Question

Van der Waals interaction. Consider two atoms a distanceapart. Because they are electrically neutral you might suppose there would be no force between them, but if they are polarizable there is in fact a weak attraction. To model this system, picture each atom as an electron (mass m , charge -e ) attached by a spring (spring constant k ) to the nucleus (charge +e ), as in Figure. We'll assume the nuclei are heavy, and essentially motionless. The Hamiltonian for the unperturbed system is

${\mathit{H}}^{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{m}}{\mathit{p}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{k}}{\mathit{x}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{m}}{\mathit{p}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{k}}{\mathit{x}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{6}\mathbf{.}\mathbf{96}\mathbf{]}$

The Coulomb interaction between the atoms is

${\mathbf{H}}^{\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi ϵ}}_{\mathbf{0}}}\mathbf{(}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{R}}\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{R}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}}\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{R}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{R}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}}$ [6.97]

(a) Explain Equation6.97. Assuming that localid="1658203563220" $\left|x_1\right|$ and $\mathbf{\left|}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right|}$are both much less than, show that

localid="1658203513972" ${\mathit{H}}^{\mathbf{\text{'}}}\mathbf{\cong }\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{2}}{\mathbf{x}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}}{\mathbf{2}\mathbf{\pi }{\mathbf{ϵ}}_{\mathbf{0}}{\mathbf{R}}^{\mathbf{3}}}$ [6.98]

(b) Show that the total Hamiltonian (Equationplus Equation) separates into two harmonic oscillator Hamiltonians:

$\mathit{H}\mathbf{=}\left[\frac{1}{2m}{p}_{+}^2+\frac{1}{2}(k-\frac{e^2}{2\pi {ϵ}_0{R}^3}{x}_{+}^2\right]\mathbf{+}\left[+\frac{1}{2m}{p}_{-}^2+\frac{1}{2}(k+\frac{e^2}{2\pi {ϵ}_0{R}^3}{x}_{-}^2\right]$ [6.99]

under the change of variables

${\mathit{x}}_{\mathbf{\pm }}\mathbf{\equiv }\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{(}{\mathit{x}}_{\mathbf{1}}\mathbf{\pm }{\mathit{x}}_{\mathbf{2}}\mathbf{)}$ Which entails ${\mathit{p}}_{\mathbf{\pm }}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{(}{\mathit{p}}_{\mathbf{1}}\mathbf{\pm }{\mathit{p}}_{\mathbf{2}}\mathbf{)}$ [6.100]

(c) The ground state energy for this Hamiltonian is evidently

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{\hslash }\mathbf{(}{\mathit{\omega }}_{\mathbf{+}}\mathbf{+}{\mathit{\omega }}_{\mathbf{-}}\mathbf{)}$ Where ${\mathit{\omega }}_{\mathbf{\pm }}\mathbf{=}\sqrt{\frac{\mathbf{k}\mathbf{\mp }\mathbf{(}{\mathbf{e}}^{\mathbf{2}}\mathbf{/}\mathbf{2}\mathbf{\pi }{\mathbf{ϵ}}_{\mathbf{0}}{\mathbf{R}}^{\mathbf{3}}\mathbf{)}}{\mathbf{m}}}$ [6.101]

Without the Coulomb interaction it would have been ${\mathit{E}}_{\mathbf{0}}\mathbf{=}\mathit{ħ}{\mathit{\omega }}_{\mathbf{0}}$, where ${\mathit{\omega }}_{\mathbf{0}}\mathbf{=}\sqrt{\mathbf{k}\mathbf{/}\mathbf{m}}$. Assuming that, show that

$\mathit{\Delta }\mathit{V}\mathbf{\equiv }\mathit{E}\mathbf{-}{\mathit{E}}_{\mathbf{0}}\mathbf{\cong }\mathbf{-}\frac{\mathbf{\hslash }}{\mathbf{8}{\mathbf{m}}^{\mathbf{2}}{\mathbf{\omega }}_{\mathbf{0}}^{\mathbf{3}}}{\left(\frac{e^2}{2\pi {ϵ}_0}\right)}^{\mathbf{2}}\frac{\mathbf{1}}{{\mathbf{R}}^{\mathbf{6}}}\mathbf{.}$ [6.102]

Conclusion: There is an attractive potential between the atoms, proportional to the inverse sixth power of their separation. This is the van der Waals interaction between two neutral atoms.

