In the quantum mechanics, special states such as eigenstate, basis states, and ground state are defined. I may be know these definitions, but I'm very confusing; this confusion occurred when I studying the Two-state quantum system. My questions are shown below.
【My questions】
Is it possible to realize a Two-state quantum system having basis states $|1>$ and $|2>$ under the Hamiltonian of (1-3)' below ? If so, why?
Here, $|1>$ and $|2>$ are expressed as follows. $$|1>:=\left(\begin{matrix}1\\0\\\end{matrix}\right) \tag{1-1}$$ $$|2>:=\left(\begin{matrix}0\\1\\\end{matrix}\right) \tag{1-2}$$
According to the Wikipedia, the Hamiltonian $\hat{H}$ of the two-state quantum system, whose basis states are $|1>$ and $|2>$ should be written in the form of (1-3). Here, ${\varepsilon_1}$ and ${\varepsilon_2}$ are real number, $\gamma$ is a complex number, and $\bar{\gamma}$ is a conjugate complex number of the $\gamma$.
$$\hat{H}={\varepsilon_1}\left|1><1|\ +{\varepsilon_2}\right|2><2|\ +\ \bar{\gamma}|1><2|\ +\ \gamma|2><1| \tag{1-3}$$
Following (1-4) is the matrix representation of the $\hat{H}$ of (1-3) $$\left(\begin{matrix}\varepsilon_1&\bar{\gamma}\\ \gamma &\varepsilon_2\\\end{matrix}\right)\tag{1-4}$$
Therefore, the Hamiltonian of equation (1-4) 'should be appropriate as the Hamiltonian of the two-state system; this is just a special case of ${\varepsilon_1} =\gamma=1$, and ${\varepsilon_2} =-1$ in the (1-3). $$\hat{H}=\left|1><1|\ -\right|2><2|\ +\ |1><2|\ +\ |2><1| \tag{1-3'}$$
Following (1-4)' is the matrix representation of the $\hat{H}$ of (1-3)'. $$\left(\begin{matrix}1&1\\1&-1\\\end{matrix}\right)\tag{1-4'}$$
From the assumption, the possible states are $|1>$ or $|2>$. Or, the $|1>$ or $|2>$ are the most stable states.
On the other hand, the $\varphi$ is an eigenstate of the hamiltonian $\hat{H}$ iff the $\varphi$ is the eigenvector of the $\hat{H}$. In this sense, eigenstates are nothing but the following $\varphi_{-\sqrt{2}}$ and $\varphi_{\sqrt{2}}$.
$$\varphi_{-\sqrt2}:=\frac{1}{\sqrt{2\left(2+\sqrt2\right)}}\left(\begin{matrix}1\\-\sqrt{2\ }\ -1\\\end{matrix}\right)$$ $$\varphi_{\sqrt2}:=\frac{1}{\sqrt{2\left(2-\sqrt2\right)}}\left(\begin{matrix}1\\ \sqrt{2\ }\ -1\\\end{matrix}\right)$$
The smallest eigenvalue is ${-\sqrt{2}}$, so $\varphi_{-\sqrt{2}}$ might be the ground state, but neither $|1>$ nor $|2>$ might be ground state; the expected value of energy of $|1>$ and $|2>$ seems higher than $-\sqrt{2}$, therefore, I have no idea why $|1>$ and $|2>$ are 'preferred' state; these are unlikely to be the stablestationary state.
