Single-Qubit Gates

An ideal single-qubit gate is a unitary operator

$$U\in U(2), \qquad U^\dagger U=UU^\dagger=I,$$

acting on a chosen two-dimensional computational basis

$$\left\{ \lvert0\rangle,\lvert1\rangle \right\}.$$

It changes amplitudes, relative phases, or both. On the Bloch sphere, every one-qubit unitary acts as a proper three-dimensional rotation. Conversely, every proper rotation of the Bloch sphere is represented by two special-unitary matrices, $U$ and $-U$, and by infinitely many $U(2)$ matrices if arbitrary global phases are retained.

The matrices below connect common gate names to rotations, phase changes and basis measurements. They describe ideal operations; the final section explains how a control Hamiltonian approximates them.

Helpful background. The Bloch sphere describes one-qubit states and measurement axes; Pauli matrices supplies the algebra; the circuit model fixes how gates are composed.

Unitary qubit gates and phase conventions

Basis and action

Unless stated otherwise, matrices below use the ordered computational basis

$$\lvert0\rangle= \begin{pmatrix} 1\\ 0 \end{pmatrix}, \qquad \lvert1\rangle= \begin{pmatrix} 0\\ 1 \end{pmatrix},$$

and act on state columns from the left. If a diagram applies $G_1$ and then $G_2$, its matrix is

$$U_{\mathrm{total}}=G_2G_1.$$

The rightmost factor acts first. A change of computational basis changes every displayed matrix by conjugation, so a gate name without a basis convention is incomplete.

Three notions of equality

For gate calculations, distinguish:

1. Exact matrix equality: $U=V$.
2. Equality up to global phase: $U=e^{i\gamma}V$.
3. Equality as an isolated quantum channel:

$$\mathcal U(\rho) = U\rho U^\dagger = V\rho V^\dagger = \mathcal V(\rho).$$

The second implies the third. It does not permit phases to be dropped blindly inside a controlled operation, an interferometer, or any larger construction in which the allegedly global phase becomes branch dependent.

A general element of $U(2)$ has four real parameters. Factoring out $e^{i\gamma}$ leaves three physically relevant parameters for an isolated one-qubit gate. Those three parameters specify a rotation axis and angle, or equivalently three Euler angles.

Pauli Gates

The Pauli gates are

$$\begin{gathered} X= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}, \\[4pt] Y= \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}, \\[4pt] Z= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}. \end{gathered}$$

Their basis-state actions are

$$\begin{array}{c|cc} &\lvert0\rangle&\lvert1\rangle\\ \hline X&\lvert1\rangle&\lvert0\rangle\\ Y&i\lvert1\rangle&-i\lvert0\rangle\\ Z&\lvert0\rangle&-\lvert1\rangle \end{array}.$$

$X$ exchanges computational-basis populations. $Z$ preserves those populations and reverses the sign of the $\lvert1\rangle$ amplitude. $Y$ combines both effects with phases fixed by the convention

$$Y= \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}.$$

The informal names “bit flip,” “phase flip,” and “bit-and-phase flip” are useful mnemonics, but they refer to the chosen computational basis. In another basis, the same operator has a different matrix and operational description.

Each Pauli gate is Hermitian, unitary, traceless, and involutory:

$$X^2=Y^2=Z^2=I.$$

Their products retain noncommutative phase information:

$$XY=iZ, \qquad YZ=iX, \qquad ZX=iY,$$

whereas reversing either factor changes the sign. Pauli matrices derives the algebra and its relation to spin.

Pauli gates as half turns

Define a rotation about a unit vector $\hat{\mathbf n}$ by

$$R_{\hat{\mathbf n}}(\theta) = \exp\left( -\frac{i\theta}{2} \hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

Then

$$\begin{aligned} R_x(\pi)&=-iX, \\ R_y(\pi)&=-iY, \\ R_z(\pi)&=-iZ. \end{aligned}$$

Thus a Pauli gate is a Bloch-sphere rotation by $\pi$ about its named axis, up to the common phase $i$. On a Bloch vector $\mathbf r=(r_x,r_y,r_z)$,

$$\begin{aligned} X:\ &(r_x,r_y,r_z) \longmapsto (r_x,-r_y,-r_z), \\ Y:\ &(r_x,r_y,r_z) \longmapsto (-r_x,r_y,-r_z), \\ Z:\ &(r_x,r_y,r_z) \longmapsto (-r_x,-r_y,r_z). \end{aligned}$$

