Controlled NOT gate

Behaviour in the Hadamard transformed basis

When viewed only in the computational basis ${\displaystyle \{|0⟩,|1⟩\}}$, the behaviour of the C_NOT appears to be like the equivalent classical gate. However, the simplicity of labelling one qubit the control and the other the target does not reflect the complexity of what happens for most input values of both qubits.

Insight can be gained by expressing the CNOT gate with respect to a Hadamard transformed basis ${\displaystyle \{|+⟩,|-⟩\}}$. The Hadamard transformed basis of a one-qubit register is given by

${\displaystyle |+⟩=\frac{1}{\sqrt{2}}(|0⟩+|1⟩),|-⟩=\frac{1}{\sqrt{2}}(|0⟩-|1⟩),}$

and the corresponding basis of a 2-qubit register is

${\displaystyle |++⟩=|+⟩\otimes |+⟩=\frac{1}{2}(|0⟩+|1⟩)\otimes (|0⟩+|1⟩)=\frac{1}{2}(|00⟩+|01⟩+|10⟩+|11⟩)}$,

etc. Viewing CNOT in this basis, the state of the second qubit remains unchanged, and the state of the first qubit is flipped, according to the state of the second bit. (For details see below.) "Thus, in this basis the sense of which bit is the control bit and which the target bit has reversed. But we have not changed the transformation at all, only the way we are thinking about it."

The "computational" basis ${\displaystyle \{|0⟩,|1⟩\}}$ is the eigenbasis for the spin in the Z-direction, whereas the Hadamard basis ${\displaystyle \{|+⟩,|-⟩\}}$ is the eigenbasis for spin in the X-direction. Switching X and Z and qubits 1 and 2, then, recovers the original transformation." This expresses a fundamental symmetry of the CNOT gate.

The observation that both qubits are (equally) affected in a C_NOT interaction is of importance when considering information flow in entangled quantum systems.

Details of the computation

We now proceed to give the details of the computation. Working through each of the Hadamard basis states, the results on the right column show that the first qubit flips between ${\displaystyle |+⟩}$ and ${\displaystyle |-⟩}$ when the second qubit is ${\displaystyle |-⟩}$:

A quantum circuit that performs a Hadamard transform followed by C_NOT then another Hadamard transform, can be described as performing the CNOT gate in the Hadamard basis (i.e. a change of basis):

(H₁ ⊗ H₁)⁻¹ . C_NOT . (H₁ ⊗ H₁)

The single-qubit Hadamard transform, H₁, is Hermitian and its own inverse. The tensor product of two Hadamard transforms operating (independently) on two qubits is labelled H₂. We can therefore write the matrices as:

H₂ . C_NOT . H₂

When multiplied out, this yields a matrix that swaps the ${\displaystyle |01⟩}$ and ${\displaystyle |11⟩}$ terms over, while leaving the ${\displaystyle |00⟩}$ and ${\displaystyle |10⟩}$ terms alone. This is equivalent to a CNOT gate where qubit 2 is the control qubit and qubit 1 is the target qubit:

$${\displaystyle \frac{1}{2}\left[\begin{array}{c}\begin{array}{rrrr}1& 1& 1& 1\\ 1& -1& 1& -1\\ 1& 1& -1& -1\\ 1& -1& -1& 1\end{array}\end{array}\right].\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right].\frac{1}{2}\left[\begin{array}{c}\begin{array}{rrrr}1& 1& 1& 1\\ 1& -1& 1& -1\\ 1& 1& -1& -1\\ 1& -1& -1& 1\end{array}\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\end{array}\right]}$$

Constructing a Bell state

A common application of the C_NOT gate is to maximally entangle two qubits into the ${\displaystyle |{\mathrm{\Phi }}^{+}⟩}$ Bell state; this forms part of the setup of the superdense coding, quantum teleportation, and entangled quantum cryptography algorithms.

To construct ${\displaystyle |{\mathrm{\Phi }}^{+}⟩}$, the inputs A (control) and B (target) to the C_NOT gate are

${\displaystyle \frac{1}{\sqrt{2}}(|0⟩+|1⟩{)}_A}$ and ${\displaystyle |0{⟩}_B}$.

After applying C_NOT, the resulting Bell state $\frac{1}{\sqrt{2}}(|00⟩+|11⟩)$ has the property that the individual qubits can be measured using any basis and will always present a 50/50 chance of resolving to each state. In effect, the individual qubits are in an undefined state. The correlation between the two qubits is the complete description of the state of the two qubits; if we both choose the same basis to measure both qubits and compare notes, the measurements will perfectly correlate.

When viewed in the computational basis, it appears that qubit A is affecting qubit B. Changing our viewpoint to the Hadamard basis demonstrates that, in a symmetrical way, qubit B is affecting qubit A.

The input state can alternately be viewed as

${\displaystyle |+{⟩}_A}$ and ${\displaystyle \frac{1}{\sqrt{2}}(|+⟩+|-⟩{)}_B}$.

In the Hadamard view, the control and target qubits have conceptually swapped and qubit A is inverted when qubit B is ${\displaystyle |-{⟩}_B}$. The output state after applying the C_NOT gate is ${\displaystyle \frac{1}{\sqrt{2}}(|++⟩+|--⟩),}$ which can be shown as follows:

${\displaystyle =\frac{1}{\sqrt{2}}(|+{⟩}_A|+{⟩}_B+|-{⟩}_A|-{⟩}_B)}$

${\displaystyle =\frac{1}{2\sqrt{2}}((|0{⟩}_A+|1{⟩}_A)(|0{⟩}_B+|1{⟩}_B)+(|0{⟩}_A-|1{⟩}_A)(|0{⟩}_B-|1{⟩}_B))}$

${\displaystyle =\frac{1}{2\sqrt{2}}((|00⟩+|01⟩+|10⟩+|11⟩)+(|00⟩-|01⟩-|10⟩+|11⟩))}$

${\displaystyle =\frac{1}{\sqrt{2}}(|00⟩+|11⟩).}$

C-ROT gate

The C-ROT gate (controlled Rabi rotation) is equivalent to a C-NOT gate except for a ${\displaystyle \pi /2}$ rotation of the nuclear spin around the z axis.

Implementations

Trapped ion quantum computers:

• Cirac–Zoller controlled-NOT gate
• Mølmer–Sørensen gate

Regulation

In May, 2024, Canada implemented export restrictions on the sale of quantum computers containing more than 34 qubits and error rates below a certain CNOT error threshold, along with restrictions for quantum computers with more qubits and higher error rates. The same restrictions quickly popped up in the UK, France, Spain and the Netherlands. They offered few explanations for this action, but all of them are Wassenaar Arrangement states, and the restrictions seem related to national security concerns potentially including quantum cryptography or protection from competition.