Exploring Quantum Physics - Week 4 Question 1

Question:

Suppose we have a potential $V(x) = ax^4+bx^2+c$, with $a,b,c$ positive, and imagine that we are examining a low energy bound state of this system. Expressing the low energy approximation of this potential as a harmonic oscillator, what is its characteristic frequency, $\omega$?

Solution:

That would be a local Taylor approximation at the lowest energy.

First, let's get a feeling on how $V(x)$ look like by computing some derivatives. We have:

$\frac{dV}{dx} = 4ax^3 + 2bx = x(4ax^2 + 2b)$

$\frac{d^2V}{dx^2} = 12ax^2 + 2b > 0$.

Therefore, the function is convex, with the minimum achieved at 0, $\sqrt{\frac{-2b}{4a}}$. So the only real minimum is at 0.

Approximating the potential there, we have $V(x) \approx V(0) + \frac{V'(0)(x - 0)}{1!} + \frac{V''(0)(x-0)^2}{2!} = c + bx^2$.

As with the typical prescription, we should have the equation

$\hat{H} = \frac{\hat{p}^2}{2m} + b\hat{x}^2 + c$

Ignoring the constant part $c$ and do some matching we have $\frac{1}{2}m\omega^2x^2 = bx^2$, that makes $\omega = \sqrt{\frac{2b}{m}}$.