I want to prove $$S^{\mu \nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu].$$ I started from $$[\gamma^\mu,S^{\alpha\beta}]=(J^{\alpha\beta})^\mu_\nu \gamma^\nu$$ Putting the value of $(J^{\alpha\beta})^\mu_\nu$ $$=i(\eta^{\alpha\mu}\delta^\beta_\nu-\eta^{\beta\mu}\delta^\alpha_\nu)\gamma^\nu$$ we get $$\gamma^\mu S^{\alpha\beta}-S^{\alpha\beta}\gamma^\mu=i(\eta^{\alpha\mu}\gamma^\beta-\eta^{\beta\mu}\gamma^\alpha)$$ Whats the next step? Also tell me if there is any other decent method. Note I am using metric $Diag(1.-1).$

  • $\begingroup$ I don't really understand what you want to do here - what is $J^{\alpha\beta}$? Where does your second equation come from? How exactly do you want to prove the first equation? $\endgroup$ – ACuriousMind May 14 '17 at 7:51
  • $\begingroup$ Well, theoretical physicists know all these basic things. $(J^{\alpha\beta})^\mu_\nu$ is defined as $(J^{\alpha\beta})^\mu_\nu=i(\eta^{\alpha\mu}\delta^\beta_\nu-\eta^{\beta\mu}\delta^\alpha_\nu)$. There is a story behind where the second equation came from. $\endgroup$ – Sami Khan May 14 '17 at 9:12
  • $\begingroup$ @ACuriousMind I think here you understand What $S^{\mu \nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]$ is. Have a look at answer you provided. $\endgroup$ – Sami Khan May 14 '17 at 9:35
  • $\begingroup$ I'm confused about what you want to do. Do you want to show that $S^{\mu \nu}$ forms a representation of the lorentz algbera? $\endgroup$ – CStarAlgebra May 29 '17 at 14:09

We do not have to guess the structure of $S^{\mu\nu}$. You are really close, just replace $$2\eta^{\mu\nu}= \{\gamma^\mu,\gamma^\nu\}.$$ Rearrange the gammas on left side you will automatically make structure like your desired one by comparison on both sides. $$\gamma^\mu S^{\alpha\beta}-S^{\alpha\beta}\gamma^\mu=i(\eta^{\alpha\mu}\gamma^\beta-\eta^{\beta\mu}\gamma^\alpha)$$ $$\gamma^\mu S^{\alpha\beta}-S^{\alpha\beta}\gamma^\mu=\frac{i}{2}( \{\gamma^\alpha,\gamma^\mu\}\gamma^\beta- \{\gamma^\beta,\gamma^\mu\}\gamma^\alpha)$$ There might be difference of some constant factor. Fix it yourself.

  • $\begingroup$ Yes it did it but there is still problem with factor of $\frac{1}{4}$. $\endgroup$ – Sami Khan May 29 '17 at 16:05
  • $\begingroup$ This is precisely a (well-motivated I admit) guess. The $1/4$ factor is tricky: use $\eta^{\alpha\mu} \gamma^{\beta}= 1/2\, (\eta^{\alpha\mu} \gamma^{\beta} + \gamma^{\beta}\eta^{\alpha\mu} )$ then develop everything and simplify. Now I still claim that one should check unicity: I don't know it for sure but one clearly sees e.g. for $\mu=0,\ \alpha, \beta$ fixed: $\gamma^{0} S^{\alpha\beta} - S^{\alpha\beta}\gamma^{0} = \gamma^{0} A - A \gamma^{0} $ **does not** imply $S^{\alpha\beta}$. It may do if one writes the condition for $\mu= 0, 1,2,3$ but someone has to check... $\endgroup$ – Noix07 May 30 '17 at 11:59
  • $\begingroup$ (does not imply $S^{\alpha\beta}= A$). Explicitly, in $2\times 2$ blocks: $\gamma^{0}= \begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix}$. If $A = \begin{pmatrix} a & b\\ c & d\end{pmatrix}$ then $\gamma^{0} A - A \gamma^{0} = \begin{pmatrix} 0 & -2 b\\ -2 c & 0 \end{pmatrix}$ so that there is no conditions on the $a$ and $d$ components... $\endgroup$ – Noix07 May 30 '17 at 12:04
  • $\begingroup$ Finally using that the off-diagonal blocks are fixed by the equality $\gamma^0 S^{\alpha\beta} - S^{\alpha\beta} \gamma^0 = \gamma^0 A - A \gamma^0$, one writes the same condition for $\gamma^i$. Doing so one obtains conditions of the form $\sigma\, b' - \sigma\, a' = \sigma\, b - \sigma\, a$ (equality of $2\times 2$ matrices) where $S^{\alpha\beta} = \begin{pmatrix}a' & b' \\ c' & d' \end{pmatrix}$. From this $a' = \sigma\, \sigma\, a = a$ with the use of $\sigma^2 = 1$. So finally there is UNICITY of the solution!!! $\endgroup$ – Noix07 May 30 '17 at 12:43

You'll probably be a little disappointed but it seems that one had to guess the answer:

Existence of a solution to your second equation: Check that your first equation does satisfy the second.

Unicity??: I first thought that it could not be shown, but as usually for linear equations, if you have a particular solution $S^{\alpha\beta}$ of a non-homogeneous equation then the others are obtained by solving the associated homogeneous ones, here $$[\gamma^\mu, T^{\alpha\beta}]= \gamma^\mu \cdot T^{\alpha\beta}-T^{\alpha\beta}\cdot \gamma^\mu=0$$ This is equivalent to $$\gamma^\mu \cdot T^{\alpha\beta} = T^{\alpha\beta}\cdot \gamma^\mu\quad \Longleftrightarrow\quad T^{\alpha\beta} = (\eta^{\mu\mu})\, \gamma^{\mu}\cdot T^{\alpha\beta}\cdot \gamma^\mu $$ I leave it to others to find possible solutions...

For those who do not know where that comes from: second equation is the linearized form of $$S(\Lambda)\cdot \gamma^{\mu}\cdot S^{-1}(\Lambda) = \Lambda^{\mu}{}_{\nu}\, \gamma^{\nu}$$ One wants to find a representation $S(\Lambda)$ of the Lorentz group that satisfies such a relation and the previous $S^{\alpha\beta}$ are the generators in this representation $$S(\Lambda)= \mathrm{Id} - \frac{i}{2 i \hbar}\, \omega_{\mu\nu}\, S^{\mu\nu} + o(\omega) $$ while the $J^{\alpha\beta}$ are the generators in the defining representation.

There are constructions in books on Clifford algebras where the $S(\Lambda)$ can explcitly be constructed without exhibithing a solution from nowhere. Sketch: Lorentz transfo can always be written as a composition of particular "symmetries" $\Lambda_0$ (of the kind $x$ mapped to $-x$ and identity for orthogonal vectors). For those, one find a simple associated $S(\Lambda_0)$...


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