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Here $H_2=\left(\begin{array}{cc}0& i\\ -i& 0\end{array}\right)$ and $H_2^2=I$. Its time evolution for an additional interval $\tau$ isAt $\tau =T=\pi \hslash /2$, $U_2(T)=-i{H}_2$. Acting on the state from the previous part givesIt differs from the initial state only by a global phase. For every observable or projection the phase cancels between bra and ket, so the final state is observationally indistinguishable from the original isolated state. A relative phase between two components or between coherently controlled alternatives would be observable; this overall phase of the entire state is not.

Take the Hadamard gate to be $H=2^{-1/2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$ and the phase gate to be $P(\varphi )=\mathrm{diag}(1,e^{i\varphi })$. This fixes the sign convention for the interference phase.

For the coherent quantum circuit, the successive quantum states areThe Born rule therefore gives the concise answer

With the stated quantum decoherence after the phase gate, the joint pure state before the last Hadamard gate isAfter that Hadamard gate, its environment coefficient at output zero is $(|e_0⟩+e^{i\varphi }|e_1⟩)/2$. Squaring its norm sums over every unobserved environment outcome, soThe Cauchy-Schwarz inequality gives $0\le v\le 1$. In this single-qubit interference with conditional environment states, $v$ is the interference visibility: orthogonal environment states erase the fringe, while a unit-modulus overlap shifts its phase without reducing visibility. The plus sign in $\varphi +\alpha$ follows from the specified overlap $⟨e_0|e_1⟩$, rather than its complex conjugate.

If the same quantum decoherence occurs before the phase gate, the joint pure state is first $(|0⟩|e_0⟩+|1⟩|e_1⟩)/\sqrt{2}$. The phase gate then produces exactly the same state used above. Equivalently, the controlled environment interaction commutes with the diagonal phase gate. The probability is unchanged.

For the Deutsch algorithm, use the reversible Boolean quantum oracleThe algebraic oracle map printed in the source has its arguments misplaced: taken literally, it sends two distinct answer inputs to the same output for constant $f$, so it cannot be a unitary operator. The corrected map above agrees with the diagram's input-control, answer-target arrangement. Put the answer qubit in $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Since the Pauli X gate satisfies $X|-⟩=-|-⟩$, quantum phase kickback givesWrite $b=f(0)\oplus f(1)$. Up to the irrelevant global phase $(-1{)}^{f(0)}$, the input qubit acquires relative phase $(-1{)}^b$, so its ideal final computational-basis measurement returns $b$: zero for a constant function and one for a balanced function.

Let the specified quantum decoherence occur in the interference portion between the two Hadamard gates. It commutes with the oracle's effective diagonal phase on the input, so the same analysis applies whether it occurs just before or just after the query. The two likelihoods areThus the diagram's usual zero-for-constant decision rule hasThis is the reliability for either promised class separately, with no prior needed. For equally likely classes, if $v\cos \alpha$ is known to be negative, reversing the interpretation gives $(1+|v\cos \alpha |)/2$. A known environment phase can instead be compensated by the phase gate $P(-\alpha )$ before the final Hadamard gate, giving success probability $(1+v)/2$. In particular, $v=0$ gives no information and $v=1$ permits certainty after phase compensation. These refinements distinguish a shifted fringe from lost quantum coherence; with the unmodified diagram a phase shift alone can reverse the apparent answer. Quantum decoherence after the final Hadamard gate leaves its computational populations unchanged, so that different placement would not spoil the measured answer.

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.

A density operator represents a pure state exactly when it has rank one, equivalently ${\rho }^2=\rho$ or $\mathrm{Tr}{\rho }^2=1$. The spectra just found giveTheir wavefunctions are defined up to a global phase. The other two are mixed states, with purity $(3/4{)}^2+(1/4{)}^2=5/8$, and have no single-wavefunction representation as rank-one projectors.

The Von Neumann equation integrates to $\rho (t)=U(t)\rho (0)U(t{)}^{†}$, so the spectrum and purity of a density operator are invariant under any allowed open-loop control. These are the kinematic restrictions.

There is no additional orbit restriction here: the generators $i{\sigma }_z$ and $i{\sigma }_x$, together with their matrix commutator, span the special unitary Lie algebra:For freely chosen durations and piecewise continuous real control values, this gives density operator controllability. The connected reachable group is $SU(2)$; its action by conjugation is the same as that of $U(2)$, because a global phase cancels. Thus exactly the pairs ${\rho }_0\leftrightarrow {\rho }_2$ and ${\rho }_1\leftrightarrow {\rho }_3$ can be interconverted. No coherent control can turn either mixed state into either pure state, even though every orientation within each spectral orbit is reachable.

