Question of the Day: Why does confinement produce discrete energies?
Learning objectives
From Notebook 02: a quantum state predicts measurement probabilities and repeated position measurements trace |ψ|². Now the boundaries themselves restrict which stationary states are possible.
Imagine a particle confined to 0 ≤ x ≤ L with V=0 inside and V=∞ outside.
Prediction: Before any calculation, can every wavelength describe a stationary state between two hard walls? What must be true at both impenetrable walls?
This notebook will make the answer visible before presenting the full box formulas.
Three investigations structure this notebook:
Each investigation follows prediction → model exploration → observation → interpretation.
We compare candidate spatial patterns $\propto \sin(q\pi x/L)$ for positive finite $q$ and ask which satisfy both hard walls. The canonical comparison uses $L=2.0$ nm and $q=1.0,1.5,2.0,2.5$.
The comparison shows that most wavelengths do not fit both boundaries simultaneously; only $q=1.0$ and $q=2.0$ satisfy both walls, while $q=1.5$ and $q=2.5$ miss the right-wall zero. Only particular standing-wave patterns survive.
Inside the box the stationary Schrödinger equation is
$$ -\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} = E\psi $$
with hard-wall conditions
$$ \psi(0)=0, \qquad \psi(L)=0 $$
which permit only
$$ k_n = \frac{n\pi}{L}, \qquad \lambda_n = \frac{2L}{n} $$
and normalized eigenfunctions
$$ \psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right), \qquad 0<x<L $$
The differential equation allows sinusoidal solutions, but the walls select the allowed subset.
For the allowed $k_n$,
$$ E_n = \frac{\hbar^2 k_n^2}{2m}, \qquad E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} $$
so $E_n \propto n^2$, $E_n \propto 1/L^2$, and $E_n \propto 1/m$. Halving $L$ at fixed $n,m$ multiplies the energy by four. Student-facing position and width use nm, energies use eV, $\psi$ uses $\mathrm{nm}^{-1/2}$, $|\psi|^2$ uses $\mathrm{nm}^{-1}$; internal calculations remain SI.
A stationary state does not mean the particle sits motionless. The ideal energy eigenvalue is definite, the spatial probability density $|\psi_n|^2$ is time independent, yet repeated position measurements still give different positions. No explicit time-dependent phase is needed here.
| Concept | Relation | |---|---| | Energy quantization | $E_n = \frac{\hbar^2 k_n^2}{2m}$ | | Energy quantization | $E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}$ | | Stationary probability | $P(a<x<b)=\int_a^b |\psi_n|^2 dx$ |
For a selected stationary eigenstate $n$, where will repeated ideal position detections accumulate? Predict how the histogram will compare with the model density, especially near an interior node.
What to change: use the prepared-state control and then increase the trial count. What to watch: the recorded-position histogram and the smooth model probability density.
The prepared state and the fixed 2.0 nm box are model/apparatus settings. Each recorded position is one detector event; the histogram is accumulated detector evidence. The smooth $|\psi_n(x)|^2$ curve is the model probability density, not another detector reading.
For the initial $n=2$ state, the model has a central node at $x=L/2=1.0$ nm: $\psi_2(L/2)=0$ and $|\psi_2(L/2)|^2=0$. A continuous coordinate is not a finite detector event, so look for strongly suppressed counts in a narrow interval around 1.0 nm, compared with neighbouring equal-width intervals.
Refresh: for a finite interval, $P(a<x<b)=\int_a^b |\psi_n(x)|^2\,dx$. Increasing $N$ should make the accumulated evidence resemble the same model density more closely; it does not change the prepared state or its density. This is an ideal detector with no resolution broadening or acquisition noise.
Quantum dots are nanoscale confined systems that can be used as optical probes and imaging labels. The one-dimensional infinite box captures the basic idea of confinement-induced quantization, but it is a minimal confinement model and not a quantitative model of a real quantum-dot device.
Real quantum-dot devices differ in at least four ways:
In two or three dimensions additional quantum numbers appear and the level structure is richer (no two-dimensional box calculation is required here).
Two nearby confined regions provide a useful qualitative bridge toward molecules: states that begin localized in separate regions can combine into states spread over both regions, with lower- and higher-energy combinations reminiscent of bonding and antibonding states. This is only an analogy, not a model of molecular bonding; real molecules involve Coulomb potentials, nuclei, and, for many-electron systems, electron–electron interaction. A quantitative treatment of molecular bonding belongs later in Notebook 09.
Consider the following four questions together before moving on:
Why can energy be discrete while position measurements remain continuous? How does confinement quantize energies without quantizing detector positions?
At fixed $n$ and $m$, what happens to the energy $E_n$ when the box width $L$ is halved, and why does $E_n \propto 1/L^2$ predict this?
How can the $n=2$ state have $\langle x \rangle = L/2$ while $|\psi_2(L/2)|^2 = 0$? What does this teach about an expectation value as an ensemble average versus the most probable individual result?
Which assumptions of the one-dimensional infinite-box model are least realistic for a real quantum dot, and how do the four limitations above bear on that judgment?
Hard-wall confinement imposes boundary conditions $\psi(0)=0$ and $\psi(L)=0$, which allow only particular standing-wave patterns $k_n = n\pi/L$. Those stationary patterns give discrete energies $E_n = n^2\pi^2\hbar^2/(2mL^2)$, a time-independent spatial probability density $|\psi_n(x)|^2$, and repeated ideal position detections whose histogram approaches that density. Stronger confinement (smaller $L$) increases the characteristic energy spacing as $1/L^2$, and the ideal one-dimensional box remains a minimal confinement model for the qualitative dot analogy, not a quantitative device model.
For any box eigenstate $\langle x \rangle = L/2$, yet for $n=2$, $|\psi_2(L/2)|^2 = 0$. An expectation value is an ensemble average over many repeated measurements; it need not be the most probable individual result. The central node suppresses counts in a finite interval around $L/2$, while the average over the symmetric distribution remains at the centre.
The infinite wall was an idealization. Real barriers are finite: the wave function can penetrate a classically forbidden region, leading to nonzero transmission, tunnelling, and the operating principle of the scanning tunnelling microscope (STM). How the penetration depends on barrier height and width, and how STM exploits it, is the subject of the next notebook.
Can a particle be detected beyond a barrier even when classical mechanics says it should not cross?