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D(T)=12∥∣h0⟩⟨h0∣−∣h1⟩⟨h1∣∥1D(T)=\frac12 \left\| |h_0\rangle\langle h_0|- |h_1\rangle\langle h_1| \right\|_1

Why this formula appears here

Assume for the moment that |h0h_0⟩\rangle and |h1h_1⟩\rangle are pure, normalized, and prepared with equal prior probability. Define their trace distinguishability D(T)=12∥∣h0⟩⟨h0∣−∣h1⟩⟨h1∣∥1D(T)=\frac12 \left\| |h_0\rangle\langle h_0|- |h_1\rangle\langle h_1| \right\|_1. For two pure states,

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Published contexts (1)

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D(T)=12∥∣h0⟩⟨h0∣−∣h1⟩⟨h1∣∥1.D(T)=\frac12 \left\| |h_0\rangle\langle h_0|- |h_1\rangle\langle h_1| \right\|_1.

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

Assume for the moment that |h0h_0⟩\rangle and |h1h_1⟩\rangle are pure, normalized, and prepared with equal prior probability. Define their trace distinguishability D(T)=12∥∣h0⟩⟨h0∣−∣h1⟩⟨h1∣∥1D(T)=\frac12 \left\| |h_0\rangle\langle h_0|- |h_1\rangle\langle h_1| \right\|_1. For two pure states,

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