I am told to show that these matrices are anticommutative (AB = -BA)$$ \beta=\begin{bmatrix}I_2&0\\0&-I_2\end{bmatrix}, \alpha_x=\begin{bmatrix}0&\sigma_1\\\sigma_1&0\end{bmatrix},\alpha_y=\begin{bmatrix}0&\sigma_2\\\sigma_2&0\end{bmatrix}, \alpha_z=\begin{bmatrix}0&\sigma_3\\\sigma_3&0\end{bmatrix}$$ where $\sigma_1=\begin{bmatrix}0&1\\1&0\end{bmatrix}, \sigma_2=\begin{bmatrix}0&-i\\i&0\end{bmatrix}, \sigma_3=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ (the pauli spin matrices). I could just use brute force and plug every matrix in for B and A, but my professor is very hard on us when it comes to proofs and only gives full credit for "elegant" proofs. The only problem is, I have no idea how I would even start to prove this; any help or hints would be greatly appreciated!!!
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$\begingroup$ In this case, I think the most elegant way is to compute. $\endgroup$– Jacky ChongOct 18, 2016 at 2:51
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$\begingroup$ Consider the effect of AB+BA on each of the two usual basis vectors. $\endgroup$– DanielWainfleetOct 18, 2016 at 4:39
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$\begingroup$ $\sigma_{i}\sigma_{j} = \delta_{ij}\ \overbrace{\sigma_{0}}^{identity} + \mathrm{i} \sum_{k}\epsilon_{ijk}\sigma_{k}$. For example: $\sigma_{2}\sigma_{1} =-\mathrm{i} \sigma_{3}$ $\endgroup$– Felix MarinOct 18, 2016 at 9:24
2 Answers
Just observe that for any (2D)matrices $\rm A,B$ (and zero matrix $\rm O$):
$$\begin{align}\begin{pmatrix}\rm A & \rm O \\ \rm O & -\rm A\end{pmatrix}\begin{pmatrix}\rm O & \rm B \\ \rm B & \rm O\end{pmatrix}~=~&\begin{pmatrix}\rm O & \rm AB \\ \rm -AB & \rm O\end{pmatrix}\\[1ex] =~&-\begin{pmatrix}\rm O & \rm -AB \\ \rm AB & \rm O\end{pmatrix}\\[1ex] =~& -\begin{pmatrix}\rm O & \rm B \\ \rm B & \rm O\end{pmatrix}\begin{pmatrix}\rm A & \rm O \\ \rm O & \rm -A\end{pmatrix}\end{align}$$
So if $\rm A=I_2$ and $\rm B\in\{\sigma_1, \sigma_2, \sigma_3\}$ you've proven three pairs are anti-commutative.
The rest follows from showing $\sigma_1, \sigma_2, \sigma_3$ are themselves anticommuative and showing if $\rm B,C$ are then $(\begin{smallmatrix}\rm O & \rm B \\ \rm B & \rm O\end{smallmatrix})$ and $(\begin{smallmatrix}\rm O & \rm C \\ \rm C & \rm O\end{smallmatrix})$ must be too.
PS: $\sigma_2, \sigma_3$ are not.
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$\begingroup$ Thanks so much, you made it really easy for me to understand! $\endgroup$ Oct 18, 2016 at 3:12
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\alpha_{i}\alpha_{j}\right\vert_{\ i\ \not=\ j} & = \pars{\begin{array}{cc} \ds{0} & \ds{\sigma_{i}} \\ \ds{\sigma_{i}} & \ds{0} \end{array}} \pars{\begin{array}{cc} \ds{0} & \ds{\sigma_{j}} \\ \ds{\sigma_{j}} & \ds{0} \end{array}} = \pars{\begin{array}{cc} \ds{\sigma_{i}\sigma_{j}} & \ds{0} \\ \ds{0} & \ds{\sigma_{i}\sigma_{j}} \end{array}} = \pars{\begin{array}{cc} \ds{-\sigma_{j}\sigma_{i}} & \ds{0} \\ \ds{0} & \ds{-\sigma_{j}\sigma_{i}} \end{array}} = \bbx{\ds{-\alpha_{j}\alpha_{i}}} \end{align}
