- #1
jk22
- 731
- 24
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.
[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.