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Define the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ and write $s=e^{-i\varphi }$. After $U_{\varphi }$ the two-qubit quantum state isApplying the Hadamard gate to the first qubit gives the normalized quantum stateFor $\varphi =n\pi$, $s=(-1{)}^n$, soThe first is a product state. The second is exactly the minus Bell state in the printed phase gate convention; its reduced density matrices are $I/2$.

For three qubits, let $U_{12}$ use the ordered pair $(1,2)$ and $U_{23}$ the ordered pair $(2,3)$. The two diagonal quantum gates commute. For $\varphi =n\pi$ their action before the final Hadamard gates isIf $n$ is even this is $|+++⟩$, and $H_1{H}_3$ gives $|0⟩|+⟩|0⟩$. If $n$ is odd, group the quantum superposition by the middle qubit:Since $H_{\mathrm{Had}}|+⟩=|0⟩$ and $H_{\mathrm{Had}}|-⟩=|1⟩$, the final states areThe odd case is a GHZ state: each individual qubit has reduced density matrix $I/2$, and it is related to $(|000⟩+|111⟩)/\sqrt{2}$ by the local Pauli gates $Z_1{X}_3$. Conversely, applying those gates converts $|G⟩$ to that standard form. They have not been applied in the printed network, so the actual output used in the next part is $|G⟩=(|001⟩-|110⟩)/\sqrt{2}$.

Define the Hadamard basis states $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. The initial two Hadamard gates prepare $|+⟩|+⟩$. For one bracket, the two control branches giveUsing the Pauli X gate $X=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, unitary conjugation by the Hadamard gate gives $H{U}_{0}H=-X$, so $G=-X{U}_f$. Keeping every sign, the target states along the control-one branch, in the exact gate order, areThe last column also includes the final target Hadamard gate, which lies outside the bracket. Writing $|{\chi }_f⟩=HG|+⟩$, applying the final top Hadamard gate gives the exact normalized two-qubit quantum stateThus the complete outputs, in top-then-bottom computational basis order, areFor the first qubit, the Pauli matrix ${\sigma }_z=Z$ has expectation value $P(0)-P(1)$ by the Born rule. The two $k=1$ cases each have $P(0)=P(1)=1/2$, while the $k=0$ and $k=2$ cases are deterministic. Equivalently, this Hadamard test measuresThis explains the approximate quantum counting interpretation: repeated preparations estimate the expectation value, and hence $k$. A single top-qubit computational-basis measurement need not identify $k$.

For $r$ brackets the control-zero branch still does nothing, and the control-one branch applies $G^r$ to the target. Immediately before the final two Hadamard gates, the quantum state isFor the constant function with $k=0$, the phase oracle is $U_f=-I$, so $G=(-X)(-I)=X$. Since $|+⟩$ is an eigenstate of the Pauli X gate with eigenvalue $+1$, $G^r|+⟩=|+⟩$ for every integer $r$. The final two Hadamard gates therefore giveIn particular, the first qubit always ends in $|0⟩$.

For the constant function with $k=2$, $U_f=I$ and $G=-X$, so $G^r|+⟩=(-1{)}^r|+⟩$. Thus the target remains in $|+⟩$, and the top qubit before its final Hadamard gate is $(|0⟩+(-1{)}^r|1⟩)/\sqrt{2}$. The sign is a relative quantum phase between the control branches. The Hadamard gate converts it into the integer parity of $r$:The first qubit is therefore $|0⟩$ for even $r$ and $|1⟩$ for odd $r$, including all requested $r>1$.

