Question: Consider a quantum system with just three linearly independent states. Suppose the Hamiltonian, in matrix form, is

$$\begin{gathered} \mathbf{H}\mathbf{=}{\mathbf{V}}_{\mathbf{0}}\left(\left(1-\dot{o}\right)00 \\ 00\dot{o} \\ 0\dot{o}2\right) \end{gathered}$$

Where${\mathbf{V}}_{\mathbf{0}}$is a constant, and$\dot{\mathrm{o}}$is some small number$\left(\in \ll 1\right)$.

(a) Write down the eigenvectors and eigenvalues of the unperturbed Hamiltonian$\mathbf{(}\dot{\mathbf{o}}\mathbf{=}\mathbf{0}\mathbf{)}$.

(b) Solve for the exact eigen values of H. Expand each of them as a power series in$\dot{\mathbf{o}}$, up to second order.

(c) Use first- and second-order non degenerate perturbation theory to find the approximate eigen value for the state that grows out of the non-degenerate eigenvector of${\mathit{H}}^{\mathbf{0}}$. Compare the exact result, from (a).

(d) Use degenerate perturbation theory to find the first-order correction to the two initially degenerate eigen values. Compare the exact results.

a)

In this problem we study the Hamiltonian

$$\begin{gathered} \mathrm{H}={\mathrm{V}}_0\left(\left(1-\dot{\mathrm{o}}\right)00 \\ 00\dot{\mathrm{o}} \\ 0\dot{\mathrm{o}}2\right) \end{gathered}$$

The unperturbed Hamiltonian is obtained by setting$\dot{\mathrm{o}}=\mathbf{0}$ and is of the form

Since it is diagonal, its Eigen values are

$V_0,V_0,2V_0$

And

the eigenvectors are

$$\begin{gathered} {\mathrm{V}}_{1=}\left(1 \\ 0 \\ 0\right)V_2=\left(0 \\ 1 \\ 0\right)V_3=\left(0 \\ 0 \\ 1\right) \end{gathered}$$

b)

We now diagonalize the total Hamiltonian by solving the equation

$$\begin{gathered} \left|{\mathrm{V}}_0(1-\dot{\mathrm{o}})-\mathrm{\lambda }00 \\ 0{\mathrm{V}}_0-\mathrm{\lambda }{\mathrm{V}}_0\dot{\mathrm{o}} \\ 0{\mathrm{V}}_0\dot{\mathrm{o}}2{\mathrm{V}}_0-\mathrm{\lambda }\right|=0 \\ \left(V_0\right(1-\dot{\mathrm{o}})-\lambda )({\mathrm{V}}_0-\mathrm{\lambda })(2{\mathrm{V}}_0-\mathrm{\lambda })-{\mathrm{V}}_0^2{\dot{\mathrm{o}}}^2\left({\mathrm{V}}_0\right(1-\dot{\mathrm{o}})-\mathrm{\lambda })=0 \\ \left({\mathrm{V}}_0\right(1-\dot{\mathrm{o}})-\mathrm{\lambda })({\mathrm{V}}_0-\mathrm{\lambda })(2{\mathrm{V}}_0-\mathrm{\lambda })-{\mathrm{V}}_0^2{\dot{\mathrm{o}}}^2=0 \end{gathered}$$

The one eigen value is

${\lambda }_1=V_0(1-\dot{\mathrm{o}})$

and the other two are obtained as

$$\begin{gathered} {\lambda }^2-3V_0\lambda +V_0(2-{\dot{\mathrm{o}}}^2)=0 \\ {\lambda }_{2,3}=\frac{1}{2}3V_0\pm \sqrt{9V_0^2-4V_0^2(2-{\dot{\mathrm{o}}}^2)} \\ =\frac{V_0}{2}3\pm \sqrt{1+4{\dot{\mathrm{o}}}^2} \end{gathered}$$

We can expand these Eigen values with respect to to obtain

$$\begin{gathered} {\lambda }_1=V_0(1-\dot{\mathrm{o}}), \\ {\lambda }_2=\frac{V_0}{2}3+\sqrt{1+4{\dot{\mathrm{o}}}^2}\approx \frac{V_0}{2}(3+1+2{\dot{\mathrm{o}}}^2) \\ =V_{0}2+{\dot{\mathrm{o}}}^2 \\ {\lambda }_3=\frac{V_0}{2} \\ 3-\sqrt{1+4{\dot{\mathrm{o}}}^2}\approx \frac{V_0}{2} \\ (3-1-2{\dot{\mathrm{o}}}^2)=V_{0}1-{\dot{\mathrm{o}}}^2 \end{gathered}$$

We have used the Taylor expansion of the square root $\sqrt{1+\dot{\mathrm{o}}}\approx 1+\dot{\mathrm{o}}/2$.

c)

We now observe the perturbed part of the Hamiltonian

$$\begin{gathered} H=\dot{o}{V}_0\left[-100 \\ 001 \\ 010\right] \end{gathered}$$

By setting$\dot{o}=0$we see that ${\lambda }_1={\lambda }_2$, so we are calculating the corrections for $E_3$. The first-order correction is

The second-order corrections are

We have

localid="1658203255352"

Therefore,

$$\begin{gathered} E_a^2=\frac{{\left(\dot{\mathrm{o}}{V}_0\right)}^2}{V_0} \\ ={\dot{\mathrm{o}}}^2{V}_0 \end{gathered}$$

The total energy is then

$$\begin{gathered} E_3=E_3^0+E_3^1+E_3^2 \\ =2V_0+0+{\dot{\mathrm{o}}}^2{V}_0 \\ =V_{0}2+{\dot{\mathrm{o}}}^2 \end{gathered}$$

which is what we obtained by expanding the exact solution.

d)

We now calculate the corrections to the degenerate energies. We have

The energies are

$$\begin{gathered} E_{\pm }^1=\frac{1}{2}{W}_{aa}+W_{bb}\pm \sqrt{{\left(W_{aa}-W_{bb}\right)}^2+4{\left|W_{ab}\right|}^2} \\ =\frac{1}{2}\left(W_{aa}\pm W_{aa}\right) \\ =\frac{1}{2}\left(-\dot{\mathrm{o}}{V}_a\pm \dot{\mathrm{o}}{V}_0\right) \end{gathered}$$

The energies corrections are

$E_{-}=-\dot{\mathrm{o}}{V}_0,E_{+}=0$

and the energies are

$E_1=V_0-\dot{\mathrm{o}}{V}_0,E_2=V_0$