6
$\begingroup$

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?

$\endgroup$

2 Answers 2

7
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.