- #1

- 103

- 4

## 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:

----------------------------------------------------------------------------------------------------------------------

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:

----------------------------------------------------------------------------------------------------------------------

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:

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##

----------------------------------------------------------------------------------------------------------------------------

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....

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##

----------------------------------------------------------------------------------------------------------------------------

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....