Question

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.

Short answer

The answers are

a)$$\begin{gathered} v_1=\left(1 \\ 0 \\ 0\right),V_0;v_2=\left(0 \\ 1 \\ 0\right),V_0;v_3=\left(0 \\ 0 \\ 1\right),2V_0 \end{gathered}$$

b)role="math" localid="1658144006650" ${\lambda }_1=V_0(1-\dot{\mathrm{o}}),{\lambda }_2=\frac{V_0}{2}\left(3-\sqrt{4{\dot{\mathrm{o}}}^2}\right)\approx V_0(1-{\dot{\mathrm{o}}}^2),{\lambda }_3=\frac{V_0}{2}\left(3+\sqrt{4{\dot{\mathrm{o}}}^2}\right)\approx V_0(2-{\dot{\mathrm{o}}}^2)$

c)$E_3^1=0,E_3^2,={\dot{\mathrm{o}}}^2{V}_0$

d)$E_1=V_0-\dot{o}{V}_0,E_2=V_{0\mathbf{1}\mathbf{+}}$

Step-by-step solution

Cigen vector and eigen values of Hamiltonian

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

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

Expand eigen values as power series of H

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 $\sqrt{1+\dot{\mathrm{o}}}\approx 1+\dot{\mathrm{o}}/2$.

Find approximate eigen values for state

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.

First order eigen values for degenerate state

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

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