Here “controllable” means density operator controllability, so all transformations on a fixed spectral orbit are available. Strict unitary operator controllability, including an independently specified global phase, is a stronger condition. Assume the usual finite-dimensional real controls and freely chosen duration.

The dimension hypothesis actually identifies the algebra, not just its size. For completeness, put the invariant inner product $⟨X,Y⟩=-\mathrm{Tr}(XY)$ on $\mathfrak{u}(N)$ and write ${\mathfrak{g}}^{\perp }=\mathbb{R}Z$. Since $\mathfrak{g}$ is closed under commutators, $[X,Z]$ lies in ${\mathfrak{g}}^{\perp }$ for every $X\in \mathfrak{g}$. The adjoint action is skew-adjoint for this inner product, so $[X,Z]$ is also orthogonal to $Z$ and must vanish. If $Z$ were nonscalar, its orthogonal eigenspaces would give a centralizer of dimension at most $(N-1{)}^2+1<N^2-1$, too small to contain $\mathfrak{g}$. Therefore $Z$ is scalar and $\mathfrak{g}=\mathfrak{su}(N)$. This proves codimension-one quantum dynamical algebra is special unitary.

a1 is true for density-state controllability; a2 is true; a3 is false. The algebra $\mathfrak{su}(N)$ generates every determinant-one unitary. For any $V\in U(N)$, choose $\phi$ with $e^{iN\phi }=\det V$; then $e^{-i\phi }V\in SU(N)$, proving that every gate is available up to global phase. But all generated Quantum Hamiltonians are traceless, andso evolution starting at the identity always has determinant one. Arbitrary determinant, requiring the additional central direction $iI$, cannot be generated. If a1 is instead read as phase-sensitive unitary-operator controllability, it is false for precisely the same reason as a3; the physical distinction is explicit here.

The stated relation places every generator, and therefore the whole Dynamical Lie algebra, inside the standard compact symplectic Lie algebra $\mathfrak{sp}(\ell )\subset \mathfrak{su}(2\ell )$. Indeed the relation is closed under linear combinations and commutators, and differentiating evolution givesThus it supplies a symplectic dynamical symmetry in quantum control, rather than a claim that $J$ commutes with every Quantum Hamiltonian.

For $\ell \ge 2$, b1 is false: even the full compact symplectic group does not act transitively on every mixed-state spectral orbit. Its algebra has dimension $\ell (2\ell +1)<(2\ell {)}^2-1$. An explicit obstruction is the symplectic invariant of a density operatorwhich is unchanged by a symplectic conjugation. For $N=4$ and the printed $J$, let $a,b>0$, $a\ne b$, $2a+2b=1$. The isospectral operators $\mathrm{diag}(a,a,b,b)$ and $\mathrm{diag}(a,b,a,b)$ have invariants $4ab$ and $2(a^2+b^2)$, respectively, and cannot be interconverted. For the exceptional case $\ell =1$, $\mathfrak{sp}(1)=\mathfrak{su}(2)$, so b1 cannot be decided from inclusion alone; equality gives density-state controllability.

b2 is not implied by the displayed relation: more information is needed. It permits the full symplectic algebra but also a single commuting generator, or even zero generators. The full $Sp(\ell )$ is pure-state controllable: identify ${\mathbb{C}}^{2\ell }$ with ${\mathbb{H}}^{\ell }$, extend any unit vector to a quaternionic orthonormal basis, and map one such basis to another. A one-axis diagonal subgroup, in contrast, cannot change level populations.

The required extra information is the generated algebra, obtained by computing its real span of iterated commutators. Within this specified symplectic inclusion, equality $\mathfrak{g}=\mathfrak{sp}(\ell )$ gives pure-state controllability; its dimension is $\ell (2\ell +1)$. The mere inclusion does not establish equality. For $N=2$, equality also answers b1 affirmatively. None of these strictly symplectic generators supplies arbitrary global phase.

c1 is true for density-state controllability. Every qubit density operator can be written $\rho ={\lambda }_{2}I+({\lambda }_1-{\lambda }_2)|\psi ⟩⟨\psi |$ using its eigenvalues. A control taking one pure orthogonal projector to any other therefore also connects every pair of qubit density operators with the same spectrum. In Lie-algebra language, $\mathfrak{sp}(1)=\mathfrak{su}(2)$; there is no larger-dimensional symplectic exception. This does not supply arbitrary global phase: $SU(2)$ alone is already a counterexample to an implication involving strict operator controllability.

c2 is false. For every even $N=2\ell \ge 4$, the defining $Sp(\ell )$ action is transitive on unit vectors, so a system generating this algebra is pure-state controllable. It still preserves additional mixed-state invariants and lacks density operator controllability, as the preceding explicit four-dimensional example demonstrates. pure states have fewer orbit requirements than general mixed states.

