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I'm attempting to figure out and solve the problems given in my Quantum Physics course below. I would like some clarification on the concept of filtering (relating to questions (ii) to (vi)).

From my understanding, if you have an equal number of particles with $S_z=\hbar/2$ and $S_z=-\hbar/2$, the first filter would permit half of the particles (only those with $S_z=\hbar/2$) through.

  1. How do I evaluate the fraction for filter 2? Do I evaluate $_z\left\langle\frac{1}{2},\frac{1}{2}|\frac{1}{2},\frac{1}{2}\right\rangle_n$?
  2. After filter 3, wouldn't the fraction transmitted be 0 since filter 1 removed particles with $S_z=-\hbar/2$? In which case, wouldn't the fraction of particles after all filters always be 0? (regardless of whether we use filter 2 or not?).

Problem

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You're correct in that you need to evaluate the braket you gave (you actually need to square it, to get a probability). But you're not correct in that nothing will make it past filter 3. The key insight is that filter 2 doesn't just filter particles, it changes their spin. After filter 2, every particle is in the state $|\frac{1}{2},\frac{1}{2}\rangle_n$. This state has a component along $|\frac{1}{2},-\frac{1}{2}\rangle_z$, and thus has some probability of being transmitted through the third filter.

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  • $\begingroup$ So presumably, if we remove filter 2, we get a fraction of 0 as filter 1 will remove particles with $\hbar/2$ and filter 3 will remove particles with $-\hbar/2$. $\endgroup$ – Akyidrian Jun 5 '16 at 4:41
  • $\begingroup$ @Akyidrian Yes. $\endgroup$ – Jahan Claes Jun 5 '16 at 14:40

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