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I would like to calculate the Pauli spin operator rotation $$ U^{\dagger } \overset{\rightharpoonup }{\sigma } U$$

where $$\overset{\rightharpoonup }{\sigma }=\sigma _x \overset{\rightharpoonup }{x}+\overset{\rightharpoonup }{y} \sigma _z+\sigma _z \overset{\rightharpoonup }{z}$$

and

$$U = e^{i t \omega \sigma _z}$$

I know that this operation can be considered a rotation of the eigenvalues of the operators:

$$ U^{\dagger } \sigma _x U = \underset{n}{\Sigma } e_n U^{\dagger } |n'\rangle \langle n'| U=\underset{n}{\Sigma } e_n |n'\rangle \langle n'|$$

and these eigenvalues represent the rotation axis of the overall matrix $U^{\dagger } \overset{\rightharpoonup }{\sigma } U$ (or Hamiltonian in my case of interest).

In other words, this should be really straightforward to compute without need a lot of algebra or Baker–Campbell–Hausdorff formulas. However I don't quite see how to do it most straightforwardly; any thoughts?

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$ {\rm U}\left(t\right) \equiv \exp\left({\rm i}\,\omega t\,\sigma_{z}\right) $ satisfies $\ddot{\rm U}\left(t\right) + \omega^{2}{\rm U}\left(t\right) = 0$ with ${\rm U}\left(0\right) = 1$ and $\dot{{\rm U}}\left(0\right) = {\rm i}\omega\sigma_{z}$. Then

\begin{align} {\rm U}\left(t\right) &= \exp\left({\rm i}\,\omega t\,\sigma_{z}\right) = \cos\left(\omega t\right) - {\rm i}\sin\left(\omega t\right)\,\sigma_{z} \\[3mm] {\rm U}^{\dagger}\left(t\right) &= \exp\left(-{\rm i}\,\omega t\,\sigma_{z}\right) = \cos\left(\omega t\right) + {\rm i}\sin\left(\omega t\right)\,\sigma_{z} \end{align}

Define $A\left(t\right) \equiv {\rm U}^{\dagger}\left(t\right)A{\rm U}\left(t\right)$ such that ${\rm i}\dot{A} = \omega\left[\sigma_{z},A\left(t\right)\right]$.

$$ {\rm i}\dot{\sigma}_{x}\left(t\right) = \omega\left[{\rm i}\sigma_{y}\left(t\right)\right]\,, \qquad {\rm i}\dot{\sigma}_{y}\left(t\right) = \omega\left[-{\rm i}\sigma_{x}\left(t\right)\right]\,; \qquad {{\rm d}\left[\sigma_{x}\left(t\right) + {\rm i}\sigma_{y}\left(t\right)\right] \over {\rm d}t} = -{\rm i}\omega \left[\sigma_{x}\left(t\right) + {\rm i}\sigma_{y}\left(t\right)\right] $$

\begin{align} \sigma_{x}\left(t\right) + {\rm i}\sigma_{y}\left(t\right) &= \left(\sigma_{x} + {\rm i}\sigma_{y}\right)\exp\left(-{\rm i}\omega t\right) \\[3mm] \sigma_{x}\left(t\right) - {\rm i}\sigma_{y}\left(t\right) &= \left(\sigma_{x} - {\rm i}\sigma_{y}\right)\exp\left({\rm i}\omega t\right) \end{align} Then $$ \sigma_{x}\left(t\right) = \sigma_{x}\cos\left(\omega t\right) + \sigma_{y}\sin\left(\omega t\right)\,, \qquad \sigma_{y}\left(t\right) = -\sigma_{x}\sin\left(\omega t\right) + \sigma_{y}\cos\left(\omega t\right) $$

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  • $\begingroup$ I'm familiar with the Euler expansion for the exponentiated pauli operator, but I dont immediately see how it follows from the differential equation statement? Or is it not a constructive derivation, in other words, are you just saying that both, exponentials and sin/cos will satisfy the 2nd Order ODE. And then you match boundary conditions for the two ways of expressing the particular solution? $\endgroup$ Commented Sep 25, 2013 at 2:02
  • $\begingroup$ @AimForClarity That works since $\sigma_{i}^{2} = 1\,, i = x, y, z$. $\endgroup$ Commented Sep 25, 2013 at 2:04
  • $\begingroup$ oh ok so you're Taylor expanding the exponential? $\endgroup$ Commented Sep 25, 2013 at 2:10

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