Controlled Gates: CNOT, CZ, SWAP

Controlled-U in block form#

\lvert a,b\rangle=\lvert a\rangle\otimes\lvert b\rangle

Definition

Controlled-U gate

U

Derivation

Block form of $C\text{-}U$

Step 1:
$P_0=\lvert0\rangle\langle0\rvert=\begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad P_1=\lvert1\rangle\langle1\rvert=\begin{pmatrix}0\\1\end{pmatrix}\begin{pmatrix}0&1\end{pmatrix}=\begin{pmatrix}0&0\\0&1\end{pmatrix}$
An outer product $\lvert u\rangle\langle v\rvert$ is the column $\lvert u\rangle$ times the row $\langle v\rvert$ .
Step 2:
$A\otimes B=\begin{pmatrix}a_{00}B&a_{01}B\\a_{10}B&a_{11}B\end{pmatrix}\ \Rightarrow\ P_0\otimes I=\begin{pmatrix}I&0\\0&0\end{pmatrix},\qquad P_1\otimes U=\begin{pmatrix}0&0\\0&U\end{pmatrix}$
The Kronecker product places a copy of $B$ , scaled by $a_{jk}$ , in block $(j,k)$ ; here each block is $2\times2$ and $0$ denotes the $2\times2$ zero block.
Step 3:
$C\text{-}U=\begin{pmatrix}I&0\\0&U\end{pmatrix}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&u_{00}&u_{01}\\0&0&u_{10}&u_{11}\end{pmatrix}$
Add the two block matrices. With qubit 0 as the most significant bit, the upper-left block acts on $\lvert00\rangle,\lvert01\rangle$ (control $0$ ) and the lower-right block on $\lvert10\rangle,\lvert11\rangle$ (control $1$ ).
Step 4:
$C\text{-}U\,\lvert0\rangle\lvert\psi\rangle=P_0\lvert0\rangle\otimes\lvert\psi\rangle+P_1\lvert0\rangle\otimes U\lvert\psi\rangle=\lvert0\rangle\lvert\psi\rangle+0=\lvert0\rangle\lvert\psi\rangle$
Apply $(A\otimes B)(\lvert u\rangle\otimes\lvert v\rangle)=A\lvert u\rangle\otimes B\lvert v\rangle$ to each term, with $P_0\lvert0\rangle=\lvert0\rangle$ and $P_1\lvert0\rangle=\lvert1\rangle\langle1\vert0\rangle=0$ .
Step 5:
$C\text{-}U\,\lvert1\rangle\lvert\psi\rangle=P_0\lvert1\rangle\otimes\lvert\psi\rangle+P_1\lvert1\rangle\otimes U\lvert\psi\rangle=0+\lvert1\rangle\otimes U\lvert\psi\rangle$
Now $P_0\lvert1\rangle=0$ and $P_1\lvert1\rangle=\lvert1\rangle$ : the target receives $U$ exactly when the control is $\lvert1\rangle$ . Compactly, $C\text{-}U\lvert a\rangle\lvert\psi\rangle=\lvert a\rangle U^a\lvert\psi\rangle$ for $a\in\{0,1\}$ .

Result

\boxed{C\text{-}U=P_0\otimes I+P_1\otimes U=\begin{pmatrix}I&0\\0&U\end{pmatrix},\qquad C\text{-}U\lvert a\rangle\lvert\psi\rangle=\lvert a\rangle\,U^a\lvert\psi\rangle}

Proposition

$C\text{-}U$ is unitary

U

Proof (7 steps)

Step 1:
$(C\text{-}U)^\dagger=P_0^\dagger\otimes I^\dagger+P_1^\dagger\otimes U^\dagger=P_0\otimes I+P_1\otimes U^\dagger$
The adjoint is additive and satisfies $(A\otimes B)^\dagger=A^\dagger\otimes B^\dagger$ ; the projectors $P_j=\lvert j\rangle\langle j\rvert$ and $I$ are Hermitian.
Step 2:
$P_jP_k=\lvert j\rangle\langle j\vert k\rangle\langle k\rvert=\delta_{jk}\lvert j\rangle\langle k\rvert=\delta_{jk}P_j,\qquad P_0+P_1=I$
Orthonormality $\langle j\vert k\rangle=\delta_{jk}$ of the computational basis gives $P_0^2=P_0$ , $P_1^2=P_1$ , $P_0P_1=P_1P_0=0$ ; completeness gives $P_0+P_1=I$ .
Step 3:
$(C\text{-}U)^\dagger(C\text{-}U)=(P_0\otimes I+P_1\otimes U^\dagger)(P_0\otimes I+P_1\otimes U)$
Substitute the adjoint from the first line.
Step 4:
$=P_0P_0\otimes I\,I+P_0P_1\otimes I\,U+P_1P_0\otimes U^\dagger I+P_1P_1\otimes U^\dagger U$
Expand the product of two sums into four terms and apply the mixed-product rule $(A\otimes B)(C\otimes D)=AC\otimes BD$ to each.
Step 5:
$=P_0\otimes I+0+0+P_1\otimes I$
$P_0P_0=P_0$ , $P_0P_1=P_1P_0=0$ , $P_1P_1=P_1$ , and $U^\dagger U=I$ because $U$ is unitary. The cross terms vanish because the two branches act on orthogonal control states.
Step 6:
$=(P_0+P_1)\otimes I=I\otimes I=I_4$
The tensor product is bilinear, and $P_0+P_1=I$ .
Step 7:
$(C\text{-}U)(C\text{-}U)^\dagger=P_0\otimes I+P_1\otimes UU^\dagger=I_4,\qquad(C\text{-}U)^\dagger=P_0\otimes I+P_1\otimes U^\dagger=C\text{-}(U^\dagger)$
The same expansion in the opposite order uses $UU^\dagger=I$ . Both products are the identity, so $C\text{-}U$ is unitary and its inverse is the controlled- $U^\dagger$ .

∎

\mathrm{CNOT}=\begin{pmatrix}I&0\\0&X\end{pmatrix}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}

CNOT = C-X (control 0, target 1)4×4

Unitary verified: $U^\dagger U = I$ (max deviation 0.0e+0)

\mathrm{CZ}=\begin{pmatrix}I&0\\0&Z\end{pmatrix}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-1\end{pmatrix}

CZ = C-Z4×4

Unitary verified: $U^\dagger U = I$ (max deviation 0.0e+0)

