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A spin quantum paradox

  1. Dec 27, 2013 #1
    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.
  2. jcsd
  3. Dec 27, 2013 #2
    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:.
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