- #1

quasar_4

- 290

- 0

## Homework Statement

I'm supposed to derive the following:

[tex] \left({\bf A} \cdot {\bf \sigma} \right) \left({\bf B }\cdot {\bf \sigma} \right) = {\bf A} \cdot {\bf B} I + i \left( {\bf A } \times {\bf B} \right) \cdot {\bf \sigma} [/tex]

using just the two following facts:

Any 2x2 matrix can be written in a basis of spin matrices:

[tex] M = \sum{m_\alpha \sigma_\alpha} [/tex]

which means that the beta-th component is given by

[tex] m_\beta = \frac{1}{2}Tr(M \sigma_\beta) [/tex]

## Homework Equations

listed above...

## The Attempt at a Solution

It should just be a left side= right side proof.

I started by saying [tex] \left({\bf A} \cdot {\bf \sigma} \right) \left({\bf B }\cdot {\bf \sigma} \right) =

\left(\sum_\alpha a_\alpha \sigma_\alpha \cdot \sigma \right) \left(\sum_\alpha b_\alpha \sigma_\alpha \cdot \sigma \right) =

\left( \sum_\beta a_\alpha \delta_{\alpha \beta} \right) \left( \sum_\gamma b_\alpha \delta_{\alpha \gamma} \right) =

\sum_\beta a_\beta \sum_\gamma b_\gamma =

\frac{1}{2} Tr(A \sigma_\beta) \frac{1}{2} Tr(B \sigma_\gamma) [/tex]

Not sure if this is even the right way to start, and I can't see at all where I would go from here to get the appropriate RHS of the identity I'm proving. Any ideas?