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The desired output is $|{\psi }_d⟩=((\alpha +\beta )|0⟩+(\alpha -\beta )|1⟩)/\sqrt{2}$. To evaluate its squared quantum fidelity, write the input Bloch vector as $(x,y,z)$, with $x=c$, $y=2\mathrm{Im}({\alpha }^{\ast }\beta )$ and $z=d$. The desired vector after the Hadamard is $(z,-y,x)$. From part (a), the actual vector isFor $k\ge 2$ this simplifies to $(1-1/k)(z,-y,x)-(x/k,0,0)$. The pure state overlap is one half of one plus the dot product with the desired unit vector, soSince $x^2+y^2+z^2=1$, $|xz|\le 1/2$. Therefore the infidelity is uniformly bounded by $3/(4k)$, and the entanglement effect on the implemented gate is negligible for large $k$, uniformly over all initial pure states. The shifted field packets then have overlaps tending to one, so they retain negligible information about the qubit transitions. A large fixed photon number without this broad number coherence would not suffice.

The printed overlap silently requires $k\ge 2$. If the allowed packet has $k=1$, the exact result instead isFor instance, $\alpha =\cos (\pi /8)$ and $\beta =\sin (\pi /8)$ give actual overlap $3/8$, whereas substituting $k=1$ into the printed expression gives $1/4$. This boundary correction follows directly from the two-shift overlap above.

Use the rotation-invariant distribution over pure initial states, namely the uniform pure-qubit average with area element $\sin \theta d\theta d\phi /(4\pi )$. Reflection symmetry gives $\mathbb{E}[xz]=0$, and rotational symmetry together with $x^2+y^2+z^2=1$ gives $\mathbb{E}[y^2]=1/3$. Averaging the squared quantum fidelity from part (b) therefore yieldsIf the single-number packet $k=1$ is included, its corrected average isUniform polar angle without the sine weight would not be the Haar-uniform average over all pure qubit states.

For a pure qubit $a|0⟩+b|1⟩$, the Bloch vector is $(2\mathrm{Re}(a^{\ast }b),2\mathrm{Im}(a^{\ast }b),|a{|}^2-|b{|}^2)$. Substitution givesBoth lie on the unit Bloch sphere, at the same azimuth and opposite heights. They are reflections across its equatorial plane, not antipodal vectors. Choose $\theta =\pi /3$ and $\phi =\pi /4$, so their coordinates are $(\sqrt{6}/4,\sqrt{6}/4,\pm 1/2)$. The following original sketch shows those vectors and their common meridian:

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.

A general single-qubit quantum gate belongs to $U(2)$. Remove its global phase to obtain $V\in SU(2)$; the displayed expression in the question is literally a special-unitary gate and represents a general physical gate up to that phase. Every $V\in SU(2)$ can be writtenand consequentlyChoose $\theta$ with $q_0=\cos (\theta /2)$ and $\Vert q\Vert =\sin (\theta /2)$, and set $n=q/\Vert q\Vert$ when $q\ne 0$. The Pauli matrix multiplication law shows $(n\cdot \sigma {)}^2=\Vert n{\Vert }^{2}I=I$. Applying part (a) provesFor $q=0$, $V=\pm I$ and any axis can be used with angle zero or $2\pi$. Thus the phase-free gate is a rotation gate by angle $\theta$ about the unit axis $n$. An arbitrary exact $U(2)$ operator also needs the scalar phase $e^{i\chi }$, which cannot in general be produced by a traceless Quantum Hamiltonian.

The claimed universality is false for the particular gate specified in part (d). Every allowed single-qubit gate is local, and $C_{\mathrm{phase}}=-Z\otimes Z$ is local too. Products of these gates remain of the form $e^{i\chi }A\otimes B$. They preserve all product states and cannot, for example, implement the CNOT gate that sends $|+⟩\otimes |0⟩$ to $(|00⟩+|11⟩)/\sqrt{2}$. Equivalently, its diagonal two-qubit phase entanglement criterion gives the alternating phase $\pi -0-0+\pi =0$ modulo $2\pi$. No construction with only the printed phase gate and local rotations implements arbitrary two-qubit gates.

