FrontierScience / 6646019d-bda1-4b1c-83ac-7953e853effd / Consider two neutral atoms separated by a distance `R`. Since they are electrically neutral, there is no force between them in the absence of perturbations. However, if one of the …

Consider two neutral atoms separated by a distance `R`. Since they are electrically neutral, there is no force between them in the absence of perturbations. However, if one of the atoms is slightly polarized, a very weak attractive force emerges between the two atoms. Here, we assume that both atoms consist of a positively charged nucleus `(+e)` fixed in place and an electron `(−e)` with mass `m`, connected to the nucleus by springs with spring constant `k`. We further assume that the electrons are constrained to move along the line connecting the two atoms, with displacements `x₁` and `x₂`, respectively. When x₁>0, the electron of the first atom is closer to atom 2 than the nucleus of the first atom is. When x₂>0, the electron of the second atom is further away from the first atom than the nucleus of the second atom is. The elastic potential energy of the two springs can be written as `(1)/(2)kx₁²` and `(1)/(2)kx₂²`.

The Hamiltonian of this system is
`H=[(p_(+)²)/(2m)+(1)/(2)k_(+)x_(+)²]+[(p_(−)²)/(2m)+(1)/(2)k_(−)x_(−)²], where \( x_{\pm}=\frac{1}{\sqrt{2}}\biggl(x_{1}\pm x_{2}\biggr) \)` and `\( p_{\pm}=\frac{1}{\sqrt{2}}\biggl(p_{1}\pm p_{2}\biggr) \)`, where \\(p_1\\) and \\(p_2\\) are the momenta of the two electrons respectively.

The system can be considered as undergoing simple harmonic oscillations with two degrees of freedom, `x_(+)` and `\( x_− \)`, each with an angular frequency `ω_(±)=√(k_(±)/m)`. The ground state energy of this system is given by:

`\( E =\frac{1}{2}\hbar\Big(\omega_{+}+\omega_{-}\Big) \)`

We define the ground state energy when the Coulomb potential energy is absent as `E₀=ℏω₀`, where `ω₀=√((k)/(m))`.

By assuming `|k_(+)−k|≪k`, and proving that:

`ΔV≡E−E₀≈−(C)/(R⁶)`

derive the expression for the constant `C using only the terms ℏ,m,ω₀,e,ϵ₀,π.`

Think step by step and solve the problem below. At the end of your response, write your final answer on a new line starting with “FINAL ANSWER”. It should be an answer to the question such as providing a number, mathematical expression, formula, or entity name, without any extra commentary or providing multiple answer attempts.

Consider two neutral atoms separated by a distance `\( R \)`. Since they are electrically neutral, there is no force between them in the absence of perturbations. However, if one of the atoms is slightly polarized, a very weak attractive force emerges between the two atoms. Here, we assume that both atoms consist of a positively charged nucleus `\( (+e) \)` fixed in place and an electron `\( (-e) \)` with mass `\( m \)`, connected to the nucleus by springs with spring constant `\( k \)`. We further assume that the electrons are constrained to move along the line connecting the two atoms, with displacements `\( x_1 \)` and `\( x_2 \)`, respectively. When $x_1 > 0,$ the electron of the first atom is closer to atom 2 than the nucleus of the first atom is. When $x_2 > 0,$ the electron of the second atom is further away from the first atom than the nucleus of the second atom is. The elastic potential energy of the two springs can be written as `\( \frac{1}{2} kx_1^2 \)` and `\( \frac{1}{2} kx_2^2 \)`.

The Hamiltonian of this system is
`\( H=\left[\frac{p_+^2}{2m}+\frac{1}{2}k_+x_+^2\right]+\left[\frac{p_-^2}{2m}+\frac{1}{2}k_-x_-^2\right] \), where \( x_{\pm}=\frac{1}{\sqrt{2}}\biggl(x_{1}\pm x_{2}\biggr) \)` and `\( p_{\pm}=\frac{1}{\sqrt{2}}\biggl(p_{1}\pm p_{2}\biggr) \)`, where \\(p_1\\) and \\(p_2\\) are the momenta of the two electrons respectively.

The system can be considered as undergoing simple harmonic oscillations with two degrees of freedom, `\( x_+ \)` and `\( x_− \)`, each with an angular frequency `\( \omega_{\pm}=\sqrt{k_{\pm}/m} \)`. The ground state energy of this system is given by:

`\( E =\frac{1}{2}\hbar\Big(\omega_{+}+\omega_{-}\Big) \)`

We define the ground state energy when the Coulomb potential energy is absent as `\( E_0=\hbar\omega_0 \)`, where `\( \omega_0=\sqrt{\frac{k}{m}} \)`.

By assuming `\( \left|k_{+}-k\right|\ll k \)`, and proving that:

`\( \Delta V\equiv E-E_0\approx-\frac{C}{R^6} \)`

derive the expression for the constant `\( C \) using only the terms \(\hbar, m, \omega_0, e, \epsilon_0, \pi\).`

Think step by step and solve the problem below. At the end of your response, write your final answer on a new line starting with “FINAL ANSWER”. It should be an answer to the question such as providing a number, mathematical expression, formula, or entity name, without any extra commentary or providing multiple answer attempts.