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

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.

• This model seems to be used in the analysis of phenomena that frequently cause transitions between $| 1>$ and $| 2>$. If we choose eigenstates as the basis, it will probably be "trapped" in eigenstates and state transitions will not occur. Am I right? Apr 19, 2020 at 2:06

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:
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.
• Thank you for your answer. I'd like to confirm if I understand your explanation. Under the Hamiltonian of (1-3'), (1) neither $|1⟩$ nor $|2⟩$ are stable state, (2) both $\varphi _{-\sqrt{2}}$ and $\varphi _{\sqrt{2}}$ are stable states, and (3) $\varphi _{-\sqrt{2}}$ is the ground state but (4) $\varphi _{\sqrt{2}}$ is not ground state, am I right? (5)The "state of definite energy" you said means stable state? Apr 19, 2020 at 1:55