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I've been reading through chapter 5 of Preskill's notes and arrived at a result I couldn't replicate (found p24 here).

The question is to find the Von-Neumann entropy of the one-qubit ensemble where $\cos^2(\pi/8)$ is the maximal fidelity that a single qubit could take, corresponding to one of two eigenstates. The Von-Neumann entropy of the one qubit ensemble according to Preskill is then:

$$S(\rho) = H (\cos^2(\pi/8) ) \simeq 0.60088$$

$S(\rho)$ is the Von-Neumann entropy, and $H$ is the Shannon entropy , $$H = \sum_{x} -p(x) \log{(p(x))}.$$

Now, I can't seem to get to 0.60088. If I try $H (\cos^2(\pi/8) ) = -\cos^2(\pi/8)\log_2{(\cos^2(\pi/8))}$, I get $0.19499$. This is using log base 2 as specified earlier in the notes by Preskill. What gives?

Also, is the Von-Neumann entropy of the one qubit ensemble only including the highest fidelity because the other is negligible?

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  • $\begingroup$ "Von-Neumann" should by "von Neumann" $\endgroup$ – phenomenon Oct 18 '17 at 6:04
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The notation $H(x)$ refers to the binary entropy $$ H(x)\equiv H(\{x,1-x\})=-x\log x - (1-x)\log (1-x)\ . $$ With this definition (base 2), the result follows.

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