Relativistic Expression for Energy of Quantum State of Harmonic Oscillator?

[ephedyn]

Homework Statement

This is supposed to be a question for high school seniors who've had instruction in introductory concepts of special relativity and non-relativistic QM. According to my TA, he isn't too certain if it can be done within these confines but nonetheless I've attempted the problem... Please let me know if I'm doing it wrong (quite sure I am); or if I'm on the right track, how to reduce r in terms of n. Thanks in advance!

A particle of mass m vibrates as a harmonic oscillator with angular frequency $$\omega$$. For this harmonic oscillator, the general expression for the energy $$E_n$$ of the state of quantum number n is

$$E_n = (n - \frac{1}{2})\hbar\omega$$

Suppose that the angular frequency $$\omega$$ is so large that the kinetic energy of the particle is comparable to $$mc^2$$. Obtain the relativistic expression for the energy $$E_n$$ of the state of quantum number n.

Relevant equations and the attempt at a solution

Considering the relativistic kinetic energy $$E_k$$ of the particle,
$$E_k = (mc^2)(\gamma - 1)$$

If $$E_k \approx mc^2$$

then $$\gamma - 1 \approx 1$$

$$\therefore \gamma \approx 2$$

Expressing angular frequency in terms of linear velocity in $$\gamma$$

$$\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}$$

$$\omega = \frac{c}{\sqrt{2}r}$$
where r is the radius of the oscillation

$$\therefore E_n = \frac{(n - \frac{1}{2}) \hbar c }{\sqrt{2}r}$$

 

[ephedyn]

Posting again to put this thread above again, since it remains unsolved.

Thanks for looking.

 

[Redbelly98]

Well, I don't know the solution, but I'll just remark that "comparable to" isn't synonymous with "approximately equal to". So the KE could easily be 1/2 of, or twice, the rest energy for example.