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The matrix contains information about the vectors(states). To see this, the matrix form can be written as $U_{ij}=\langle i\mid U\mid j\rangle$ or the total form of $U$ mentioned by @Quantum Mechanic, i.e., $U=\sum_{ij}U_{ij}\mid i\rangle\langle j\mid$. To show it more vividly, the stabilizer code will be a good example. Another easier example is that: when ...


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The columns of a unitary matrix must be orthonormal! So, in this case, since you know the image of $|0\rangle$ under $U$, it must be that $U|1\rangle$ is the unit vector orthogonal to $U|0\rangle=a|0\rangle+b|1\rangle$ in $\mathbb{C}^2$.


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Here's a solution that follows a slightly different idea, but provides the precise number of rows that need to be added to get a unitary matrix. Let $M$ be an arbitrary $n\times m$ matrix. As a first observation, note that the task of adding rows/columns can be reduced to that of adding rows to get $m$ orthonormal columns. Indeed, once this is done, we get ...


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I didn't go through the attached pdf. But if you want to find a unitary matrix $U$ that maps a quantum state $|\psi \rangle$ to $|\phi\rangle$ then you can use the Householder transformation as I commented. Here the two vectors have the same length (they are unit vectors) because we are thinking of them as a quantum state, so there will always exist a ...


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$U-I$ is a normal matrix so $||U-I||_{op}$ is its eigenvalue with the largest magnitude. The eigenvalue equation for this matrix is $$(U-I)|\psi\rangle=\lambda|\psi\rangle,$$ so $$|\lambda|^2=(\cos\phi-1)^2+\sin^2\phi=4\sin^2\frac{\phi}{2}\Rightarrow |\lambda|=2\left|\sin\frac{\phi}{2}\right|,$$ where $e^{i\phi}$ is some eigenvalue of $U$. Now, the ...


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If you want to be more precise about it, quantum (pure, ket) states are elements of complex projective spaces, $\mathbb{CP}^n$. This is the set of equivalence classes of elements of $\mathbb C^{n+1}$ modulo multiplication by complex scalars. So "gates" should really be described as maps between such equivalence classes. This is the projective ...


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Two states that are the same up to a global phase are physically indistinguishable; some even go as far as to say that they are exactly the same state. Any $U_{1}$ and $U_{2} = e^{i\alpha}U_{1}$ will thus give physically indistinguishable states, meaning that there is nothing that $U_{2}$ can do that $U_{1}$ cannot. Would a quantum processor restricted to ...


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