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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

\begin{equation}\label{eqn:reflection:20}

\Bx = (\Bx \cdot \ncap) \ncap + \lr{ \Bx – (\Bx \cdot \ncap) \ncap }.

\end{equation}

A reflection about the plane normal to \( \ncap \) just flips the component in the direction of \( \ncap \), leaving the rest unchanged. That is

\begin{equation}\label{eqn:reflection:40}

-(\Bx \cdot \ncap) \ncap + \lr{ \Bx – (\Bx \cdot \ncap) \ncap }

=

\Bx – 2 (\Bx \cdot \ncap) \ncap.

\end{equation}

We may write this in \( \Bsigma \) notation as

\begin{equation}\label{eqn:reflection:60}

\Bsigma \cdot \Bx – 2 \Bx \cdot \ncap \Bsigma \cdot \ncap.

\end{equation}

We also know that

\begin{equation}\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}

\end{equation}

or

\begin{equation}\label{eqn:reflection:100}

a \cdot b = \inv{2} \symmetric{\Bsigma \cdot \Ba}{\Bsigma \cdot \Bb},

\end{equation}

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

\begin{equation}\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}

\end{equation}

which completes the proof.

When we expand \( (\Bsigma \cdot \Bx)^\T \) and find

\begin{equation}\label{eqn:reflection:n}

(\Bsigma \cdot \Bx)^\T

=

\sigma^1 x^1 – \sigma^2 x^2 + \sigma^3 x^3,

\end{equation}

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.

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