A1  ^?? What could the state vector be for the ground state? Remember that the wave function is definedup to the phase factor $e^{i \varphi}$, where $\varphi$ is a real number.

The ground state wave function is
\[\Psi = e^{i\varphi} \Psi_g.\]Then the state vector could be written just with the deffinition:

A2  ^?? Write down the Hamiltonian $\hat{\mathcal{H}}$ as a $2 \times 2$ matrix, if the energies of the ground state and excited state are equal to $E_g$ and $E_e$.

Hamiltonian should act on the state vector as it acts on the wave function:
\[ \hat{\mathcal{H}} \Psi = \hat{\mathcal{H}} (\alpha \Psi_e + \beta \Psi_g) = \alpha E_e \Psi_e + \beta E_g \Psi_g.\]Thus, we can write the Hamiltonian as the matrix:
\[ \hat{\mathcal{H}}\begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \alpha E_e \\ \beta E_g \end{pmatrix}.\]

A3  ^?? Let us assume that at the initial moment of time $| \psi(t=0) \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ 1\end{pmatrix}$. Using Schrödinger's equation
\[ i \hbar \frac{d |\psi(t) \rangle}{dt} = \hat{\mathcal{H}} |\psi(t) \rangle,\]express $|\psi(t) \rangle$ in terms of $E_g$ and $E_e$.

Let's rewrite Schrödinger's equation component by component:
\[i\hbar \begin{pmatrix} \dot{\alpha} \\ \dot{\beta} \end{pmatrix} = \begin{pmatrix}
E_e \alpha \\ E_g \beta \end{pmatrix}.\]
We got the system of two independent equations for $\alpha$ and $\beta$, which have the complex exponent solutions:
\[ \alpha(t) = \alpha(0) e^{-\frac{iE_et}{\hbar}}, \quad \beta(t) = \beta(0) e^{-\frac{iE_gt}{\hbar}}.\]

A4  ^?? Find the dependence of the average observed $\langle x(t) \rangle$ value of the coordinate of the TLS on time if its initial state is $| \psi(t=0) \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ 1\end{pmatrix}$. Express your answer in terms of $a$, $E_g$ and $E_e$.

Let's calculate straght forward:
\[ \langle x(t) \rangle = \frac{1}{2} \begin{pmatrix} e^{\frac{iE_et}{\hbar}} & e^{\frac{iE_gt}{\hbar}} \end{pmatrix} \begin{pmatrix} 0 & a \\ a & 0 \end{pmatrix} \begin{pmatrix} e^{-\frac{iE_et}{\hbar}} \\ e^{-\frac{iE_gt}{\hbar}} \end{pmatrix} = \frac{a}{2} \left( e^{\frac{it}{\hbar} (E_e - E_g) } + e^{-\frac{it}{\hbar} (E_e - E_g) } \right) = a \cos \frac{(E_e-E_g)t}{\hbar} \]