Time-evolution with a time-dependent Hamiltonian [closed]

Consider a quantum mechanical system whose Hilbert space of states is $\mathbb{C}^2$, and has Hamiltonian $$\hat{H}= \begin{pmatrix} E_0e^{t/w_0} & E_1 \\ E_1 & E_0e^{t/w_0} \end{pmatrix}$$ (a) Describe the time evolution of the system.

I am not exactly sure how exactly to start-off here, and guessing that the time-evolution operator, $U$ is required. I know that $\hat{U} = e^{- \frac{i}{\hbar} \hat{H} t}$ but not sure how to proceed further.

closed as off-topic by AccidentalFourierTransform, Kyle Kanos, Jon Custer, rob♦Sep 6 '17 at 1:22

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The identity $\hat{U} = e^{- \frac{i}{\hbar} \hat{H} t}$ holds only for a time-independent hamiltonian, which does not apply here. Instead, the propagator here is given by the time-ordered exponential $\hat{U}(t_2,t_1) = \mathcal T \left[e^{- \frac{i}{\hbar} \int_{t_1}^{t_2} \hat{H}(t) \mathrm d t}\right]$, which is not particularly useful in this situation.
In your situation, you've got very few options left to you other than direct solution of the coupled Schrödinger equations, \begin{align} i\hbar \dot a(t) & = E_0e^{t/w_0} a(t) + E_1 b(t), \\ i\hbar \dot b(t) & = E_1 a(t) + E_0e^{t/w_0} b(t). \end{align} This one is hard to solve but you can make a start by setting $E_1=0$, in which case both $a$ and $b$ obey the differential equation $$i\hbar \dot c(t) = E_0e^{t/w_0} c(t),$$ whose solution is $$c(t)= c(0)e^{-i e^{t/w_0}E_0 w_0/\hbar},$$ so you can ride on that and define $a(t) = \alpha(t) e^{-i e^{t/w_0}E_0 w_0/\hbar}$ and $b(t) = \beta(t) e^{-i e^{t/w_0}E_0 w_0/\hbar}$, for which the Schrödinger equation simplifies to \begin{align} i\hbar \dot \alpha(t) & =E_1 \beta(t) , \\ i\hbar \dot \beta(t) & = E_1 \alpha(t) , \end{align} whose solutions are \begin{align} \alpha(t) & = \alpha(0) \cos(E_1t/\hbar) -i \beta(0) \sin(E_1t/\hbar)\\ \beta(t) & = -i\alpha(0) \sin(E_1t/\hbar) + \beta(0) \cos(E_1t/\hbar). \end{align} Anything beyond that will depend on exactly what you want to do with those solutions.
Actually in this specific case you are helped a bit because the time dependence is contained in a term proportional to the unit matrix $$\hat H = E_0e^{t/\omega_0}\hat I + E_1\left(\begin{array}{cc} 0&1 \\ 1&0 \end{array}\right) \, .$$ You can then make the time independent change of basis defined by $$U=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1&1 \\ 1&-1\end{array}\right)$$ that will take $\hat H$ to $$\hat h= U^{-1}\cdot \hat H \cdot U= \left(\begin{array}{cc} e^{t/\omega_0}E_0+E_1& 0 \\ 0&e^{t/\omega_0}E_0-E_1 \end{array}\right)$$ The Schrodinger equation for each component in this basis is uncoupled and of the form: $$i\hbar \frac{d}{dt}\vert\phi_{\pm}(t)\rangle = \left(e^{t/\omega_0}E_0\pm E_1\right)\vert\phi_{\pm}(t)\rangle$$ with solution $$\vert \phi_{\pm}(t)\rangle=C_\pm e^{-i(\pm E_1 t +e^{t/\omega_0}E_0\omega_0)/\hbar}$$ and you can go back to the original basis. You can verify that this solution is NOT of the form $e^{-i t H/\hbar}$ precisely because, as noted by @EmilioPisanty, $U(t)=e^{-i t H/\hbar}$ is only valid for time-independent Hamiltonians.