Harmonic oscillator - rotating molecule?

[jeebs]

I have a planar molecule with a torsional oscillation mode where it twists around a C-C bond by an angle $$\theta$$ from some equilibrium position.
The restoring force is a function of theta, and the potential energy involved is given by $$V(\theta) = V_0(1-cos(2\theta))$$

I need to "use a Taylor expansion to approximate the system as a harmonic oscillator about the equilibrium point $$\theta = 0$$. Write down the Hamiltonian for the harmonic oscillator and find the frequency of oscillation $$\omega$$."

I am also told that the kinetic energy is $$T = \frac{L^2}{2I}$$ where L is angular momentum conjugate to $$\theta$$ and I is the relevant moment of inertia.

So, I have my Taylor (well Maclaurin specifically) series expression, $$f(x) = \Sigma \frac{f^(^n^)(0)}{n!}x^n$$.

I know in the case of a mass on an ideal spring we have simple harmonic motion described by $$\frac{d^2x}{dt^2} = - k^2x$$, which says that the mass' accelleration is directly proportional to the displacement from equilibrium, and acting in the opposition direction to the displacement. I assume that for angular displacement I can write exactly the same type of equation, $$\frac{d^2\theta}{dt^2} = - k^2\theta$$. This has the solution $$\theta(t) = e^i^\omega^t$$.

So, I'm not sure where to go from here. It wants an approximation to a harmonic oscillator, so I have stuck my harmonic oscillator equation $$\theta(t) = e^i^\omega^t$$ into the Taylor expansion, and truncated it after the third term, where I have $$\theta(t) = 1 + i\omega t - \omega^2 t^2 + ...$$

However, I'm a bit suspicious that I haven't done what I am being asked here, because the question does not explicitly mention anything with a t dependence, and the question is talking about oscillations about $$\theta(t) = 0$$. That to me suggests that I should in fact be putting $$V(\theta)$$ into the Taylor series $$f(x) = \Sigma \frac{f^(^n^)(a)}{n!}(x-a)^n$$, where a would be the angular displacement from equilibrium, ie. $$a = \theta_0 = 0$$ in this case.

Having done that I get $$V(\theta) = 4V_0\theta^2 - 16V_0\theta^4 +...$$ if I truncate it to the first two non-zero terms. Moving on to the Hamiltonian part, I have said $$H = T + V = \frac{L^2}{2I} +4V_0\theta^2 - 16V_0\theta^4$$. I don't think there's anything more I can do with this, so I want to then try and find the frequency of oscillation. I thought about just substituting my $$\theta(t) = e^i^\omega^t$$ into that Hamitonian but couldn't see a way to proceed after that.

So, am I approaching this the right way?
Can anybody suggest a way forward for me?

Thanks.

 

[mark726]

Hi there,

You are on the right track with your approach! Let's break it down step by step.

Firstly, you are correct in using the Taylor series for the potential energy function V(\theta) around the equilibrium point \theta_0 = 0. This will give you a polynomial approximation for V(\theta) in terms of \theta, where the first non-zero term will be the quadratic term 4V_0\theta^2. This is because the first derivative of V(\theta) at \theta_0 = 0 is zero, so the first non-zero term in the Taylor series will be the second derivative.

Next, you need to write down the Hamiltonian for this system. The Hamiltonian is simply the sum of the kinetic energy T and the potential energy V. In this case, the kinetic energy is given by T = \frac{L^2}{2I}, where L is the angular momentum conjugate to \theta and I is the relevant moment of inertia. So, the Hamiltonian will be H = \frac{L^2}{2I} + V(\theta).

Now, to find the frequency of oscillation, we need to use the equations of motion for a harmonic oscillator. These are given by \ddot{\theta} = -\omega^2\theta, where \omega is the frequency of oscillation. So, we need to find a way to write our Hamiltonian in terms of \theta and its derivatives. This is where the Taylor series approximation comes in handy.

Substituting the Taylor series for V(\theta) into the Hamiltonian, we get:

H = \frac{L^2}{2I} + 4V_0\theta^2 - 16V_0\theta^4 + ...

Now, we can use the equations of motion to write \frac{L^2}{2I} in terms of \theta and its derivatives. Since the angular momentum L is conjugate to \theta, we have L = I\dot{\theta}. So, substituting this into \frac{L^2}{2I}, we get:

\frac{L^2}{2I} = \frac{I^2}{2I}\dot{\theta}^2 = \frac{I}{2}\dot{\theta}^2

Now, we can rewrite our Hamiltonian as:

H = \frac{I}{2}\