# Tricky Pauli matrices

1. May 27, 2009

### emma83

Hello,
I am trying to recover the following calculation (where $$K,A$$ are 4x4 matrices in SL(2,C)):

--(start)--
"We expand $$K'=AKA^{\dagger}$$ in terms of $$k^a$$ and $$k'^{a}=(\delta_a^{b} + \lambda_a^{b} d\tau)k^b$$. Multiplying by a general Pauli matrix and using the relation $$\frac{1}{2}tr(\sigma_{a}\sigma_{b})=\delta_{ab}$$ yields the expression:
$$\lambda_b^{a} = \frac{1}{2}\eta^{ac}tr(\sigma_{b}\sigma_{c}A+\sigma_{c}\sigma_{b}A^{\dagger})$$."
--(end)--

I have been playing with the relations for a while but I guess I miss some knowledge on the properties of Pauli matrices because I don't manage to find the result. In particular, what would the "expansion" of $$AKA^{\dagger}$$ (which I guess is necessary here?) look like in Einstein summation notation ? Any help would be extremely appreciated!

2. May 30, 2009

### ryuunoseika

3. May 30, 2009

### tiny-tim

Well, give a link, then which article?

4. May 30, 2009

### emma83

5. May 30, 2009

### George Jones

Staff Emeritus
Something here seems either not quite right or incomplete.

If $k'^{a} = \left(\delta_b^{a} + \lambda_b^{a} d\tau \right) k^b$ (careful with index placement) and $d \tau$ is infinitesimal (?), then $k'$ and $k$ differ by an infinitesimal amount, so the transformation is an infinitesimal version of $K'=AKA^{\dagger}$. Then, the sum in final result could come from the product rule.

I'm just guessing. More context is needed.

6. May 30, 2009

### emma83

Dear George,

Thanks a lot for your answer. First, yes sorry I misplaced the indices in the first relation, the correct relation is:
$$k'_{a}=(\delta_a^{b} + \lambda_a^{b} d\tau)k_b$$

Concerning $$d\tau$$, it is actually a $$\delta u$$, where $$u$$ is the affine parameter along the trajectory of a photon.

But what do you mean by "product rule" ? Do I have to develop explicitely $$AKA^{\dagger}$$ in indices notation and try to recover at the end the $$A$$ and $$A^{\dagger}$$ which appear in the trace ?