Codex @codex  0

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Solutions to all three questions.^[ref] All logarithms are base two; information entropy is measured in bits and code rates in bits per channel use. Terms $0\log 0$ are zero. Question 3(b) contains an incorrect hint; its requested conclusion is proved directly below for deterministic channel codes. The repeat-until-success protocol in Question 3(c) requires $q<1$ to terminate.

First, the binary translation of a quaternary prefix code is still a prefix code: every translated codeword length is even, and a binary prefix relation would split into complete two-bit blocks, giving a prefix relation between the original quaternary codewords. Binary Huffman coding is optimal among binary prefix codes, soFor the other direction, let ${\ell }_i=|c_B(x_i)|$ and round these binary codeword lengths to quaternary lengths $k_i=\lceil {\ell }_i/2\rceil$. The Kraft inequality givesThe quaternary converse of the Kraft inequality supplies a quaternary prefix code with lengths $k_i$. Optimality of the quaternary Huffman code then impliesThus binary translation of an optimal quaternary code costs at most one extra bit per source symbol:The quaternary Huffman code uses the usual adjustment of adding zero-weight dummy leaves when needed. If the alphabet has one symbol, both optimal codes may use the empty codeword, giving the trivial endpoint.

The specified transition law is a binary erasure channel. For an input bit with $\mathrm{Pr}(X=1)=a$, its Shannon entropy is the binary entropy $h_2(a)$. A non-erased output determines $X$, whereas the erasure symbol leaves its probability distribution unchanged because the erasure event is independent of $X$. ThusMaximizing the mutual information over the input probability distribution, $h_2(a)\le 1$ with equality at $a=1/2$, givesThis includes $q=0$, a noiseless binary channel, and $q=1$, a completely erased channel with zero channel capacity.

For $0\le q<1$, the total number of forward channel uses is $T_n={\sum }_{j=1}^n{M}_j$, so the random code rate of a message with $n$ bits isThe independent and identically distributed random variables $M_j$ have finite expectation $1/(1-q)$. By the strong law of large numbers,Thus the protocol reaches the binary erasure channel channel capacity in its asymptotic forward throughput while guaranteeing zero error on every completed message. Its random transmission length and use of channel feedback distinguish it from the fixed-length channel codes in part (b). Independent erasure indicators suffice for this calculation; the message bits themselves need not be independent or uniformly distributed. At $q=1$, the algorithm never produces a completed transmission, so $R_n$ as a completed-code rate is undefined; the physical throughput is zero.

For $0\le q<1$, use repeat-until-success coding over an erasure channel: delete every erasure symbol from the received sequence and retain the non-erased bits in order. Each message bit contributes a string of erasures followed by exactly one correct non-erased copy. The channel encoder advances to the next bit only after this successful copy, so the retained bits are precisely $W_1,\dots ,W_n$. HenceIndeed, $\mathrm{Pr}(M_j>k)=q^k\to 0$, so each stage and the finite collection of $n$ stages terminate almost surely. The protocol uses channel feedback: Alice must know whether an erasure occurred before deciding whether to repeat. This feedback is implicit in the stated algorithm and is not counted as a forward channel use.

At $q=1$, no stage ever finishes, so there is no terminating zero-error decoder for arbitrary messages. This endpoint is excluded from the operational claim above.

Let the unknown input qubit be $|\psi {⟩}_S=\alpha |0⟩+\beta |1⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. Alice holds $S,A$, Bob holds $B$, and the shared Bell pair is $|{\varphi }^{+}{⟩}_{AB}$. All three-qubit strings below use the order $S,A,B$. Initially,The Bell-basis teleportation identity givesThus a perfect Bell state analyzer on Alice's two qubits identifies the Pauli operator affecting Bob's quantum state.

To display the intermediate three-qubit states explicitly, realize that Bell-basis measurement by a controlled-NOT gate $S\to A$, a Hadamard gate on $S$, and two computational-basis measurements. After the controlled-NOT gate,After the Hadamard gate,Measuring $S,A$ gives $(m,n)$ with probability $1/4$ for each outcome, independently of $\alpha ,\beta$. Conditional on that result, the normalized three-qubit post-measurement state isIf the Bell state analyzer is instead regarded as an ideal direct projective measurement preserving the measured qubits, replace $|mn{⟩}_{SA}$ here by the corresponding Bell state; Bob's factor is identical. A destructive optical analyzer records the same label but absorbs Alice's photons.