Hadamard Gate

The Hadamard gate is

$$H= \frac{1}{\sqrt2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}.$$

It exchanges the computational and $X$ eigenbases:

$$\begin{aligned} H\lvert0\rangle&=\lvert+\rangle, & H\lvert1\rangle&=\lvert-\rangle, \\ H\lvert+\rangle&=\lvert0\rangle, & H\lvert-\rangle&=\lvert1\rangle. \end{aligned}$$

This is why $H$ appears before a computational-basis measurement when one wants to measure in the $X$ basis. It is better understood as a basis-changing unitary than as a device that “creates randomness.” Applied to $\lvert0\rangle$, it creates a coherent superposition. Randomness appears only when that state is measured in an incompatible basis.

Conjugation by $H$ exchanges $X$ and $Z$ while reversing $Y$:

$$\begin{aligned} HXH&=Z, \\ HZH&=X, \\ HYH&=-Y. \end{aligned}$$

Consequently, its Bloch action is

$$(r_x,r_y,r_z) \longmapsto (r_z,-r_y,r_x).$$

Geometrically, $H$ is a rotation by $\pi$ about

$$\hat{\mathbf n}_H = \frac{\hat{\mathbf x}+\hat{\mathbf z}}{\sqrt2}$$

up to phase:

$$H=iR_{\hat{\mathbf n}_H}(\pi).$$

The exact Hadamard matrix has determinant $-1$ and lies in $U(2)$ rather than $SU(2)$; $-iH$ and $iH$ are its two special-unitary representatives up to phase.

Phase, S, and T Gates

The continuous phase-gate family is

$$P(\phi) = \begin{pmatrix} 1&0\\ 0&e^{i\phi} \end{pmatrix}.$$

It leaves computational-basis probabilities unchanged but changes the relative phase:

$$P(\phi) \left( a\lvert0\rangle+b\lvert1\rangle \right) = a\lvert0\rangle +e^{i\phi}b\lvert1\rangle.$$

The common discrete gates are

$$\begin{aligned} S &= P\left(\frac{\pi}{2}\right) = \begin{pmatrix} 1&0\\ 0&i \end{pmatrix}, \\ T &= P\left(\frac{\pi}{4}\right) = \begin{pmatrix} 1&0\\ 0&e^{i\pi/4} \end{pmatrix}. \end{aligned}$$

They satisfy

$$\begin{aligned} S^2&=Z, & T^2&=S, \\ T^4&=Z, & T^8&=I. \end{aligned}$$

$T$ is often called the π/8 gate because, up to global phase,

$$T = e^{i\pi/8} \exp\left(-\frac{i\pi Z}{8}\right).$$

Its relative phase is $\pi/4$, while the exponent contains $\pi/8$. Stating the matrix removes the naming ambiguity.

Phase gates versus z rotations

The $z$ rotation is

$$R_z(\phi) = \begin{pmatrix} e^{-i\phi/2}&0\\ 0&e^{i\phi/2} \end{pmatrix}.$$

Therefore

$$P(\phi)=e^{i\phi/2}R_z(\phi).$$

For an isolated qubit, $P(\phi)$ and $R_z(\phi)$ produce the same density-operator transformation and the same Bloch rotation. They are not the same matrix. The distinction matters when the operation is controlled and the matrix phase becomes relative between branches.

$H$, $S$, and the Pauli gates belong to the one-qubit Clifford group: they map Pauli operators to Pauli operators under conjugation. $T$ does not. Universal Gate Sets develops the computational consequences of that distinction; it should not be confused with a claim that one physical $T$ pulse is intrinsically more difficult on every hardware platform.

Continuous Rotation Gates

For $j\in\{x,y,z\}$,

$$\begin{aligned} R_j(\theta) &= \exp\left( -\frac{i\theta}{2}\sigma_j \right) \\ &= \cos\frac{\theta}{2}\,I -i\sin\frac{\theta}{2}\,\sigma_j. \end{aligned}$$

With the convention above,

$$R_x(\theta) = \begin{pmatrix} \cos\frac{\theta}{2} & -i\sin\frac{\theta}{2} \\ -i\sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix},$$ $$R_y(\theta) = \begin{pmatrix} \cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix},$$ $$R_z(\theta) = \begin{pmatrix} e^{-i\theta/2}&0\\ 0&e^{i\theta/2} \end{pmatrix}.$$

Positive $\theta$ gives the active right-hand-rule rotation of the Bloch vector when states transform as $\rho\mapsto R_j\rho R_j^\dagger$. A source using $e^{+i\theta\sigma_j/2}$, a passive coordinate rotation, or a different $Y$ convention will reverse signs. Never compare pulse angles without comparing definitions.

