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What physical reason is there for a mass on a spring to have discrete energy levels? And why are those energy levels equally spaced, i.e. why is $E \ \alpha \ f$?

Personal background and clarification:
I'm a 2nd year college student. I've taken a QM class, so I know why things like "particle-in-a-box" are quantized (or at least, I can accept that they are).

Now I'm in thermo, and we keep talking about oscillators having discrete, equally spaced energy levels. From the comments, I gather that there's no classical explanation for a mass on a spring to have discrete energy levels.

My question then becomes, starting from "QM is right", how do we get to "oscillators have discrete, equally spaced energy levels? Might it have to do with how the potential well for an oscillator is quadratic (as opposed to that of a particle in a box, which is square)?

LastStar007
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1 Answers1

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For simplicity, look at the case where $m=\omega=1$ (where $m$ is mass and $\omega$ is frequency) so that the hamiltonian is $$H={P^2\over2}+{Q^2\over 2}$$ Put $$A={1\over\sqrt2}(P+iQ)\qquad B={1\over\sqrt2}(P-iQ)$$ so that $$AB=H-\hbar/2\qquad BA=H+\hbar/2$$

From this, check that if $H\phi=\lambda\phi$ ($\lambda$ a scalar) then $$H(A\phi)=(\lambda+\hbar)\phi\qquad H(B\phi)=(\lambda-\hbar)\phi$$

Thus if $\lambda$ is an eigenvalue, so are $\lambda+\hbar$ (unless $A\phi=0$) and $\lambda-\hbar$ (unless $B\phi=0$).

Next, check that if $\lambda$ is an eigenvalue then $\lambda+\hbar/2$ is non-negative, and is zero if and only if $A\phi=0$ ($\phi$ being a corresponding eigenvector). (Hint: Use $$0\le (A\phi,A\phi)=(\phi,BA\phi)$$ and use the formula above for $BA$). Simliarly, check that if $\lambda$ is an eigenvalue then $\lambda-\hbar/2$ is non-negative, and is zero if and only if $B\phi=0$. (**)

From here, it's easy to conclude that if you choose any eigenvalue $\lambda$, then every $$\lambda+k\hbar\ge 0$$ is also an eigenvalue ($k$ an integer), that the smallest eigenvalue is $\hbar/2$, that therefore every $\hbar/2+k\hbar$ is an eigenvalue ($k$ a positive integer).

It remains to show that these are the only eigenvalues. Suppose $\mu$ were an eigenvalue not on the list, with corresponding eigenvector $\psi$. Then (by (**)) $B^k\psi$ is never zero ($k$ a positive integer), so you get an infinite decreasing sequence of evenly spaced eigenvalues, hence negative eigenvalues, contradiction.

WillO
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