# geometric intepretation of Helstrom formula

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?