Use the Hadamard basis states $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. The Hadamard gate satisfies $H|0⟩=|+⟩$, $H|1⟩=|-⟩$, and $H^2=I$. After the first three Hadamard gates, the full quantum state isThe Pauli X gate obeys $X|-⟩=-|-⟩$. Consequently the Boolean quantum oracle produces phase kickback:Factoring out the ancilla qubit, the four states immediately after the function oracle areThe final Hadamard gates send $|+⟩$ to $|0⟩$ and $|-⟩$ to $|1⟩$. In particular, the printed quantum circuit includes a final Hadamard gate on the ancilla qubit too. Its full outputs, ordered as $x_1,x_2,y$, areThus one oracle call identifies $a$ with certainty: a computational-basis measurement of the first two qubits returns $a_1{a}_2$. The ancilla qubit returns to $|1⟩$ and carries no additional information about $a$. The Bernstein-Vazirani algorithm turns the pattern of relative quantum phases into a definite computational-basis state.

Use $a|n⟩=\sqrt{n}|n-1⟩$, $a^{†}|n⟩=\sqrt{n+1}|n+1⟩$, ${\sigma }_z|e⟩=|e⟩$ and ${\sigma }_z|g⟩=-|g⟩$. Only one interaction term acts on each of the two quantum states. HenceThe interaction exchanges one atomic excitation with one field quantum. The square-root factor is the photon-number enhancement of the transition matrix element.

Subtract the common sector energy and call the remaining eigenvalue $\hslash \lambda$. Its characteristic equation isThe dressed-state energies are consequentlyOn resonance their splitting is $2\hslash |g|\sqrt{n+1}$; far off resonance the eigenstates approach the uncoupled quantum states with small dispersive energy shifts. The eigenvalue splitting is an angular-frequency difference ${\Omega }_n$, so an ordinary frequency would be ${\Omega }_n/(2\pi )$.

Write $ϵ=0.08$ and $v=1-2ϵ=0.84$. The ensemble is the mixture $(1-ϵ)|\mathrm{GHZ}⟩⟨\mathrm{GHZ}|+ϵ|{\mathrm{GHZ}}^{\prime }⟩⟨{\mathrm{GHZ}}^{\prime }|$, rather than a coherent superposition of the two phases. The ZZZ probabilities stay unchanged. Reversing the relative phase interchanges the allowed parity sector in every equatorial basis, so for XXY, XYX and YXX,In YYY the quantum prediction is the opposite:For a concrete continuation of the ideal local analysis, apply the same phase-flip event to a predetermined local assignment: it reverses X and Y at the affected photon. All three mixed-basis products and the YYY product then reverse together. With unbiased marginals in each parity sector this shared-error local model predicts negative-product YYY outcomes at $0.23$ each and positive-product outcomes at $0.02$ each. The histograms compare this specified model with the quantum result.

Imperfect mixed-basis statistics alone do not force that particular local histogram. A model-independent constraint follows because a positive YYY product requires at least one of the three ideal mixed-product constraints to fail. Each fails with probability $ϵ$, so the union bound givesEquivalently, local assignments satisfy $|⟨XXY⟩+⟨XYX⟩+⟨YXX⟩-⟨YYY⟩|\le 2$, whereas this noisy quantum state gives $4v=3.36$. Thus the violation survives the specified phase error without assuming a unique noisy local histogram or independent errors in the three measurement contexts.

For the stated Hamiltonian operator, put $\mathbf{\Omega }=(V,0,-\delta )$ and $\Omega =\sqrt{V^2+{\delta }^2}$. The Bloch vector obeysThus it rotates right-handedly about the fixed axis $\mathbf{\Omega }$, preserving its length and its projection on that axis. For example, on resonance $\delta =0$, a north-pole Bloch vector follows $(0,-\sin Vt,\cos Vt)$ and reaches the south pole at $t=\pi /V$.

Take $V>0$, with the field phase chosen accordingly. The Hadamard gate is $H=({\sigma }_x+{\sigma }_z)/\sqrt{2}$. Align the axis with $(1,0,1)$ and rotate through $\pi$:The factor $-i$ is an irrelevant common global phase. This realizes the whole quantum gate, rather than merely its action on one input quantum state.

