# Solving for SHM Diatomic Energy Levels

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## Main Question or Discussion Point

So I'm trying to figure out how we got the allowed vibrational energy levels for a diatomic molecule by approximating it with simple harmonic motion.

I do know how to use the uncertainty principle to get the zero-point energy:

We know that the potential function is $V(x) = \frac{1}{2}mx^2$ where x is the distance away from the equilibrium:

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1) Knowing that $k \Delta X^2 = \frac{\Delta P^2}{m}$
$\Delta X = \frac{\Delta P}{\sqrt{km}}$

So that's one Delta X, the other one you can get from the uncertainty principle:​

$\Delta X \Delta P \geq \frac{\hbar}{2}$ therefore $\Delta X = \frac{\hbar}{2 \Delta P}$ at a minimum

These two values for delta x and inserted back into the original energy equation:

$V(x) = \frac{1}{2}m\frac{\Delta P}{\sqrt{km}}\frac{\hbar}{2 \Delta P}$

Delta P cancel out, multiply both numerator and denominator by k squared and you get the zero point energy:

$E_0 = \frac{\hbar}{2} \omega$
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So that's one energy level, how do I find the rest? If I start with just the Schrodinger equation:

$(-\frac{\hbar}{2m} \frac{d}{dx^2} + \frac{1}{2}kx^2)\psi(x) = E \psi(x)$

I get that, but it's not so easy to solve, I know the answer should be: $E = \hbar \omega(n + 1/2), n = 0,1,2...$

Also, how would I find which energy level the system is in based on temperature, the only thing I can think of right now is Botlzman $E = \frac{3}{2}kT$, but I understand that much of this energy can be in the form of translational, rotational or oscillation, so....