Quantum Processing – The Quantum Gates: Single QBit.

Posted on Thu 10 July 2025 in misc

Like the classical AND, OR, NOT, etc... gates for classical bit, there are Quantum Gates for QBits. We will see some examples on Quantum gates acting on a single QBit.

Small reminder of classical gates

A boolean gate is a gate taking one or several boolean values as input, and return a boolean value. For example, the NOT gate return 0 if input is 1 and return 1 if input is 0. Its Truth Table is as below:

In Out
0 1
1 0

The AND gate takes 2 input bits and return one value:

In1 In2 Out
0 0 0
0 1 0
1 0 0
1 1 1

QBit Vector Representation

In previous article, wa have seen that a qbit can be represented using cartesian coordinates with \(\left| \Psi \right> = \alpha \left| 0 \right> + \beta \left| 1 \right>\):

and using the Bloch Sphere and some spherical coordinates with \( \left| \Psi \right> = cos(\frac{\theta}{2}) \left| 0 \right>+ sin(\frac{\theta}{2}) e^{i\phi} \left| 0 \right>\)

Bloch Sphere

A qbit can have an infinite number of values, we cannot represent it through a Truth Table. However, they can be represented as a matrix.

There are an uncountable infinite number of quantum gates, but here are the most interesting one.

Identity

$$ \left< I | \Psi \right> = \left| \Psi \right> $$

Or, using matrix representation:

$$ \begin{bmatrix} 1 & 0 \cr 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} \alpha \cr \beta \cr \end{bmatrix} = \begin{bmatrix} \alpha \cr \beta \cr \end{bmatrix} $$

Pauli X

Equivalent of NOT gate, noted \(\sigma_x\):

$$ \left< \sigma_x | 0 \right> = \left| 1 \right> \\ \left< \sigma_x | 1 \right> = \left| 0 \right> $$

Using matrix representation:

$$ \begin{bmatrix} 0 & 1 \cr 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} \alpha \cr \beta \cr \end{bmatrix} = \begin{bmatrix} \beta \cr \alpha \cr \end{bmatrix} $$

PauliX Applied to Phi QBit using cartesian representation

In Bloch Sphere:

PauliX Applied to Phi QBit on the Bloch Sphere

It apply a 180º rotation around X axis, swap α and β probabilities and keep the same phase ϕ.

$$ \begin{align} \begin{bmatrix} 0 & 1 \cr 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} cos( \frac{\theta}{2}) \cr e^{i\phi} sin(\frac{\theta}{2}) \cr \end{bmatrix} &= \begin{bmatrix} e^{i\phi} sin(\frac{\theta}{2}) \cr cos( \frac{\theta}{2}) \cr \end{bmatrix} \\ &= \begin{bmatrix} e^{i\phi} cos(\frac{\theta}{2} + \frac{\pi}{2}) \cr sin(\frac{\theta}{2} + \frac{\pi}{2}) \cr \end{bmatrix} \\ &\cong \begin{bmatrix} cos(\frac{\theta + \pi}{2}) \cr e^{-i\phi}sin( \frac{\theta + \pi}{2}) \cr \end{bmatrix} \end{align} $$

Pauli Y

Noted \(\sigma_y\):

$$ \begin{align} &\left< \sigma_y | 0 \right> = i\left| 1 \right> \\ &\left< \sigma_y | 1 \right> = -i\left| 0 \right> \end{align} $$

Using matrix representation:

$$ \begin{bmatrix} 0 & -i \cr i & 0 \end{bmatrix} \cdot \begin{bmatrix} \alpha \cr \beta \cr \end{bmatrix} = \begin{bmatrix} -i\beta \cr i\alpha \cr \end{bmatrix} $$

PauliY Applied to Phi QBit using cartesian representation

In Bloch Sphere:

PauliY Applied to Phi QBit on the Bloch Sphere

It apply a 180º rotation around Y axis, swap α and β probabilities and shift the phase ϕ. ,

$$ \begin{align} \begin{bmatrix} 0 & -i \cr i & 0 \end{bmatrix} \cdot \begin{bmatrix} cos( \frac{\theta}{2}) \cr e^{i\phi} sin(\frac{\theta}{2}) \cr \end{bmatrix} &= \begin{bmatrix} -i \cdot e^{i\phi} sin(\frac{\theta}{2}) \cr i \cdot cos( \frac{\theta}{2}) \cr \end{bmatrix} \\ &= \begin{bmatrix} e^{i\frac{-\pi}{2}}e^{i\phi} cos(\frac{\pi}{2} - \frac{\theta}{2}) \cr e^{i\frac{\pi}{2}}sin(\frac{\pi}{2} - \frac{\theta}{2}) \cr \end{bmatrix} \\ &= \begin{bmatrix} e^{i (\phi - \frac{\pi}{2})}cos(\frac{\pi - \theta}{2}) \cr e^{i\frac{\pi}{2}}sin(\frac{\pi - \theta}{2}) \cr \end{bmatrix} \\ &\cong \begin{bmatrix} cos(\frac{\pi - \theta}{2}) \cr e^{i(\frac{\pi}{2} - (\phi - \frac{\pi}{2}))}sin(\frac{\pi - \theta}{2}) \cr \end{bmatrix} \\ &= \begin{bmatrix} cos(\frac{\pi - \theta}{2}) \cr e^{i(\pi - \phi)}sin(\frac{\pi - \theta}{2}) \cr \end{bmatrix} \end{align} $$

