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If I have a quantum system described by a time-independent Hamiltonian $\hat{H}$, then the Liouville-von Neumann equation is

\begin{align} i\hbar\frac{\partial\hat{\rho}}{\partial t}=[\hat{H},\hat{\rho}]\,, \end{align}

where $\hat{\rho}$ is the density matrix. What happens if the Hamiltonian is explicitly time-dependent, such that $\hat{H}=\hat{H}(t)$? Is the Liouville-von Neumann equation the same?

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If the Hamiltonian is time-dependent the evolution of a pure state is $$ |\psi(t)\rangle = {\mathcal T} exp\left[ -\frac{i}{\hbar} \int_{t_0}^t d\tau H(\tau)\right] |\psi(0)\rangle = U(t; t_0) |\psi(0)\rangle $$ where $\mathcal T$ is the time-ordering operator and $$ i\hbar\dot{U} = H(t) U(t, t_0) $$ Then a density matrix evolves according to $$ \rho(t) = U(t; t_0) \rho(0) U^\dagger (t; t_0) $$ and you can check that taking the time derivative gives $$ i\hbar\dot{\rho} = \left[ H(t), \rho(t)\right] $$

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  • $\begingroup$ I have $\frac{\partial}{\partial t}\exp\left[-\frac{i}{\hbar}\int_{t_0}^tH(\tau)d\tau\right]=\frac{1}{i\hbar}H(t)\exp\left[-\frac{i}{\hbar}\int_{t_0}^tH(\tau)d\tau\right]$, but how do I treat the time-ordering operator? Is it true that $\frac{\partial}{\partial t}\left(\mathcal{T}\exp\left[-\frac{i}{\hbar}\int_{t_0}^tH(\tau)d\tau\right]\right)=\mathcal{T}\left(\frac{\partial}{\partial t}\exp\left[-\frac{i}{\hbar}\int_{t_0}^tH(\tau)d\tau\right]\right)$? $\endgroup$
    – user85503
    Mar 1, 2017 at 20:27
  • $\begingroup$ If this is so, then why would $\mathcal{T}\left(H(t)\exp\left[-\frac{i}{\hbar}\int_{t_0}^tH(\tau)d\tau\right]\right)=\left(H(t)\mathcal{T}\exp\left[-\frac{i}{\hbar}\int_{t_0}^tH(\tau)d\tau\right]\right)$? $\endgroup$
    – user85503
    Mar 1, 2017 at 20:27
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    $\begingroup$ No, that's not how it works. You may want to take a look at Sec.5.2.1 in ocw.mit.edu/courses/nuclear-engineering/…. The time ordered exponential ${\mathcal T}\exp\left[-(i/\hbar)\int_{t_0}^t{d\tau\; H(\tau)}\right]$ is a compact way of writing a series where each term is a nested integral of the form $$\int_{t_0}^t{d\tau_1\int_{t_0}^{t_1}{d\tau_2\; \dots \int_{t_0}^{t_{n-1}}{d\tau_n H(\tau_1) H(\tau_2)\dots H(\tau_n)}}}$$. $\endgroup$
    – udrv
    Mar 2, 2017 at 5:59

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