The eigenstates, basis states, and the ground state on two-state quantum system

Question

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 stationary state.

Answer 1

Your Hamiltonian $\hat H$ in (1-4') is not diagonal so $\vert 1\rangle$ and $\vert 2\rangle$ are not eigenstates of $\hat H$. Thus, neither $\vert 1\rangle$ nor $\vert 2\rangle$ are the most stable, but some linear combination of them is. Alternatively, neither or them is "favoured" in any ways: they are just basis states useful in computing matrix elements (here of $\hat H$). The linear combination you have found to be the most stable is the one you write as $\varphi_{-\sqrt{2}}$.

Since neither $\vert 1\rangle$ nor $\vert 2\rangle$ are eigenstates, the expectation values of $\hat H$ in those states will not correspond to the lowest eigenvalue of the system. However, if you compute the expectation value of $\hat H$ for the state $\varphi_{-\sqrt{2}}$ then you should find that:

1. this expectation value is the lowest eigenvalue,
2. The fluctuation $\Delta E=0$, meaning this state has definite energy (the energy does not fluctuate). This is in contradistinction with - say - computing $\Delta E$ for the system in state $\vert 1\rangle$, which will give you a non-zero value, indicating that $\vert 1\rangle$ is not a state of definite energy.