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An exact two-level calculation shows how far this can depart from the intuitive picture. Preselect a spin-one-half system along x , |$\psi$$\rangle$ = $\tfrac{1}{\sqrt2}(|{\uparrow}\rangle+|{\downarrow}\rangle)$ , and postselect along y , |f$\rangle$ = $\tfrac{1}{\sqrt2}(|{\uparrow}\rangle+i|{\downarrow}\rangle)$ . Then $\langle$ f|$\psi$$\rangle$ = $\tfrac12(1-i)$ and $\langle$ f|$\sigma_z$|$\psi$$\rangle$ = $\tfrac12(1+i)$ , so the weak value of $\sigma_z$ is

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