Pauli Z

Noted \(\sigma_z\):

$$ \begin{align} &\left< \sigma_z | 0 \right> = \left| 0 \right> \\ &\left< \sigma_z | 1 \right> = -\left| 1 \right> \end{align} $$

Using matrix representation:

$$ \begin{bmatrix} 1 & 0 \cr 0 & -1 \end{bmatrix} \cdot \begin{bmatrix} \alpha \cr \beta \cr \end{bmatrix} = \begin{bmatrix} \alpha \cr -\beta \cr \end{bmatrix} $$

PauliZ Applied to Phi QBit using cartesian representation

In Bloch Sphere:

PauliZ Applied to Phi QBit on the Bloch Sphere

It apply a 180º rotation around Z axis, keep α and β probabilities and flip the phase ϕ.

$$ \begin{align} \begin{bmatrix} 1 & 0 \cr 0 & -1 \end{bmatrix} \cdot \begin{bmatrix} cos( \frac{\theta}{2}) \cr e^{i\phi} sin(\frac{\theta}{2}) \cr \end{bmatrix} &= \begin{bmatrix} cos( \frac{\theta}{2}) \cr -e^{i\phi} sin(\frac{\theta}{2}) \cr \end{bmatrix} \\ &= \begin{bmatrix} cos( \frac{\theta}{2}) \cr e^{i\pi}e^{i\phi} sin(\frac{\theta}{2}) \cr \end{bmatrix} \\ &= \begin{bmatrix} cos( \frac{\theta}{2}) \cr e^{i(\phi + \pi)} sin(\frac{\theta}{2}) \cr \end{bmatrix} \end{align} $$

Hadamard

Hadamard gate is used to create a superposition state given a basis state.

Noted \( H :\)

$$ \begin{align} &\left< H | 0 \right> = \frac{\left| 0 \right> + \left| 1 \right>}{\sqrt{2}} = \left|+\right> \\ &\left< H | 1 \right> = \frac{\left| 0 \right> - \left| 1 \right>}{\sqrt{2}} = \left|-\right> \end{align} $$

Using matrix representation:

$$ \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \cr 1 & -1 \end{bmatrix} \cdot \begin{bmatrix} \alpha \cr \beta \cr \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} \alpha + \beta \cr \alpha - \beta \cr \end{bmatrix} $$

Let's note that:

$$ \begin{align} &H = XY^{\frac{1}{2}} = R_x(\pi)R_y(\frac{\pi}{2}) \\ &H = Y^{-\frac{1}{2}}X = R_y(-\frac{\pi}{2})R_x(\pi) \\ &H = Y^{\frac{1}{2}}Z = R_y(\frac{\pi}{2})R_z(\pi) \\ &H = ZY^{-\frac{1}{2}} = R_z(\pi)R_y(-\frac{\pi}{2}) \\ &H = H^{-1} \end{align} $$

In Bloch Sphere:

Hadamard Applied to Phi QBit on the Bloch Sphere

It apply a 90º rotation around the Y-axis, followed by a 180º rotation around the X-axis.

$$ \begin{align} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \cr 1 & -1 \end{bmatrix} \cdot \begin{bmatrix} cos( \frac{\theta}{2}) \cr e^{i\phi} sin(\frac{\theta}{2}) \cr \end{bmatrix} &= \begin{bmatrix} cos( \frac{\theta}{2}) + e^{i\phi} sin(\frac{\theta}{2}) \cr cos( \frac{\theta}{2}) - e^{i\phi} sin(\frac{\theta}{2}) \cr \end{bmatrix} \\ \end{align} $$

Takeaways

  • Single Qbit Quantum gates change the orientation of QBit on the Bloch sphere. They act on \(\alpha\) and \(\beta\) probability (through acting on \(\theta\) angle) and on phase \(\phi\).
  • When there are only 2 logic gates that can act on a single bit (Identity and NOT), there is an infinite number of Quantum gates that can act on a single qbit.
  • PauliX gate is the equivalent of NOT gate.
  • Hadamard gate is classically used to create superposition state.