# What is the quantum Fourier transform of $\alpha|0\rangle+\beta|1\rangle$?

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 $$QFT(|\psi\rangle) = |\psi\rangle$$?

$$QFT |\psi\rangle = \alpha H|0\rangle + \beta H|1\rangle = \frac{\alpha+\beta}{\sqrt2}|0\rangle + \frac{\alpha-\beta}{\sqrt2}|1\rangle$$
And $$QFT(|\psi\rangle) = |\psi\rangle$$ is possible if $$|\psi\rangle$$ is an eigenvector of $$H$$.
• @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. $-|\psi\rangle = |\psi\rangle$. Apr 27 at 10:34