5.5 Reflections
Given any unit vector a, we can create the unitary operator Ref a, which reflects any other unit vector b around a. Geometrically, this is done by dropping a line from the tip of b that hits the body of a in a right angle and continuing the line the same distance further to a point b' . Then b' likewise lies on the unit sphere of the Hilbert space. The operation mapping b to b' preserves the unit sphere and is its own inverse, so it is unitary. In geometrical terms, the point on the body of a is the projection of b onto a and is given by a' = a$\langle$a, b$\rangle$. Thus, b' = b − 2(b − a$\langle$a, b$\rangle$) = (2Pa − I)b
where Pa is the operator doing the projection: for all b, $P_ab$ = a$\langle$a, b$\rangle$.
Question: I don't understand why b' = b − 2(b − a$\langle$a, b$\rangle$) = (2Pa − I)b?
If you break up $b$ into two components, one along the vector $a$ and the other orthogonal to $a$, we can then write $b$ as
$$ b \;\; = \;\; \underbrace{\left (b - \langle b,a\rangle a\right )}_{\perp} + \underbrace{\langle b,a\rangle a}_{||}. $$
Now notice that $b'$ is obtained by reflecting the part of $b$ which is perpendicular to $a$, hence we get that
\begin{eqnarray*} b' & = & -\left (b - \langle b,a\rangle a\right ) + \langle b,a\rangle a \\ & = & -b + \langle b,a\rangle a+ \langle b,a\rangle a \\ & = & -b + 2\langle b,a\rangle a \\ & = & b - 2b + 2\langle b,a\rangle a \\ & = & b - 2\left (b - \langle b,a\rangle a\right ). \end{eqnarray*}
We get the identity by acknowledging that $P_ab = \langle b, a\rangle a$ and seeing that \begin{eqnarray*} b' & = & b - 2\left (b - \langle b,a\rangle a\right ) \\ & = & b - 2b + 2P_ab \\ & = & 2P_ab - b \\ & = & (2P_a - I)b \end{eqnarray*}
where we factor out $b$ and $I$ is the identity matrix/mapping.