There is a basis-ordering convention to make explicit: this $H$ is the Hadamard gate matrix in the ordered basis $(|1⟩,|0⟩)$ chosen in part (a). If instead its definition is the usual ket rules $H_{01}|0⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $H_{01}|1⟩=(|0⟩-|1⟩)/\sqrt{2}$, its matrix in our order is $({\sigma }_x-{\sigma }_z)/\sqrt{2}$. For that definition choose $\boxed{\delta =+V,\tau =\pi /(\sqrt{2}V)}$. Both conventions use the same pulse duration; the sign of the Rabi detuning selects the desired axis.

A bipartite pure state is entangled when it cannot be written $|\alpha {⟩}_a\otimes |\beta {⟩}_b$. For mixed states, the corresponding definition excludes every convex mixture of product states, not just a single product state.

Apply a Hadamard gate to ion $a$, followed by the CNOT gate of part (d):The lower network in the figure shows this Bell pair preparation. Either ion alone has reduced density matrix $I/2$, whereas the joint quantum state is pure; hence the joint quantum state cannot be a product state. The entangling operation can be expanded into $H_b,G,Z_a,H_b$ exactly as above.

Use the ordered polarization basis $(|H⟩,|V⟩)$ and define the Bell statesRegrouping the four-photon quantum state by channel pairs $(1,4)$ and $(2,3)$ givesFor instance, projecting channels 2 and 3 on $⟨{\Psi }^{-}|$ retains the two original terms with inner polarizations $HV$ and $VH$, with opposite signs, leaving $(|HV⟩-|VH⟩{)}_{14}/(2\sqrt{2})$. Its squared norm is $1/4$.

By the same orthonormality argument, each Bell-basis measurement outcome prepares the identically named Bell state in channels 1 and 4, withBefore communication of the outcome, Alice and Bob's density matrix is $I_4/4$; the usable conditional entanglement is established by the announced Bell-basis measurement result. This assumes an ideal complete Bell-basis measurement; ordinary passive linear optics alone does not deterministically distinguish all four polarization Bell states.

An ideal projective measurement of the field's photon number selects the coefficient of $|1⟩$. ThereforeWhenever this probability is nonzero, normalization of the projected joint quantum state gives the atomic Bell stateup to a common global phase. The maximum heralding probability is $1/3$, attained at $\sqrt{6}gt=\pi /2+n\pi$. At a time with $P_1=0$, that conditioning event is impossible, so a conditional state is not defined there. The interpretation is a photon number measurement at the selected time; continuous monitoring of escaping photons would instead change the dynamics and require a different model.

The five DiVincenzo criteria are: a scalable collection of well-characterized qubits; preparation in a known fiducial quantum state; relevant coherence times long compared with quantum gate times; a universal quantum gate set; and qubit-specific quantum measurement. The criterion is useful control of the whole computation, including errors, rather than merely observing two resolved levels.^[ref]

For trapped ions, the implementation is:

For trapped neutral atoms, the corresponding implementation is:

Thus neither platform is forbidden from meeting a criterion in principle. Weak neutral-ground-state interactions make entangling control harder, but claiming that trapped neutral atoms cannot implement a universal quantum gate set would be false. Gate-model quantum algorithms using Rydberg blockade had already been demonstrated in a programmable neutral-atom processor by 2022.^[ref]

Assume $|\alpha {|}^2+|\beta {|}^2=1$. Let Alice's first recorded quantum measurement outcome be $a\in \{0,1\}$. Immediately afterward she assigns the conditional pure quantum state $|a⟩$, with probabilities $p_0=|\alpha {|}^2$, $p_1=|\beta {|}^2$. Bob, who has no record, assigns the density operatorThis is a nonselective quantum measurement, which removes the initial computational basis coherence. It is not generally the original pure quantum state.

After the Hadamard gate, Alice assigns $H|a⟩=|+⟩$ for $a=0$, or $|-⟩$ for $a=1$. Bob assignsBoth $|+⟩$ and $|-⟩$ give either final computational basis outcome with probability $1/2$. Thus, after Alice records the second outcome $b$, she assigns $|b⟩$, whereas Bob assignsAlice's conditional quantum states are pure; Bob's are averages over her unobserved records. They agree on all probabilities when supplied the same information. Their first descriptions can coincide if the initial quantum state is a computational basis state; after the second quantum measurement, Bob's maximally mixed state differs from each conditional pure quantum state.

