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The three traceless Pauli matrices $\sigma_x,\sigma_y,\sigma_z$ are arbitrary in the sense that any three operators with the appropriate commutation relations can be represented with those matrices. This is expressed by the action of the group SU(2) on $\mathbb{C}^2$ and hence also on the space of linear operators $\mathcal{L}(\mathbb{C}^2)$.

I want to prove the following: Let $A,B$ two anti-commuting Hermitian unitary traceless operators on a spin. Then there exists unitary U such that $U^\dagger AU=\sigma_z$ and $U^\dagger B U=\sigma_x $

This can be proven quite easily by a solving a system of linear equations on the entries of U. In fact there is even a simpler way. But I am more interested in a simple abstract argument which is more insightful. Perhaps some argument that will have to do with the action of the group $SU(3)\times SU(2) $ where $SU(3)$ expresses the freedom in choosing $A$ to be represented by $Z$ while $SU(2)$ expressed the freedom of representing $B$ as $X$ while keeping the representation of $A$ fixed.

As the argument above expressed an intuition regarding the symmetry of Pauli group, it is informal as a proof.

Is there a way to formalize this idea in a way which proves the claim above?

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  • $\begingroup$ This seems like a math question. $\endgroup$ – Omry Sep 17 '17 at 15:00
  • $\begingroup$ It is, though often physicists bring the best insights in those matters. And after all, this question came from quantum information $\endgroup$ – Itamar Vigi Sep 17 '17 at 16:17
  • $\begingroup$ maybe check out Patera, J., and H. Zassenhaus. "The Pauli matrices in n dimensions and finest gradings of simple Lie algebras of type A n− 1." Journal of Mathematical Physics 29.3 (1988): 665-673 although not quite what you want. $\endgroup$ – ZeroTheHero Sep 17 '17 at 16:31
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Yes, there even exists a $U\in SU(2)$ such that the claim $UAU^{-1}=\sigma_z$ and $UBU^{-1}=\sigma_x$ is true.

Sketched proof: Define $C:=\frac{1}{2i}[A,B]$. Then the triples $(\sigma_x,\sigma_y,\sigma_z)$ and $(A,B,C)$ can be viewed as two orthonormal dreibeins on 3D space $su(2)\cong \mathbb{R}^3$ with the same orientation. So there exists a 3D rotation $R\in SO(3)$ that takes one dreibein into the other. Moreover, the Lie algebra $su(2)$ is the vector space for the adjoint representation ${\rm Ad}(U)(\cdot):=U(\cdot)U^{-1}$ of the Lie group $SU(2)$, which is the double cover of $SO(3)$, so this induces an $SO(3)$ representation that turns out to be not projective. This is of course the aforementioned 3D $SO(3)$ representation. Hence we can find a $U\in SU(2)$ that corresponds to $R\in SO(3)$. $\Box$

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  • $\begingroup$ Thanks. Regarding your proof, it seems like a good direction however I could not manage to exactly understand it. For example, I am not familiar with the term dreibeins and could not find anything on it online in a quick search. By the way, the A and B in your example are not unitary, so this is not a counter-example. $\endgroup$ – Itamar Vigi Sep 17 '17 at 19:34
  • $\begingroup$ Ups, I first missed the unitary condition. Corrected. $\endgroup$ – Qmechanic Sep 17 '17 at 19:43

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