For a general axis,

$$\hat{\mathbf n} = (n_x,n_y,n_z), \qquad \lVert\hat{\mathbf n}\rVert=1,$$

the matrix is

$$\begin{gathered} R_{\hat{\mathbf n}}(\theta) = \\[2pt] \begin{pmatrix} c-in_zs & (-in_x-n_y)s \\ (-in_x+n_y)s & c+in_zs \end{pmatrix}, \end{gathered}$$

where

$$c=\cos\frac{\theta}{2}, \qquad s=\sin\frac{\theta}{2}.$$

This formula is useful for checking signs, but the exponential form better exposes the generator and composition structure.

Bloch-Sphere Action

Write

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right).$$

Under a unitary $U$,

$$\rho' = U\rho U^\dagger = \frac12 \left[ I+(R_U\mathbf r)\cdot\boldsymbol\sigma \right],$$

where $R_U\in SO(3)$. One convenient definition is

$$(R_U)_{ij} = \frac12 \operatorname{Tr} \left( \sigma_iU\sigma_jU^\dagger \right).$$

This representation makes several facts immediate:

• unitaries preserve $\lVert\mathbf r\rVert$, purity, and angles between Bloch vectors;
• no isolated one-qubit unitary can turn a mixed state into a pure state;
• $U$ and $e^{i\gamma}U$ give the same $R_U$;
• after removing a global phase from a general $U(2)$ matrix, the resulting special-unitary matrix $V$ and $-V$ represent the same spatial rotation;
• gate composition maps to rotation composition in the same operator order.

The Bloch sphere is exact for one qubit, but it does not scale into a complete picture of a multi-qubit state. A local one-qubit gate acting on one part of an entangled state still has a $2\times2$ matrix on that subsystem, yet the joint state lives in a tensor-product space. Local unitaries preserve bipartite entanglement measures even while they change local measurement statistics and correlation axes.

Global Phase Can Become Relative Phase

State-vector phase

The state vectors

$$\lvert\psi\rangle \quad\text{and}\quad e^{i\gamma}\lvert\psi\rangle$$

represent the same ray. By contrast,

$$\frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2} \quad\text{and}\quad \frac{ \lvert0\rangle+e^{i\gamma}\lvert1\rangle }{\sqrt2}$$

are generally different rays because only one branch was rephased. Their relative phase changes interference and equatorial measurement probabilities.

Gate phase

Let

$$V=e^{i\gamma}U.$$

Then the isolated channels agree, $\mathcal V=\mathcal U$. Now define a controlled operation with the control as the first tensor factor:

$$C(U) = \lvert0\rangle\langle0\rvert\otimes I + \lvert1\rangle\langle1\rvert\otimes U.$$

Then

$$\begin{aligned} C(e^{i\gamma}U) &= \left[ P(\gamma)\otimes I \right] C(U), \end{aligned}$$

so the two controlled gates differ by a phase gate on the control. This is why replacing $P(\phi)$ by $R_z(\phi)$ is harmless for a standalone target but not automatically harmless after adding a control. Exact phase conventions are part of a reusable gate definition.

Composition and Useful Identities

The most reliable way to simplify a short gate sequence is to choose one representation and keep the order explicit:

• multiply exact matrices when phase matters;
• conjugate Pauli operators for basis-change questions;
• compose $SO(3)$ rotations for Bloch-vector questions;
• act on basis states for simple circuit semantics.

For example, $SH\lvert0\rangle$ means $H$ acts first:

$$SH\lvert0\rangle = S\lvert+\rangle = \frac{ \lvert0\rangle+i\lvert1\rangle }{\sqrt2} = \lvert+i\rangle.$$

But

$$HS\lvert0\rangle = H\lvert0\rangle = \lvert+\rangle.$$

The two sequences differ because $H$ and $S$ do not commute.

