0
$\begingroup$

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 $Diag(1.-1).$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $S(\Lambda_0)$...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.