Question

What do you think about the energy levels for a particle moving in a potential field of the shape? V(x) = {m omega^2 x^2/2 x < 0 infinity, x Greaterthanorequalto 0 Write an expression for the energy levels. Explain your answer.

Answer 1

The Potential of a particle in Simple Harmonic Motion is given by: $\small V(x) = \frac{1}{2}kx^2$

where k is the spring's constant. It is related to the angular frequency of the oscillations and the mass of the particle by the expression: $\small \omega = \sqrt{\frac{k}{m}}$

So $\small \omega^2 = \frac{k}{m}$

=> $\small k = m\omega^2$

therefore the Potential Energy will be $\small V(x) = \frac{1}{2} m\omega^2 x^2$ [which corresponds to the potential value for x<0].

Hence our particle behaves as a simple harmonic oscillator for x < 0 and since we know the allowed Energy levels for a particle in simple harmonic motion the same energy levels will also correspond for this particle for x < 0.

So the Energy levels for the particle for x < 0 will be

$\small E_n = (n + \frac{1}{2})\frac{h}{2\pi}\omega$

for all n = 0, 1, 2 ,....

Now the Potential Energy is infinite beyond x = 0 and thus the particle cannot exist beyond this point. So the particle exists only as a simple harmonic oscillator with Energy levels given by $\small E_n = (n + \frac{1}{2})\frac{h}{2\pi}\omega$ .