Derivation

$\mathrm{CNOT}$ and $\mathrm{CZ}$ on basis states

Step 1:
$X^a\lvert b\rangle=\lvert a\oplus b\rangle,\qquad a,b\in\{0,1\}$
For $a=0$ , $X^0=I$ and $0\oplus b=b$ ; for $a=1$ , $X\lvert0\rangle=\lvert1\rangle$ and $X\lvert1\rangle=\lvert0\rangle$ , i.e. $X\lvert b\rangle=\lvert1\oplus b\rangle$ . Here $\oplus$ is addition modulo $2$ (XOR).
Step 2:
$\mathrm{CNOT}\lvert a,b\rangle=\lvert a\rangle\,X^a\lvert b\rangle=\lvert a,\,a\oplus b\rangle$
Use $C\text{-}U\lvert a\rangle\lvert\psi\rangle=\lvert a\rangle U^a\lvert\psi\rangle$ with $U=X$ : the target bit is XORed with the control bit.
Step 3:
$\lvert00\rangle\mapsto\lvert00\rangle,\quad\lvert01\rangle\mapsto\lvert01\rangle,\quad\lvert10\rangle\mapsto\lvert11\rangle,\quad\lvert11\rangle\mapsto\lvert10\rangle$
The four cases of the previous line. Column $k$ of the matrix is the image of basis state $k$ , which reproduces the permutation matrix shown above.
Step 4:
$Z^a\lvert b\rangle=(-1)^{ab}\lvert b\rangle$
$Z\lvert b\rangle=(-1)^b\lvert b\rangle$ , so $Z^a\lvert b\rangle=(-1)^{ab}\lvert b\rangle$ for $a\in\{0,1\}$ .
Step 5:
$\mathrm{CZ}\lvert a,b\rangle=\lvert a\rangle\,Z^a\lvert b\rangle=(-1)^{ab}\lvert a,b\rangle$
The same rule with $U=Z$ : only $\lvert11\rangle$ , where $ab=1$ , acquires the sign $-1$ , which is the diagonal $(1,1,1,-1)$ .

Result

\boxed{\mathrm{CNOT}\lvert a,b\rangle=\lvert a,a\oplus b\rangle,\qquad\mathrm{CZ}\lvert a,b\rangle=(-1)^{ab}\lvert a,b\rangle}

Circuit · 2 qubits

\lvert10\rangle

\lvert\psi\rangle = 1\lvert 11\rangle

H

Derivation

The state after each operation of $H_0$ , $\mathrm{CNOT}_{01}$

Step 1:
$k=0:\quad\lvert00\rangle\ \leftrightarrow\ (1,0,0,0)^{\mathsf T}$
Every qubit starts in $\lvert0\rangle$ ; amplitudes are listed in the order $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$ .
Step 2:
$k=1:\quad(H\otimes I)\lvert00\rangle=\tfrac1{\sqrt2}\big(\lvert0\rangle+\lvert1\rangle\big)\otimes\lvert0\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert10\rangle\big)\ \leftrightarrow\ \big(\tfrac1{\sqrt2},0,\tfrac1{\sqrt2},0\big)^{\mathsf T}$
$H$ acts on qubit 0, the most significant bit, so the weight moves to indices $0$ and $2$ .
Step 3:
$k=2:\quad\mathrm{CNOT}\,\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert10\rangle\big)=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)\ \leftrightarrow\ \big(\tfrac1{\sqrt2},0,0,\tfrac1{\sqrt2}\big)^{\mathsf T}$
Linearity with $\mathrm{CNOT}\lvert00\rangle=\lvert00\rangle$ and $\mathrm{CNOT}\lvert10\rangle=\lvert11\rangle$ : the branch with control $1$ flips its target, the other is left alone.
Step 4:
$(a\lvert0\rangle+b\lvert1\rangle)\otimes(c\lvert0\rangle+d\lvert1\rangle)=ac\lvert00\rangle+ad\lvert01\rangle+bc\lvert10\rangle+bd\lvert11\rangle$
A product state would need $ad=bc=0$ while $ac=bd=\tfrac1{\sqrt2}$ ; but $ad=0$ forces $a=0$ (then $ac=0$ ) or $d=0$ (then $bd=0$ ). The output is therefore not a product state: it is entangled, the Bell state $\lvert\Phi^+\rangle$ of the next lesson.

Result

\boxed{\lvert00\rangle\ \xmapsto{H_0}\ \tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert10\rangle\big)\ \xmapsto{\mathrm{CNOT}_{01}}\ \tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)}

Statevector viewStatevectorView

Lists every basis amplitude of the state with its magnitude, phase and Born probability.

Initial state |00⟩ (no op applied yet)

\lvert\psi_{0}\rangle = \lvert 00\rangle

2 qubits · 4 amplitudes · 1 non-zero · ‖ψ‖² = 1.000000

Show phase

Phase arg α:0π/2π−π/2

Statevector amplitudes
Basis	Amplitude α	\|α\|	Phase arg α	Probability
\|00⟩	1	1.0000	0	100.00%
\|01⟩	0	0.0000	—	0.00%
\|10⟩	0	0.0000	—	0.00%
\|11⟩	0	0.0000	—	0.00%

Quick check

\mathrm{CNOT}_{01}\lvert11\rangle

CZ is symmetric; CNOT from CZ#

Proposition

$\mathrm{CZ}$ does not distinguish control from target

\mathrm{CZ}=P_0\otimes I+P_1\otimes Z=I\otimes P_0+Z\otimes P_1=\mathrm{diag}(1,1,1,-1)

Proof (5 steps)

Step 1:
$P_0\otimes I=\mathrm{diag}(1,1,0,0),\qquad P_1\otimes Z=\mathrm{diag}(0,0,1,-1)$
Kronecker products of diagonal matrices are diagonal: $\mathrm{diag}(p_0,p_1)\otimes\mathrm{diag}(q_0,q_1)=\mathrm{diag}(p_0q_0,p_0q_1,p_1q_0,p_1q_1)$ , with $P_0=\mathrm{diag}(1,0)$ , $P_1=\mathrm{diag}(0,1)$ , $Z=\mathrm{diag}(1,-1)$ .
Step 2:
$P_0\otimes I+P_1\otimes Z=\mathrm{diag}(1,1,1,-1)$
Add the two diagonals entry by entry.
Step 3:
$I\otimes P_0=\mathrm{diag}(1,0,1,0),\qquad Z\otimes P_1=\mathrm{diag}(0,1,0,-1)$
The same rule with the factors in the other order: $\mathrm{diag}(1,1)\otimes\mathrm{diag}(1,0)$ and $\mathrm{diag}(1,-1)\otimes\mathrm{diag}(0,1)$ .
Step 4:
$I\otimes P_0+Z\otimes P_1=\mathrm{diag}(1,1,1,-1)=\mathrm{CZ}$
Add the diagonals. The operator $I\otimes P_0+Z\otimes P_1$ is, by definition, the controlled- $Z$ with control on qubit 1 and target on qubit 0, so both orientations give the same matrix.
Step 5:
$\mathrm{SWAP}\,\mathrm{CZ}\,\mathrm{SWAP}\lvert a,b\rangle=\mathrm{SWAP}\,\mathrm{CZ}\lvert b,a\rangle=(-1)^{ba}\,\mathrm{SWAP}\lvert b,a\rangle=(-1)^{ab}\lvert a,b\rangle=\mathrm{CZ}\lvert a,b\rangle$
Equivalent formulation with basis labels: $\mathrm{SWAP}\lvert a,b\rangle=\lvert b,a\rangle$ , $\mathrm{CZ}\lvert a,b\rangle=(-1)^{ab}\lvert a,b\rangle$ , and $ba=ab$ . The two operators agree on a basis, hence everywhere by linearity.