The original Figure 1 has the allowed chain $1\leftrightarrow 2\leftrightarrow 3\leftrightarrow 4$. Its positive resonance frequencies are ${\omega }_1=E_2-E_1$, ${\omega }_2=E_2-E_3$ and ${\omega }_3=E_4-E_3$ in units $\hslash =1$; level 2 lies above level 3. Use a pulse at ${\omega }_1$ to transfer level population from 1 to 2, then one at ${\omega }_2$ to transfer it from 2 to 3, and finally one at ${\omega }_3$ to transfer it from 3 to 4.

For a selected pair, choose the optical phase so the effective Quantum Hamiltonian has a fixed off-diagonal rotation axis. If its coupling coefficient is $g_{jk}(t)=A_{jk}(t)|d_{jk}|/2$ in this normalization, the active two-dimensional propagator iswhere $K_{jk}^2=I_{jk}$. A full level population transfer has $\beta =\pi /2$. Equivalently, in the Rabi-frequency convention ${\Omega }_R=2g_{jk}$, the required area is $\int {\Omega }_{R}dt=\pi$. On spectator levels the embedded two-level quantum rotation is the identity.

Thus three selective $\pi$ pulses at ${\omega }_1,{\omega }_2,{\omega }_3$, in that order, transfer $|1⟩$ to $|4⟩$ up to global phase. A strongly regular quantum Hamiltonian separates the resonances; envelopes long enough that their bandwidth and coupling strength are small relative to unwanted frequency gaps make off-resonant excitation arbitrarily small within the selective-pulse limit. The phases accumulated along this path do not affect the final level population.

Start by turning on ${\Omega }_{12}$ first, while ${\Omega }_{23}=0$. The dark state is then exactly $|3⟩$, which is uncoupled even while the first pulse rises. Next turn on ${\Omega }_{23}$ while ${\Omega }_{12}$ remains appreciable, and then turn off ${\Omega }_{12}$ while ${\Omega }_{23}$ remains on. The ratio ${\Omega }_{23}/{\Omega }_{12}$ rises from zero to infinity, soThe final minus sign is an irrelevant global phase for level population transfer. This is the counter-intuitive sequence of stimulated Raman adiabatic passage: the laser coupling the target state to the excited state precedes the laser coupling the initial state to the excited state.

The quantum adiabatic theorem requires the dark state to rotate slowly compared with its separation from the bright states. Here $\dot{D}=-\dot{\theta }B$, so its coupling to either bright eigenstate has magnitude $|\dot{\theta }|/\sqrt{2}$. In the printed convention the entries ${\Omega }_{jk}=A_{jk}{d}_{jk}/(2\hslash )$ are angular-frequency couplings, and the condition isIf one instead uses an energy-valued Quantum Hamiltonian, its spectral gap is $\hslash \Omega$ and the same condition is $\hslash |\dot{\theta }|\ll \hslash \Omega$. Maintain nonzero overlap and gap while the angle changes. The pulses can vanish at the beginning and end, where the dark state has stopped rotating and is an exactly decoupled basis state.

An explicit smooth sequence has ${\Omega }_{12}={\Omega }_0{\sin }^2(\pi t/(6T))$ on $0<t<6T$ and zero otherwise, while ${\Omega }_{23}={\Omega }_0{\sin }^2(\pi (t-2T)/(6T))$ on $2T<t<8T$ and zero otherwise. Choose ${\Omega }_{0}T\gg 1$. The first pulse comes earlier, both overlap during the rotation, and the second persists after the first has ended. The following original sketch shows the envelopes and instantaneous dark-state level populations.

Write $a=⟨\psi |w⟩=2^{-N/2}$, which is real and satisfies $0<a<1$ for $N\ge 1$. The invariant two-dimensional span has orthonormal eigenvectorswith eigenvalues $\Delta (1+a)$ and $\Delta (1-a)$. In this basis,Transfer therefore requires the two evolved eigenphases to differ by an odd multiple of $\pi$:The first positive time isEquivalently, exponentiating on the invariant span gives the full evolutionAt $t_0$ this is $-i{e}^{-i\Delta t_0}|w⟩$, so the requested global phase can be chosen as $\alpha =-\Delta t_0-\pi /2$ modulo $2\pi$. Since the two vectors are linearly independent, their relative coefficient cannot produce exact transfer at an earlier positive time. This is continuous-time quantum search, with time proportional to the square root of the search-space dimension for fixed $\Delta$. Restoring units multiplies the stated time by $\hslash$. The degenerate zero-qubit case would already have $\psi =w$, so there would be no smallest positive transfer time; the formula concerns the nontrivial $N\ge 1$ search.

