# Confused regarding explanation of Schumachers compression in N&C

On page 547 of N&C, for $$|\psi_{0}\rangle=|0\rangle$$ and $$|\psi_{1}\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$$ and for $$|\tilde{0}\rangle=\cos(\pi/8)|0\rangle+\sin(\pi/8)|1\rangle$$ and $$|\tilde{1}\rangle=-\sin(\pi/8)|0\rangle + \cos(\pi/8)|1\rangle$$, that $$|\langle\tilde{0}|\psi_{k}\rangle| = \cos(\pi/8)$$ and $$|\langle\tilde{1}|\psi_{k}\rangle| = \sin(\pi/8)$$ for k = $$\{0,1\}$$. I just don't see how this can be the case for either of them.

I get $$\cos(\pi/8)$$ and $$\cos(\pi/8)/\sqrt{2}+\sin(\pi/8)\sqrt{2}$$ for $$|\langle\tilde{0}|\psi_{k}\rangle|$$ and for $$|\langle\tilde{1}|\psi_{k}\rangle|$$ I get $$-\sin(\pi/8)$$ and $$-\sin(\pi/8)/\sqrt{2}+\cos(\pi/8)/\sqrt{2}$$.

The inner product only produce these for $$|\psi_{0}\rangle$$. Is this a typo, and what they mean to say is that the inner product $$|\langle\tilde{0}|\psi_{k}\rangle|$$ is much larger than $$|\langle\tilde{1}|\psi_{k}\rangle|$$. But even in this case, that isn't completely true, as it's only just larger in the case of $$|\langle\tilde{1}|\psi_{k}\rangle|$$

For context, $$|\tilde{0}\rangle$$ and $$|\tilde{1}\rangle$$ come from the spectral decomposition of the density operator representing the source that generates $$|\psi_{0}\rangle$$ and $$|\psi_{1}\rangle$$ with probability a half for each.

What am I missing here? It seems like a simple inner product should be used but I can't get their results.

You're missing a bit of algebraic trickery. Remember that $$\frac{1}{\sqrt{2}}=\sin(\pi/4)=\cos(\pi/4)$$. Thus, $$\cos(\pi/8)/\sqrt{2}+\sin(\pi/8)/\sqrt{2}=\cos(\pi/8)\cos(\pi/4)+\sin(\pi/8)\sin(\pi/4)=\cos(\pi/4-\pi/8)=\cos(\pi/8)$$ by the double angle formula.
Also, be careful of signs. It might be an amplitude is $$\pm\sin(\pi/8)$$, but when you take the modulus, that becomes $$\sin(\pi/8)$$.