∎

\mathrm{CZ}

Matrix displayMatrixDisplay

Renders a complex matrix entry by entry, with its Hermitian and unitary properties checked numerically.

3 decimals

Exact constantsBasis labels

\begin{array}{rrc} & & \begin{matrix*}[r] \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 00\rangle}} \\ \hphantom{1} \\ \hphantom{0} \\ \hphantom{0} \\ \hphantom{0}\end{matrix}}\mathllap{{\scriptstyle \lvert 00\rangle}} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 01\rangle}} \\ \hphantom{0} \\ \hphantom{1} \\ \hphantom{0} \\ \hphantom{0}\end{matrix}}\mathllap{{\scriptstyle \lvert 01\rangle}} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 10\rangle}} \\ \hphantom{0} \\ \hphantom{0} \\ \hphantom{1} \\ \hphantom{0}\end{matrix}}\mathllap{{\scriptstyle \lvert 10\rangle}} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 11\rangle}} \\ \hphantom{0} \\ \hphantom{0} \\ \hphantom{0} \\ \hphantom{-1}\end{matrix}}\mathllap{{\scriptstyle \lvert 11\rangle}} \end{matrix*} \\ \mathrm{CZ} = & \begin{matrix} \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}{\scriptstyle \lvert 00\rangle} \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}{\scriptstyle \lvert 01\rangle} \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}{\scriptstyle \lvert 10\rangle} \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}{\scriptstyle \lvert 11\rangle} \end{matrix} & \begin{pmatrix*}[r] \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}\smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 00\rangle}} \\ \hphantom{1}\end{matrix}}\mathllap{1} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 01\rangle}} \\ \hphantom{0}\end{matrix}}\mathllap{0} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 10\rangle}} \\ \hphantom{0}\end{matrix}}\mathllap{0} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 11\rangle}} \\ \hphantom{0}\end{matrix}}\mathllap{0} \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}0 & 1 & 0 & 0 \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}0 & 0 & 1 & 0 \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}0 & 0 & 0 & -1 \end{pmatrix*} \end{array}

Row i, column j holds ⟨i|A|j⟩, so column j is the image of the input |j⟩. Basis order |00⟩, |01⟩, |10⟩, |11⟩ (qubit 0 = most significant bit).

RowColumn

\langle 00\rvert\, \mathrm{CZ} \,\lvert 00\rangle = 1

|·| = 1.000, arg = 0

Properties (tolerance 1e-9)

Hermitian A = A†: yes max|A − A†| = 0.0e+0
Unitary A†A = I: yes max|A†A − I| = 0.0e+0
Normal AA† = A†A: yes max|AA† − A†A| = 0.0e+0
Trace: $\operatorname{Tr} \mathrm{CZ} = 2.000$
Determinant: $\det \mathrm{CZ} = -1$
Eigenvalues: $\lambda = -1,\; 1,\; 1,\; 1$

Derivation

$\mathrm{CNOT}=(I\otimes H)\,\mathrm{CZ}\,(I\otimes H)$ , block by block

Step 1:
$I\otimes H=\begin{pmatrix}H&0\\0&H\end{pmatrix}=\frac1{\sqrt2}\begin{pmatrix}1&1&0&0\\1&-1&0&0\\0&0&1&1\\0&0&1&-1\end{pmatrix}$
The Kronecker product $I\otimes H$ has $1\cdot H$ in both diagonal blocks and $0\cdot H$ off the diagonal.
Step 2:
$(I\otimes H)\,\mathrm{CZ}\,(I\otimes H)=\begin{pmatrix}H&0\\0&H\end{pmatrix}\begin{pmatrix}I&0\\0&Z\end{pmatrix}\begin{pmatrix}H&0\\0&H\end{pmatrix}=\begin{pmatrix}HIH&0\\0&HZH\end{pmatrix}$
Block-diagonal matrices multiply block by block: the $(0,0)$ block is $H\cdot I\cdot H$ and the $(1,1)$ block is $H\cdot Z\cdot H$ ; every off-diagonal block contains a factor $0$ .
Step 3:
$HIH=H^2=I,\qquad HZH=X$
Both identities were proved in the Hadamard lesson: $H^2=I$ follows from $\{X,Z\}=0$ , and $HZH=\tfrac12(X+Z)Z(X+Z)=\tfrac12(I-iY)(X+Z)=X$ .
Step 4:
$(I\otimes H)\,\mathrm{CZ}\,(I\otimes H)=\begin{pmatrix}I&0\\0&X\end{pmatrix}=\mathrm{CNOT}$
Insert the two blocks; this is the block form of $C\text{-}X$ .
Step 5:
$\mathrm{CZ}\,(I\otimes H)=\frac1{\sqrt2}\begin{pmatrix}1&1&0&0\\1&-1&0&0\\0&0&1&1\\0&0&-1&1\end{pmatrix},\qquad\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\cdot\frac1{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}=\frac12\begin{pmatrix}0&2\\2&0\end{pmatrix}=X$
Entry-level check of the lower-right block: left multiplication by the diagonal $\mathrm{CZ}$ negates the last row of $I\otimes H$ ; then $H$ times the block $ZH=\tfrac1{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}$ gives $\tfrac12\begin{pmatrix}1-1&1+1\\1+1&1-1\end{pmatrix}=X$ . The upper-left block is $H\cdot H=I$ .
Step 6:
$(I\otimes H)(P_0\otimes I+P_1\otimes Z)(I\otimes H)=P_0\otimes HIH+P_1\otimes HZH=P_0\otimes I+P_1\otimes X$
The same computation in projector form, using the mixed-product rule: conjugating the target of a controlled gate conjugates $U$ , $C\text{-}U\mapsto C\text{-}(VUV^\dagger)$ for $V=H=H^\dagger$ .
Step 7:
$(I\otimes H)\,\mathrm{CNOT}\,(I\otimes H)=(I\otimes H)^2\,\mathrm{CZ}\,(I\otimes H)^2=\mathrm{CZ}$
Conversely, multiply both sides of $\mathrm{CNOT}=(I\otimes H)\mathrm{CZ}(I\otimes H)$ by $I\otimes H$ on the left and right, and use $(I\otimes H)^2=I\otimes H^2=I_4$ .

Result

\boxed{\mathrm{CNOT}=(I\otimes H)\,\mathrm{CZ}\,(I\otimes H),\qquad\mathrm{CZ}=(I\otimes H)\,\mathrm{CNOT}\,(I\otimes H)}

Circuit · 2 qubits

H

U = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}

Unitary heatmapUnitaryHeatmap

Magnitude/phase heatmap of the full unitary matrix implemented by a circuit.