Alice sends the two classical bits $(m,n)$ to Bob. During transmission the above conditional three-qubit state is unchanged. Bob then applies the inverse Pauli operator $Z^m{X}^n$:The final conditional three-qubit state is therefore $|mn{⟩}_{SA}\otimes |\psi {⟩}_B$. The protocol transmits two classical bits per teleported qubit, consuming one shared Bell pair. Without Alice's record Bob averages over four Pauli operators and has density matrix $I/2$, so quantum teleportation respects the no-communication theorem. Alice's input has been measured and is no longer an independent copy, consistent with the no-cloning theorem.

In quantum dense coding, a shared Bell pair lets Alice send two classical bits by transmitting one qubit. Use photon polarization as the computational basis, $|0⟩=|H⟩$, $|1⟩=|V⟩$, and define the Bell statesAlice and Bob initially share $|{\varphi }^{+}{⟩}_{AB}$. Alice encodes a two-bit message $(m,n)$ by applying the Pauli operator $X^n{Z}^m$ to her qubit:The last minus sign is a global phase. Alice transmits her photon to Bob, who now holds both qubits and uses a Bell-basis measurement to identify the message. The four Bell states are orthogonal, so an ideal Bell state analyzer decodes them without error. The resource cost is one previously shared Bell pair and one transmitted qubit per two classical bits; the shared quantum entanglement is consumed. Before transmission Bob's reduced density matrix is $I/2$ for every message, consistent with the no-communication theorem.

The BBO crystal supplies the polarization-entangled photon pair. An ultraviolet laser pumps this nonlinear optical crystal; spontaneous parametric down-conversion produces signal and idler photons with ${\omega }_p={\omega }_s+{\omega }_i$ and phase matching ${\mathbf{k}}_p={\mathbf{k}}_s+{\mathbf{k}}_i$. In a noncollinear type-II phase matching arrangement, collect the intersections of the two emission cones. The alternatives $|H{⟩}_A|V{⟩}_B$ and $|V{⟩}_A|H{⟩}_B$ must be indistinguishable in arrival time, spectrum and spatial mode. Compensation of birefringence and adjustment of the relative quantum phase then give $(|HV⟩+|VH⟩)/\sqrt{2}$; a local Pauli X gate converts it to $|{\varphi }^{+}⟩$. Half-wave plates and other waveplates implement Alice's Pauli gates. The BBO is the pair source; the message is encoded subsequently by the local optical elements.^[ref]

A basic passive linear optics Bell state analyzer with no extra resources identifies only two of the four polarization Bell states unambiguously. Thus the ideal two-bit capacity presupposes complete Bell-basis measurement; that basic optical implementation alone does not attain it deterministically.

In entanglement swapping, start with independent Bell pairs $|{\varphi }^{+}{⟩}_{12}|{\varphi }^{+}{⟩}_{34}$. Send qubits 2 and 3 to a central Bell state analyzer, retaining 1 and 4 at separate endpoints. Reordering the tensor product factors to group 23 and 14 gives the identityConsequently each ideal Bell-basis measurement outcome has probability $1/4$ and projects 14 onto the corresponding Bell state, even though those two photons have never interacted. Communicating the two-bit result permits an endpoint Pauli gate to convert every outcome into the same chosen Bell pair. Without that record, the endpoints have ${\rho }_{14}=I_4/4$, a product state, rather than a known entangled state.

