- #1
jk22
- 732
- 25
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.
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.
.