Compare up to a global phase

U

T

|U − T|

Opacity = |U_ij| (1 = opaque), hue = arg U_ij (red 0, yellow-green π/2, cyan π, violet −π/2). Rows and columns are labelled |q₀q₁…⟩ with qubit 0 the most significant bit. Click a cell for its exact value.

U=\begin{pmatrix} 1.0000 & 0.0000 & 0.0000 & 0.0000 \\ 0.0000 & 1.0000 & 0.0000 & 0.0000 \\ 0.0000 & 0.0000 & 0.0000 & 1.0000 \\ 0.0000 & 0.0000 & 1.0000 & 0.0000 \end{pmatrix}

U = T exactly: the circuit implements the target.

target from matrix; process fidelity |Tr(T†U)|/d = 1.000000; global phase φ = arg Tr(T†U) = 0.000000; max |U − T| = 2.22e-16 (tolerance 1e-9)

F=\frac{\lvert\mathrm{Tr}(T^\dagger U)\rvert}{4}=1.000000,\qquad \mathrm{Tr}(T^\dagger U)=4.0000+0.0000i

Quick check

\mathrm{CZ}

Exchanging control and target#

\mathrm{CNOT}_{10}=I\otimes P_0+X\otimes P_1

\mathrm{CNOT}_{10}=I\otimes P_0+X\otimes P_1=\begin{pmatrix}1&0&0&0\\0&0&0&1\\0&0&1&0\\0&1&0&0\end{pmatrix}

CNOT10 (control 1, target 0)4×4

Unitary verified: $U^\dagger U = I$ (max deviation 0.0e+0)

Proposition

Reversing a CNOT with four Hadamards

(H\otimes H)\,\mathrm{CNOT}_{01}\,(H\otimes H)=\mathrm{CNOT}_{10}

Proof (6 steps)

Step 1:
$H\otimes H=(H\otimes I)(I\otimes H)=(I\otimes H)(H\otimes I)$
Mixed-product rule: $(H\otimes I)(I\otimes H)=HI\otimes IH=H\otimes H$ , and likewise in the other order. Gates on different qubits commute.
Step 2:
$(H\otimes H)\,\mathrm{CNOT}_{01}\,(H\otimes H)=(H\otimes I)(I\otimes H)\,(I\otimes H)\,\mathrm{CZ}\,(I\otimes H)\,(I\otimes H)(H\otimes I)$
Substitute $\mathrm{CNOT}_{01}=(I\otimes H)\mathrm{CZ}(I\otimes H)$ from the previous section, and split the outer $H\otimes H$ as in the first line (left factor $(H\otimes I)(I\otimes H)$ , right factor $(I\otimes H)(H\otimes I)$ ).
Step 3:
$=(H\otimes I)\,\mathrm{CZ}\,(H\otimes I)$
$(I\otimes H)(I\otimes H)=I\otimes H^2=I_4$ on both sides of $\mathrm{CZ}$ .
Step 4:
$=(H\otimes I)(I\otimes P_0+Z\otimes P_1)(H\otimes I)=HIH\otimes P_0+HZH\otimes P_1$
Use the symmetric form $\mathrm{CZ}=I\otimes P_0+Z\otimes P_1$ proved above, then the mixed-product rule on each term.
Step 5:
$=I\otimes P_0+X\otimes P_1=\mathrm{CNOT}_{10}$
$H^2=I$ and $HZH=X$ . The result applies $X$ to qubit 0 exactly when qubit 1 is $\lvert1\rangle$ .
Step 6:
$(H\otimes H)\,\mathrm{CNOT}_{01}\,(H\otimes H)\,\lvert01\rangle=\mathrm{CNOT}_{10}\lvert01\rangle=\lvert0\oplus1,\,1\rangle=\lvert11\rangle$
Spot check on a basis state with $a=0$ , $b=1$ : the target qubit 0 is flipped because the control qubit 1 is set, matching the second column of the $\mathrm{CNOT}_{10}$ matrix.

∎

Circuit · 2 qubits

H\otimes H

U = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}

\mathrm{CNOT}_{10}

Matrix displayMatrixDisplay

Renders a complex matrix entry by entry, with its Hermitian and unitary properties checked numerically.

3 decimals

Exact constantsBasis labels

\begin{array}{rrc} & & \begin{matrix*}[r] \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 00\rangle}} \\ \hphantom{1} \\ \hphantom{0} \\ \hphantom{0} \\ \hphantom{0}\end{matrix}}\mathllap{{\scriptstyle \lvert 00\rangle}} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 01\rangle}} \\ \hphantom{0} \\ \hphantom{0} \\ \hphantom{0} \\ \hphantom{1}\end{matrix}}\mathllap{{\scriptstyle \lvert 01\rangle}} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 10\rangle}} \\ \hphantom{0} \\ \hphantom{0} \\ \hphantom{1} \\ \hphantom{0}\end{matrix}}\mathllap{{\scriptstyle \lvert 10\rangle}} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 11\rangle}} \\ \hphantom{0} \\ \hphantom{1} \\ \hphantom{0} \\ \hphantom{0}\end{matrix}}\mathllap{{\scriptstyle \lvert 11\rangle}} \end{matrix*} \\ \mathrm{CNOT}_{10} = & \begin{matrix} \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}{\scriptstyle \lvert 00\rangle} \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}{\scriptstyle \lvert 01\rangle} \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}{\scriptstyle \lvert 10\rangle} \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}{\scriptstyle \lvert 11\rangle} \end{matrix} & \begin{pmatrix*}[r] \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}\smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 00\rangle}} \\ \hphantom{1}\end{matrix}}\mathllap{1} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 01\rangle}} \\ \hphantom{0}\end{matrix}}\mathllap{0} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 10\rangle}} \\ \hphantom{0}\end{matrix}}\mathllap{0} & \smash{\begin{matrix}\hphantom{{\scriptstyle \lvert 11\rangle}} \\ \hphantom{0}\end{matrix}}\mathllap{0} \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}0 & 0 & 0 & 1 \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}0 & 0 & 1 & 0 \\ \vphantom{\tfrac{1}{\sqrt{2}}e^{-3i\pi/4}}0 & 1 & 0 & 0 \end{pmatrix*} \end{array}

Row i, column j holds ⟨i|A|j⟩, so column j is the image of the input |j⟩. Basis order |00⟩, |01⟩, |10⟩, |11⟩ (qubit 0 = most significant bit).