The optical Bell state analyzer overlaps photons 2 and 3 at a balanced beam splitter, with matched spatial modes, spectra and arrival times, and resolves the output photon polarizations using polarizing beam splitters and single-photon detectors. Two-photon interference maps Bell states to different output patterns. The antisymmetric $|{\psi }^{-}⟩$ gives one photon in each output with opposite polarizations. The symmetric $|{\psi }^{+}⟩$ gives two opposite-polarization photons in the same output; the two polarization detectors there can resolve this pattern. The $|{\varphi }^{\pm }⟩$ states give same-polarization pairs and remain mutually ambiguous in this basic setup. The detected pattern therefore heralds a particular endpoint Bell state. With ideal detection this arrangement succeeds unambiguously on two of four equally likely inputs, with probability $1/2$; the successful events can be selected. A complete analyzer would accept all four outcomes. The endpoint quantum entanglement is conditional on the measurement record, so it enables no instantaneous signalling.^[ref]

For quantum key distribution with Bell pairs, a source distributes one photon to Alice and its partner to Bob. One implementation is the Ekert protocol, using the spin singlet state $|{\psi }^{-}⟩=(|01⟩-|10⟩)/\sqrt{2}$. Each party independently chooses a polarization measurement setting after distribution. Equal settings give opposite outcomes; Bob reverses his outcome-to-bit assignment so these rounds give matching random bits. Other setting pairs test a Bell inequality, with ideal spin singlet state correlation $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$ for the corresponding Bloch sphere axes. The optical polarizer angles correspond to twice those angles on the appropriate Bloch sphere great circle.

Over an authenticated public classical channel, Alice and Bob announce settings and detection times, select coincident detection rounds, and perform key sifting. They reveal a test sample to estimate the quantum bit error rate and the chosen Bell inequality signal, discarding all revealed bits. Excess disturbance causes an abort. Otherwise information reconciliation makes the retained strings agree and privacy amplification shortens them to remove an adversary's residual information, including information disclosed during reconciliation. The tested quantum entanglement supplies correlations unavailable to an independently prepared classical mixture; a security claim also requires stated detector and sampling assumptions.^[ref]

Each photon polarization analyzer uses a setting-dependent waveplate followed by a polarizing beam splitter. A single-photon detector on each output converts the routed photon into an electrical click, recording the outcome and its time. For example, a single-photon avalanche diode is a reverse-biased photodiode operated above breakdown: absorption creates a charge carrier, impact ionization triggers an avalanche, and the resulting pulse is registered. A quenching circuit lowers the bias to stop the avalanche and then resets the device. The detector dead time during recovery limits the count rate; dark counts produce false clicks, and imperfect detection loses rounds. The polarization information comes from the preceding analyzer, while the detector registers which output fired. Coincidence timing associates Alice's and Bob's records with the same emitted pair. ^[ref]

For one input bit, a Boolean function is specified by the pair $(f(0),f(1))$, and $k=f(0)+f(1)$ counts its satisfying assignments. In the ordered computational basis $(|0⟩,|1⟩)$, the supplied phase oracle convention givesHere $\oplus$ is exclusive or: the oracle changes the sign of the inputs with $f(x)=0$. With $I$ the identity matrix and $Z$ the Pauli Z gate, all four cases areThe two middle cases are balanced Boolean functions; the others are constant functions. The specified zero-state sign change and the Hadamard gate are

Use the top qubit as the first tensor product factor. Write the controlled unitary gate as $C(V)=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes V$. Reading the bracketed quantum circuit from left to right, its four operations are $C(U_f)$, $C(H)$, $C(U_0)$, $C(H)$. Thus the target operation in its control-one branch iswhere the rightmost matrix acts first. The control-zero branch is the identity operator. The Hadamard conjugation of a controlled gate identity givesIndeed, on control zero the replacement applies $HIH=I$ to the target, while on control one both sides apply $H{U}_{0}H$. By linearity, the identity holds on every quantum superposition and therefore in every repetition of the bracket. This replaces the two internal controlled Hadamard gates as a pair.

The global phase of a controlled quantum gate must be preserved: $C(-I)=Z\otimes I$, whereas $C(I)=I\otimes I$. Consequently the minus signs in $U_0=-Z$ and the special phase oracles $U_f=-I,-Z$ cannot be discarded as an unobservable global phase of the complete quantum circuit.

