A speaker from a talk/lecture I attended recently quoted the following inequality without proof, and for no good reason I have taken it upon myself to prove it:
Let $A\in SL(2,\mathbb{C})$, so $\det\left(A\right) = 1$. Also, let $\not x \equiv x^\mu \sigma_\mu$ where $x^\mu=(x^0,x^1,x^2,x^3)$ is an arbitrary real-valued four-vector (tuple of 4 real numbers) and $\sigma_\mu=(\sigma_0,\sigma_1,\sigma_2,\sigma_3)$ are the usual Pauli matrices with:
$$\begin{align}
\sigma_0&=-\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}\\
&\\
\sigma_1=\sigma_x\hspace{0.7cm}\,\,\,\sigma_2&=\sigma_y\hspace{1cm}\,\,\,\sigma_3=\sigma_z
\end{align}
$$
I would like to prove the following harmless looking inequality:
$$\text{Tr} [A\not x A ^{\dagger}\not x]\geq0$$
My progress:
I know that the trace is definitely real, and so the inequality is well-defined:
$$\begin{align}
\text{Tr}[A\not x A ^{\dagger}\not x]^*&=\text{Tr}[(A\not x A ^{\dagger} \not x )^\dagger]\\
&=\text{Tr}[(\not x )^{\dagger}(A^{\dagger})^{\dagger}(\not x )^{\dagger}(A)^{\dagger}]\\
&=\text{Tr}[\not x A \not x A ^{\dagger} ]\\
&=\text{Tr}[A\not x A ^{\dagger} \not x]
\end{align}$$
Honestly, apart from this I don't know what I could do. Hints and/or suggestions would be much appreciated.
Best Answer
Let $x=\sigma_3$, $A=\mathrm{i}\sigma_2$. Then $\det A = 1$, so $A\in\mathrm{SL}_2\mathbb{C}$. Nevertheless $$\begin{split}AxA^{\dagger}x&=(\mathrm{i}\sigma_2)\sigma_3(\mathrm{i}\sigma_2)^{\dagger}\sigma_3\\ &=\sigma_2\sigma_3\sigma_2\sigma_3\\ &=-\sigma_2^2\sigma_3^2\\ &=\sigma_0\end{split}$$ whence $$\mathrm{Tr}\,AxA^{\dagger}x=-2\text{.}$$