Potential Energy Surface

[jgens]

Homework Statement

Estimate the ground-state potential energy surface for H₂⁺ using the first-order perturbative change in the energy.

Homework Equations

N/A

The Attempt at a Solution

I can calculate the first-order correction to the energy using the fact that $E^1_0 = \langle \mathrm{1s}_A |V| \mathrm{1s}_A \rangle$. In particular,

$$E_0^1 = \int_{-\infty}^{\infty}\overline{\mathrm{1s}}_A V \mathrm{1s}_A\mathrm{d}\mathbf{r} = \int_{-\infty}^{\infty}\overline{\mathrm{1s}}_A\left( \frac{1}{R} - \frac{1}{r_B}\right)\mathrm{1s}_A = e^{-2R}\left(1+\frac{1}{R}\right)$$

However, I'm having trouble getting from the first-order correction in the energy to obtaining a potential energy surface. Can anyone help with this?

 

[nate808]

Hi there! To obtain the potential energy surface for H2+, we can use the relation V(R) = E_0^1(R) - E_0^0, where E_0^1(R) is the first-order correction to the energy at a given internuclear distance R and E_0^0 is the energy of the unperturbed system.

To calculate E_0^0, we can use the fact that the energy of the unperturbed system is given by the sum of the energies of the individual hydrogen atoms, which is -1/R - 1/r_B. Therefore, E_0^0 = -2/R - 1/r_B.

Substituting this into the equation for V(R), we get:

V(R) = E_0^1(R) - E_0^0 = e^{-2R}\left(1+\frac{1}{R}\right) + \frac{2}{R} + \frac{1}{r_B}

This is the potential energy surface for H2+ as a function of internuclear distance R. I hope this helps! Let me know if you have any further questions.