RowColumn

\langle 00\rvert\, \mathrm{CNOT}_{10} \,\lvert 00\rangle = 1

|·| = 1.000, arg = 0

Properties (tolerance 1e-9)

Hermitian A = A†: yes max|A − A†| = 0.0e+0
Unitary A†A = I: yes max|A†A − I| = 0.0e+0
Normal AA† = A†A: yes max|AA† − A†A| = 0.0e+0
Trace: $\operatorname{Tr} \mathrm{CNOT}_{10} = 2.000$
Determinant: $\det \mathrm{CNOT}_{10} = -1$
Eigenvalues: $\lambda = -1,\; 1,\; 1,\; 1$

SWAP from three CNOTs#

Definition

SWAP gate

\mathrm{SWAP}\lvert a,b\rangle=\lvert b,a\rangle

\mathrm{SWAP}=\lvert00\rangle\langle00\rvert+\lvert01\rangle\langle10\rvert+\lvert10\rangle\langle01\rvert+\lvert11\rangle\langle11\rvert=\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}

SWAP4×4

Unitary verified: $U^\dagger U = I$ (max deviation 0.0e+0)

Theorem

$\mathrm{SWAP}=\mathrm{CNOT}_{01}\,\mathrm{CNOT}_{10}\,\mathrm{CNOT}_{01}$

\mathrm{CNOT}_{01}

Proof (7 steps)

Step 1:
$x\oplus x=0,\qquad x\oplus0=x,\qquad x\oplus y=y\oplus x,\qquad(x\oplus y)\oplus z=x\oplus(y\oplus z)$
Addition modulo $2$ on $\{0,1\}$ is commutative and associative, $0$ is neutral, and every bit is its own inverse.
Step 2:
$\mathrm{CNOT}_{01}\lvert a,b\rangle=\lvert a,\,a\oplus b\rangle$
First gate: control qubit 0, target qubit 1.
Step 3:
$\mathrm{CNOT}_{10}\lvert a,\,a\oplus b\rangle=\lvert a\oplus(a\oplus b),\,a\oplus b\rangle=\lvert b,\,a\oplus b\rangle$
Second gate: control qubit 1 (value $a\oplus b$ ), target qubit 0, so qubit 0 becomes $(a\oplus b)\oplus a$ ; by associativity and $a\oplus a=0$ this is $0\oplus b=b$ .
Step 4:
$\mathrm{CNOT}_{01}\lvert b,\,a\oplus b\rangle=\lvert b,\,b\oplus(a\oplus b)\rangle=\lvert b,a\rangle$
Third gate: control qubit 0 (now $b$ ), target qubit 1, which becomes $b\oplus a\oplus b=a\oplus(b\oplus b)=a$ .
Step 5:
$M\lvert a,b\rangle=\lvert b,a\rangle=\mathrm{SWAP}\lvert a,b\rangle\quad\text{for all }a,b\in\{0,1\}$
The operator for the wire order $\mathrm{CNOT}_{01};\mathrm{CNOT}_{10};\mathrm{CNOT}_{01}$ is the product $M=\mathrm{CNOT}_{01}\mathrm{CNOT}_{10}\mathrm{CNOT}_{01}$ (the first gate acts first); the three lines above evaluate it on every basis state.
Step 6:
$M\lvert\psi\rangle=M\textstyle\sum_{a,b}c_{ab}\lvert a,b\rangle=\sum_{a,b}c_{ab}M\lvert a,b\rangle=\sum_{a,b}c_{ab}\lvert b,a\rangle=\mathrm{SWAP}\lvert\psi\rangle$
Every two-qubit state is a linear combination $\lvert\psi\rangle=\sum_{a,b}c_{ab}\lvert a,b\rangle$ ; both $M$ and $\mathrm{SWAP}$ are linear and agree on the basis, so they agree everywhere: $M=\mathrm{SWAP}$ .
Step 7:
$\mathrm{SWAP}(\lvert\varphi\rangle\otimes\lvert\chi\rangle)=\textstyle\sum_{a,b}\varphi_a\chi_b\lvert b,a\rangle=\Big(\sum_b\chi_b\lvert b\rangle\Big)\otimes\Big(\sum_a\varphi_a\lvert a\rangle\Big)=\lvert\chi\rangle\otimes\lvert\varphi\rangle$
For a product state $\lvert\varphi\rangle\otimes\lvert\chi\rangle=\sum_{a,b}\varphi_a\chi_b\lvert a,b\rangle$ , regrouping the sum by bilinearity of $\otimes$ shows that superpositions, not only basis labels, are exchanged.

∎

Derivation

The three CNOTs on the four basis states

Step 1:
$\lvert00\rangle\xmapsto{\mathrm{CNOT}_{01}}\lvert00\rangle\xmapsto{\mathrm{CNOT}_{10}}\lvert00\rangle\xmapsto{\mathrm{CNOT}_{01}}\lvert00\rangle$
Every control is $0$ , so nothing happens.
Step 2:
$\lvert01\rangle\xmapsto{\mathrm{CNOT}_{01}}\lvert01\rangle\xmapsto{\mathrm{CNOT}_{10}}\lvert11\rangle\xmapsto{\mathrm{CNOT}_{01}}\lvert10\rangle$
The first control (qubit 0) is $0$ ; then qubit 1 is $1$ and flips qubit 0; then qubit 0 is $1$ and flips qubit 1.
Step 3:
$\lvert10\rangle\xmapsto{\mathrm{CNOT}_{01}}\lvert11\rangle\xmapsto{\mathrm{CNOT}_{10}}\lvert01\rangle\xmapsto{\mathrm{CNOT}_{01}}\lvert01\rangle$
Qubit 0 is $1$ and flips qubit 1; then qubit 1 is $1$ and flips qubit 0 back to $0$ ; finally qubit 0 is $0$ , so the last CNOT does nothing.
Step 4:
$\lvert11\rangle\xmapsto{\mathrm{CNOT}_{01}}\lvert10\rangle\xmapsto{\mathrm{CNOT}_{10}}\lvert10\rangle\xmapsto{\mathrm{CNOT}_{01}}\lvert11\rangle$
Qubit 1 is flipped to $0$ ; then the control qubit 1 is $0$ ; finally qubit 0 is $1$ and flips qubit 1 back to $1$ .

Result

\boxed{\lvert00\rangle\mapsto\lvert00\rangle,\quad\lvert01\rangle\mapsto\lvert10\rangle,\quad\lvert10\rangle\mapsto\lvert01\rangle,\quad\lvert11\rangle\mapsto\lvert11\rangle}

Circuit · 2 qubits

\lvert01\rangle

U = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}

Circuit · 2 qubits

\lvert10\rangle

\lvert\psi\rangle = 1\lvert 01\rangle

Circuit builderCircuitBuilder

Grid of qubit wires on which gates are placed column by column; the circuit is simulated live after every edit.

Activate a gate in the palette, then a cell to place it, or drag a gate onto the grid. Activate a placed gate to select it.

