 # I am asked to show that the matrix S = 1 &#x2212;<!-- − --> i 4 </mfr lurtzslikgtgjd 2022-05-18 Answered
I am asked to show that the matrix
$S=1-\frac{i}{4}{\sigma }_{\mu \nu }{\epsilon }^{\mu \nu }$
represents the infinitesimal Lorentz transformation
${\mathrm{\Lambda }}^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }+{\epsilon }^{\mu }{}_{\nu },$
in the sense that
${S}^{-1}{\gamma }^{\mu }S={\mathrm{\Lambda }}^{\mu }{}_{\nu }{\gamma }^{\nu }.$
I have already proven that ${S}^{-1}={\gamma }^{0}{S}^{†}{\gamma }^{0}$, so I can begin with the left-hand side of this third equation:
$\begin{array}{rl}{S}^{-1}{\gamma }^{\mu }S& ={\gamma }^{0}{S}^{†}{\gamma }^{0}{\gamma }^{\mu }S\\ & ={\gamma }^{0}\left(1+\frac{i}{4}{\sigma }_{\mu \nu }{\epsilon }^{\mu \nu }\right){\gamma }^{0}{\gamma }^{\mu }\left(1-\frac{i}{4}{\sigma }_{\mu \nu }{\epsilon }^{\mu \nu }\right)\\ & ={\gamma }^{0}{\gamma }^{0}{\gamma }^{\mu }-{\gamma }^{0}{\gamma }^{0}{\gamma }^{\mu }\frac{i}{4}{\sigma }_{\mu \nu }{\epsilon }^{\mu \nu }+{\gamma }^{0}\frac{i}{4}{\sigma }_{\mu \nu }{\epsilon }^{\mu \nu }{\gamma }^{0}{\gamma }^{\mu }-{\gamma }^{0}\frac{i}{4}{\sigma }_{\rho \tau }{\epsilon }^{\rho \tau }{\gamma }^{0}{\gamma }^{\mu }\frac{i}{4}{\sigma }_{\mu \nu }{\epsilon }^{\mu \nu }\\ & ={\gamma }^{\mu }+\frac{1}{16}{\gamma }^{0}{\sigma }_{\rho \tau }{\epsilon }^{\rho \tau }{\gamma }^{0}{\gamma }^{\mu }{\sigma }_{\mu \nu }{\epsilon }^{\mu \nu }\\ & ={\gamma }^{\mu }+\frac{1}{16}\left({\sigma }_{\rho \tau }{\epsilon }^{\rho \tau }{\sigma }_{\mu \nu }{\epsilon }^{\mu \nu }\right){\gamma }^{\mu }.\end{array}$
Remember, we want this to be equal to
${\mathrm{\Lambda }}^{\mu }{}_{\nu }{\gamma }^{\nu }=\left({\delta }^{\mu }{}_{\nu }+{\epsilon }^{\mu }{}_{\nu }\right){\gamma }^{\nu }={\gamma }^{\mu }+{\epsilon }^{\mu }{}_{\nu }{\gamma }^{\nu }.$
The first term is there already, but I have no idea how that second term is going to work out to ${\epsilon }^{\mu }{}_{\nu }{\gamma }^{\nu }$. Can anyone give me a hint, or tell me what I’ve done wrong?
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Raiden Williamson
To order ${S}^{-1}$, isn't ${S}^{-1}$ just $1+\frac{i}{4}{\sigma }_{\mu \nu }{ϵ}^{\mu \nu }$? Then try expanding out ${S}^{-1}{\gamma }^{\mu }S$ keeping terms of order $ϵ$ and using the Clifford algebra relations when you have to commute $\gamma$s through ${\sigma }_{\mu \nu }$ terms.