Expand the possibly unnormalized second-qubit kets as $|\alpha ⟩=a_0|0⟩+a_1|1⟩$ and $|\beta ⟩=b_0|0⟩+b_1|1⟩$. The joint quantum state isThe Born rule in the orthonormal computational basis gives $P_{00}=|a_0{|}^2$, $P_{01}=|a_1{|}^2$, and similarly for the two outcomes beginning with one. Ignoring the second outcome therefore givesEquivalently these are the squared norms of $(|0⟩⟨0|\otimes I)|\psi ⟩$ and $(|1⟩⟨1|\otimes I)|\psi ⟩$. Orthogonality of the first qubit's basis gives $P_0+P_1=⟨\psi |\psi ⟩=1$, even if $⟨\alpha |\beta ⟩$ is nonzero. For a nonzero-probability outcome, the conditional second-qubit quantum state is the corresponding ket divided by $\sqrt{P_i}$.

Read the source network as a Hadamard test: a Hadamard gate on the first qubit, a controlled unitary gate with the first qubit as control, and another Hadamard gate on that control. Starting from $|0⟩|\varphi ⟩$, the successive quantum states areApply b to the last line. Since $U^{†}U=I$ by unitarity,For the Pauli Z gate observable on the first qubit, the two eigenvalues are $+1$ and $-1$. ThusAs a check, $U=I$ returns outcome zero certainly, while $U=-I$ returns one certainly. The controlled unitary gate makes this relative phase measurable even though multiplying an isolated system's quantum state by $-1$ is only a global phase.

Yes. Replace the target qubit by a register of $k$ qubits and implement the controlled unitary gateThe algebra in c uses only normalization and unitarity, so for an arbitrary pure register quantum state $|\varphi ⟩$ it still gives $⟨Z⟩=\mathrm{Re}⟨\varphi |U|\varphi ⟩$. For a mixed register quantum state $\rho$, linearity gives $\mathrm{Re}\mathrm{Tr}(\rho U)$. With $\rho =I/2^k$,Only the control qubit needs to be pure. A uniformly randomized computational basis preparation supplies the target maximally mixed state. This is the principle of trace estimation by a maximally mixed Hadamard test. The extension is algebraic; efficiently implementing controlled-$U$ still requires a known suitable gate construction. Moreover, estimating the unnormalized trace to fixed additive precision requires increasingly precise estimation of the normalized signal as $k$ grows. There is no automatic efficient reconstruction of all entries of $U$.

Let $P$ be the orthogonal projector onto the quantum code $\mathcal{X}$, and choose an orthonormal basis $|i_L⟩$ for it. The necessary and sufficient Knill--Laflamme condition is that one matrix $C$, independent of the logical labels, satisfiesEquivalently, $⟨i_L|E_a^{†}{E}_b|j_L⟩=C_{ab}{\delta }_{ij}$. This makes the information in the error label independent of the encoded quantum state. It applies to the whole linear span of the errors, and hence to every noise quantum channel with Kraus operators in that span.

Here is a proof of both directions. For necessity of the Knill-Laflamme condition, let $R_{\mu }$ be Kraus operators of a common recovery, so ${\sum }_{\mu }{R}_{\mu }^{†}{R}_{\mu }=I$. Since recovery after $E_a$ must return the pure input state, every vector $R_{\mu }{E}_a|\psi ⟩$ is parallel to $|\psi ⟩$. A linear operator on the code which sends every vector to a multiple of itself is scalar: apply it to two basis vectors and their sum to force the two multipliers to agree. Therefore $R_{\mu }{E}_{a}P=c_{\mu a}P$ with no dependence on $|\psi ⟩$. It follows thatThis proves necessity without assuming in advance that the physical errors have orthogonal syndromes.

