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For a general time dependent Hamiltonian, if the Hamiltonian at two different times $t_1,\,t_2$ satisfies $$\left[ \hat{H}(t_1),\hat{H}(t_2) \right]=0,$$ then the time evolution operator is $\hat{U}(t)= \exp\left( \frac{-i}{\hbar}\int_0^{t} \hat{H}(\tau) d\tau \right).$

The following is the effective Hamiltonian of a composite system of a two level system and a harmonic oscillator in the interaction picture. Here $\hat{\sigma}_+= |e\rangle\langle g|$ and $\hat{\sigma}_-= |g\rangle\langle e|$ where $|g\rangle,|e\rangle$ are the ground and excited state of the two level system respectively.

\begin{equation*} \hat{H}_{\text{eff}}(t)= \frac{\hbar g}{2}(\hat{\sigma}_+ +\hat{\sigma}_-)\otimes(\hat{a} e^{-i\delta t} + \hat{a}^\dagger e^{i\delta t} ). \end{equation*} We now find $\left[ \hat{H}_{\text{eff}}(t_1),\hat{H}_{\text{eff}}(t_2) \right]\neq 0.$ \begin{equation} \left[ \hat{H}_{\text{eff}}(t_1),\hat{H}_{\text{eff}}(t_2) \right]= -i\frac{\hbar^2 g^2}{8}\sin(\delta(t_2-t_1))\hat{\mathbb{I}}_{2}\otimes\hat{\mathbb{I}}_{QHO} \end{equation}

In the paper I'm following, the author simply proceeds and obtains a result for time evolution as if this $\hat{U}(t)= \exp\left( \frac{-i}{\hbar}\int_0^{t} \hat{H}(\tau) d\tau \right)$ holds true.

My first question would be, is it possible to actually treat this Hamiltonian this way and simply exponentiate without timeordering? My answer based on standard QM knowledge through textbooks is no you cannot do that, cf. e.g. this Phys.SE.

I'm currently reading on how one treats Hamiltonians periodic in time but have not so far found that to give any credibility to simply integrating and exponentiating.

This Hamiltonian of interest is clearly time periodic with period $T=\frac{2\pi}{\delta}$.

Any help is appreciated.

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2 Answers 2

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You are right that in general for a time-dependent Hamiltonian, one can not just exponentiate the time integral to get time evolution operator. However, this example is more of a special case for a particular initial state. Note that the Hamiltonian is a tensor product: $H=g\sigma^x(ae^{-i\delta t}+a^\dagger e^{i\delta t})$. The initial state studied right below Eq (5) can be written $|+\rangle |0\rangle + |-\rangle|0\rangle$. Let us consider the first term. Since $|+\rangle$ is an eigenstate of $\sigma^x$, we can basically replace $\sigma^x$ in the Hamiltonian by the eigenvalue $1$, so what remains is to figure out how $|0\rangle$ of the harmonic oscillator evolves according to $H=+g(ae^{-i\delta t}+a^\dagger e^{i\delta t})$. I believe it will indeed involve to the coherent state $|\alpha\rangle$ (I can check it, but will take more time). The key is that the Hamiltonian and the initial state are both tensor products.

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  • $\begingroup$ For it to evolve into a coherent state is not a problem since yes the this Hamiltonian IF (integrated then) exponentiated does give the displacement operator! But the problem still lies in the explicit time dependence, how/why is it justified to do that? I have no issue with the resulting state, but rather with the way it is arrived it (implicitly just integrating and exponentiating). $\endgroup$ May 10, 2022 at 6:41
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I think this answer will help you. The interesting part is equation (S) and (T). So the non-vanishing commutator only amounts to a global phase factor that can be neglected (since it's just an imaginary number). To get rid of the $\sigma_x$ operator (to match the definition of the Hamiltonian in the link above) you have to consider the evolution of the states $|\pm>|\psi>$ separatly, for which you can replace $\sigma_x$ by $\pm 1$. Of course, this was already pointed out in the answer by Meng Cheng.

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