Useful inverse and power relations include

$$\begin{gathered} H^\dagger=H, H^2=I, \\ S^\dagger=P\left(-\frac{\pi}{2}\right), \\ T^\dagger=P\left(-\frac{\pi}{4}\right), \\ R_j(\theta)^\dagger = R_j(-\theta), \\ R_j(\alpha)R_j(\beta) = R_j(\alpha+\beta). \end{gathered}$$

The last identity holds without a Baker–Campbell–Hausdorff correction because both rotations have the same generator. Rotations about different axes generally do not commute.

Basis Changes for Preparation and Readout

A basis change can move complexity from the measurement to a preceding gate. Suppose hardware measures $Z$ after a unitary $B$. The effective observable on the input is

$$B^\dagger ZB.$$

With the conventions on this page:

Desired input basis	Gates before $Z$ readout	Total pre-rotation $B$	Effective observable
$Z$	none	$I$	$Z$
$X$	$H$	$H$	$X$
$Y$	$S^\dagger$, then $H$	$HS^\dagger$	$Y$

For the last row,

$$\begin{aligned} B^\dagger ZB &= SHZH S^\dagger \\ &= SXS^\dagger = Y. \end{aligned}$$

This table specifies the temporal order as well as the matrix product. Writing only “use $H$ and $S^\dagger$” is ambiguous because the gates do not commute.

The reverse viewpoint prepares cardinal states. Starting from $\lvert0\rangle$,

$$\begin{aligned} H\lvert0\rangle&=\lvert+\rangle, \\ SH\lvert0\rangle&=\lvert+i\rangle, \\ X\lvert0\rangle&=\lvert1\rangle. \end{aligned}$$

Preparation and readout circuits are related by adjoints, but a real experiment may assign different errors to preparation pulses, frame updates, and measurement. Gate identities alone do not certify state-preparation-and-measurement performance.

Arbitrary One-Qubit Gates

Euler decomposition

Every one-qubit unitary can be written

$$U = e^{i\alpha} R_z(\beta) R_y(\gamma) R_z(\delta).$$

To extract angles, first choose $\alpha$ so that $e^{2i\alpha}=\det U$. Then

$$e^{-i\alpha}U=\begin{pmatrix}u&-v^*\\v&u^*\end{pmatrix}, \qquad |u|^2+|v|^2=1.$$

When $u,v\ne0$, one choice is

$$\begin{aligned} \gamma&=2\operatorname{atan2}(|v|,|u|),\\ \beta&=\arg v-\arg u,\\ \delta&=-\arg v-\arg u. \end{aligned}$$

Indeed, the product has $u=e^{-i(\beta+\delta)/2}\cos(\gamma/2)$ and $v=e^{i(\beta-\delta)/2}\sin(\gamma/2)$. If $v=0$, only $\beta+\delta$ matters; if $u=0$, only $\beta-\delta$ matters. Set one redundant angle to zero rather than taking the argument of a zero matrix entry.

The angles are not unique at coordinate singularities or after periodic shifts. This is a coordinate decomposition, not a claim that a device literally executes three pulses in that form. A compiler may cancel neighboring rotations, absorb $z$ rotations into a frame, or choose another native basis. Gate Decomposition uses Euler factors as one-qubit leaves inside controlled, Cartan/KAK, dense-unitary, and finite-alphabet synthesis.

The decomposition also gives a direct state-preparation rule. For

$$\lvert\psi\rangle = \cos\frac{\theta}{2}\lvert0\rangle + e^{i\phi} \sin\frac{\theta}{2}\lvert1\rangle,$$

one has

$$R_z(\phi)R_y(\theta)\lvert0\rangle = e^{-i\phi/2} \lvert\psi\rangle.$$

Thus two rotations prepare any pure qubit state from $\lvert0\rangle$ up to global phase. A general gate needs a third rotation because it must specify the action on an entire basis, not only on one input ray.

Software parameterizations

Circuit languages often expose a three-angle universal gate, but names, angle order, signs, and global phases are not universal. The OpenQASM 3 specification, for example, defines a built-in three-parameter $U$ together with an explicit global-phase operation; compatibility gates inherited from earlier versions use related but phase-shifted conventions.