The intended universality argument works after replacing that local gate by an entangling controlled phase. The same tunable Ising coupling suppliesThus this quarter-area Ising pulse, with local corrections, realizes the Controlled-Z gate. Conjugating it by the Hadamard gate on the second qubit gives $\mathrm{CNOT}=(I\otimes H)\mathrm{CZ}(I\otimes H)$, with harmless scalar phases if the Hadamards are implemented using traceless control Hamiltonians.

The Cartan decomposition of a two-qubit gate writes a target aswith local $SU(2)$ factors. The three central Pauli-product operators commute, so their exponential is a product of three variable interaction gates. From two fixed CNOT gates and a variable local rotation about the z-axis,because $\mathrm{CNOT}(I\otimes Z)\mathrm{CNOT}=Z\otimes Z$. Conjugating this interaction by $R_y(\pi /2)$ on both qubits yields $e^{-icX\otimes X}$; conjugating by $R_x(-\pi /2)$ on both yields $e^{-icY\otimes Y}$. Part (c) implements all outer local factors. This gives the full requested Cartan construction with an entangling phase gate, and pinpoints the defect in the printed version. Exact arbitrary scalar phase requires an identity Hamiltonian term; the physical gate construction is up to global phase.

An always-on Ising coupling produces evolution during every nominally local pulse, and generally does not commute with the applied $x,y$ Quantum Hamiltonians. A naive product of single-qubit rotations therefore accumulates unwanted conditional phases and can entangle the qubits. Idling is also no longer an identity operation.

For a pulse of duration $\tau$, neglecting the coupling requires $|J_{12}|\tau \ll 1$ in units $\hslash =1$, or $|J_{12}|\tau /\hslash \ll 1$ in ordinary units. For order-one rotation angles, this is the strong-local-drive regime $|B|\gg |J_{12}|$. Large drive amplitude alone does not justify neglect over a long total sequence: the accumulated interaction time or an appropriate error bound must also be small.

If it cannot be neglected, include it in the quantum optimal control model, or refocus it. A $\pi$ pulse $P=R_x^{(1)}(\pi )$ obeys $P(Z\otimes Z)P^{†}=-Z\otimes Z$, giving the ideal Ising spin echoThis cancellation assumes the refocusing pulses are instantaneous relative to $1/|J_{12}|$, or that the coupling during finite pulses is compensated. It explains how fast local controls can suppress the fixed coupling without mistaking it for a controllable zero interaction.

Write $N=2k+1$ and abbreviate the Dicke states by $|D_j^n⟩=|{\Psi }_j^n⟩$. Alice's shared resource has decompositionFollow the actual circuit order: a controlled-NOT gate from the input to $q_0$, a Hadamard gate on the input and a Pauli X gate on $q_0$, followed by computational basis measurements. Just before measurement, the joint vector isThe Dicke states in each branch are orthogonal, so each branch has squared norm $1/4$. Conditioning on outcome $00$ and normalizing therefore gives the state of the other $N$ parties:This is the encoding used in Dicke-resource telecloning. The extra Pauli X gate on Alice's resource qubit is essential to the displayed ordering of the amplitudes.

The state in part (b) is symmetric in the receiving qubits. Single out Bob using the Dicke state recursion. For $k\ge 1$, the two vectors areDistinct excitation counts give orthogonal states of the other recipients. Taking their partial trace, the only surviving off-diagonal term comes from the common rest vector $|D_k^{N-1}⟩$. The one-qubit reduction of Dicke-state superpositions is consequentlyPut $p=|\alpha {|}^2$ and $q=|\beta {|}^2$, so $p+q=1$. The overlap with the input isFor $N=1$ there is no system to trace out, and the same matrix is $|\psi ⟩⟨\psi |$, with overlap one. For every $N>1$ the overlap is below one, even at its maximum $p=q=1/2$, so these are imperfect copies.