For the normalized vector defining the orthogonal projection $P=|\varphi ⟩⟨\varphi |$, $P^k=P$ for every $k\ge 1$. Thuswhich proves the projector phase rotation identity in this case:It changes the sign of the component along $\varphi$ and fixes the orthogonal complement, so it is a reflection operator.

For a marked computational basis vector $w$, $U_{P_w}=I-2|w⟩⟨w|$ is the marked-state phase oracle. With $|\psi ⟩=H^{\otimes N}|0^N⟩$, the other reflection isThe Hadamard gates thus implement its change of basis from the known all-zero state. The usual Grover diffusion operator is $D=2|\psi ⟩⟨\psi |-I=-U_{P_{\psi }}$. Consequently a Grover search algorithm iteration isThe minus sign is an irrelevant global phase. These two reflections rotate the search state in the span of the marked vector and the unmarked uniform vector, with the Grover rotation angle controlled by the initial marked amplitude. Iterating them approximately $\pi \sqrt{2^N}/4$ times gives high-probability search. The same pair of projectors appears additively in the continuous-time Hamiltonian, but the product of full reflections and evolution under their sum are distinct dynamics.

Use the PDF's Bell state labels throughout this question: its ${\Phi }_{-}$ is the antisymmetric spin singlet state, whereas ${\Psi }_{\pm }$ are the even-parity states. WriteExpanding the antisymmetric combination cancels its $00$ and $11$ components and givesThe two normalized orthogonal vectors are the columns of a unitary matrix $U$. Its determinant $a_0{b}_1-a_1{b}_0$ has modulus one because $\det (U^{†}U)=|\det U{|}^2=1$. ThereforeThis proves collective-unitary covariance of the two-qubit singlet: a common basis change alters only its global phase, and leaves its density projector unchanged.

By orthonormality in the Generalized Bloch representation,A pure state has density operator $\rho =|\psi ⟩⟨\psi |$ with $⟨\psi |\psi ⟩=1$, so ${\rho }^2=\rho$ and its purity of a density operator is one. ConsequentlyEvery pure state therefore maps to the specified radius-$\sqrt{1-1/N}$ sphere in ${\mathbb{R}}^{N^2-1}$, of dimension $N^2-2$. The expansion is unique, so it distinguishes pure-state density operators, while vectors differing only by global phase describe the same point. This proves that pure states lie on the sphere, without asserting that they fill it.

A real Lie algebra is a real vector space with a bilinear antisymmetric bracket satisfying the Jacobi identity. For matrices, the bracket is the commutator $[X,Y]=XY-YX$. The Dynamical Lie algebra isIt is found by repeatedly adjoining commutators and taking real linear spans until no new independent generators appear. Let $G$ be its connected dynamical Lie group. Under the standard assumptions of freely timed controls with unrestricted real amplitudes, the following are the relevant Lie criteria.

For pure-state controllability, $G$ must act transitively on normalized state vectors, or on their rays when global phase is disregarded. The standard finite-dimensional classification is, up to a unitary change of basis,Here $\mathfrak{sp}(n)$ is the compact symplectic Lie algebra in its defining complex $2n$-dimensional representation. Equivalently, the infinitesimal action spans the tangent directions to the whole pure-state orbit. The symplectic cases are important because pure-state reachability alone need not give complete mixed-state reachability.

For density operator controllability, the action must be transitive on every unitary orbit of a density operator, meaning on every fixed-spectrum class. The criterion is $\mathfrak{g}=\mathfrak{su}(N)$ or $\mathfrak{u}(N)$. This concerns all density operators, not merely one unusually degenerate spectrum.

For exact unitary operator controllability, the criterion is $\mathfrak{g}=\mathfrak{u}(N)$. If only the physical gate up to global phase matters, it is enough that the image after removing the scalar direction is $\mathfrak{su}(N)$, equivalently that $\mathfrak{g}+\mathbb{R}iI=\mathfrak{u}(N)$. A traceless system can implement all special-unitary gates while lacking independent control of the overall phase.

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.

Write $X={\sigma }_x$, $Y={\sigma }_y$, $Z={\sigma }_z$. The specified diagonal two-qubit gate has the exact factorizationThus one exact implementation is to set the controllable Ising coupling to zero and perform the two local rotations about the z-axis, synthesized from the available $x,y$ pulses as in part (c). Alternatively, turn off the local fields and choose the interaction area $\int J_{12}(t)dt=\pi /2$. Thenwhich implements the same physical gate up to global phase. The printed gate is local, not entangling; the pulse construction does not change that fact.