How to show the trace relation: $\text{Tr} [A\not x A ^{\dagger} \not x]\geq0$

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.

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 $$\mathrm{Tr}\,AxA^{\dagger}x=-2\text{.}$$
• It's true when $x$ is a positive-semidefinite (psd) matrix, which is the case whenever $x^{\mu}$ is a spacelike or lightlike vector. For then $x$ has a psd square root $y$ and the trace is of the form $\mathrm{Tr}\,B B^{\dagger}$ where $B= y A y$. – K B Dave Oct 19 '18 at 21:25
• "..which is the case whenever $x^\mu$ is a spacelike or lightlike vector." how did you arrive at this? And what convention are you using when you say "spacelike"? I think the 4-vector $x^\mu=(0,0,0,1)$ is definitely spacelike, but you have shown that the trace is negative with that choice. – Arturo don Juan Oct 19 '18 at 23:29
• Sorry, I was hasty and mixed up spacelike and timelike :/. By direct diagonalization, the eigenvalues of $x$ are $-x^0 \pm \sqrt{(x^1)^2+(x^2)^2+(x^3)^2}$. For $x$ to be psd, we need the eigenvalues to both be nonnegative, whence $x^0\leq 0$ and $\lvert x^0\rvert\geq \sqrt{(x^1)^2+(x^2)^2+(x^3)^2}$—a timelike vector. – K B Dave Oct 19 '18 at 23:45