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Part three (going by N&C page 544) states that $$tr(S(n)\rho^{\otimes n})=tr(S(n)\rho^{\otimes n}P(n,\epsilon))+tr(S(n)\rho^{\otimes n}(I-P(n,\epsilon))).$$ Now I understand how the term on the left of + goes to 0 as n $\to \infty$. However, I am confused how the term on the right does. N&I states that you can set $$0 \le tr(S(n)\rho^{\otimes n}(I-P(n,\epsilon))) \le tr(\rho^{\otimes n}(I-P(n,\epsilon))) \rightarrow 0\,\,\text{ as } n\to \infty.$$

I don't quite understand why this is the case. My only assumption is that the eigenvalues of $\rho^{\otimes n}(I-P(n,\epsilon))$ are bounded in such a way that as $n \to \infty$ it will go to zero. However, I am unsure how to go about calculating this bound, though I assume it is of a similar form to the eigenvalues of $\rho^{\otimes n}P(n,\epsilon)), 2^{-n(S(\rho)-\epsilon)}$

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By part 1, we have that for any $\delta > 0$, then for sufficiently large $n$, $tr( \rho ^ {\otimes n} P(n, \epsilon)) \geq 1 - \delta$.

This means that $tr( \rho ^ {\otimes n} P(n, \epsilon)) \rightarrow 1$ as $n \rightarrow \infty$, since it is at most 1.

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    $\begingroup$ So it goes to 0 not because of the projector acting on it, but because the trace of it's orthogonal complement goes to 1? Can this be viewed as simply, for large enough n, there aren't and states that will be atypical, so the trace of the projected state is 0, or am I way off? $\endgroup$ – GaussStrife Jul 8 at 18:36

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