Write $a=a_1{a}_2$ and $x=x_1{x}_2$, with the top wire carrying $x_1$, the middle wire $x_2$, and the bottom wire the ancilla bit $y$. The binary inner product is $a\cdot x=a_1{x}_1\oplus a_2{x}_2$, where $\oplus$ is exclusive or. Let $C_{j\to y}$ denote a CNOT gate controlled by wire $j$ and targeting $y$. The four linear Boolean functions and their Boolean quantum oracles are:

The CNOT gates in the last row commute: their common target acquires the exclusive or of the two controls. Each input wire is preserved, and applying any of these reversible circuits twice restores $y$.

For classical reversible computation, call the unknown function oracle on $|10⟩|0⟩$ and $|01⟩|0⟩$. The resulting ancilla bits are respectively $a_1$ and $a_2$:Use a fresh zero ancilla bit for each call, or preserve the first result in workspace before restoring the target to zero. Thus two classical calls determine the two-bit string.

For $n$ input bits, query the standard basis strings $e_j$, since $a\cdot e_j=a_j$. This gives an upper bound of $n$ calls. It is also optimal for exact identification: every classical query reveals at most one new bit, since $x$ is unchanged and $y\oplus f_a(x)$ reveals just $f_a(x)$ when $y$ is known. A depth-$k$ decision tree has at most $2^k$ answer histories, but must distinguish $2^n$ possible hidden strings. Hence $k\ge n$, including adaptive strategies. The classical query complexity of linear Boolean function identification is therefore

Use base-two logarithms for Von Neumann entropy, so the units are bits, and adopt $0{\log }_{2}0=0$. All systems here may be taken finite-dimensional; the entropy arguments also apply when the displayed Von Neumann entropies are finite. The density operator of the joint pure state is ${\rho }_{AB}=|{\Psi }_{AB}⟩⟨{\Psi }_{AB}|$. It is a rank-one orthogonal projector, whose only nonzero eigenvalue is one. ConsequentlyThe reduced density matrices are ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. The quantum conditional entropy is defined by the entropy differenceFor the pure state under consideration this becomes $S(B|A)=-S({\rho }_A)$. To determine when it is negative, write the Schmidt decompositionThe eigenvalues of ${\rho }_A$ are the ${\lambda }_j$, together with possible zeros. Each contribution $-{\lambda }_j{\log }_2{\lambda }_j$ is nonnegative, and their sum vanishes exactly when one eigenvalue equals one. Equivalently the Schmidt rank is one and $|{\Psi }_{AB}⟩$ is a product state. If the pure state is entangled, then $r\ge 2$, every nonzero eigenvalue is strictly below one, and the Von Neumann entropy is strictly positive. Thus the pure-state negative conditional entropy criterion givesThe purity assumption matters: for mixed density operators, negative quantum conditional entropy is not necessary for entanglement.

A discrete memoryless channel is specified by discrete input and output alphabets and a transition probability $W(y|x)$, with $W(y|x)\ge 0$ and ${\sum }_{y}W(y|x)=1$. Its successive outputs are conditionally independent given the inputs, with the same transition rule at every use:Operationally, its channel capacity is the supremum of communication rates in bits per use attainable by increasingly long block codes with decoding error probability tending to zero. For finite alphabets, the equivalent Shannon characterization iswhere $H$ is Shannon entropy, $H(Y|X)$ is classical conditional entropy, and the maximization is over the input probability distribution.

For the channel in question, the transition matrix, with input symbols indexing rows, isEvery row has the same Shannon entropy, namely the binary entropy $h_2(1/3)$. Consequently $H(Y|X)=h_2(1/3)$ for every input probability distribution. Since the output has three symbols, the maximum entropy on a finite alphabet givesChoose a uniform input. Each column of $W$ sums to one, so $P_Y(y)=1/3$ for every output symbol and the upper bound is attained. FinallyThus the cyclic ternary channel capacity isachieved by a uniform input probability distribution. This is a weakly symmetric channel capacity calculation: the matrix rows are permutations and the column sums agree. The channel is not the usual ternary symmetric-error channel, since an error has only one possible changed output for each input.

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.

