Is There a Mistake in the Spin Quantum Paradox Calculation?

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SUMMARY

The discussion centers on the spin 1/2 measurement matrices B and A, where B is defined as B=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1\\1&-1\end{array}\right) and A=diag(1,-1). It is established that B^2 equals A, and a normalized eigenstate |\Psi\rangle with eigenvalue 1 leads to the conclusion that \langle B^2\rangle equals 1. However, the expectation value \langle A\rangle results in a value less than 1, specifically a^2-b^2<1, indicating a contradiction. The core issue lies in the misunderstanding of the eigenvector alignment with respect to the measurement matrices.

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jk22
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Suppose we consider the spin 1/2 measurement matrices

[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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jk22 said:
it's easy to show that [tex]B^2=A[/tex]

I didn't read the whole post thoroughly, so there might be other mistakes, but is it? It would seem to me that B2=1 :-p.
 

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