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I need help with a mathematical physics question. I have given the following system:

A spin is coupled to a single harmonic oscillator mode with Hamiltonian $$H=(\epsilon/2) \sigma_{z} + \omega\, a^{\star} a + (\lambda/2) \sigma_{z} \otimes (a^{\star} + a) $$

which acts on the Hilbert space

$$ \mathbb C^{2} \otimes \mathcal H_{\text{osc}} $$ and $a^{\star} a \equiv 1 \otimes a^{\star} a , \sigma_{z}\equiv \sigma_{z} \otimes 1 $.

Task: Calculate explicitly the reduced density spin density matrix

$$\rho(t)= \operatorname{Tr}_{R} (e^{-itH} (\rho(0)\otimes\rho_{R}) e^{itH} ) $$

with the following properties:

$$ \operatorname{Tr}_{R}\rho_{R} e^{(i(a^{\star}+a))}= e^{-1/2 \coth(\beta\omega /2)}$$ and $\rho_{R}=1/Z_{\beta}\times e^{-\beta\omega a^{\star}a}, Z_{\beta}=(1-e^{-\beta\omega})^{-1}$.

Use that $(P_{1}\otimes 1)H=P_{1}\otimes H_{1}$ and $(P_{2}\otimes 1)H=P_{2}\otimes H_{2}$ with

$$H_{1}=(\epsilon/2) + \omega a^{\star} a + (\lambda/2) (a^{\star} + a) $$

and

$$H_{2}=-(\epsilon/2) + \omega a^{\star} a - (\lambda/2) (a^{\star} + a) $$

and one can use that

$$\rho(t) = \sum _{i,j\in \{1,2\}} P_{i} \rho (0) P_{j} c_{ij}$$ with $c_{ij}= \operatorname{Tr}_{R}e^{-itH_{i}}\rho_{R}e^{itH_{j}}$.


So now when I calculate $c_{ij}$ I get for $i=j$ since $\operatorname{Tr}(A^{-1}BA=\operatorname{Tr}(B))$

$$c_{ii}=\operatorname{Tr}(\rho_{R})=1$$ as $\rho_R$ is a density matrix.

Also if I take $i\neq j$ I can't seem to find an answer. I have so far

$$ c_{21}= \operatorname{Tr}_{R}e^{-itH_{2}}\rho_{R}e^{itH_{1}} = \operatorname{Tr}_{R}e^{itH_{1}}e^{-itH_{2}}\rho_{R}$$

I realize I should be using property 1 here but can't seem how. Any help would be greatly appreciated.

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  • $\begingroup$ You surely mean $\sigma_z\equiv \sigma_z\otimes 1$ in your explication above, right? $\endgroup$ – Cosmas Zachos Sep 18 '17 at 16:24

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