Proving Identity of S12^2 in Two Particle System

[evilcman]

I am trying to prove the identity
S_{12} ^ 2 = 4S^2-2S_{12}
where S12 is the tensor operator:
S_{12} = 3(\vec{\sigma_1} \vec{r})(\vec{\sigma_2} \vec{r}) / r^2 - (\vec{\sigma_1} \vec{\sigma_2})
where sigmas are vectors made of the Pauli matrices in the space of particle 1 and 2, and
\vec{S} = (\vec{\sigma_1} + \vec{\sigma_2})/2
the spin of the two particle system, and I am using the identity:
(\vec{a} \vec{\sigma})(\vec{b} \vec{\sigma}) = \vec{a}\vec{b} + i \sigma (\vec{a} X \vec{b})
to match the terms in each sides, however, i get a term like:
(\vec{\sigma_1} \vec{n})(\vec{\sigma_2} \vec{n})(\vec{\sigma_1} \vec{\sigma_2}) = 1 + i (\vec{\sigma_1} \vec{n}) (\vec{n}(\vec{\sigma_1} X \vec{\sigma_2}))
and I don't know how to further simplify this, but if the identity really holds, then this should be a linear combination of 1, \vec{\sigma_1} \vec{\sigma_2} and (\vec{\sigma_1} \vec{n}) (\vec{\sigma_2}\vec{n}), where n is a vec{r} / r. So, how do I further simplify this? Or is there an easier way to prove this without tedious computation?

 

[azntoon]

The identity you are trying to prove can be proven by using the definition of the S12 tensor operator. Since S12 is defined as 3(\vec{\sigma_1} \vec{r})(\vec{\sigma_2} \vec{r}) / r^2 - (\vec{\sigma_1} \vec{\sigma_2}), we can simply expand the left side of the equation and substitute the definition of S12 for the right side. On the left side, we have S12 squared: S_{12} ^ 2 = (3(\vec{\sigma_1} \vec{r})(\vec{\sigma_2} \vec{r}) / r^2 - (\vec{\sigma_1} \vec{\sigma_2}))^2 = 9(\vec{\sigma_1} \vec{r})^2 (\vec{\sigma_2} \vec{r})^2 /r^4 - 6(\vec{\sigma_1} \vec{r})(\vec{\sigma_2} \vec{r})(\vec{\sigma_1} \vec{\sigma_2}) / r^2 + (\vec{\sigma_1} \vec{\sigma_2})^2 On the right side, we have 4S^2-2S_{12}: 4S^2-2S_{12} = 4((\vec{\sigma_1} + \vec{\sigma_2})^2/4) - 2(3(\vec{\sigma_1} \vec{r})(\vec{\sigma_2} \vec{r}) / r^2 - (\vec{\sigma_1} \vec{\sigma_2})) = 3(\vec{\sigma_1} \vec{r})^2 (\vec{\sigma_2} \vec{r})^2 /r^4 - (\vec{\sigma_1} \vec{\sigma_2})^2 As you can see, both sides of the equation have the same terms (up to a factor of 9/4 on the