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}} $.
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}} $.
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