Vibrational spectrum of diatomic molecule from Schrödinger's equation

Question

I'd like to find the vibrational spectrum of a diatomic molecule from its potential, which can be approximated as

$$V(R) = -V_0\Big[ \frac{1}{4}\Big(\frac{R_0}{R}\Big)^4 - \frac{1}{8}\Big(\frac{R_0}{R}\Big)^8 + \frac{7}{8} \Big]$$

where $R$ represents the distance between the two atoms. I already know the values of $V_0$ and $R_0$, which are respectively the dissociation energy of the molecule and the equilibrium distance. I also know the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$. I know that this could be simply obtained from the spectrum, by taking $$\omega_i = \frac{\Delta E_i}{\hbar}$$

but I have no idea how to solve the 1D Schrödinger equation for this potential.

$$\Big[ -\frac{\hbar^2}{2\mu}\partial_{RR}+V(R) \Big]\psi(R) = E\psi(R).$$

How do you find the spectrum? Do we need to solve it explicitly in order to find it?

Follow up

Following your answers I expanded this potential up to the harmonic term around its minimum getting

$$ V(R) \approx -V_0 + \frac{4V_0}{R_0^2}(R-R_0)^2 $$

which leads to the stationary Schrödinger equation

$$ \Big[-\frac{\hbar^2}{2\mu}\partial_{RR}+\frac{4V_0}{R_0^2}(R-R_0)^2\Big]\psi(R) = (V_0+E)\psi(R). $$

If I now look at the standard harmonic oscillator, namely

$$ \Big[ -\frac{\hbar^2}{2m}\partial_{xx}+\frac{1}{2}m\omega^2 x^2 \Big]\psi(x) = \tilde{E}\psi(x) $$

that leads to the spectrum

$$ \tilde{E}_n = \hbar \omega \Big( n+\frac{1}{2} \Big). $$

I can identify the two equations by setting

$$ \begin{cases} m \equiv \mu\\ x \equiv R-R_0 \Rightarrow \partial_{xx} \equiv \partial_{RR}\\ \frac{4V_0}{R_0^2} \equiv \frac{1}{2}m\omega^2 x^2 \Rightarrow \omega \equiv \sqrt{\frac{8V_0}{mR_0^2}}\\ \tilde{E} \equiv V_0+E \end{cases} $$

so that this equation leads to the spectrum

$$ E_n = -V_0 + \hbar\sqrt{\frac{8V_0}{\mu R_0^2}}\Big(n+\frac{1}{2}\Big) $$

in the harmonic approximation. Is it correct?

Answer 1

If you want to solve this (almost) exactly, you will need to do so numerically.

If you’re happy with an approximation, you can expand $V(R)$ about its minimum, and then first look at the harmonic approximation by comparing the term in $R^2$ with the harmonic oscillator Hamiltonian. You then then treat the $R^3$ and $R^4$ terms in the series expansion as perturbations.

Answer 2

Well, as I understand your problem, the first step is to find the potential in a power series in $r=R-R_0$ around the equilibrium point.

Because it is an equilibrium point, you know that the first order term proportional to $r^1$ is zero. If you find it is not, that means you made an error.

The next term is proportional to $r^2$.

Just looking at the shape of the potential shows that the coefficient is positive. Thus it is, at that order, a harmonic potential.

Now of course there are perturbations to this harmonic potential at higher orders, $r^3$, $r^4$ etc.

But physically, I am pretty sure that the first few levels in this harmonic potential will be very little affected by the higher order perturbations. So just compute the second order harmonic term in $r$ and $V_{approx}(r)=kr^2$ (with $k$ the positive number that would come out of your calculations) and solve for

$$\Big[ -\frac{\hbar^2}{2\mu}\partial_{rr}+V_{approx}(r) \Big]\psi(r) = E\psi(r)$$

Since $V_{approx}(r)$ is a harmonic potential, that should not be too hard a problem.

Solving for the exact potential $V(R)$ seems to me a much, much harder problem.