Solving Algebraic Problem to Calculate Vector Operator Rotation

[IsNoGood]

Homework Statement

I'm trying to comprehend
\hat{P}_t^{-1} \left( \vec{\sigma} \cdot \vec{A} \right) \hat{P}_t = \<br /> \cos{\Psi\left(t\right)}\left( \vec{\sigma} \cdot \vec{A} \right) - \sin{\Psi\left(t\right)} \sigma \cdot \left[ \hat{a}\left(t\right) \times \vec{A} \right] + 2\sin^2{\frac{\Psi\left(t\right)}{2}} \left[ \hat{a}\left(t\right)\cdot\vec{A} \right]\left[\vec{\sigma}\cdot\hat{a}\left(t\right)\right]<br />
with \vec{\sigma} as the usual vector of pauli matrices, \vec{A} as an (more or less) arbitrary operator vector and \hat{a} as the axis of the rotation represented by \hat{P}_t.

Homework Equations

I already know \left[ \vec{\sigma},\vec{A} \right]_- = \left[ \vec{\sigma},\hat{a} \right]_- = \left[ \hat{a},\vec{A} \right]_- = 0.

Further on, the following identities are given (time dependencies \left(t\right) omitted):
(I) <br /> \hat{P}_t = \cos{\frac{\Psi}{2}} - i\left(\vec{\sigma}\cdot\hat{a}\right) \sin{\frac{\Psi}{2}}<br />
(II) <br /> \left( \vec{m}\cdot\vec{\sigma} \right) \left( \vec{n}\cdot\vec{\sigma} \right) = \<br /> \vec{m}\cdot\vec{n} + i\vec{\sigma} \cdot \left( \vec{m} \times \vec{n} \right)<br />
(III) <br /> \vec{m}\times\left(\vec{n}\times\vec{l}\right) = \vec{n}\left(\vec{m}\vec{l}\right) - \vec{l}\left(\vec{m}\vec{n}\right)<br />

Just in case I forgot something important, the problem appears in Physical Review A 80, 022328, page 3 (http://pra.aps.org/abstract/PRA/v80/i2/e022328" ).

The Attempt at a Solution

I desperately reproduced the following steps over and over again (so I'm relatively sure they are correct). But I just don't know where to go from there:

\hat{P}_t^{-1} \left( \vec{\sigma} \cdot \vec{A} \right) \hat{P}_t<br />

using (I), i obtain:
\left[\cos{\frac{\Psi}{2}} + i\left(\vec{\sigma}\cdot\hat{a}\right) \sin{\frac{\Psi}{2}}\right]\cdot\<br /> \left( \vec{\sigma} \cdot \vec{A} \right)\cdot\<br /> \left[\cos{\frac{\Psi}{2}} - i\left(\vec{\sigma}\cdot\hat{a}\right) \sin{\frac{\Psi}{2}}\right]<br />

expanding, using \sin{\frac{\Psi}{2}}\cdot \cos{\frac{\Psi}{2}} = \frac{1}{2} \sin{\Psi} yields:
<br /> \cos^2{\frac{\Psi}{2}} \left(\vec{\sigma}\vec{A}\right) +\<br /> \frac{i}{2} \sin{\Psi} \left[ \left( \vec{\sigma} \hat{a} \right) \left( \vec{\sigma} \vec{A} \right) - \left( \vec{\sigma} \vec{A} \right) \left( \vec{\sigma}\hat{a} \right) \right] +\<br /> \sin^2{\frac{\Psi}{2}} \left( \vec{\sigma}\hat{a} \right) \left( \vec{\sigma}\vec{A}\right) \left(\vec{\sigma}\hat{a}\right)<br />

using (II) two times on \left( \vec{\sigma} \hat{a} \right) \left( \vec{\sigma} \vec{A} \right) - \left( \vec{\sigma} \vec{A} \right) \left( \vec{\sigma}\hat{a} \right) together with \left[\hat{a},\vec{A}\right]_- = 0 yields:
<br /> \cos^2{\frac{\Psi}{2}} \left(\vec{\sigma}\vec{A}\right) -\<br /> \sin{\Psi} \vec{\sigma} \left( \hat{a} \times \vec{A} \right) +\<br /> \sin^2{\frac{\Psi}{2}} \left( \vec{\sigma}\hat{a} \right) \left( \vec{\sigma}\vec{A}\right) \left( \vec{\sigma}\hat{a} \right)<br />

I'm reasonably sure so far, especially as -\sin{\Psi} \vec{\sigma} \left( \hat{a} \times \vec{A} \right) is a part of the solution. However, I can't see how (III) comes into play. The best i tried further on is again using (II) yielding:
<br /> \cos^2{\frac{\Psi}{2}} \left(\vec{\sigma}\vec{A}\right) -\<br /> \sin{\Psi} \vec{\sigma} \left( \hat{a} \times \vec{A} \right) +\<br /> \sin^2{\frac{\Psi}{2}} \left[ \hat{a}\vec{A} + i\vec{\sigma} \left(\hat{a} \times \vec{A} \right) \right] \left( \vec{\sigma}\hat{a} \right)<br />

However, this yet leaves me without any good idea how to go on.
I guess there is "just" some nifty algebra trick I constantly fail to see ... so every help is greatly appreciated.

Thank you in advance!

 

[IsNoGood]

It's astonishing how long one can stare at an expression without the slightest idea until suddenly out of nowhere it seems absolutely clear where to go.
I'm not done yet because I've got something different to do, but I think I finally got the "nifty trick".
Will post again if it turns out to be correct!

 

[IsNoGood]

OK, did the calculation, everything is fine now.