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Given $|\psi\rangle = \alpha |0\rangle + \beta |1\rangle$ and $|\alpha|^2 + |\beta|^2 = 1$, what would the quantum Fourier transform of $|\psi\rangle$ be? I know it is of the form $\frac{1}{\sqrt{2}}(x |0\rangle + y |1\rangle)$, but how would I determine $x$ and $y$?

Is it possible that $QFT(|\psi\rangle) = |\psi\rangle$?

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Quantum Fourier transform for a single-qubit case is just the Hadamard gate, so

$QFT |\psi\rangle = \alpha H|0\rangle + \beta H|1\rangle = \frac{\alpha+\beta}{\sqrt2}|0\rangle + \frac{\alpha-\beta}{\sqrt2}|1\rangle$

And $QFT(|\psi\rangle) = |\psi\rangle$ is possible if $|\psi\rangle$ is an eigenvector of $H$.

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  • $\begingroup$ Thank you for the quick response, though I'm not sure I understand why an eigenvector of H would remain unchanged when put through the QFT. $\endgroup$ Apr 26 at 22:46
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    $\begingroup$ An eigenvector of a unitary operator remains unchanged by definition. I'm not sure if you want exactly the same vector or the same up to a global phase, in the first case you need the eigenvector that corresponds to eigenvalue of 1, in the second case either one will do. $\endgroup$ Apr 26 at 23:08
  • $\begingroup$ Can you provide a source for that definition? I'm sorry, I'm pretty new to all of this and am not completely solid on a lot of the math. I really appreciate your help. $\endgroup$ Apr 26 at 23:14
  • $\begingroup$ I would recommend you to take a look at github.com/microsoft/QuantumKatas/#learning-path- - the first two tutorials in the learning path offer all definitions you need to get started with the math for quantum computing, including eigenvectors and eigenvalues. Alternatively, any source that defines an eigenvector will work :-) $\endgroup$ Apr 26 at 23:27
  • $\begingroup$ @QuantumLearner: Concerning eigenvectors, if $|\psi\rangle$ is eigenvector of $H$ then $H|\psi\rangle = \lambda|\psi\rangle$, where $\lambda$ is respective eigenvalue. Since $H$ is unitary and hermitian, $\lambda = \pm 1$, hence $H|\psi\rangle = \pm |\psi\rangle$. Minus one is just global phase and can be neglected, i.e. $-|\psi\rangle = |\psi\rangle$. $\endgroup$ Apr 27 at 10:34

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