The link between discrete energy level in quantum mechanics and harmonic series in Acoustics

Question

Consider a quantum square potential well with infinite depth: $$ V(x)=\begin{cases} 0, &|x|<a \\ +\infty, &\text{otherwise}. \end{cases}$$ Solving the Schodinger equation of a particle with mass $m$ in it gives the energy levels $$ E_n=\frac{n^2}{8m}\left(\frac{\hbar \pi}{a}\right)^2. $$ You can see that the energies are not equally spaced.

Now consider a string with both ends fixed. Standing waves on the string can vibrate at frequencies $nf$ only, where $f$ is the fundamental frequency, and $n$ is a positive integer. This time the "energy levels" ($\propto$ frequency) is equally spaced.

(We also get equally spaced energy levels in quantum harmonic oscillators.)

The similarity of the two situations above is that the value of a quantity (such as energy) has to be on discrete "levels" (cannot change continuously) due to the boundary conditions. However, the "levels" produced are not exactly the same: one of them $\propto n^2$, the other $\propto n$.

Questions:

1. What difference between the Schrodinger equation for a potential well and the wave equation for a string is responsible for the difference in the spacing of energy levels? Intuitively, both equations are describing waves, so why one produce levels $\propto n^2$ while the other produce levels $\propto n$? What's really different?
2. Is there a system in classical mechanics that behave very similar to the quantum potential well? I want to find a classical system that also has levels of $\propto n^2$?

Answer 1

1. The square well (resp. harmonic) potential with spectrum $E_n \propto n^2$ (resp. $E_n \propto n$) can be viewed as a power law potential $|x|^p$, where the power $p$ is $p=\infty$ (resp. $p=2$). See e.g. this related Phys.SE post.

2. Although the classical standing wave with spectrum $E_n \propto n$ may literally take place in a square well, it is not produced by a single particle but instead a collective phenomenon of many particles. Their effective potential is in fact harmonic, see e.g. phonons.

Answer 2

In the classical case, the energies are related to the eigenvalues of the square root of the 1D Laplace operator, so that they are proportional to the numbers $E_n^{1/2}$ which are equally spaced as functions of the natural $n$. In the quantum case, they are instead related to the eigenvalues $E_n$ of Laplace operator. These differnces can be traced back to the order of the time derivative appearing in the relevant equations: second order for Klein-Gordon (D'Alembert if $m=0$ below) equation $$-\frac{1}{v^2} \frac{\partial^2 u}{\partial t^2}+ \Delta u -m^2 u = 0 \tag{1}$$ and first order for Schroedinger equation $$-i\hbar \frac{\partial \psi }{\partial t} + \frac{\hbar^2}{2m} \Delta \psi + V\psi =0.\tag{2}$$ D'Alembert equation is similar to (a pair of) Schroedinger equation(s) with spatial Laplacian replaced by its square root (with both signs) and without potential $V$, here replaced by boundary conditions.

As a matter of fact, (1) can be rewritten (for $m=0$), $$\left(\partial_t + v \sqrt{-\Delta} \right)\left(\partial_t - v \sqrt{-\Delta} \right) u =0\:.$$ Regarding your second question, I do not know. What I know is that vibrations of some rigid structures satisfy an equation where the Laplace operator is replaced by its square. Maybe a rigid linear elastic stick has proper frequencies which are quadratically spaced...

The eigenvalues enter the game because one search for a solution $$u=u(t,x),\quad \psi =\psi(t,x)$$ which is a linear combination of eigenvectors of the Laplacian or its square root and the coefficients of the combination are function of time to be determined.

Let us consider the two cases with a simplified choice of the constants for the sake of simplicity: $m=0$, $v=1$ (D'Alemebert equation) and $V=0$, $\hbar =1$, $2m =1$ (in the Schroedinger equation) and assume that we have constructed a basis of (time independent) eigenfunctions, respectively, $$(\sqrt{-\Delta} u_n)(x) = \sqrt{-E_n} u_n(x)\tag{3}$$ or $$-(\Delta\psi_n)(x) = E_n \psi(x)\tag{4}\:,$$ where boundary conditions have been taken into account.

Observe that, since we used the same boundary conditions we can assume $$u_n(x) = \psi_n(x)$$ but I use a different notation just to distinguish between the two equations.

The solution of the corresponding time dependent equation can be decomposed as a liner combination $$u(t,x) = \sum_n c_n(t) u_n(x)$$ and $$\psi(t,x) = \sum _n d_n(t) \psi_n(x)\:.$$ Inserting these expansions in (1) and (2), taking (3) and (4) into account, the first case leads to the differential equation $$\frac{d^2c_n}{dt^2} = -E_nc_n$$ so that $$c_n(t) = C_n^{(+)} e^{i\sqrt{E_n}t} + C_n^{(-)} e^{-i\sqrt{E_n}t}\:,$$ whereas the second case yields $$\frac{dd_n}{dt} = -i E_n d_n$$ so that $$d_n(t) = D_n e^{-iE_nt}\:.$$ The constants $C_n^{(\pm)}$ and $D_n$ are fixed by the initial conditions imposed on the solutions $u(t,x)$ and $\psi(t,x)$ respectively.