Statevector amplitudes
Basis	Amplitude α	\|α\|	Phase arg α	Probability
\|00⟩	1	1.0000	0	100.00%

3 basis states with zero amplitude hidden

Measurement probabilities

Exact P(k) = |⟨k|ψ⟩|²

Quick check

\mathrm{CNOT}_{01}

Phase kickback#

Proposition

Phase kickback

x\in\{0,1\}

Proof (6 steps)

Step 1:
$X\lvert-\rangle=\tfrac1{\sqrt2}\big(X\lvert0\rangle-X\lvert1\rangle\big)=\tfrac1{\sqrt2}\big(\lvert1\rangle-\lvert0\rangle\big)=-\lvert-\rangle$
$X$ is linear and exchanges $\lvert0\rangle$ and $\lvert1\rangle$ ; so $\lvert-\rangle$ is an eigenvector of $X$ with eigenvalue $-1$ .
Step 2:
$\mathrm{CNOT}\big(\lvert x\rangle\otimes\lvert-\rangle\big)=\lvert x\rangle\otimes X^x\lvert-\rangle$
Block-form rule $C\text{-}U\lvert a\rangle\lvert\psi\rangle=\lvert a\rangle U^a\lvert\psi\rangle$ with $U=X$ and $\lvert\psi\rangle=\lvert-\rangle$ .
Step 3:
$X^x\lvert-\rangle=(-1)^x\lvert-\rangle$
For $x=0$ , $X^0=I$ ; for $x=1$ , the first line.
Step 4:
$\mathrm{CNOT}\lvert x\rangle\lvert-\rangle=\lvert x\rangle\otimes(-1)^x\lvert-\rangle=(-1)^x\lvert x\rangle\lvert-\rangle$
A scalar can be moved across the tensor product: $u\otimes(cv)=c(u\otimes v)=(cu)\otimes v$ . The sign can be read as belonging to the control.
Step 5:
$\mathrm{CNOT}\lvert1\rangle\lvert-\rangle=\tfrac1{\sqrt2}\big(\mathrm{CNOT}\lvert10\rangle-\mathrm{CNOT}\lvert11\rangle\big)=\tfrac1{\sqrt2}\big(\lvert11\rangle-\lvert10\rangle\big)=-\lvert1\rangle\lvert-\rangle$
Direct check of the case $x=1$ with the basis table $\lvert10\rangle\mapsto\lvert11\rangle$ , $\lvert11\rangle\mapsto\lvert10\rangle$ .
Step 6:
$C\text{-}U\lvert x\rangle\lvert u\rangle=\lvert x\rangle\otimes U^x\lvert u\rangle=\lvert x\rangle\otimes e^{ix\varphi}\lvert u\rangle=e^{ix\varphi}\lvert x\rangle\lvert u\rangle$
The same three steps for any controlled- $U$ and eigenvector $U\lvert u\rangle=e^{i\varphi}\lvert u\rangle$ , using $U^x\lvert u\rangle=e^{ix\varphi}\lvert u\rangle$ . The CNOT case is $U=X$ , $\lvert u\rangle=\lvert-\rangle$ , $e^{i\varphi}=-1$ .

∎

Derivation

$\mathrm{CNOT}\lvert+\rangle\lvert-\rangle=\lvert-\rangle\lvert-\rangle$

Step 1:
$\lvert+\rangle\lvert-\rangle=\tfrac1{\sqrt2}(\lvert0\rangle+\lvert1\rangle)\otimes\tfrac1{\sqrt2}(\lvert0\rangle-\lvert1\rangle)=\tfrac12\big(\lvert00\rangle-\lvert01\rangle+\lvert10\rangle-\lvert11\rangle\big)$
Expand the tensor product by bilinearity; the amplitude vector is $\tfrac12(1,-1,1,-1)^T$ .
Step 2:
$\mathrm{CNOT}\lvert+\rangle\lvert-\rangle=\tfrac12\big(\lvert00\rangle-\lvert01\rangle+\lvert11\rangle-\lvert10\rangle\big)$
Apply $\mathrm{CNOT}$ term by term: $\lvert00\rangle$ and $\lvert01\rangle$ are unchanged, $\lvert10\rangle\mapsto\lvert11\rangle$ and $\lvert11\rangle\mapsto\lvert10\rangle$ .
Step 3:
$=\tfrac12\big(\lvert00\rangle-\lvert01\rangle-\lvert10\rangle+\lvert11\rangle\big)=\tfrac1{\sqrt2}(\lvert0\rangle-\lvert1\rangle)\otimes\tfrac1{\sqrt2}(\lvert0\rangle-\lvert1\rangle)=\lvert-\rangle\lvert-\rangle$
Reorder the terms by index; the coefficients $\tfrac12(1,-1,-1,1)$ factor as $\tfrac1{\sqrt2}(1,-1)\otimes\tfrac1{\sqrt2}(1,-1)$ .
Step 4:
$\mathrm{CNOT}\lvert+\rangle\lvert-\rangle=\tfrac1{\sqrt2}\big(\mathrm{CNOT}\lvert0\rangle\lvert-\rangle+\mathrm{CNOT}\lvert1\rangle\lvert-\rangle\big)=\tfrac1{\sqrt2}\big(\lvert0\rangle-\lvert1\rangle\big)\lvert-\rangle=\lvert-\rangle\lvert-\rangle$
The same result from the kickback proposition and linearity: the $\lvert1\rangle$ branch picks up $(-1)^1$ .
Step 5:
$\mathrm{CNOT}\big(\lvert+\rangle\otimes\lvert-\rangle\big)=\big(Z\lvert+\rangle\big)\otimes\lvert-\rangle,\qquad Z\lvert+\rangle=\tfrac1{\sqrt2}\big(\lvert0\rangle-\lvert1\rangle\big)=\lvert-\rangle$
The target qubit is untouched, while the control has been flipped in the $X$ basis, exactly as if $Z$ had acted on it. This is the reversed-CNOT theorem seen on one input.

Result

\boxed{\mathrm{CNOT}\lvert+\rangle\lvert-\rangle=\lvert-\rangle\lvert-\rangle=\tfrac12\big(\lvert00\rangle-\lvert01\rangle-\lvert10\rangle+\lvert11\rangle\big)}

Circuit · 2 qubits

H

\lvert\psi\rangle = \tfrac{1}{2}\lvert 00\rangle - \tfrac{1}{2}\lvert 01\rangle - \tfrac{1}{2}\lvert 10\rangle + \tfrac{1}{2}\lvert 11\rangle

Quick check

\mathrm{CNOT}

Any controlled-U from two CNOTs#

U=e^{i\alpha}R_Z(\beta)R_Y(\gamma)R_Z(\delta)

Lemma

$XR_Y(\theta)X=R_Y(-\theta)$ and $XR_Z(\theta)X=R_Z(-\theta)$

\theta

Proof (6 steps)

