# Why are harmonic oscillators quantized? [closed]

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)?

## closed as too broad by DanielSank, John Rennie, Kyle Kanos, Alfred Centauri, GertNov 26 '15 at 14:44

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• no need classically. quantum mechanically, i.e. dimensions commensurate to h, the postulates of quantum mechanics require it . – anna v Nov 26 '15 at 6:39
• Welcome to Physics Stack Exchange! This is a very broad question which essentially asks "why is quantum mechanics right?" or at least "please derive the results of the harmonic oscillator in quantum mechanics". We generally require questions to be less broad than this, particularly in cases where the answer is discussed in common references, such as textbooks. – DanielSank Nov 26 '15 at 6:44
• A macroscopic mass on a macroscopic spring won't have quantised energy levels, even in principle, because it won't remain coherent for long enough. Quantum oscillators are quantised and describing them is a routine part of any quantum physics course. If you're interested you need to go off and read up about it. Describing the details here would require a review length answer. – John Rennie Nov 26 '15 at 7:50
• Related: physics.stackexchange.com/q/39208/2451 and links therein. – Qmechanic Nov 26 '15 at 8:32
• Edited. Is it narrow enough to be taken off hold? If not, what more is needed? – LastStar007 Nov 27 '15 at 19:23

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.