Use $\oplus$ for exclusive or and regard all bits as elements of ${\mathbb{F}}_2$. For every public label $z\in \{0,1{\}}^m$, Alice computes the indicator functionwhile Bob computes $v_z(y)=f(z,y)$ from the known truth table. Exactly one of the $u_z(x)$ is one, so the separated Boolean decomposition is

For each $z$, they invoke an independent PR box with respective input bits $u_z(x),v_z(y)$ and receive outputs $a_z,b_z$. These satisfy $a_z\oplus b_z=u_z(x)v_z(y)$. Alice forms $A={⨁}_z{a}_z$, and Bob forms $B={⨁}_z{b}_z$. Taking the exclusive or of all box constraints provesAlice sends the single bit $A$. Bob combines it with his locally known $B$, obtainingThis is one-bit computation using Popescu–Rohrlich boxes. It uses $2^m$ boxes; decomposing by Bob's input instead gives $2^n$, so the smaller truth-table factorization uses at most $2^{\min (m,n)}$. No efficiency bound on local calculation or box use is needed for the stated unlimited resource.

Each PR box has uniform local output distributions regardless of the other party's input. Independent invocations therefore give Bob a uniform output string before the message, carrying no information about $x$. The useful correlation becomes accessible through Alice's final bit; the construction respects the no-signalling property of the boxes.

The target states here are qubit states, as appropriate to sharing singlet states of two qubits. Write $|\psi ⟩=\alpha |0⟩+\beta |1⟩$ and choose $|{\psi }^{\perp }⟩=-{\beta }^{\ast }|0⟩+{\alpha }^{\ast }|1⟩$. These form an orthonormal basis, and the shared Bell state can be expressed asAlice knows the classical description of $|\psi ⟩$, so she can perform the projective measurement in this basis. The outcome $|{\psi }^{\perp }⟩$ has probability $1/2$ and leaves Bob in $|\psi ⟩$, up to an irrelevant overall phase. Alice sends a success bit: one for this outcome and zero for the other. Bob keeps his qubit only on success. This proves heralded remote preparation of an arbitrary qubit:The other outcome leaves $|{\psi }^{\perp }⟩$ at Bob's side; the protocol makes no claim of success in that branch.

For a block of $n$ targets, share $n$ independent singlet states. Alice measures her $j$th qubit in the basis $|{\psi }_j⟩,|{\psi }_j^{\perp }⟩$. She sends one bit indicating whether all $n$ outcomes were of the successful kind. Because the pairs are independent, all-success has probability $2^{-n}$; conditioned on it, Bob's joint state is precisely ${⨂}_{j=1}^n|{\psi }_j⟩$. Thus one-bit heralding of a remote state preparation block givesOn failure Bob discards the block, so he does not need a list of which component states failed.

For asymptotic remote state preparation by block indexing, prelabel a supply of independent candidate blocks, each containing $n$ singlet states. Alice measures blocks until she finds the first all-success block, of index $J$. She does not send a flag after each trial. Since $q=2^{-n}$ is the success probability of each block,This geometric distribution is independent of the target states. The probability of never succeeding is ${\lim }_{k\to \infty }(1-q{)}^k=0$. Alice needs to communicate only $J$; Bob retrieves the labelled qubits in that block, and thereby knows that every requested state was prepared exactly.

Here is an explicit prefix code for a rare geometric success. Set $M=2^n$ and divide $J-1=MQ+R$, where $0\le R<M$. Encode $Q$ as $Q$ ones followed by zero, then encode $R$ in exactly $n$ bits. Bob can decode this without knowing any target-state description. The message length is $L=n+Q+1$. With $r=(1-q{)}^M$,Since $(1-1/M{)}^M\le e^{-1}$,This version succeeds exactly with probability one and has the stated expected classical communication cost. It consumes $n\mathbb{E}J=n{2}^n$ singlet states on average, compatible with the unlimited entanglement assumption.

