Hadamard Quantum Gate

quantum-computing

Definition

Hadamard Quantum Gate

The Hadamard quantum gate is

$$H=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right).$$

It acts on the basis states as

$$\begin{array}{rl}H|0⟩& =\frac{1}{\sqrt{2}}(|0⟩+|1⟩)\\ H|1⟩& =\frac{1}{\sqrt{2}}(|0⟩-|1⟩)\end{array},$$

i.e., $H$ turns basis state into superpositions.

Applying it to an arbitrary quantum state

$$|\psi ⟩=\alpha |0⟩+\beta |1⟩$$

leads to

$$\begin{array}{rl}H|\psi ⟩& =\alpha H|0⟩+\beta H|1⟩\\ & =\frac{\alpha 1}{\sqrt{2}}(|0⟩+|1⟩)+\frac{\beta 1}{\sqrt{2}}(|0⟩-|1⟩)\\ & =\frac{\alpha +\beta }{\sqrt{2}}|0⟩+\frac{\alpha -\beta }{\sqrt{2}}|1⟩.\end{array}$$

We can infer the formula:

$$H:\left(\begin{array}{c}\alpha \\ \beta \end{array}\right)\mapsto \frac{2}{\sqrt{2}}\left(\begin{array}{c}\alpha +\beta \\ \alpha -\beta \end{array}\right)$$

Destructive Inference

The output amplitudes are sums and differences of the input amplitudes.

If $\alpha =\beta$, the second amplitude cancels, i.e.:

$$\alpha -\beta =0.$$

If $\alpha =-\beta$, the second amplitude cancels:

$$\alpha +\beta =0.$$

That is destructive interference.