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I want to find a similarity transformation $T$ between the Weyl representation and the Dirac representation of the gamma matrices: $\gamma_W^\mu=T \gamma_D^\mu T^{-1}$. It turns out that I can look at the zero component and get $$\gamma_D^0=(A^{-1} \sigma^0 A) \otimes(B^{-1} \sigma^1 B)=\sigma^0 \otimes\sigma^3.$$ Hence, $A =\mathbb{1}$. For B we have $\sigma^1 B = B \sigma^3.$ For the other components: $\sigma^2 B= B \sigma^2$. Now I am interested in the form of $B$. If we think about the commutator relation of the Pauli matrices we get: $B=a \sigma^0 + i b \sigma^2$, by using the other equation above, we directly get $a=-b.$ So far so good.

My question is now how this could be seen from a point of view of linear algebra, i.e. directly using basis transformation properties of the 2x2 matrices and NOT making use of the commutation relations.

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  • $\begingroup$ just form a matrix having its columns as the eigenvectors of $sigm^1$ then form another matrix having its columns as the eigenvectors $sigm^2$ then divide on to another. Here it is your $B$. $\endgroup$ – physshyp Oct 26 '17 at 18:50
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Try (A-26) of Itzykson & Zuber, or any decent QFT text. $$ T= \frac {1}{\sqrt 2} (I\otimes I + i\sigma^2\otimes I), $$ which leaves $$ \vec{\gamma}= i\sigma^2\otimes \vec{\sigma} $$ invariant under a similarity transformation and maps $\gamma^0$: $$ T\gamma_D^0T^{-1}=\frac {1}{\sqrt 2} (I\otimes I + i\sigma^2\otimes I) (\sigma^3\otimes I ) \frac {1}{\sqrt 2} (I\otimes I - i\sigma^2\otimes I) = -\sigma^1\otimes I=\gamma_W^0 ~. $$ The roles of $\sigma ^1$ and $\sigma^3$ are essentially reversed between the W and D representations for $\gamma^5$ , so the inverse T here transforms $\gamma^5$, and the above equation alternatively amounts to $$ T\gamma_D^5 T^{-1}=\frac {1}{\sqrt 2} (I\otimes I + i\sigma^2\otimes I) (\sigma^1\otimes I ) \frac {1}{\sqrt 2} (I\otimes I - i\sigma^2\otimes I) = \sigma^3\otimes I=\gamma_W^5 ~. $$

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