Analyzer projections of a two-dimensional polarisation space and their overlaps

[The rotated pairs are orthonormal] By [F1] and [F2], $⟨e_{\theta },e_{\theta }⟩={\cos }^2\theta +{\sin }^2\theta =1$, $⟨e_{\theta }^{\perp },e_{\theta }^{\perp }⟩={\sin }^2\theta +{\cos }^2\theta =1$ and $⟨e_{\theta },e_{\theta }^{\perp }⟩=-\cos \theta \sin \theta +\sin \theta \cos \theta =0$; hence $(e_{\theta },e_{\theta }^{\perp })$ is an orthonormal family of the two-dimensional space $V$, therefore an orthonormal basis of $V$.

[Overlaps and the cosine identity] By [F1] and [F2], $⟨e_{\theta },{\psi }_{\phi }⟩=\cos \theta \cos \phi +\sin \theta \sin \phi =\cos (\theta -\phi )$ and $⟨e_{\theta }^{\perp },{\psi }_{\phi }⟩=-\sin \theta \cos \phi +\cos \theta \sin \phi =\sin (\phi -\theta )$. Then $⟨{\psi }_{\phi },P_{\theta }{\psi }_{\phi }⟩=⟨{\psi }_{\phi },⟨{\psi }_{\phi },e_{\theta }⟩e_{\theta }⟩=\mathrm{\mid }⟨{\psi }_{\phi },e_{\theta }⟩{\mathrm{\mid }}^2={\cos }^2(\theta -\phi )$ because $\cos (\theta -\phi )$ is real, and $\mathrm{\parallel }{P}_{\theta }{\psi }_{\phi }{\mathrm{\parallel }}^2=\mathrm{\mid }\cos (\theta -\phi ){\mathrm{\mid }}^2\mathrm{\parallel }{e}_{\theta }{\mathrm{\parallel }}^2={\cos }^2(\theta -\phi )$; similarly $⟨{\psi }_{\phi },P_{\theta }^{\perp }{\psi }_{\phi }⟩={\sin }^2(\theta -\phi )$. This is (b).

[The projections are orthogonal projections] For $x\in V$, $P_{\theta }x=⟨x,e_{\theta }⟩e_{\theta }$, so $P_{\theta }(P_{\theta }x)=⟨⟨x,e_{\theta }⟩e_{\theta },e_{\theta }⟩e_{\theta }=⟨x,e_{\theta }⟩⟨e_{\theta },e_{\theta }⟩e_{\theta }=P_{\theta }x$ by [F1] and step 1.1: $P_{\theta }^2=P_{\theta }$. For the adjoint, $⟨P_{\theta }x,y⟩=⟨⟨x,e_{\theta }⟩e_{\theta },y⟩=⟨x,e_{\theta }⟩⟨e_{\theta },y⟩=⟨x,⟨y,e_{\theta }⟩e_{\theta }⟩=⟨x,P_{\theta }y⟩$ for all $x,y$, so $P_{\theta }^{\ast }=P_{\theta }$; the same computation with $e_{\theta }^{\perp }$ in place of $e_{\theta }$ gives $P_{\theta }^{\perp 2}=P_{\theta }^{\perp }=P_{\theta }^{\perp \ast }$. For either unit vector $u=e_{\theta }$ or $u=e_{\theta }^{\perp }$, $P_{u}x=⟨x,u⟩u$ lies in $\mathbb{C}u$ and $⟨x-P_{u}x,u⟩=0$, so the residual is orthogonal to every multiple of $u$. Hence [F4] identifies $P_u$ as the orthogonal projection onto $\mathbb{C}u$; its range is exactly that line because $P_{u}u=u$. By the finite norm identity in [F3], $\mathrm{\parallel }{P}_{u}x{\mathrm{\parallel }}^2=\mathrm{\mid }⟨x,u⟩{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}^2$, so it is bounded. The line is closed: if $a_{r}u\to v$, then $a_r=⟨a_{r}u,u⟩\to ⟨v,u⟩$ (use the finite coordinate norm identity), and therefore $v=⟨v,u⟩u$.

[Complementarity] By step 1.1 and [F3] applied to the orthonormal basis $(e_{\theta },e_{\theta }^{\perp })$, every $x\in V$ satisfies $x=⟨x,e_{\theta }⟩e_{\theta }+⟨x,e_{\theta }^{\perp }⟩e_{\theta }^{\perp }=P_{\theta }x+P_{\theta }^{\perp }x$; hence $P_{\theta }+P_{\theta }^{\perp }=I$. Applying $P_{\theta }$ to the same expansion and using $P_{\theta }{e}_{\theta }^{\perp }=⟨e_{\theta }^{\perp },e_{\theta }⟩e_{\theta }=0$ gives $P_{\theta }{P}_{\theta }^{\perp }=0$, and likewise $P_{\theta }^{\perp }{P}_{\theta }=0$. Since $P_{\theta }=I-P_{\theta }^{\perp }$ and $\mathbb{C}{e}_{\theta }\cap \mathbb{C}{e}_{\theta }^{\perp }=\{0\}$ by orthogonality of the nonzero vectors $e_{\theta },e_{\theta }^{\perp }$, the eigenspace of $P_{\theta }$ at eigenvalue $1$ is $\mathbb{C}{e}_{\theta }$ and at eigenvalue $0$ is $\mathbb{C}{e}_{\theta }^{\perp }$; this proves (a) and (d).

[Periodicity and phase invariance] Since $e_{\theta +\pi }=-e_{\theta }$ and $e_{\theta +\pi \mathrm{/}2}=e_{\theta }^{\perp }$, the identities $P_{\theta +\pi }=P_{\theta }$ and $P_{\theta +\pi \mathrm{/}2}=P_{\theta }^{\perp }$ follow from the definition of $P_{\theta }$ and [F1]; in particular $P_{\theta }$ is $2\pi$-periodic. For $z\mathrm{\ne }0$ and the unit vector ${\psi }_{\phi }$ one has $\mathrm{\parallel }z{\psi }_{\phi }{\mathrm{\parallel }}^2=\mathrm{\mid }z{\mathrm{\mid }}^2$, and $P_{\theta }(z{\psi }_{\phi })=z{P}_{\theta }{\psi }_{\phi }$ by linearity of $x\mapsto ⟨x,e_{\theta }⟩$ in the first argument of [F1]; hence $\mathrm{\parallel }{P}_{\theta }(z{\psi }_{\phi }){\mathrm{\parallel }}^2\mathrm{/}\mathrm{\parallel }z{\psi }_{\phi }{\mathrm{\parallel }}^2=\mathrm{\parallel }{P}_{\theta }{\psi }_{\phi }{\mathrm{\parallel }}^2={\cos }^2(\theta -\phi )$ by step 1.2. This proves (c) and completes the proof.