, one additionally requires ${\sum }_k{A}_k^{†}{A}_k=I$; a physical outcome branch instead obeys ${\sum }_k{A}_k^{†}{A}_k\le I$.

Transposition is positive on one system: for any positive $\rho$ and any $v$, $v^{†}{\rho }^{T}v$ is the complex conjugate of ${\bar{v}}^{†}\rho \bar{v}$, hence is real and nonnegative. It fails complete positivity already for a qubit. Apply partial transposition to the second subsystem of the Bell state $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. The resulting operator isOn the antisymmetric vector $(|01⟩-|10⟩)/\sqrt{2}$, its eigenvalue is $-1/2$. Thus the partial transpose is not positive on this entangled input, andThe same witness embeds into any larger input dimension; in the exceptional one-dimensional case transposition is simply the identity.

Use most-significant-bit-first ordering, so $l={\sum }_{j=1}^n{b}_j{2}^{n-j}$ and $|l⟩=|b_1⟩\otimes \cdots \otimes |b_n⟩$. Then the coefficient factors as $e^{i{\varphi }_{m}l}={\prod }_j{e}^{i{\varphi }_m{b}_j{2}^{n-j}}$. Expanding a tensor product over all bit strings proves the product decomposition of a Fourier phase state:Each factor is a normalized one-qubit state. The whole register is therefore a product state, hence a separable state; no claim that arbitrary outputs of the Fourier transform are separable is needed. Reversing the bit convention reverses the order of these factors.

The PDF circuit has its upper wire controlling both CNOT gates. Its successive states areThe first arrow is the upper Hadamard gate; the middle phase arises because both ones receive a phase. The second CNOT disentangles and resets the lower wire. Hence the output isThis implements doubled logical phase using two parallel phase gates, rather than two serial applications on one qubit.

Write the promised phase as $\varphi =2\pi m/8$. Prepare three logical qubits in $|+⟩$ and attach zero ancillas in three blocks of sizes four, two and one. A block of size $r$ is prepared by fanout CNOT gates asApply one copy of $U_{\varphi }$ to every physical qubit in every block, all in the same time step. Undo each fanout. The logical block output is $(|0⟩+e^{ir\varphi }|1⟩)/\sqrt{2}$ and every ancillary wire returns to zero. This is parallel phase multiplication by coherent fanout; its intermediate entanglement copies basis labels, and does not clone an arbitrary quantum state.

For block weights $4,2,1$, the three logical wires now containApply the specified negative-sign quantum Fourier transform and measure. Part (a) gives label $m$ deterministically. Thus seven parallel phase-gate applications, on three logical wires and four ancillas, identify the phase in one oracle time step. Preparation, uncomputation and the Fourier transform cost no time under the stated model; using serial powers would not meet the time requirement.

Denote the normalized initial field by $|F⟩$, and its shifted packets byHere the photon-number kets are orthogonal Fock states. Counting shared number labels givesAfter the interaction, linearity gives the joint stateTake the partial trace over the field. Put $d=|\alpha {|}^2-|\beta {|}^2$ and $c=\alpha {\beta }^{\ast }+{\alpha }^{\ast }\beta$. In the ordered qubit basis $(|0⟩,|1⟩)$, the reduced density matrix isFor example, its upper off-diagonal entry is half the overlap of the field accompanying $|1⟩$ with the field accompanying $|0⟩$, fixing the conjugation order. The trace is one, and positivity follows because this is the partial trace of a normalized pure state. For $k\ge 2$, $s_k=1-2/k$; at $k=1$ it is zero, not the continuation $-1$. This is the photon-number reference for a Hadamard gate channel, not an actual coherent-state packet.