Hamiltonian after transformation to interaction picture

[lutherblissett]

Dear all,

I am encoutering some difficulties while calculating the Hamiltonian after the transformation to the interaction picture. I am following the tutorial by Sasura and Buzek:
https://arxiv.org/abs/quant-ph/0112041

Previous:
I already know that the Hamiltonian for the j-th ion is given by two terms:

$$\hat{H} = \hat{H}_{0j} + \hat{V}_j$$

where the two terms are the free Hamiltonian and the interaction hamiltonian
$$\hat{H}_{0j} = \dfrac{1}{2} \hbar \omega_0 \sigma_{zj} + \nu \hat{a}^\dagger \hat{a}$$
$$\hat{V}_j = -q_e\left [ \left (\mathbf{r}_{eg}\right )_j \hat{\sigma}_{+j} + \left (\mathbf{r}_{eg}\right )^*_j \hat{\sigma}_{-j} \right ] \cdot \dfrac{E_0 \varepsilon }{2} \left \{ e^{ -i \left [ \omega_L t - \eta_j \left ( \hat{a}^\dagger + \hat{a} \right ) + \phi_j \right ] } + \text{ h.c.} \right \}$$

I don't understand how one can obtain the transformed Hamiltonian $\hat{\mathcal{H}} = \hat{U}^\dagger_0 \hat{V} \hat{U}_0$ where $\hat{U}_0 = exp \left \{ - \dfrac{i}{\hbar} \hat{H}_0 t \right \}$.

Could someone explain me the steps to obtain it, please? I will be very grateful :D

thank you in advance

LB

 

[Paul Colby]

I recommend taking as much advantage of commutativity as possible. The unperturbed operator is the sum of two operator which commute. This allows $U_0$ to be factored into two commuting unitary operators. One then commutes the various terms such that the spin factors and mode operators are each transformed by their respective unitary transforms. Also recall that $U^\dagger f(\hat{o}) U = f(U^\dagger \hat{o} U)$ where $f$ is a c-number function and $\hat{o}$ is some unsuspecting operator.

 

[lutherblissett]

Thank you Paul, I've done it! Your hint on c-number functions has been fundamental.

Thank you.

LB