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Let's suppose Alice transmit to Bob either one of the two states:

$$|\psi_{\pm}\rangle = \cos(\theta)|H\rangle \pm \sin(\theta)|V\rangle, \quad \theta \in [-\frac{\pi}{4}, +\frac{\pi}{4}]$$ The Helstrom formula is given as $$p(\text{error}) = |\psi_{\pm}\rangle = \frac{1}{2}[1-\sqrt(1-|\langle \psi_{+}|\psi_{-} \rangle|^{2})]$$

From the above: $\langle \psi_{+}|\psi_{-} \rangle$ is the probability amplitude associated with the projection of $|\psi_{+}\rangle$ onto $|\psi_{-}\rangle$. Then, $|\langle \psi_{+}|\psi_{-} \rangle|^{2}$ is the probability associated with the projection of $|\psi_{+}\rangle$ onto $|\psi_{-}\rangle$.

Now, the probability that $|\psi_{+}\rangle$ onto $|\psi_{-}\rangle$ add to the probability that $|\psi_{+}\rangle$ not projected onto $|\psi_{-}\rangle$ must equal 1 by law of total probability. From this: $1-|\langle \psi_{+}|\psi_{-} \rangle|^{2}$ must be the probability that $|\psi_{+}\rangle$ not projected onto $|\psi_{-}\rangle$.

Then, $1-|\langle \psi_{+}|\psi_{-} \rangle|^{2}$ is a probability so $\sqrt(1-|\langle \psi_{+}|\psi_{-} \rangle|^{2})$ must be the probability amplitude that $|\psi_{+}\rangle$ not projected onto $|\psi_{-}\rangle$.

But what does the 1 in $1-\sqrt(1-|\langle \psi_{+}|\psi_{-} \rangle|^{2})$ represents? What does the $\frac{1}{2}$ physically represents?

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