Step 1:
$XYX=X(YX)=X(-iZ)=-i\,XZ=-i(-iY)=-Y$
Pauli products $YX=-iZ$ and $XZ=-iY$ , and $(-i)(-i)=i^2=-1$ .
Step 2:
$XZX=X(ZX)=X(iY)=i\,XY=i(iZ)=-Z$
Pauli products $ZX=iY$ and $XY=iZ$ , and $i\cdot i=-1$ .
Step 3:
$XR_Y(\theta)X=X\big(\cos\tfrac\theta2\,I-i\sin\tfrac\theta2\,Y\big)X=\cos\tfrac\theta2\,X^2-i\sin\tfrac\theta2\,XYX=\cos\tfrac\theta2\,I+i\sin\tfrac\theta2\,Y$
Conjugation is linear; $X^2=I$ and $XYX=-Y$ from the first line.
Step 4:
$=\cos\big(-\tfrac\theta2\big)I-i\sin\big(-\tfrac\theta2\big)Y=R_Y(-\theta)$
Cosine is even and sine is odd: $\cos(-t)=\cos t$ , $\sin(-t)=-\sin t$ .
Step 5:
$XR_Z(\theta)X=\cos\tfrac\theta2\,I-i\sin\tfrac\theta2\,XZX=\cos\tfrac\theta2\,I+i\sin\tfrac\theta2\,Z=R_Z(-\theta)$
The same two steps with $XZX=-Z$ .
Step 6:
$XR_Z(\theta)X=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}e^{-i\theta/2}&0\\0&e^{i\theta/2}\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}e^{i\theta/2}&0\\0&e^{-i\theta/2}\end{pmatrix}=R_Z(-\theta)$
Matrix check: conjugating a diagonal matrix by $X$ exchanges its two diagonal entries.

∎

Theorem

The $ABC$ decomposition

U=e^{i\alpha}R_Z(\beta)R_Y(\gamma)R_Z(\delta)

Proof (7 steps)

Step 1:
$ABC=R_Z(\beta)R_Y(\tfrac\gamma2)\,R_Y(-\tfrac\gamma2)R_Z(-\tfrac{\delta+\beta}2)\,R_Z(\tfrac{\delta-\beta}2)$
Write out the three definitions in order.
Step 2:
$=R_Z(\beta)\,R_Y(0)\,R_Z\big(-\tfrac{\delta+\beta}2+\tfrac{\delta-\beta}2\big)=R_Z(\beta)\,R_Z(-\beta)$
Adjacent rotations about the same axis add: $\tfrac\gamma2-\tfrac\gamma2=0$ gives $R_Y(0)=I$ , and $-\tfrac{\delta+\beta}2+\tfrac{\delta-\beta}2=\tfrac{-\delta-\beta+\delta-\beta}2=-\beta$ .
Step 3:
$=R_Z(\beta-\beta)=R_Z(0)=I$
Add the two $z$ angles once more.
Step 4:
$XBX=X\,R_Y(-\tfrac\gamma2)\,X\,X\,R_Z(-\tfrac{\delta+\beta}2)\,X=R_Y(\tfrac\gamma2)\,R_Z(\tfrac{\delta+\beta}2)$
Insert $X^2=I$ between the two factors of $B$ , then apply the lemma to each: $XR_Y(-\tfrac\gamma2)X=R_Y(\tfrac\gamma2)$ and $XR_Z(-\tfrac{\delta+\beta}2)X=R_Z(\tfrac{\delta+\beta}2)$ .
Step 5:
$AXBXC=R_Z(\beta)R_Y(\tfrac\gamma2)\cdot R_Y(\tfrac\gamma2)R_Z(\tfrac{\delta+\beta}2)\cdot R_Z(\tfrac{\delta-\beta}2)$
Substitute $A$ , the previous line for $XBX$ , and $C$ .
Step 6:
$=R_Z(\beta)\,R_Y\big(\tfrac\gamma2+\tfrac\gamma2\big)\,R_Z\big(\tfrac{\delta+\beta}2+\tfrac{\delta-\beta}2\big)=R_Z(\beta)R_Y(\gamma)R_Z(\delta)$
Add adjacent same-axis angles: $\tfrac\gamma2+\tfrac\gamma2=\gamma$ and $\tfrac{\delta+\beta+\delta-\beta}2=\delta$ .
Step 7:
$e^{i\alpha}AXBXC=e^{i\alpha}R_Z(\beta)R_Y(\gamma)R_Z(\delta)=U$
Multiply by the scalar $e^{i\alpha}$ and compare with the ZYZ form of $U$ .

∎

Corollary

Controlled- $U$ from two CNOTs

A,B,C

Proof (6 steps)

Step 1:
$W=(P(\alpha)\otimes I)(I\otimes A)\,\mathrm{CNOT}\,(I\otimes B)\,\mathrm{CNOT}\,(I\otimes C)$
The operator is the product of the gates in reverse wire order. $P(\alpha)$ acts on the control only and is diagonal there, so it commutes with both CNOTs and may be placed anywhere on the control wire.
Step 2:
$(I\otimes A)(P_0\otimes I+P_1\otimes X)(I\otimes B)=P_0\otimes AB+P_1\otimes AXB$
Insert $\mathrm{CNOT}=P_0\otimes I+P_1\otimes X$ and use the mixed-product rule on each term: $(I\otimes A)(P_j\otimes X^j)(I\otimes B)=P_j\otimes AX^jB$ .
Step 3:
$(P_0\otimes AB+P_1\otimes AXB)(P_0\otimes I+P_1\otimes X)(I\otimes C)=P_0\otimes ABC+P_1\otimes AXBXC$
Expand the product: the cross terms contain $P_0P_1=P_1P_0=0$ and vanish, while $P_0P_0=P_0$ and $P_1P_1=P_1$ ; the last factor appends $C$ to both branches.
Step 4:
$=P_0\otimes I+P_1\otimes AXBXC$
The theorem gives $ABC=I$ : when the control is $\lvert0\rangle$ the target sees $C$ , then $B$ , then $A$ , whose product is the identity.
Step 5:
$W=(P(\alpha)\otimes I)(P_0\otimes I+P_1\otimes AXBXC)=P_0\otimes I+e^{i\alpha}P_1\otimes AXBXC$
$P(\alpha)P_0=P_0$ and $P(\alpha)P_1=e^{i\alpha}P_1$ , because $P(\alpha)$ is diagonal with entries $1$ and $e^{i\alpha}$ .
Step 6:
$=P_0\otimes I+P_1\otimes U=C\text{-}U$
The theorem gives $e^{i\alpha}AXBXC=U$ . Without $P(\alpha)$ the circuit would implement $C\text{-}(e^{-i\alpha}U)$ , which differs from $C\text{-}U$ by a relative, observable phase between the two control branches.

