- #1
IsNoGood
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Homework Statement
I'm trying to comprehend
[itex]\hat{P}_t^{-1} \left( \vec{\sigma} \cdot \vec{A} \right) \hat{P}_t = \
\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]
[/itex]
with [itex] \vec{\sigma} [/itex] as the usual vector of pauli matrices, [itex] \vec{A} [/itex] as an (more or less) arbitrary operator vector and [itex] \hat{a} [/itex] as the axis of the rotation represented by [itex]\hat{P}_t[/itex].
Homework Equations
I already know [itex] \left[ \vec{\sigma},\vec{A} \right]_- = \left[ \vec{\sigma},\hat{a} \right]_- = \left[ \hat{a},\vec{A} \right]_- = 0[/itex].
Further on, the following identities are given (time dependencies [itex]\left(t\right)[/itex] omitted):
(I) [itex]
\hat{P}_t = \cos{\frac{\Psi}{2}} - i\left(\vec{\sigma}\cdot\hat{a}\right) \sin{\frac{\Psi}{2}}
[/itex]
(II) [itex]
\left( \vec{m}\cdot\vec{\sigma} \right) \left( \vec{n}\cdot\vec{\sigma} \right) = \
\vec{m}\cdot\vec{n} + i\vec{\sigma} \cdot \left( \vec{m} \times \vec{n} \right)
[/itex]
(III) [itex]
\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)
[/itex]
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:
[itex]\hat{P}_t^{-1} \left( \vec{\sigma} \cdot \vec{A} \right) \hat{P}_t
[/itex]
using (I), i obtain:
[itex] \left[\cos{\frac{\Psi}{2}} + i\left(\vec{\sigma}\cdot\hat{a}\right) \sin{\frac{\Psi}{2}}\right]\cdot\
\left( \vec{\sigma} \cdot \vec{A} \right)\cdot\
\left[\cos{\frac{\Psi}{2}} - i\left(\vec{\sigma}\cdot\hat{a}\right) \sin{\frac{\Psi}{2}}\right]
[/itex]
expanding, using [itex] \sin{\frac{\Psi}{2}}\cdot \cos{\frac{\Psi}{2}} = \frac{1}{2} \sin{\Psi} [/itex] yields:
[itex]
\cos^2{\frac{\Psi}{2}} \left(\vec{\sigma}\vec{A}\right) +\
\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] +\
\sin^2{\frac{\Psi}{2}} \left( \vec{\sigma}\hat{a} \right) \left( \vec{\sigma}\vec{A}\right) \left(\vec{\sigma}\hat{a}\right)
[/itex]
using (II) two times on [itex] \left( \vec{\sigma} \hat{a} \right) \left( \vec{\sigma} \vec{A} \right) - \left( \vec{\sigma} \vec{A} \right) \left( \vec{\sigma}\hat{a} \right) [/itex] together with [itex] \left[\hat{a},\vec{A}\right]_- = 0 [/itex] yields:
[itex]
\cos^2{\frac{\Psi}{2}} \left(\vec{\sigma}\vec{A}\right) -\
\sin{\Psi} \vec{\sigma} \left( \hat{a} \times \vec{A} \right) +\
\sin^2{\frac{\Psi}{2}} \left( \vec{\sigma}\hat{a} \right) \left( \vec{\sigma}\vec{A}\right) \left( \vec{\sigma}\hat{a} \right)
[/itex]
I'm reasonably sure so far, especially as [itex] -\sin{\Psi} \vec{\sigma} \left( \hat{a} \times \vec{A} \right) [/itex] 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:
[itex]
\cos^2{\frac{\Psi}{2}} \left(\vec{\sigma}\vec{A}\right) -\
\sin{\Psi} \vec{\sigma} \left( \hat{a} \times \vec{A} \right) +\
\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)
[/itex]
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!
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