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Suppose you have a pure state $\vert\psi\rangle$. Consider the following operation.

For unitaries $U_1$ and $U_2$, one can take complex numbers $\alpha, \beta$ where $|\alpha|^2 + |\beta|^2 = 1$ and construct the operator $\alpha U_1 + \beta U_2$. By linearity, applying this on $\vert\psi\rangle$ gives us $\alpha U_1\vert\psi\rangle + \beta U_2\vert\psi\rangle$. Since $U_i\vert\psi\rangle$ is a pure quantum state, we have by linearity that $(\alpha U_1 + \beta U_2)\vert\psi\rangle$ is a pure quantum state.

However, $\alpha U_1 + \beta U_2$ is not unitary itself. So what is the channel that corresponds to this operation?

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$(\alpha U_1 + \beta U_2)|\psi\rangle$ is not in general a pure quantum state because pure quantum states need to be normalized. A matrix is unitary if and only if it maps pure quantum states to pure quantum states. But, as you say, $\alpha U_1 + \beta U_2$ is not unitary.

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  • $\begingroup$ I see, so for any $\alpha$ with $|\alpha|\leq 1$, there exist $\beta$ that satisfy the normalization requirement for all $\vert\psi\rangle$ - is this true or does the $\beta$ depend on the choice of $\vert\psi\rangle$? $\endgroup$ Dec 16, 2023 at 12:56
  • $\begingroup$ I unfortunately don't understand what you're asking here -- can you please clarify? $\endgroup$ Dec 17, 2023 at 2:51
  • $\begingroup$ Sorry, I was asking if for every choice of $\alpha, U_1, U_2$, there exists a fixed $\beta$ that makes $\alpha U_1 + \beta U_2$ a unitary. But having thought about it, I think this claim is false. $\endgroup$ Dec 17, 2023 at 13:06

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