Before translating a matrix into software:

1. read the versioned gate definition;
2. check whether the equality is exact or only up to phase;
3. verify basis and qubit ordering;
4. test the action on both $\lvert0\rangle$ and $\lvert1\rangle$;
5. retain the documented phase when the gate may later be controlled.

From Ideal Matrix to Physical Control

An ideal gate symbol names a target unitary. A physical device realizes time evolution generated by a control Hamiltonian, often modeled in an appropriate rotating frame as

$$H_{\mathrm c}(t) = \frac{\hbar}{2} \left[ \Omega_x(t)X + \Omega_y(t)Y + \Omega_z(t)Z \right].$$

The resulting propagator is

$$U(t_f,t_i) = \mathcal T \exp\left[ -\frac{i}{\hbar} \int_{t_i}^{t_f} H_{\mathrm c}(t)\,dt \right].$$

For a constant direction $\hat{\mathbf n}$ and pulse area

$$\theta = \int_{t_i}^{t_f} \Omega(t)\,dt,$$

this reduces to $R_{\hat{\mathbf n}}(\theta)$ in the ideal rotating-frame model. Detuning, leakage, crosstalk, finite bandwidth, decoherence, and calibration drift turn the implemented operation into a channel $\mathcal E$ that only approximates the target.

A named $Z$ gate need not correspond to a dedicated physical pulse. In many driven-qubit platforms it can be implemented as a software frame update that changes the phases of later drives. Such a virtual Z gate can be effectively instantaneous in the control schedule, but it still changes the interpretation of subsequent axes and must be tracked consistently. Other platforms may implement the same logical matrix by different controls.

Keep four layers separate:

Layer	Object	Question
Logical	ideal matrix $U$	What transformation is intended?
Compiled	sequence in a native gate set	How is $U$ decomposed for this target?
Scheduled	pulses, frame updates, and timing	What controls are issued and when?
Implemented	noisy channel $\mathcal E$	What transformation did the device realize?

Rabi and Ramsey control develops the driven dynamics; pulse-level control discusses calibrated frames and waveforms; noise in quantum information describes implementation errors.

Comparing two gate matrices

For two ideal matrices $U,V\in U(2)$, a quick test for equality up to phase is

$$\frac12 \left| \operatorname{Tr}(U^\dagger V) \right| = 1.$$

This criterion assumes both matrices are unitary. It says their isolated channels agree; it does not establish that two noisy implementations have the same error process.

Exercises

1. Pauli action with phases

For

$$\lvert\psi\rangle = a\lvert0\rangle+b\lvert1\rangle,$$

compute $X\lvert\psi\rangle$, $Y\lvert\psi\rangle$, and $Z\lvert\psi\rangle$. Verify $XY\lvert\psi\rangle=iZ\lvert\psi\rangle$.

Solution

Direct matrix action gives

$$\begin{aligned} X\lvert\psi\rangle &= b\lvert0\rangle+a\lvert1\rangle, \\ Y\lvert\psi\rangle &= -ib\lvert0\rangle+ia\lvert1\rangle, \\ Z\lvert\psi\rangle &= a\lvert0\rangle-b\lvert1\rangle. \end{aligned}$$

Applying $Y$ first and then $X$,

$$XY\lvert\psi\rangle = ia\lvert0\rangle-ib\lvert1\rangle = iZ\lvert\psi\rangle.$$

The phase $i$ is part of the exact operator identity.

2. Phase gate versus z rotation

Show that $P(\phi)=e^{i\phi/2}R_z(\phi)$. Then show that their controlled versions differ by $P(\phi/2)$ on the control qubit.

Solution

Multiplying $R_z(\phi)$ by $e^{i\phi/2}$ gives

$$e^{i\phi/2}R_z(\phi) = \begin{pmatrix} 1&0\\ 0&e^{i\phi} \end{pmatrix} = P(\phi).$$

Using

$$C(e^{i\gamma}U) = \left[ P(\gamma)\otimes I \right] C(U),$$

with $\gamma=\phi/2$ and $U=R_z(\phi)$ gives

$$C(P(\phi)) = \left[ P(\phi/2)\otimes I \right] C(R_z(\phi)).$$

The factor is a relative phase between the control branches and cannot generally be discarded.

3. Gate order on a cardinal state

Starting from $\lvert0\rangle$, compare $SH\lvert0\rangle$ with $HS\lvert0\rangle$. Give both final Bloch vectors.

