# How to calculate the overlap of the orthogonal state?

This is probably a very obvious question, but I am going through this problem set and I don't understand why in 1b) it says that it is obvious that $$|\langle\psi_1^\perp|\psi_2\rangle|=\sin\theta$$ given that $$|\langle\psi_1|\psi_2\rangle| = \cos\theta$$.

TL;DR: These inner products are equal to the amplitudes and therefore the squares of their magnitudes sum to one. By Pythagoras theorem $$\sin^2\theta + \cos^2\theta = 1$$, so if one of the amplitudes is $$\cos\theta$$ then the magnitude of the other must be $$|\sin\theta|$$.

Since $$\{|\psi_1\rangle, |\psi_1^\perp\rangle\}$$ is a basis, we can expand $$|\psi_2\rangle$$ as

$$|\psi_2\rangle = \alpha |\psi_1\rangle + \beta |\psi_1^\perp\rangle\tag1$$

where $$|\alpha|^2 + |\beta|^2=1$$. Moreover, since the basis is orthonormal, we can compute $$\alpha$$ and $$\beta$$ in terms of inner products

$$\alpha = \langle \psi_1|\psi_2\rangle \\ \beta = \langle \psi_1^\perp|\psi_2\rangle$$

as is easy to check by taking the inner product of $$(1)$$ with the elements of the dual basis. Now, from $$|\alpha|^2 + |\beta|^2=1$$ we see that

$$|\langle \psi_1^\perp|\psi_2\rangle| = |\beta| = \sqrt{1 - |\alpha|^2} = \sqrt{1 - |\langle \psi_1|\psi_2\rangle|^2} = \sqrt{1 - \cos^2\theta} = |\sin\theta|$$

but $$\theta \in (0, \pi)$$ so $$|\langle \psi_1^\perp|\psi_2\rangle|=\sin\theta$$.