91影视

The expression for the total Hamiltonian is given as follows:

$$\begin{gathered} H=H^0+H\text{'} \\ H=\frac{p_1^2}{2m}+\frac{k{x}_1^2}{2}+\frac{p_2^2}{2m}+\frac{k{x}_2^2}{2}+\frac{e^2}{4蟺{蔚}_0}\left[\frac{1}{R}-\frac{1}{R-x_1}-\frac{1}{R+x_2}+\frac{1}{R-x_1+x_2}\right] \end{gathered}$$

Write equation 6.97that is the Coulomb interaction between the atoms.

$H^{\text{'}}=\frac{1}{4蟺{系}_0}\frac{e^2}{R}-\frac{e^2}{R-x_1})-\frac{e^2}{R+x_2}+\frac{e^2}{R-x_1+x_2}$

The Coulomb interaction between nuclei is the first term in the perturbed Hamiltonian. The second and third terms are the interactions between the nuclei of one atom and the

electrons of another, while the last term is the interaction between the electrons of two separate atoms.

Consider that $\left|x_1\right|,\left|x_2\right|<R$,and expand interaction terms in Taylor series.

$$\begin{gathered} \frac{1}{R卤x}=\frac{1}{R}.\frac{1}{1卤{\displaystyle \frac{x}{R}}} \\ 鈮�\frac{1}{R}\left(1鈭�\frac{x}{R}+\frac{x^2}{R^2}+...\right)\frac{1}{R-x_1+x_2} \\ =\frac{1}{R}.\frac{1}{1-{\displaystyle \frac{\left(x_1-x_2\right)}{R}}} \\ 鈮�\frac{1}{R}\left(1+\frac{x_1{x}_2}{R}+\frac{{\left(x_1{x}_2\right)}^2}{R^2}+...\right) \end{gathered}$$

Evaluate the above expression further.

role="math" localid="1658206133753" $$\begin{gathered} H^{\text{'}}鈮�\frac{e^2}{4{\mathrm{蟺蔚}}_0}\left[\frac{1}{R}-\frac{1}{R}\left(1+\frac{x_1}{R}\frac{{x^2}_1}{R}\right)-\frac{1}{R}\left(1-\frac{x_2}{R}+\frac{{x^2}_2}{R}\right)+\frac{1}{R}\left(1+\frac{x_1{x}_2}{R}+\frac{{\left(x_1-x_2\right)}^2}{R}\right)\right] \\ =\frac{e^2{x}_1{x}_2}{2{\mathrm{蟺蔚}}_0{\mathrm{R}}^3} \end{gathered}$$

Thus, Equation 6.97 is explained and the equation $H^{\text{'}}=\frac{e^2{x}_1{x}_2}{2{\mathrm{蟺蔚}}_0{\mathrm{R}}^3}$is true.

Use the expressions for and into new Hamiltonian.

$$\begin{gathered} H=\frac{1}{2m}\left[\frac{\left(p_1^2+p_2^2\right)}{2}+\frac{\left(p_2^1-p_2^2\right)}{2}\right]+\frac{k}{2}\left[\frac{\left(x_1^2+x_2^2\right)}{2}+\frac{\left(x_2^1-x_2^2\right)}{2}\right]-\frac{e^2}{4{\mathrm{蟺蔚}}_0{\mathrm{R}}^3}\left[\frac{\left(x_1^2+x_2^2\right)}{2}+\frac{\left(x_2^1-x_2^2\right)}{2}\right] \\ =\frac{1}{2m}\left(p_1^2+p_2^2\right)+\frac{k}{2}\left(x_1^2+x_2^2\right)-\frac{e^2}{4{\mathrm{蟺蔚}}_0{\mathrm{R}}^3} \\ =H^0+H^{\text{'}} \end{gathered}$$

Thus, it is clearly seen that the total Halmintonian is a combination of two harmonic oscillator Hamiltonians.

Consider $E=\frac{鈩�}{2}({蝇}_{+}+{蝇}_{-})$and${蝇}_{卤}=\sqrt{\frac{k}{m}鈭�\frac{e^2}{2蟺{蔚}_{0}m{R}^3}}$.

Expand frequency${蝇}_{p}m$in Taylor series.

$$\begin{gathered} {蝇}_{卤}=\sqrt{\frac{k}{m}}.\sqrt{1鈭�\frac{e^2}{2{\mathrm{蟺蔚}}_0{\mathrm{R}}^3}} \\ ={蝇}_0\sqrt{1鈭�\frac{e^2}{2{\mathrm{蟺蔚}}_0{\mathrm{R}}^3}} \\ 鈮�{蝇}_0\left[1鈭�\frac{1}{2}.\frac{e^2}{2{\mathrm{蟺蔚}}_0{\mathrm{R}}^3}-\frac{1}{8}{\left(\frac{e^2}{2{\mathrm{蟺蔚}}_0{\mathrm{R}}^3}\right)}^2\right] \end{gathered}$$

Here,$k={蝇}_0^{2}m$.

