Swap-test measurement and pure-state outcome law

On the orthonormal product basis in [F2], $F(e_i\otimes e_j)=e_j\otimes e_i$. This basis permutation is unitary and is its own inverse, so $F^{\ast }=F$ and $F^2=I$. Direct expansion now gives $P_{\pm }^{\ast }=P_{\pm }$, $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$, and $P_{+}+P_{-}=I$. Thus they form a complementary projective two-outcome measurement.

Put $v=\psi \otimes \varphi$, which is a unit vector by the tensor inner product. By [F3], $\mathrm{\mid }v⟩⟨v\mathrm{\mid }$ sends $z$ to $⟨z,v⟩v$, and [F4]–[F5] make this a density operator. In an orthonormal basis $(e_k)$, the diagonal coefficient of $\mathrm{\mid }v⟩⟨v\mathrm{\mid }{P}_{-}$ at $e_k$ is $⟨P_{-}{e}_k,v⟩⟨v,e_k⟩$. If $[P_{-}{]}_{ik}$ and $v_i$ are the matrix and vector coordinates, this coefficient is ${\sum }_i[P_{-}{]}_{ik}\overline{v_i}{v}_k$; summing over $k$ by [F7] gives ${\sum }_{i,k}[P_{-}{]}_{ik}{v}_k\overline{v_i}=⟨P_{-}v,v⟩$. This is the coordinate expansion asserted by [F8], and yields $\mathrm{tr}(\mathrm{\mid }v⟩⟨v\mathrm{\mid }{P}_{-})={\sum }_k⟨P_{-}{e}_k,v⟩⟨v,e_k⟩=⟨P_{-}v,v⟩,$ so [F6] makes this the outcome probability. Also $⟨Fv,v⟩=⟨\varphi ,\psi ⟩⟨\psi ,\varphi ⟩=\mathrm{\mid }⟨\psi ,\varphi ⟩{\mathrm{\mid }}^2$ by the tensor inner product and [F1]. Hence $\mathrm{Pr}[-]=⟨P_{-}v,v⟩=\frac{1-\mathrm{\mid }⟨\psi ,\varphi ⟩{\mathrm{\mid }}^2}{2}\mathrm{.}$ The complementary outcome has the remaining probability.