91影视

In order to find Eigen values of H, calculate:

$$\begin{gathered} det\left(H-位.l\right)=0 \\ \left(E_a-位\right)\left(E_b-位\right)-h^2=0 \\ {位}^2-位\left(E_a+E_b\right)+E_a{E}_b-h^2=0 \end{gathered}$$

$$\begin{gathered} 位=\frac{1}{2}\left(E_a+E_b卤\sqrt{E_a^2+2E_a{E}_b+E_b^2-4E_a{E}_b+4h^2}\right) \\ 鈬�E_{卤}=\frac{1}{2}\left[E_a+E_b卤\sqrt{{\left(E_a-E_b\right)}^2+4h^2}\right] \end{gathered}$$

Zero order: $E_a^0=E_a,E_b^0=E_b$

First order:

Second order:

Here estimating the ground state energy.

So,

$$\begin{gathered} =1-\frac{{\stackrel{藱}{o}}^2}{1+{\stackrel{藱}{o}}^2}=\frac{1}{1+{\stackrel{藱}{o}}^2}; \\ \cos 2蠒=\frac{\overline{+}1}{\sqrt{1+{\stackrel{藱}{o}}^2}}\left(signdictatedby\tan 2蠒=\frac{\sin 2蠒}{\cos 2蠒}=-\stackrel{藱}{o}\right).{\cos }^2蠒 \\ =\frac{1}{2}\left(1+\mathrm{co}\left(\right)s2蠒\right) \\ =\frac{1}{2}\left(1\overline{+}\frac{1}{\sqrt{1+{\stackrel{藱}{o}}^2}}\right); \\ {\sin }^2蠒=\frac{1}{2}\left(1-\cos 2蠒\right) \\ =\frac{1}{2}\left(1卤\frac{1}{\sqrt{1+{\stackrel{藱}{o}}^2}}\right) \end{gathered}$$

But

We can obtain results from task (b) if we expand in Taylor series exact energies from task (a). we use limit where h is very small, which has to be if we want to use perturbation theory:

$$\begin{gathered} E_{卤}=\frac{1}{2}[\left(E_a+E_b\right)卤\left(E_b+E_a\right)\sqrt{1+\frac{4h^2}{\left(E_b+E_a\right)}} \\ {E}_{卤}鈮�\frac{1}{2}\left\{E_a+E_b卤\left(E_b+E_a\right)\left[1+\frac{2h^2}{{\left(E_b+E_a\right)}^2}\right]\right\} \\ =\frac{1}{2}\left[E_a+E_b卤\left(E_b+E_a\right)卤\frac{2h^2}{\left(E_b+E_a\right)}\right] \end{gathered}$$

so localid="1658396154675" $$\begin{gathered} E_{+}鈮�E_b+\frac{h^2}{\left(E_b+E_a\right)}, \\ {E}_{-}鈮�E_a-\frac{h^2}{\left(E_b+E_a\right)}, \end{gathered}$$

Confirming the perturbation theory results in (b). The variation principle (c) gets the ground state(E鈭�) exactly right-not too surprising since the trial wave function Eq. 7.75 is almost the most general state (there could be a relative phase factor $e^{i胃}$ .