There is also a fixed-message-length version. Try at most $K=n{2}^n$ candidate blocks and send the first successful index, or a zero codeword reporting that none succeeded. Encoding the $K+1$ possibilities uses $\lceil {\log }_2(K+1)\rceil =n+{\log }_{2}n+O(1)$ bits. Its failure probability isThus fixed-length communication approaches one bit per state with success probability tending to one; variable-length communication achieves exact eventual success with expected cost approaching one bit per state. Before the index arrives, Bob's unconditional state remains independent of the target descriptions, in agreement with the no-signalling theorem.

For each hidden variable $\lambda$, a deterministic local hidden-variable model assigns values $a_i(\lambda ),b_j(\lambda )\in \{0,\dots ,d-1\}$ to all settings. The responses depend only on the local setting; their averaging distribution is setting independent, as required by measurement independence.

Let $T(\lambda )$ be the sum of the nonnegative modular representatives in the chained modular Bell inequality. Before reduction modulo $d$, the chain telescopes:Reduction changes this integer only by multiples of $d$. Since all reduced terms are nonnegative,The smallest possible nonnegative value in this residue class is $d-1$. Averaging over the hidden variable provesAdding local random seeds to $\lambda$ gives the same result for stochastic local hidden-variable theories.

For the coplanar qubit realization of the chained Bell inequality, shareAt angle $\theta$ in the $xz$ plane, use the projective measurement of the observableHere ${\sigma }_x,{\sigma }_z$ are Pauli matrices, and the classical bit $a$ records eigenvalue $(-1{)}^a$. Since $⟨{\sigma }_z\otimes {\sigma }_z⟩=⟨{\sigma }_x\otimes {\sigma }_x⟩=1$ and both crossed expected values vanish in this Bell state,The one-party expected values vanish, so the complete joint probabilities areFor binary outcomes, $[a-b]$ is one exactly when $a\ne b$, while $[b-a-1]$ is one exactly when $a=b$. Consequently

For $N\ge 2$, set $\delta =\pi /(2N)$ and choose Alice's angle ${\theta }_i=2(i-1)\delta$ and Bob's angle ${\varphi }_i=(2i-1)\delta$. Every adjacent pair $A_i,B_i$ or $B_i,A_{i+1}$ differs by $\delta$, giving expectation ${\sin }^2(\delta /2)$. The closing pair has angle difference $(2N-1)\delta =\pi -\delta$; its agreement probability is also ${\sin }^2(\delta /2)$. Thus every one of the $2N$ terms contributes equally:Indeed $\sin x<x$ for $x>0$, so this value is below ${\pi }^2/(8N)\le {\pi }^2/16<1$. For an explicit two-setting instance, Alice measures ${\sigma }_z,{\sigma }_x$ and Bob measures $({\sigma }_z+{\sigma }_x)/\sqrt{2},(-{\sigma }_z+{\sigma }_x)/\sqrt{2}$. The four relevant expected values sum toviolating the local bound one. At $N=1$ there is no violation; the two terms always sum to one. As $N\to \infty$, the displayed quantum value is asymptotic to ${\pi }^2/(8N)$ and approaches the algebraic lower bound zero.

The violation rules out a local hidden-variable theory with setting-independent preassigned local responses reproducing these quantum correlations. It does not permit faster-than-light communication: the joint formula gives each local outcome probability $1/2$, independent of the remote setting, as required by the no-signalling theorem. The contradiction concerns the joint correlations and the Bell assumptions, not the ability to send a signal.

Digital acquisition converts a continuous voltage signal into timed numerical samples. A practical chain is a suitably buffered and amplified sensor signal, an analogue anti-aliasing filter, a sampling or sample-and-hold stage, an analog-to-digital converter, and a clocked data-storage system. Choose the gain so that the interesting variation uses the available input range without clipping. Finite input impedance can load a sensor, so an appropriate buffer or operational amplifier may be needed.

Sampling makes time discrete. If the conditioned signal is confined below frequency $f_{\max }$, choose a sampling rate $f_s>2f_{\max }$, allowing a transition band for the physical filter. The Nyquist–Shannon sampling theorem permits ideal reconstruction of a suitably bandlimited function from exact samples. Without filtering, components separated by integer multiples of $f_s$ have identical samples; in particular real sinusoids at $f$ and $f_s-f$ can be indistinguishable. This aliasing cannot generally be undone by later digital filtering. Oversampling can ease filter design, but does not replace the requirement to suppress unwanted high-frequency content before sampling.

