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I'm reading a paper[1] in which the propagator is calculated for this kind of Hamiltonian

\begin{align} \hat{H}(t) = \omega(t)\hat{J}_3 + \Omega^*(t)\hat{J}_{+} + \Omega(t)\hat{J}_-. \end{align}

The operators $\hat{J}_i$ satisfy an $SU(2)$ algebra:

\begin{align} \left[\hat{J}_3,\hat{J}_{\pm} \right] &= \pm\hat{J}_{\pm},\\ \left[\hat{J}_+,\hat{J}_- \right] & = -2\hat{J}_3. \end{align}

Then, it is stated that the time evolution operator can be written as a time ordered product of exponentials

\begin{align} \hat{U}(t) &= \exp\left[ \left( h(t) - i\int_0^t\omega(t')dt' \right)\hat{J}_3\right]\\ &\times\exp\left[ g(t)\hat{J}_+ \right]\exp\left[f(t)\hat{J}_-\right], \end{align}

which is the Wei-Norman form.

I have doubts whether this propagator is unitary. Taking the hermitian conjugate doesn't seem to give the inverse of it, but it must be unitary since the system is closed. Can anybody help me whith this?

[1] Phys. Rev. A 35, 1582 (1987).

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