There is something I don't understand about measurement in other basis than the Z-Pauli Basis. If measurement fixes the state of a quantum system thus destroying superposition, how can we get a superposition state $|+\rangle$ in the $X$ basis after measurement which is a superposition state of $|0\rangle$ and $|1\rangle$?
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$\begingroup$ Similarly some might argue that a z-measurement yields a superposition of |-> and |+>. However, taking a measurement in any orientation 'destroys' the information from any basis but that which it was measured in. If you are measuring in z, you are left with basis elements |0> or |1>. If you measure in x, you are left with basis elements |+> or |->. The superposition you are noting above is the result of rewriting the |+> |-> basis in terms of the computational basis rather than leaving them in the measured basis. (Oversimplification, but I hope it helps) $\endgroup$– PGibbonCommented Mar 28, 2023 at 12:37
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2 Answers
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If measurement fixes the state of a quantum system thus destroying superposition, how can we get a superposition state |+> in the X basis after measurement
Measurement doesn't destroy superpositions. Measurements are projective operators. What they create or destroy or change depends on the projection you do.
If you project onto non-superposed states, you get non-superposed states. If you project onto superposed states, you get superposed states. If you project onto entangled states, you get entangled states. For example, take any two qubit state, then measure the $X \otimes X$ operator getting result $x$ and the $Z \otimes Z$ operator getting result $z$. The two qubits are now entangled; their state is $(X^zZ^x \otimes I) \cdot (|00\rangle + |11\rangle)$. The measurements created that entanglement. And the ability to do that is extremely important to error correcting codes.
Also keep in mind that the whole notion of "superposed" is basis dependent. It's clearly impossible to have a well defined unconditional "destroy all superpositions" operation, because from another basis' perspective it's a "create superposition" operation.
answered Mar 28, 2023 at 16:47
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For measuring in the $X$ basis one needs to apply a $H$ gate followed by a (standard) $Z$ basis measurement. Then an original $|+\rangle$ transforms to $|0\rangle$ and yields $0$, while an original $|-\rangle$ transforms to $|1\rangle$ and yields $1$. So you know which final output corresponds to which initial state.
If you would have a machine that knows how to measure in the $X$ basis ($|+\rangle$, $|-\rangle$) itself, then in a similar manner to a $Z$ basis ($|0\rangle$, $|1\rangle$) measurement - a projection of the eigenstate $|+\rangle$ onto the basis yields a $+1$ eigenvalue (binary $0$), and a projection of the eigenstate $|-\rangle$ onto the basis yields a $-1$ eigenvalue (binary $1$).
