Working with an uncoupled harmonic oscillator Hamiltonian: $H = H_A + H_B$, where $H_A = \hbar \omega (a_+ a_{-} + 1/2 )$ and $H_B = B \sigma_z$.

I'm trying to calculate the density matrix $\rho (t)$, assuming that the spin state is $| \uparrow_x \rangle$ at $t=0$.

I know that $\rho (0) = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}$ (when expressed in terms of the $| \uparrow, \downarrow \rangle$-basis) and that $\rho (t) = U^{\dagger} \rho (0) U$. However, once I plug in:

$e^{-i (H_A + H_B) t /\hbar} \rho (0) e^{i (H_A + H_B) t /\hbar}$ I have no idea how to simplify this any further. Any hints?

  • $\begingroup$ Hey, thanks for the reponse. No, that's where I am stuck. Once I write up $e^{i \hbar \omega (a_+ + a_{-} )t/ \hbar} e^{i B \sigma_z t / \hbar} \rho (0) e^{-i \hbar \omega (a_+ + a_{-} )t/ \hbar} e^{-i B \sigma_z t / \hbar}$, I don't know how to proceeed any further? I assume there is some identity that could help. $\endgroup$ – user103984 Sep 30 '17 at 10:36
  • $\begingroup$ Well... I think part of the problem you have is that the action of $a_+a_-$ on your $\vert \uparrow_x\rangle$ and $\vert \downarrow_x\rangle$ states is not clear. if you can figure out what’s mean by this you can work out $H_A+H_B$ as a $2\times 2$ matrix and then work on getting the exponential. $\endgroup$ – ZeroTheHero Oct 1 '17 at 0:14

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