Solution

Since $H\lvert0\rangle=\lvert+\rangle$,

$$SH\lvert0\rangle = \frac{ \lvert0\rangle+i\lvert1\rangle }{\sqrt2} = \lvert+i\rangle,$$

whose Bloch vector is $(0,1,0)$. On the other hand, $S\lvert0\rangle=\lvert0\rangle$, so

$$HS\lvert0\rangle = H\lvert0\rangle = \lvert+\rangle,$$

whose Bloch vector is $(1,0,0)$. The rightmost gate acts first.

4. Prepare an arbitrary pure qubit

Prove that

$$R_z(\phi)R_y(\theta)\lvert0\rangle$$

prepares the Bloch-sphere state with polar angle $\theta$ and azimuth $\phi$, up to global phase.

Solution

First,

$$R_y(\theta)\lvert0\rangle = \cos\frac{\theta}{2}\lvert0\rangle + \sin\frac{\theta}{2}\lvert1\rangle.$$

Applying $R_z(\phi)$ gives

$$\begin{aligned} R_z(\phi)R_y(\theta)\lvert0\rangle &= e^{-i\phi/2} \cos\frac{\theta}{2}\lvert0\rangle + e^{i\phi/2} \sin\frac{\theta}{2}\lvert1\rangle\\ &= e^{-i\phi/2} \left( \cos\frac{\theta}{2}\lvert0\rangle + e^{i\phi} \sin\frac{\theta}{2}\lvert1\rangle \right). \end{aligned}$$

The prefactor is global. The remaining state has the required polar and azimuthal angles.

5. Hadamard conjugation

Verify $HXH=Z$ and $HZH=X$ by matrix multiplication. Use the result to explain why applying $H$, measuring $Z$, and forgetting the final state realizes an $X$-basis measurement of the input.

Solution

Direct multiplication gives

$$HXH = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} = Z,$$

and

$$HZH = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} = X.$$

The probability of outcome $z=\pm1$ after applying $H$ is

$$\operatorname{Tr} \left[ \rho\, H^\dagger \frac{I+zZ}{2} H \right] = \operatorname{Tr} \left[ \rho\, \frac{I+zX}{2} \right].$$

These are exactly the $X$-measurement probabilities of the original state.

6. Pulse area and sign check

Suppose a rotating-frame control Hamiltonian is constant:

$$H_{\mathrm c} = \frac{\hbar\Omega}{2}X$$

for a duration $\tau$, with real nonzero $\Omega$. Find the ideal gate and the earliest positive duration that produces $X$ up to global phase. How does that phase depend on the sign of $\Omega$?

Solution

The propagator is

$$U(\tau) = \exp\left( -\frac{i\Omega\tau}{2}X \right) = R_x(\Omega\tau).$$

Choose $\tau=\pi/|\Omega|$. Then the pulse area is $\operatorname{sgn}(\Omega)\pi$, and

$$U(\tau)=-i\operatorname{sgn}(\Omega)X.$$

For positive $\Omega$ the phase is $-i$; for negative $\Omega$ it is $+i$. Both give the isolated channel of $X$. If a coherent controlled implementation is later constructed, its relative branch phase must follow the actual controlled Hamiltonian, including terms omitted as global phases in the isolated-qubit model.

References

• A. Barenco, C. H. Bennett, R. Cleve, D. P. DiVincenzo, N. Margolus, P. Shor, T. Sleator, J. A. Smolin, and H. Weinfurter, “Elementary gates for quantum computation,” Physical Review A 52, 3457–3467, 1995, doi:10.1103/PhysRevA.52.3457.
• D. C. McKay, C. J. Wood, S. Sheldon, J. M. Chow, and J. M. Gambetta, “Efficient $Z$ gates for quantum computing,” Physical Review A 96, 022330, 2017, doi:10.1103/PhysRevA.96.022330.
• N. D. Mermin, Quantum Computer Science: An Introduction, Cambridge University Press, 2007, doi:10.1017/CBO9780511813870.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, doi:10.1017/CBO9780511976667.
• OpenQASM contributors, OpenQASM 3 Specification, “Gates” and “Standard library,” versioned specification.
• J. Preskill, Lecture Notes for Physics 219/Computer Science 219: Quantum Computation, Chapter 5, California Institute of Technology, course materials.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, doi:10.1017/9781108587280.