In class yesterday (lecture 19, notes not yet posted) we used $$\Bsigma^\T = -\sigma_2 \Bsigma \sigma_2$$, which implicitly shows that $$(\Bsigma \cdot \Bx)^\T$$ is a reflection about the y-axis.
This form of reflection will be familiar to a student of geometric algebra (see [1] — a great book, one copy of which is in the physics library). I can’t recall any mention of the geometrical reflection identity from when I took QM. It’s a fun exercise to demonstrate the reflection identity when constrained to the Pauli matrix notation.

## Theorem: Reflection about a normal.

Given a unit vector $$\ncap \in \mathbb{R}^3$$ and a vector $$\Bx \in \mathbb{R}^3$$ the reflection of $$\Bx$$ about a plane with normal $$\ncap$$ can be represented in Pauli notation as
\begin{equation*}
-\Bsigma \cdot \ncap \Bsigma \cdot \Bx \Bsigma \cdot \ncap.
\end{equation*}

To prove this, first note that in standard vector notation, we can decompose a vector into its projective and rejective components
\label{eqn:reflection:20}
\Bx = (\Bx \cdot \ncap) \ncap + \lr{ \Bx – (\Bx \cdot \ncap) \ncap }.

A reflection about the plane normal to $$\ncap$$ just flips the component in the direction of $$\ncap$$, leaving the rest unchanged. That is
\label{eqn:reflection:40}
-(\Bx \cdot \ncap) \ncap + \lr{ \Bx – (\Bx \cdot \ncap) \ncap }
=
\Bx – 2 (\Bx \cdot \ncap) \ncap.

We may write this in $$\Bsigma$$ notation as
\label{eqn:reflection:60}
\Bsigma \cdot \Bx – 2 \Bx \cdot \ncap \Bsigma \cdot \ncap.

We also know that
\label{eqn:reflection:80}
\begin{aligned}
\Bsigma \cdot \Ba \Bsigma \cdot \Bb &= a \cdot b + i \Bsigma \cdot (\Ba \cross \Bb) \\
\Bsigma \cdot \Bb \Bsigma \cdot \Ba &= a \cdot b – i \Bsigma \cdot (\Ba \cross \Bb),
\end{aligned}

or
\label{eqn:reflection:100}
a \cdot b = \inv{2} \symmetric{\Bsigma \cdot \Ba}{\Bsigma \cdot \Bb},

where $$\symmetric{\Ba}{\Bb}$$ is the anticommutator of $$\Ba, \Bb$$.
Inserting \ref{eqn:reflection:100} into \ref{eqn:reflection:60} we find that the reflection is
\label{eqn:reflection:120}
\begin{aligned}
\Bsigma \cdot \Bx –
\symmetric{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap
&=
\Bsigma \cdot \Bx –
{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap

{\Bsigma \cdot \Bx}{\Bsigma \cdot \ncap}
\Bsigma \cdot \ncap \\
&=
\Bsigma \cdot \Bx –
{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap

{\Bsigma \cdot \Bx} \\
&=

{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap,
\end{aligned}

which completes the proof.

When we expand $$(\Bsigma \cdot \Bx)^\T$$ and find
\label{eqn:reflection:n}
(\Bsigma \cdot \Bx)^\T
=
\sigma^1 x^1 – \sigma^2 x^2 + \sigma^3 x^3,

it is clear that this coordinate expansion is a reflection about the y-axis. Knowing the reflection formula above provides a rationale for why we might want to write this in the compact form $$-\sigma^2 (\Bsigma \cdot \Bx) \sigma^2$$, which might not be obvious otherwise.

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.