10. Gates: Hadamard

The Hadamard Gate ($H$) is the indispensable single-qubit tool. It is the gate that generates balanced superposition and enables the ability to measure the qubit in a new basis, which is necessary for algorithms like the Quantum Fourier Transform.

1. The Matrix and the Operation

The Hadamard gate is a $2 imes 2$ unitary matrix:

$$H = rac{1}{\sqrt{2}} egin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix}$$

Its action is to map the computational basis states ($|0 angle, |1 angle$) to the new basis states, often called the Hadamard Basis or the X-Basis states ($|+ angle, |- angle$):

$$egin{array}{l} H|0 angle = |+ angle = rac{1}{\sqrt{2}}|0 angle + rac{1}{\sqrt{2}}|1 angle \ H|1 angle = |- angle = rac{1}{\sqrt{2}}|0 angle - rac{1}{\sqrt{2}}|1 angle \end{array}$$

2. Physical Significance: Basis Change

On the Bloch sphere, the Hadamard gate represents a specific sequence of rotations (a $90^\circ$ rotation about the Y-axis followed by a $180^\circ$ rotation about the X-axis) that effectively swaps the Z-axis with the X-axis.

When you apply $H$ to $|0 angle$, the state vector moves from the North Pole (Z-axis) to the positive X-axis ($|+ angle$).
If you measure the state $|+ angle$ in the Z-basis, the result is random (50/50).
However, if you apply $H$ again, the state returns to $|0 angle$. Now, if you measure, the result is deterministic ($100\%$ probability of $|0 angle$).

This ability to switch bases is what allows quantum computation to function. Interference effects (which rely on the relative phase $\phi$) are only visible when the state is measured in the correct basis.

Your Task: Proving Reversibility

Since $H$ is a valid quantum gate, it must be unitary (Postulate 2). The simplest way to show this is to prove that the gate is its own inverse, meaning applying it twice returns the original state.

Prove that:

$$H^2 = I$$

where $I = egin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$ is the Identity matrix.

Show the matrix multiplication $H \cdot H$ step-by-step.

10. Gates: Hadamard