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Why can the QFT be replaced by Hadamard gates?

While the QFT and Hadamard transforms are different, their action on the input state $|00\ldots 0\rangle$ is identical; both produce the uniform superposition of all states. So, if you've got a choice ...
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• 13.8k
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How do I prove that the Hadamard satisfies $H\equiv e^{i\pi H/2}$?

First of all, note that the statement, as written, is wrong (or rather, it is correct only as long as the "$\equiv$" symbol is taken to mean "equal up to a phase"). An easy way to see it is by ...
• 24.9k
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• 58.1k

What are the $|+\rangle$ and $|-\rangle$ states?

The $|+⟩$ and $|-⟩$ are states given by the following decomposition in the Z-basis: \begin{aligned} |+⟩ &= \frac{1}{\sqrt{2}} \Big(|0⟩ + |1⟩\Big)\\ |-⟩ &= \frac{1}{\sqrt{2}} ...
• 51

Could the Hadamard gate have been constructed differently with similar characteristics?

You can easily check that $H^1=XHX$, which is another way to say that $H^1$ is the same as $H$ modulo swapping $|0\rangle$ and $|1\rangle$. This means that it is a different gate. On the other hand, ...
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