## If the eigenvalues of $Z$ are $\pm1$, why are the computational basis states labeled with "$0$" and "$1$"?

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The computational basis is also known as the $$Z$$-basis as the kets $$|0\rangle,|1\rangle$$ are chosen as the eigenstates of the Pauli gate $$\begin{equation} Z=\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}. \end{equation}$$ I've got a quick question. As $$|0\rangle$$ has eigenvalue $$+1$$ with respect to $$Z$$ and $$|1\rangle$$ has eigenvalue $$-1$$, according to the postulates of quantum mechanics $$\pm 1$$ are the results of the measurement. So why is it that in every text the results are written as $$"0"$$ or $$"1"$$? Is it simply because of a conventional correspondence $$"0"\leftrightarrow +1,"1"\leftrightarrow -1$$ that no one ever even bothers to write explicitly?

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The reason is when you make measurement on the state $$|\psi \rangle = \alpha |0\rangle + \beta |1\rangle$$, the state $$|\psi\rangle$$ will collapsed onto whether the state $$|0\rangle$$ or $$|1\rangle$$. If it collapsed to the state $$|0\rangle$$ then it is indeed belongs to the $$+1$$ eigenspace, and you record it as an event occurred in that space. If instead it collapsed to the state $$|1\rangle$$ then it is belonging to the $$-1$$ eigenspace. In any case, the idea is upon measurement in the computational basis, your state will always be in the $$|0\rangle$$ state or the $$|1\rangle$$ state, hence the statement "0" or "1".

Note that after many measurements, you will have a statistical distribution of how many times your prepared state has collapsed into the $$|0\rangle$$ or the $$|1\rangle$$ state.

1I feel like this sort of misses the fundamental aspect of the matter. "0" and "1" in this context are nothing but conventional labels. We use them because they are reminiscent of how we denote bits, but we could equivalently use any other pair of symbols. – glS – 2021-02-02T10:40:08.680

1Sure. We can denote the states $|0\rangle$ and $|1\rangle$ as $| :) \ \rangle$ and $| :( \ \rangle$. – KAJ226 – 2021-02-02T11:16:07.273

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Is it simply because of a conventional correspondence "0"↔+1,"1"↔−1 that no one ever even bothers to write explicitly?

Yes. Labels can be chosen arbitrary, so long as they are clear. This particular convention is convenient simply because it makes the labels of the eigenstates look like classical binary strings. Note that the labels can be seen as powers of (-1) in the Z eigenvalues: $$(-1)^0 = 1$$, and $$(-1)^1 = -1$$.