Jimmy Snyder
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Homework Statement
[tex](1 - \frac{i}{2}\mathbf{\sigma\cdot\theta})\mathbf{\sigma}(1 + \frac{i}{2}\mathbf{\sigma\cdot\theta}) = \mathbf{\sigma - \theta\times\sigma}[/tex]
Homework Equations
The Attempt at a Solution
At one point in this, I temporarily ignore the y and z components. I hope the notation is not too confusing.
[tex](1 - \frac{i}{2}\mathbf{\sigma\cdot\theta})\mathbf{\sigma}(1 + \frac{i}{2}\mathbf{\sigma\cdot\theta}) = \mathbf{\sigma} + \frac{i}{2}(\mathbf{\sigma}(\mathbf{\sigma\cdot\theta}) - (\mathbf{\sigma\cdot\theta})\mathbf{\sigma})[/tex]
[tex]= \mathbf{\sigma} + \frac{i}{2}(\sigma_x(\sigma_x\theta_x + \sigma_y\theta_y + \sigma_z\theta_z) - (\sigma_x\theta_x + \sigma_y\theta_y + \sigma_z\theta_z)\sigma_x) + ()[/tex]
[tex]= \mathbf{\sigma} + \frac{i}{2}(\theta_x + 2i\sigma_z\theta_y - 2i\sigma_y\theta_z) - (\theta_x - 2i\sigma_z\theta_y + 2i\sigma_y\theta_z)) + ()[/tex]
[tex]= \mathbf{\sigma} - 2(\sigma_z\theta_y - \sigma_y\theta_z) + ()[/tex]
and finally, putting the y and z compenents back in:
[tex](1 - \frac{i}{2}\mathbf{\sigma\cdot\theta})\sigma(1 + \frac{i}{2}\mathbf{\sigma\cdot\theta}) = \mathbf{\sigma - 2\theta\times\sigma}[/tex]
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