Write the expression for the difference in energy.

$$\begin{gathered} 鈭�V=E-魔{蝇}_0 \\ =\frac{魔}{2}\left({蝇}_{+}{蝇}_{-}\right)-魔{蝇}_0 \\ =\frac{魔{蝇}_0}{2}\left[1-\frac{1}{2}.\frac{e^2}{2{\mathrm{蟺蔚}}_0{\mathrm{m蝇}}_0^2{\mathrm{R}}^3}-\frac{1}{8}{\left(\frac{e^2}{2{\mathrm{蟺蔚}}_0{\mathrm{m蝇}}_0^2{\mathrm{R}}^3}\right)}^2+1\frac{1}{2}\frac{e^2}{2{\mathrm{蟺蔚}}_0{\mathrm{m蝇}}_0^2{\mathrm{R}}^3}-\frac{1}{8}{\left(\frac{e^2}{2{\mathrm{蟺蔚}}_0{\mathrm{m蝇}}_0^2{\mathrm{R}}^3}\right)}^2\right]-魔{蝇}_0 \\ =\frac{魔}{8m^2{蝇}_0^3{\mathrm{R}}^6}\left(\frac{e^2}{2{\mathrm{蟺蔚}}_0}\right) \end{gathered}$$Therefore, it is shown thatlocalid="1658208887907" $鈭�V=-\frac{魔}{8m^2{蝇}_0^3{\mathrm{R}}^6}\left(\frac{e^2}{2{\mathrm{蟺蔚}}_0}\right)$.

Consider the first order correction of ground state (n=0) .

$$\begin{gathered} E_0^1=\left(0\left|H^{\text{'}}\right|0\right) \\ =-\frac{e^2}{2{\mathrm{蟺蔚}}_0}\left({蠄}_0\left(x_1\right){蠄}_0\left(x_2\right)\left|x_1{x}_2\right|{蠄}_0\left(x_1\right){蠄}_0\left(x_2\right)\right) \\ =0 \end{gathered}$$

Here,$H^{\text{'}}=-\frac{e^2}{2{\mathrm{蟺蔚}}_0}{x}_1{x}_2$

The expectation value of position operator of a harmonic oscillator is zero.

Write the expression for the second-order correction of ground state.

$$\begin{gathered} E_0^2=\underset{n=1}{\overset{鈭�}{鈭�}}\frac{{\left|\left({蠄}_n\left|H^{\text{'}}\right|{蠄}_0\right)\right|}^2}{E_0-E_n} \\ ={\left(\frac{e^2}{2{\mathrm{蟺蔚}}_0}\right)}^2\underset{n_1=1}{\overset{鈭�}{鈭�}}\underset{n_2=1}{\overset{鈭�}{鈭�}}\frac{{\left|\left(n_1\left|x_1\right|0\right)\right|}^2{\left|\left(n_1\left|x_2\right|0\right)\right|}^2}{E_{0,0}-E_{n_1{n}_2}} \end{gathered}$$

Here,$\overline{){蠄}_0>=\overline{)0>\overline{)0>\overline{){蠄}_n>=\overline{)n_1>\overline{)n_2>}}}}}}$.

Use $n_1=n_2=1,{\left|\left(n_1\left|x\right|0\right)\right|}^2=0$.

$$\begin{gathered} E_0^2={\left(\frac{e^2}{2{\mathrm{蟺蔚}}_0{\mathrm{R}}^3}\right)}^2\frac{{\left|\left(1\left|x\right|0\right)\right|}^2{\left|\left(1\left|x\right|0\right)\right|}^2}{\left({\displaystyle \frac{魔{蝇}_0}{2}}+\frac{魔{蝇}_0}{2}\right)-\left(\frac{3魔{蝇}_0}{2}+\frac{3魔{蝇}_0}{2}\right)}{\left|\left(1\left|x\right|0\right)\right|}^2 \\ =\frac{魔}{2m{蝇}_0}={\left(\frac{e^2}{2{\mathrm{蟺蔚}}_0{\mathrm{R}}^3}\right)}^2\left(\frac{1}{魔{蝇}_0}\right){\left(\frac{魔}{2m{蝇}_0}\right)}^2 \\ =\frac{魔}{8m^2{{蝇}^3}_0}{\left(\frac{e^2}{2{\mathrm{蟺蔚}}_0}\right)}^2\frac{1}{R^6} \end{gathered}$$

Thus, the second-order perturbation theory is applied.