(d) Now do the same calculation using second-order perturbation theory. Hint: The unperturbed states are of the form ${\mathit{\psi }}_{\mathbf{n}\mathbf{1}}\left(x_1\right){\mathit{\psi }}_{\mathbf{n}\mathbf{2}}\left(x_2\right)$, where ${\mathit{\psi }}_{\mathbf{n}}\left(x\right)$is a one-particle oscillator wave function with mass mand spring constant $\mathit{k}\mathbf{;}\mathit{\Delta }\mathit{V}$is the second-order correction to the ground state energy, for the perturbation in Equation 6.98 (notice that the first-order correction is zero).

Short answer

(a) Equation 6.97 is elaborated and the equation$H^{\text{'}}=-\frac{e^2{x}_1{x}_2}{2\pi {\epsilon }_0{R}^3}$ holds true.

(b)The total Hamiltonian is separated intotwo harmonic oscillator Hamiltonians.

(c) It is shownthat $\Delta V=-\frac{\hslash }{8m^2{\omega }_0^3{R}^6}{\left(\frac{e^2}{2\pi {\epsilon }_0}\right)}^2$.

(d) The same calculation with the second-order perturbation theory is done.

Step-by-step solution

Expression for the total Halmintonian

The expression for the total Hamiltonian is given as follows:

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

(a) Explanation of equation 6.97

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

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

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.

$$\begin{gathered} \frac{1}{R\pm x}=\frac{1}{R}.\frac{1}{1\pm {\displaystyle \frac{x}{R}}} \\ \approx \frac{1}{R}\left(1\mp \frac{x}{R}+\frac{x^2}{R^2}+...\right)\frac{1}{R-x_1+x_2} \\ =\frac{1}{R}.\frac{1}{1-{\displaystyle \frac{\left(x_1-x_2\right)}{R}}} \\ \approx \frac{1}{R}\left(1+\frac{x_1{x}_2}{R}+\frac{{\left(x_1{x}_2\right)}^2}{R^2}+...\right) \end{gathered}$$

Evaluate the above expression further.

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

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

(b) Verification of the fact that the total Halmintonian is two harmonic oscillator Hamiltonians

Use the expressions for and into new Hamiltonian.

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

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

(c) Verification of the relation given

Consider $E=\frac{\hslash }{2}({\omega }_{+}+{\omega }_{-})$and${\omega }_{\pm }=\sqrt{\frac{k}{m}\mp \frac{e^2}{2\pi {\epsilon }_{0}m{R}^3}}$.

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

$$\begin{gathered} {\omega }_{\pm }=\sqrt{\frac{k}{m}}.\sqrt{1\mp \frac{e^2}{2{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^3}} \\ ={\omega }_0\sqrt{1\mp \frac{e^2}{2{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^3}} \\ \approx {\omega }_0\left[1\mp \frac{1}{2}.\frac{e^2}{2{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^3}-\frac{1}{8}{\left(\frac{e^2}{2{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^3}\right)}^2\right] \end{gathered}$$

Here,$k={\omega }_0^{2}m$.

Write the expression for the difference in energy.

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

(d) Application of the second-order perturbation theory

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

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

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

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

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

$$\begin{gathered} E_0^2=\sum _{n=1}^{\infty }\frac{{\left|\left({\psi }_n\left|H^{\text{'}}\right|{\psi }_0\right)\right|}^2}{E_0-E_n} \\ ={\left(\frac{e^2}{2{\mathrm{\pi \epsilon }}_0}\right)}^2\sum _{n_1=1}^{\infty }\sum _{n_2=1}^{\infty }\frac{{\left|\left(n_1\left|x_1\right|0\right)\right|}^2{\left|\left(n_1\left|x_2\right|0\right)\right|}^2}{E_{0,0}-E_{n_1{n}_2}} \end{gathered}$$

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

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

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

Thus, the second-order perturbation theory is applied.