91Ó°ÊÓ

a)

In this problem we study the Hamiltonian

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

The unperturbed Hamiltonian is obtained by setting$\stackrel{Ë™}{\mathrm{o}}=\mathbf{0}$ 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

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

The one eigen value is

${\mathrm{λ}}_1=V_0(1-\stackrel{˙}{\mathrm{o}})$

and the other two are obtained as

$$\begin{gathered} {\mathrm{λ}}^2-3V_0\mathrm{λ}+V_0(2-{\stackrel{Ë™}{\mathrm{o}}}^2)=0 \\ {\mathrm{λ}}_{2,3}=\frac{1}{2}3V_0Â\pm \sqrt{9V_0^2-4V_0^2(2-{\stackrel{Ë™}{\mathrm{o}}}^2)} \\ =\frac{V_0}{2}3Â\pm \sqrt{1+4{\stackrel{Ë™}{\mathrm{o}}}^2} \end{gathered}$$

We can expand these Eigen values with respect to to obtain

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

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

c)

We now observe the perturbed part of the Hamiltonian

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

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

The second-order corrections are

We have

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Therefore,

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

The total energy is then

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

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

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

The energies corrections are

$E_{-}=-\stackrel{Ë™}{\mathrm{o}}{V}_0,E_{+}=0$

and the energies are

$E_1=V_0-\stackrel{Ë™}{\mathrm{o}}{V}_0,E_2=V_0$