1. Dec 27, 2013

### jk22

Suppose we consider the spin 1/2 measurement matrices

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

it's easy to show that $$B^2=A$$
and a normalized eigenstate of B $$|\Psi\rangle=\left(\begin{array}{c}a\\b\end{array}\right)$$ with eigenvalue 1 : $$B|\Psi\rangle=|\Psi\rangle$$

then we obvisouly have $$\langle B^2\rangle=\langle\Psi|BB|\Psi\rangle=1$$

But $$\langle A\rangle=a^2-b^2<1$$ 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.

2. Dec 27, 2013

### DeIdeal

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 :tongue:.