Initialize the register inThis can be prepared by Pauli X gates on the one-bits of $y$, starting from $|0^N⟩$, followed by Hadamard gates on all qubits. Alternatively, prepare the original uniform state and apply Pauli Z gates at those same bit positions. The conjugation calculation givesThe final diagonal sign operator changes no computational basis outcome probabilities. Thus the success probability and the iteration count in part (b) are unchanged. No final correction is needed when the objective is to measure a marked string.

Define orthonormal symmetric vectorsThe symmetric-arm reduction of an exchange Hamiltonian here has two arms. To verify it directly, the link between radial distances $j$ and $j+1$ has strength $J_j$ on either side, while the two central links have strength $J_0/\sqrt{2}$. The calculation of part (a) therefore givesIn particular this subspace is invariant and contains the initial vector. Define the linear isometry $F|j⟩=|s_j⟩$. The displayed matrix entries prove $H_{e}F=F{H}_T$, or $F^{†}{H}_{e}F=H_T$. Induction gives $H_e^{m}F=F{H}_T^m$ for every nonnegative integer $m$; hence the convergent exponential series proves $e^{-i{H}_{e}t}F=F{e}^{-i{H}_{T}t}$. This establishes the mapping for the entire evolution, rather than just its first two powers. The supplied endpoint transfer also forces every $J_j$ to be nonzero, since a broken link would disconnect the two endpoints. Successive powers applied to $|s_0⟩$ then generate all $|s_j⟩$, so this is exactly the cyclic subspace governing the initial state.

Using the stated perfect quantum state transfer identity at time $\pi /2$,The two terms have the excitation at sites $0$ and $2N$, respectively, and all intermediate qubits are zero in both terms. The full state thus factors into a Bell state at the endpoints and the vacuum on the interior. Its partial trace isThe coherent cross terms survive the trace because the interior vectors coincide. We have used the transfer phase convention supplied in the question; a common transfer phase would leave this density operator unchanged.

Replace the two arms by three identical arms of length $N$ attached to one central qubit. Label the hub $c$ and the sites on arm $a$ by $(a,j)$, $a=1,2,3$, $j=1,\dots ,N$. Use XY exchange interactions with coupling $J_0/\sqrt{3}$ on each hub-to-first-site link, and coupling $J_j$ on each arm link $(a,j)$ to $(a,j+1)$. Initially excite only the hub: this is the product state $|1{⟩}_c$ with every arm qubit in $|0⟩$.

In the symmetric single-excitation subspace, define $|s_0⟩=|c⟩$ and $|s_j⟩=3^{-1/2}{\sum }_{a=1}^3|a,j⟩$. The hub coupling isand the reverse coupling has the same coefficient. All other radial links have coefficients $J_j$, just as in part (c). Thus the symmetric-arm reduction of an exchange Hamiltonian is again exactly $H_T$. At time $\pi /2$ its evolution isEvery nontip qubit is zero, so tracing them out leaves the three tips in the pure W stateThe essential change is the additional identical arm and the normalization of each hub coupling by $\sqrt{3}$ instead of $\sqrt{2}$.

Label the control qubits by binary weight: the bit $x_n$ contributes $2^n$ to $x={\sum }_{n=0}^{N-1}{2}^n{x}_n$. For each $n$, use that qubit as the control of the supplied controlled-$V^{2^n}$, with the same target eigenstate $|v_j⟩$. The controlled unitary gate acts bysince $V^{2^n}|v_j⟩=e^{i{2}^n{\varphi }_j}|v_j⟩$. The target remains unchanged after every application. Multiplying these factors givesThe ordering of the tensor factors is chosen to match the binary place values. Discarding the unchanged target leaves the required phase register. This uses quantum phase kickback in the phase accumulation step of quantum phase estimation.