Householder reflection

The projector $\ket u\bra u$ is self-adjoint, so $\hat S$ is as well. Expanding and using $\braket uu=1$ gives

$$\hat S^2=\mathbf{1}-4\ket u\bra u+4\ket u\braket uu\bra u=\mathbf{1}-4\ket u\bra u+4\ket u\bra u=\mathbf{1}.$$

Thus $\hat S^\dagger\hat S=\hat S^2=\mathbf{1}$: $\hat S$ is unitary. An operator that is both self-adjoint and unitary is its own inverse.

We have $\hat S\ket u=\ket u-2\ket u=-\ket u$: $\ket u$ is an eigenvector with eigenvalue $-1$. For every $\ket v$ orthogonal to $\ket u$, $\hat S\ket v=\ket v$: the hyperplane $\{\ket u\}^\perp$, of dimension $n-1$, is the eigenspace associated with the eigenvalue $+1$. This confirms that the eigenvalues of an operator that is both unitary and self-adjoint, being real and of unit modulus, are $\pm1$. It follows that $\tr\hat S=(n-1)-1=n-2$ and $\det\hat S=-1$. We may also write $\hat S=\hat P_\perp-\ket u\bra u$, where $\hat P_\perp=\mathbf{1}-\ket u\bra u$ is the projector onto the hyperplane. In $\R^3$, $\hat S$ leaves the vectors in the plane orthogonal to $\ket u$ unchanged and reverses the sign of the component along $\ket u$: it is the orthogonal reflection in this plane.

We have $\ket u\bra u=\frac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$, hence

$$\begin{aligned} S=\begin{pmatrix}1&0\\0&1\end{pmatrix}-\begin{pmatrix}1&-1\\-1&1\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}=\sigma_x . \end{aligned}$$

The Pauli matrix $\sigma_x$ is the reflection that leaves the vector $\frac{1}{\sqrt2}(1,1)$ invariant and reverses the sign of $\frac{1}{\sqrt2}(1,-1)$. For a photon, it may be interpreted as a plate that interchanges horizontal and vertical polarisations.

The required reflection interchanges $\ket x$ and $\ket y$; it therefore reverses the sign of their difference and preserves their sum. Take $\ket u$ collinear with $\ket x-\ket y=(-2,4,0)$, whose norm is $\sqrt{20}$: $\ket u=\frac{1}{\sqrt{20}}(-2,4,0)$. Then $\braket ux=\frac{1}{\sqrt{20}}(-6+16)=\frac{10}{\sqrt{20}}$, and

$$\hat S\ket x=\ket x-2\braket ux\ket u=(3,4,0)-2\cdot\frac{10}{20}(-2,4,0)=(3,4,0)-(-2,4,0)=(5,0,0)=\ket y .$$

This construction works whenever $\norm x=\norm y$ and $\braket xy$ is real: one then checks that $\ket x+\ket y$ is orthogonal to $\ket x-\ket y$, and is therefore invariant under the reflection. It is widely used in numerical analysis to reduce a matrix to triangular form.