Eigenstates of the Two-Level System

In the previous section we had seen that the total energy of the two tunnel-coupled wells is represented by the total Hamiltonian $\hat{H}$,

$$\hat{H}=\left( \begin{array}{cc} \varepsilon_L & T\\ T^* & \varepsilon_R \end{array}\right).$$ (203)

If we measure the energy, the possible outcomes are the eigenvalues of the corresponding observable, that is the total Hamiltonian $\hat{H}$. We therefore have to find the two eigenvectors $\vert{i}\rangle$ and eigenvalues $\varepsilon_{i}$ of $\hat{H}$, that is the solutions of

$$\hat{H}\vert{i}\rangle = \varepsilon_{i} \vert{i}\rangle,\quad i=1,2.$$ (204)

The result is

$$\vert 1\rangle = \frac{1}{N_1}\left[-2T \vert L\rangle + (\Delta +\varepsilon) \ve... ...quad \varepsilon_1=\frac{1}{2}\left(\varepsilon_L+\varepsilon_R - \Delta\right)$$

$$\vert 2\rangle = \frac{1}{N_2}\left[\phantom{-}2T \vert L\rangle + (\Delta -\varep... ...quad \varepsilon_2=\frac{1}{2}\left(\varepsilon_L+\varepsilon_R + \Delta\right)$$

$$\varepsilon := \varepsilon_L-\varepsilon_{R},\quad \Delta:=\varepsilon_2-\varepsilon_1=\sqrt{\varepsilon^2+4\vert T\vert^2}$$

$$N_{1,2} := \sqrt{4\vert T\vert^2+(\Delta\pm \varepsilon)^2}.$$ (205)

Figure: New hybridized basis states of the double well potential.

Discussion:

1. The eigenvectors $\vert 1\rangle$ and $\vert 2\rangle$ form a new orthonormal basis of the Hilbert space ${\cal H} = C^2$ (the $N_{1,2}$ are normalization factors).

2. The level splitting $\Delta$ gives the energy difference between the new eigenenergies. It increases with increasing $\vert T\vert$.

3. For $\varepsilon=0$, we find

$$\vert 1\rangle = \frac{1}{\sqrt{2}}\left[-(T/\vert T\vert) \vert L\rangle + \vert R\rangle \right]$$

$$\vert 2\rangle = \frac{1}{\sqrt{2}}\left[(T/\vert T\vert) \vert L\rangle + \vert R\rangle \right].$$ (206)

Remember that $T$ is a complex quantity. If we chose $T$ real and negative, i.e. $T=-\vert T\vert$, we find

$$\varepsilon_L=\varepsilon_R=:\varepsilon_0,\quad T=-\vert T\vert\leadsto \vert 1\rangle = \frac{1}{\sqrt{2}}\left[\vert L\rangle + \vert R\rangle \right],\quad \varepsilon_1=\varepsilon_0 - \vert T\vert$$

$$\vert 2\rangle = \frac{1}{\sqrt{2}}\left[ -\vert L\rangle + \vert R\rangle \right],\quad \varepsilon_2= \varepsilon_0 + \vert T\vert.$$ (207)

In particular, the symmetric linear combination $\vert 1\rangle$ now has a lower energy than the antisymmetric combination $\vert 2\rangle$: this is what we had found in the original double quantum well problem.