Suppose we consider the spin 1/2 measurement matrices(adsbygoogle = window.adsbygoogle || []).push({});

[tex]B=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1\\1&-1\end{array}\right)[/tex] and A=diag(1,-1)

it's easy to show that [tex]B^2=A[/tex]

and a normalized eigenstate of B [tex]|\Psi\rangle=\left(\begin{array}{c}a\\b\end{array}\right)[/tex] with eigenvalue 1 : [tex]B|\Psi\rangle=|\Psi\rangle[/tex]

then we obvisouly have [tex]\langle B^2\rangle=\langle\Psi|BB|\Psi\rangle=1[/tex]

But [tex]\langle A\rangle=a^2-b^2<1[/tex] since the eigenvector of A are not along x.

This implies that 1<1 ?? which is wrong, but I can't understand where the mistake hides.

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# A spin quantum paradox

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