On standard basis, the sum of the probability of a vector $\newcommand\ket[1]{\left|#1\right\rangle}\ket{v} = a \ket{0} + b \ket{1}$ is $a^2 + b^2 = 1$, right?

What about the two states of the basis are not orthogonal? like $\ket{b_1} = ( \ket{0} + \sqrt 3 \ket{1}) / 2$ and $\ket{b_2} = ( \ket{0} - \sqrt 3 \ket{1}) / 2$? Is the sum still 1? I got 3/4 but I'm not too sure.

  • $\begingroup$ I thought the preview were fine with the \ket things.. Why there are like this??? $\endgroup$
    – Gabe Ebag
    Commented Nov 16, 2022 at 23:32
  • $\begingroup$ They probably require some package to render properly. You can use |v\rangle instead. $\endgroup$ Commented Nov 17, 2022 at 0:19

1 Answer 1


In Quantum Computing, a measurement projects a vector onto an eigenspace of an observable $A$, which is an Hermitian matrix.

Since it is Hermitian, its eigenvectors form an orthonormal basis of the Hilbert space they live in. As such, it isn't allowed/doesn't make sense to talk about measuring in a basis which is not orthonormal.

When talking about a probability, one talk about the potential outcome of a measurement. Thus, it also doesn't make sense to talk about a probability in a basis which is not orthonormal.


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