Editing Hadamard transform (section)

===Hadamard gate operations===

<math display="block">\begin{align}
  H(|0\rangle) &= \frac{1}{\sqrt 2}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle =: |+\rangle\\
  H(|1\rangle) &= \frac{1}{\sqrt 2}|0\rangle - \frac{1}{\sqrt{2}}|1\rangle =: |-\rangle\\
  H(|+\rangle) &= H\left( \frac{1}{\sqrt 2}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle \right) = \frac{1}{2}\Big( |0\rangle + |1\rangle\Big) + \frac{1}{2}\Big(|0\rangle - |1\rangle\Big) = |0\rangle\\
  H(|-\rangle) &= H\left( \frac{1}{\sqrt 2}|0\rangle - \frac{1}{\sqrt{2}}|1\rangle \right) = \frac{1}{2}\Big( |0\rangle + |1\rangle\Big) - \frac{1}{2}\Big(|0\rangle - |1\rangle\Big) = |1\rangle
\end{align}</math>

One application of the Hadamard gate to either a 0 or 1 qubit will produce a [[quantum state]] that, if observed, will be a 0 or 1 with equal probability (as seen in the first two operations). This is exactly like flipping a fair coin in the standard [[Probabilistic Turing machine|probabilistic model of computation]].  However, if the Hadamard gate is applied twice in succession (as is effectively being done in the last two operations), then the final state is always the same as the initial state.