A quantum mechanical system has two states $\chi_{0}$ and $\chi_{1}$, which are normalised energy eigenstates of a Hamiltonian $\mathrm{H}_{3}$, with

$$H_{3} \chi_{0}=-\chi_{0}, \quad H_{3} \chi_{1}=+\chi_{1} .$$

A general time-dependent state may be written

$$\Psi(t)=a_{0}(t) \chi_{0}+a_{1}(t) \chi_{1},$$

where $a_{0}(t)$ and $a_{1}(t)$ are complex numbers obeying $\left|a_{0}(t)\right|^{2}+\left|a_{1}(t)\right|^{2}=1$.

(a) Write down the time-dependent Schrödinger equation for $\Psi(t)$, and show that if the Hamiltonian is $\mathrm{H}_{3}$, then

$$i \hbar \frac{d a_{0}}{d t}=-a_{0}, \quad i \hbar \frac{d a_{1}}{d t}=+a_{1} .$$

For the general state given in equation (1) above, write down the probability to observe the system, at time $t$, in a state $\alpha \chi_{0}+\beta \chi_{1}$, properly normalised so that $|\alpha|^{2}+|\beta|^{2}=1$.

(b) Now consider starting the system in the state $\chi_{0}$ at time $t=0$, and evolving it with a different Hamiltonian $H_{1}$, which acts on the states $\chi_{0}$ and $\chi_{1}$ as follows:

$$H_{1} \chi_{0}=\chi_{1}, \quad H_{1} \chi_{1}=\chi_{0}$$

By solving the time-dependent Schrödinger equation for the Hamiltonian $H_{1}$, find $a_{0}(t)$ and $a_{1}(t)$ in this case. Hence determine the shortest time $T>0$ such that $\Psi(T)$ is an eigenstate of $H_{3}$ with eigenvalue $+1$.

(c) Now consider taking the state $\Psi(T)$ from part (b), and evolving it for further length of time $T$, with Hamiltonian $H_{2}$, which acts on the states $\chi_{0}$ and $\chi_{1}$ as follows:

$$H_{2} \chi_{0}=-i \chi_{1}, \quad H_{2} \chi_{1}=i \chi_{0}$$

What is the final state of the system? Is this state observationally distinguishable from the original state $\chi_{0}$ ?