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 $|\langle\psi_1|\psi_2\rangle| = \cos\theta$.


1 Answer 1


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 $|\langle \psi_1^\perp|\psi_2\rangle|=\sin\theta$.


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