∎

Derivation

Worked example: controlled- $H$

Step 1:
$H=e^{i\pi/2}R_Y(\tfrac\pi2)R_Z(\pi)=e^{i\pi/2}R_Z(0)R_Y(\tfrac\pi2)R_Z(\pi)\ \Rightarrow\ \alpha=\tfrac\pi2,\ \beta=0,\ \gamma=\tfrac\pi2,\ \delta=\pi$
The Hadamard lesson computed $R_Y(\tfrac\pi2)R_Z(\pi)=-iH$ , so $H=iR_Y(\tfrac\pi2)R_Z(\pi)$ ; inserting $R_Z(0)=I$ puts it in ZYZ form.
Step 2:
$A=R_Z(0)R_Y(\tfrac\pi4)=R_Y(\tfrac\pi4),\qquad B=R_Y(-\tfrac\pi4)R_Z(-\tfrac\pi2),\qquad C=R_Z(\tfrac\pi2)$
Substitute into the definitions: $\tfrac\gamma2=\tfrac\pi4$ , $-\tfrac{\delta+\beta}2=-\tfrac\pi2$ , $\tfrac{\delta-\beta}2=\tfrac\pi2$ .
Step 3:
$ABC=R_Y(\tfrac\pi4)R_Y(-\tfrac\pi4)\,R_Z(-\tfrac\pi2)R_Z(\tfrac\pi2)=R_Y(0)R_Z(0)=I$
Add the same-axis angles.
Step 4:
$AXBXC=R_Y(\tfrac\pi4)\,R_Y(\tfrac\pi4)R_Z(\tfrac\pi2)\,R_Z(\tfrac\pi2)=R_Y(\tfrac\pi2)R_Z(\pi)=-\frac{i}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}=-iH$
The lemma turns $XBX$ into $R_Y(\tfrac\pi4)R_Z(\tfrac\pi2)$ ; then add the angles and use the matrix product from the Hadamard lesson.
Step 5:
$e^{i\pi/2}AXBXC=i\cdot(-iH)=H,\qquad P(\tfrac\pi2)=\begin{pmatrix}1&0\\0&i\end{pmatrix}=S$
The control phase gate for $\alpha=\tfrac\pi2$ is the $S$ gate; without it the circuit would give $C\text{-}(-iH)$ .
Step 6:
$C\text{-}H=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&\frac1{\sqrt2}&\frac1{\sqrt2}\\0&0&\frac1{\sqrt2}&-\frac1{\sqrt2}\end{pmatrix}$
The circuit $R_Z(\tfrac\pi2)$ ; $\mathrm{CNOT}$ ; $R_Z(-\tfrac\pi2)$ , $R_Y(-\tfrac\pi4)$ ; $\mathrm{CNOT}$ ; $R_Y(\tfrac\pi4)$ on the target, with $S$ on the control, has block form $\begin{pmatrix}I&0\\0&H\end{pmatrix}$ by the corollary. On the wire $B$ is applied as $R_Z(-\tfrac\pi2)$ first, because $B=R_Y(-\tfrac\pi4)R_Z(-\tfrac\pi2)$ .

Result

\boxed{C\text{-}H=(S\otimes I)(I\otimes A)\,\mathrm{CNOT}\,(I\otimes B)\,\mathrm{CNOT}\,(I\otimes C),\quad A=R_Y(\tfrac\pi4),\ B=R_Y(-\tfrac\pi4)R_Z(-\tfrac\pi2),\ C=R_Z(\tfrac\pi2)}

Circuit · 2 qubits

ABC

U = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \\ 0 & 0 & \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{pmatrix}

Quick check

ABC

Numerical verification and lab#

10^{-12}

TypeScriptrunnable

import { HADAMARD, I2, PAULI_X, SWAP_GATE, circuitUnitary, matrix, ryMatrix, rzMatrix } from '@/lib/quantum-simulator';

const tol = 1e-12;
const P0 = matrix.fromRows([[1, 0], [0, 0]]);
const P1 = matrix.fromRows([[0, 0], [0, 1]]);
// Engine unitaries of single two-qubit gates (qubit 0 = control = most significant bit)
const CNOT = circuitUnitary({ version: 1, numQubits: 2, numClbits: 0, ops: [{ kind: 'gate', gate: 'CNOT', targets: [0, 1] }] });
const CZ = circuitUnitary({ version: 1, numQubits: 2, numClbits: 0, ops: [{ kind: 'gate', gate: 'CZ', targets: [0, 1] }] });

// Block form C-U = |0><0| (x) I + |1><1| (x) U
console.log('CNOT = P0 x I + P1 x X :', matrix.approxEqual(matrix.add(matrix.kron(P0, I2), matrix.kron(P1, PAULI_X)), CNOT, tol));
console.log('CNOT is unitary        :', matrix.isUnitary(CNOT, tol));

// CNOT from CZ, and the reversed CNOT
const IH = matrix.kron(I2, HADAMARD);
const HH = matrix.kron(HADAMARD, HADAMARD);
const CNOT10 = matrix.add(matrix.kron(I2, P0), matrix.kron(PAULI_X, P1));
console.log('(I x H) CZ (I x H) = CNOT     :', matrix.approxEqual(matrix.mulAll(IH, CZ, IH), CNOT, tol));
console.log('(H x H) CNOT (H x H) = CNOT10 :', matrix.approxEqual(matrix.mulAll(HH, CNOT, HH), CNOT10, tol));

// SWAP from three CNOTs, and the two-CNOT near miss
console.log('CNOT CNOT10 CNOT = SWAP :', matrix.approxEqual(matrix.mulAll(CNOT, CNOT10, CNOT), SWAP_GATE, tol));
const V = matrix.mul(CNOT10, CNOT);
const tr = matrix.trace(matrix.mul(matrix.adjoint(SWAP_GATE), V));
console.log('two CNOTs: |Tr(SWAP^dag V)| / 4 =', (Math.hypot(tr.re, tr.im) / 4).toFixed(4));

// ABC construction for beta = 0.3, gamma = 1.1, delta = -0.7
const beta = 0.3, gamma = 1.1, delta = -0.7;
const A = matrix.mul(rzMatrix(beta), ryMatrix(gamma / 2));
const B = matrix.mul(ryMatrix(-gamma / 2), rzMatrix(-(delta + beta) / 2));
const C = rzMatrix((delta - beta) / 2);
const zyz = matrix.mulAll(rzMatrix(beta), ryMatrix(gamma), rzMatrix(delta));
console.log('ABC = I                   :', matrix.approxEqual(matrix.mulAll(A, B, C), I2, tol));
console.log('AXBXC = RZ(b) RY(g) RZ(d) :', matrix.approxEqual(matrix.mulAll(A, PAULI_X, B, PAULI_X, C), zyz, tol));

Expected output

CNOT = P0 x I + P1 x X : true
CNOT is unitary        : true
(I x H) CZ (I x H) = CNOT     : true
(H x H) CNOT (H x H) = CNOT10 : true
CNOT CNOT10 CNOT = SWAP : true
two CNOTs: |Tr(SWAP^dag V)| / 4 = 0.5000
ABC = I                   : true
AXBXC = RZ(b) RY(g) RZ(d) : true

References

M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), §4.3 (controlled operations, Corollary 4.2 and the CNOT identities of Exercises 4.17 and 4.20).doi:10.1017/CBO9780511976667
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

Controlled-U in block form#

Controlled-U gate

CZ is symmetric; CNOT from CZ#

Exchanging control and target#

SWAP from three CNOTs#

SWAP gate

Phase kickback#

Any controlled-U from two CNOTs#

Numerical verification and lab#

References

Hands-on labs

SWAP from three CNOTs