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D(T)=1−V(T)2,V(T)2+D(T)2=1D(T)=\sqrt{1-V(T)^2}, \qquad V(T)^2+D(T)^2=1

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For two pure states, D(T)=1−V(T)2,V(T)2+D(T)2=1D(T)=\sqrt{1-V(T)^2}, \qquad V(T)^2+D(T)^2=1. This is the first translation. A visibility of 0.6 corresponds to a trace distinguishability of 0.8; neither number is a bit count.

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Published contexts (1)

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D(T)=1−V(T)2,V(T)2+D(T)2=1.D(T)=\sqrt{1-V(T)^2}, \qquad V(T)^2+D(T)^2=1.

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This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

For two pure states, D(T)=1−V(T)2,V(T)2+D(T)2=1D(T)=\sqrt{1-V(T)^2}, \qquad V(T)^2+D(T)^2=1. This is the first translation. A visibility of 0.6 corresponds to a trace distinguishability of 0.8; neither number is a bit count.

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