For sufficiency, $C$ is positive semidefinite: for any normalized code state and coefficients $v_a$, $v^{†}Cv=\Vert {\sum }_a{v}_a{E}_a|\psi ⟩{\Vert }^2\ge 0$. Diagonalize $C$ by a unitary change of the error basis to get operators $F_j$ with $P{F}_j^{†}{F}_{l}P=c_j{\delta }_{jl}P$. If $c_j>0$, $V_j=F_{j}P/\sqrt{c_j}$ is a linear isometry from the code to an error-image subspace, and distinct such images are orthogonal. If $c_j=0$, then $F_{j}P=0$. A projective measurement of these image subspaces followed by $V_j^{†}$ recovers every logical state. Complete the recovery on the unused orthogonal complement in any trace-preserving manner. For a coherent error ${\sum }_j{v}_j{F}_j$, the measurement branch $j$ contains $v_j\sqrt{c_j}|\psi ⟩$; the branch probability is independent of the logical state, and discarding the syndrome leaves the original state. This is constructive recovery from the Knill-Laflamme condition.

For the usual family of distinct unitary Pauli errors modulo scalar phase, nondegenerate quantum code correction means that different physical errors have orthogonal image subspaces. The precise condition isFor nonunitary errors, replace the unit diagonal by positive norm factors. In an arbitrary error basis the Gram matrix need not initially be diagonal; its positive-eigenvalue modes give the independent syndrome spaces just constructed. A kernel of $C$ exhibits a linear combination of errors acting as zero on the code. In particular two distinct physical errors with the same action on the code cannot be distinguished and give degenerate correction, as in part (c).

Use the normalized three-qubit states $|G_{\pm }⟩=(|000⟩\pm |111⟩)/\sqrt{2}$. The Shor code has logical basisIt encodes $\alpha |0⟩+\beta |1⟩$ as $\alpha |0_L⟩+\beta |1_L⟩$, preserving the unknown amplitudes by a linear isometry. It concatenates a three-block phase-flip repetition code with the three-qubit bit-flip repetition code inside each block. Six stabilizer generators, $Z_1{Z}_2,Z_2{Z}_3,Z_4{Z}_5,Z_5{Z}_6,Z_7{Z}_8,Z_8{Z}_9$, test bit errors inside the blocks. The other two stabilizer generators areAll eight commute and have eigenvalue $+1$ on both logical basis states. They define the $[[9,1,3]]$ stabilizer code. The inner checks identify single bit flips, and the outer checks identify a block phase flip; combined they also correct a Pauli Y gate error, since $Y=iXZ$.

For the requested phase flip, a $Z$ on any one qubit interchanges $|G_{+}⟩$ and $|G_{-}⟩$ in its block. It commutes with the six $Z$-pair checks. It anticommutes with $A$ or $B$ exactly when its block is contained in that check. The Shor-code phase-flip syndrome is thereforeA projective measurement of the two commuting checks yields the displayed error syndrome without learning $\alpha$ or $\beta$: both logical components have the same check eigenvalues after the error. Applying the indicated Pauli Z gate restores every encoded quantum state.

The correction is genuinely degenerate. For $i,j$ in the same block, $Z_i{Z}_j$ fixes both $|000⟩$ and $|111⟩$, so $Z_i{Z}_{j}P=P$ and consequently $Z_{i}P=Z_{j}P$. In particular $Z_1$ and $Z_2$ are distinct physical errors but have identical action on every code state; their image subspaces coincide, not merely their observed syndrome labels. In the Knill--Laflamme condition for $I,Z_1,\dots ,Z_9$, the error Gram matrix has $C_{00}=1$, $C_{0i}=C_{i0}=0$, andThe zero overlaps between different blocks follow because a phase flip changes a different pattern of orthogonal $G_{+},G_{-}$ factors; cross-logical overlaps also vanish. Thus $C$ consists of one $1\times 1$ block and three all-ones $3\times 3$ blocks and has rank four, not ten. This both proves correctability of the entire single-phase-error span and exhibits its degeneracy. A coherent sum of phase errors in one block has the same logical action with its coefficients summed, so inability to locate the individual qubit loses no logical information. The two block checks suffice to correct every single phase flip, even though the within-block errors are indistinguishable.

