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5

It's true that the $|\pm\rangle=\tfrac{1}{\sqrt{2}}(|0\rangle\pm|1\rangle$ states are orthonormal eigenvectors of the $X$ gate (meaning they are "unneffected" by $X$), but this does not mean that $X$ doesn't have any effect on states in other superpositions. For instance, consider the state $|\psi\rangle=\frac{1}{\sqrt{4}}|0\rangle+\frac{\sqrt{3}}{...


5

This depends on what superposition you're in. Recall that the not gate can be written in the form $$ \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) $$ so it has eigenvectors $(1,\pm 1)$. The whole point of an eigenvector is that when the matrix acts on it, it's unchanged. So, the state $H|0\rangle$ (Hadamard acting on a 0 input state) is $(...


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