Old Quantum Theory and the Linear Harmonic Oscillator

[Mark Spearman]

Homework Statement

Calculate the quantized energy levels of a linear harmonic oscillator of angular frequency $\omega$ in the old quantum theory.

Homework Equations

\[
\oint p_i dq_i = n h
\]

The Attempt at a Solution

This is supposed to be a simple "exercise" (the first in Merzbacher). I believe the
answer is is $E_n = n \hbar \omega$. But I don't see how. The classical total energy of the LHO is
\[
H = m q_0^2 \omega^2/2
\]
so the energy would not be linear in omega. The solution using the Schroedinger equation is (n + 1/2) \hbar.

Thanks,

MLS

 

[javierR]

In "old quantum theory" people had the uncertainty relationship which you can take to have the lower bound $$\Delta x \Delta p ~ \hbar/2$$ (other people don't care about the 1/2 factor in there since h-bar is a tiny order of magnitude anyway). The total energy of a particle in SHM is $$E=p^2/2m+\frac{1}{2}mx^{2}\omega^{2}$$, where x is the displacement of the particle. Put in the estimate of the *uncertainty* in p as the "value of p" in the first energy term and you can take the uncertainty in x to be the "particle displacement". Then you have energy as a function of this characteristic x, for which you want to find the extremum in order to find the amplitude of minimal energy. So solve dE/dA=0 for A. Now put this back into the expression for E(A) to get the energy in the desired form. As you noted, the energy levels do not end up being just multiples of this ground state estimate, but odd multiples.

 

[Mark Spearman]

Thank you, JavairR for your prompt response. The method you outlined is excellent and very clear. It results in the minimum energy state for the LHO which is, hbar omega/2.

However, my problem is different and since the last post, I was able to arrive at the answer which is think is correct (I was having problems with generalized coordinates and
INTEGRATION! I have attached a better problem description and solution as a pdf.

Thank you all for a fine forum.

MLS