Amplitude quantization is separate from sampling. A uniform $b$-bit converter covering voltage span $V_{\mathrm{span}}$ has step $q\simeq V_{\mathrm{span}}/2^b$. Rounding an in-range input gives the uniform-quantization error bound $|ϵ|\le q/2$. If the rounding errors can additionally be modeled as uniform and uncorrelated, their standard deviation is $q/\sqrt{12}$. This statistical estimate is not a guarantee for a constant input or a coherent periodic signal. More bits increase resolution, but accuracy also depends on voltage-reference calibration, offset, gain error, nonlinearity and electronic noise. Out-of-range signals clip and invalidate the rounding bound.

Clock quality matters too. Timing jitter $\delta t$ gives a voltage error approximately $(dV/dt)\delta t$; for a sinusoid its scale grows with $2\pi f{V}_p{\sigma }_t$. A clock calibration error biases measured frequencies and time intervals. Simultaneous multichannel signals need known relative timing, and the sampling and analogue settling times must be appropriate to the signal bandwidth.

For $n_c$ channels the raw recording rate is approximately $n_c{f}_{s}b$ bits per second, before overhead, so storage and sustained transfer speed must prevent missing data. Store units, channel identities, timestamps, sampling rate, input ranges and calibration information along with the readings. Preserve the original samples when applying digital filters or other processing. Averaging can improve precision for independent fluctuating errors, but cannot restore aliased frequencies, clipped peaks or a fixed rounding bias. Digital recording makes reproducible analysis and transmission possible when its sampling, amplitude and timing limitations are all controlled.

Quantum cryptography includes quantum key distribution, which uses quantum transmissions and authenticated public discussion to establish a secret random key. In BB84, the sender independently chooses a random bit and either the computational or Hadamard basis, sending one of $|0⟩,|1⟩,|+⟩,|-⟩$. A receiver independently chooses one of those two bases. They announce bases after receipt, retain matching-basis positions and reveal a random sample of those bits to estimate noise or eavesdropping. Ideal uniform independent basis choices retain about half of the detected signals before testing and further processing.^[ref]

The no-cloning theorem prevents an adversary from copying arbitrary unknown nonorthogonal signal states perfectly. In a simple intercept–resend attack, the adversary chooses a random protocol basis. On half of the sifted positions it is the wrong basis, and the resent state then yields an erroneous receiver bit with probability one half. Thus the BB84 intercept-resend error rate isif every signal is intercepted, or $Q=q/4$ when an independently selected fraction $q$ is attacked this way. This is an illustrative attack calculation, not a security proof against all possible quantum attacks.

After parameter estimation, classical error reconciliation aligns the remaining keys; privacy amplification compresses them to remove the adversary's bounded information, including information leaked during public reconciliation. The classical channel requires message authentication, otherwise an attacker could impersonate each party to the other. Real sources, losses, multiphoton pulses and detector behavior must fit the security model. A low observed error rate alone is insufficient to certify arbitrary hardware. Finally the protocol distributes a key; encrypting a later message is a separate step.

For the exact difference, the representability condition becomesWithin the same operand square $[-H,U{]}^2$, the closed diagonal band between $y=x-U$ and $y=x+H$ is representable. Above the upper line, $x-y<-H$ gives negative overflow; below the lower line, $x-y>U$ gives positive overflow. The right-hand panel of the figure shows these regions, including their representable boundary points.

Positive overflow can occur only when $x\ge 0$ and $y<0$; negative overflow can occur only when $x<0$ and $y\ge 0$. The corresponding exact inequalities are still required: operands with different signs do not always overflow. Same-sign subtraction is always representable. In sign-bit terms, subtraction overflows exactly when the operands have different signs and the wrapped result's sign differs from that of $x$.

One must test subtraction directly or use its correct overflow rule: forming -y in the same $n$ bits fails at $y=-H$, even though some differences with that operand are representable. For example $(-H)-(-H)=0$ is valid.