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.

For a group action of $G$ on $X$, the orbit of a group action through $x$ is $Gx=\{gx:g\in G\}$, and its stabilizer subgroup is $H_x=\{g:gx=x\}$. A transitive group action has one orbit. A simply transitive group action has exactly one element carrying any given point to any other; equivalently it is transitive and every stabilizer subgroup is trivial.

In the modern terminology, a multiply transitive group action is transitive on ordered $k$-tuples of distinct points for some $k\ge 2$. In older geometric usage, the phrase can instead mean transitive but not simply transitive, so that there are several elements carrying one point to another. Under either interpretation the action is transitive, and that alone is what the following argument needs. These meanings should not be confused: a transitive action with nontrivial stabilizers need not be two-transitive.

Fix $x_0\in X$ and put $H=H_{x_0}$. DefineThis is well-defined because every element of $H$ fixes $x_0$. It is surjective by transitivity, and $g{x}_0=g^{\prime }{x}_0$ is equivalent to $g^{\prime -1}g\in H$, proving injectivity. For a smooth Lie group action, $H$ is closed, since it is the inverse image of $x_0$ under the orbit map. The orbit map has constant rank: translations in $G$ and the action diffeomorphisms of $X$ intertwine its differentials. Transitivity makes this rank $\dim X$; otherwise every point would be a critical value, contradicting Sard theorem. Its local submersion charts give smooth local sections and hence a smooth inverse for $F$. ThereforeChoosing $x_1=a{x}_0$ replaces $H$ by $aH{a}^{-1}$. Thus the stabilizer subgroup is determined by the base point, and intrinsically only up to conjugacy. A simply transitive group action gives $H=\{e\}$.

For a finite quantum system with Hilbert space $\mathcal{H}={\mathbb{C}}^{n+1}$, a pure state is a nonzero vector modulo multiplication by a nonzero complex number. Equivalently, it is a unit vector modulo its physically irrelevant phase. Consequently the space of pure states is the space of complex lines,The qualification “pure” is essential: arbitrary quantum states include mixed density matrices and do not form Complex projective space.

The unitary group $U(n+1)$ acts transitively on these lines: choose unit vectors in two given lines, complete each to an orthonormal basis, and map one basis to the other by a unitary transformation. The stabilizer of the line $\mathbb{C}{e}_0$ consists exactly of block matrices $\mathrm{diag}(z,V)$ with $z\in U(1)$ and $V\in U(n)$. Indeed a unitary transformation preserving the line also preserves its orthogonal complement. The homogeneous space description is thereforeIts real dimension is $(n+1{)}^2-1-n^2=2n$. This action is generally not two-transitive: the absolute inner product between unit representatives of two lines is invariant.

The Hopf fibration is the projection $\pi :S^{2n+1}\to {\mathbb{CP}}^n$, $z\mapsto [z]$. The right action $z\cdot \lambda =z\lambda$, $\lambda \in S^1$, is free, and its orbits are exactly the fibres. On the chart $U_i$ where $z_i\ne 0$, write $w_j=z_j/z_i$ for $j\ne i$. A smooth local section isEvery unit vector over this chart has the unique expression $z={\sigma }_i([z])\lambda$, where $\lambda =z_i/|z_i|$. These formulas provide local trivializations ${\pi }^{-1}(U_i)\cong U_i\times S^1$, proving that this is a principal bundle with fibre $S^1$. For $n\ge 1$ it has no global section: a global section would give $S^{2n+1}\cong {\mathbb{CP}}^n\times S^1$, contradicting their fundamental groups. For $n=0$ the base is a point and the bundle is trivial.