A Particle in a Box II · Quantum Mechanics (Phillips)

A box is the shortest possible route from a boundary condition to a spectrum. Two walls and nothing else produce energies that come in a discrete list.

This is the section the book has been building toward since page one:

One of the key features of quantum physics is that the possible energies of a confined particle are quantized.

And the mechanism is now completely unmysterious, because §4.2 already showed it on a guitar string. The claim Phillips makes here is stronger than it sounds — that the familiar quantized levels of atomic, nuclear and particle physics are all manifestations of confinement. Not of anything exotic. Of walls.

A one-dimensional box

The potential is a wall on each side and nothing in between:

V(x) = \begin{cases} 0 & \text{if } 0 < x < a\\ \infty & \text{elsewhere}\end{cases}\tag{4.30}

Each state obeys the one-dimensional Schrödinger equation

i\hbar\frac{\partial\Psi}{\partial t} = \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right]\Psi\tag{4.31}

and for a state of definite energy, §4.3 tells us the form:

\Psi(x,t) = \psi(x)\,e^{-iEt/\hbar}\tag{4.32}

with $\psi$ satisfying the energy eigenvalue equation

\left[-\frac{\hbar^2}{2m}\frac{\mathrm d^2}{\mathrm dx^2} + V(x)\right]\psi(x) = E\psi(x)\tag{4.33}

V(x)
energy level Eₙ
ψₙ(x)
classically forbidden (V > E)

selected level: E₀ = 4.9348
nodes in ψ: 0
bound states: infinitely many
E₁ − E₀: 14.8043

A flat box with impenetrable walls. Every state is bound, the levels go up as n², and ψ must vanish at both walls — which is the entire reason the energies are quantized.

The same eigenvalue problem as the string

step 1 of 4

Three moves, and then the work is done — because it was done in §4.2.

1Outside the box V is infinite, so ψ must vanish there — anything else would give infinite energy. Inside, V = 0 and the equation loses its potential term entirely.
$- \frac{ℏ ^{2}}{2 m} \frac{d ^{2} ψ}{d x ^{2}} = E ψ for 0 < x < a$
2Rename the energy in terms of a wave number. This is a definition, not an assumption — it just packages the constants.
$E = \frac{ℏ ^{2} k ^{2}}{2 m}$
Equation (4.34).
3Substituting turns the equation into one you have seen before.
$\frac{d ^{2} ψ}{d x ^{2}} = - k^{2} ψ$
Equation (4.35) — and this is EXACTLY Eq. (4.12), the string.
4The boundary conditions come from ψ having to be continuous at the walls, since the position probability density cannot jump.
$ψ (0) = ψ (a) = 0$
Equation (4.36) — and this is exactly Eq. (4.13), the string again.

Phillips states the conclusion plainly, and it is worth pausing on: the energy eigenvalue problem for a particle in a box is identical to the eigenvalue problem for a vibrating string. Same differential equation, same boundary conditions. So the answer is already known:

\psi_n(x) = N\sin k_nx,\quad\text{with}\quad k_n = \frac{n\pi}{a}\tag{4.37}

These are the functions of Fig. 4.1. In classical physics they are the shapes of a vibrating string; in quantum physics they are the wave functions of a particle in a box with definite energy, labelled by the quantum number $n = 1, 2, 3, \dots$

Quantized energy

Now feed $k_n$ back through Eq. (4.34), and the discreteness transfers from $k$ to $E$ :

E_n = \frac{n^2\pi^2\hbar^2}{2ma^2},\quad\text{with}\quad n = 1, 2, 3, \dots\tag{4.38}

\Psi_n(x,t) = N\sin k_nx\;e^{-iE_nt/\hbar}\tag{4.39}

Three things worth noticing

The fractional spacing shrinks. The gaps grow with $n$ , but relative to the energy they shrink:

\frac{E_{n+1} - E_n}{E_n} \to \frac{2}{n}\quad\text{as}\quad n \to \infty

So the discrete nature of the levels becomes less important at high energy — which is why nobody noticed quantization for two centuries. This is the correspondence principle showing up unprompted.

The lowest energy is not zero. Setting $n = 1$ ,

E_1 = \frac{\hbar^2\pi^2}{2ma^2}

a zero-point energy that the classical picture flatly denies. Phillips explains it from the uncertainty principle: a particle confined to a region of size $a$ has $\Delta x \le a$ , hence $\Delta p \gtrsim \hbar/2a$ ; since the average magnitude of the momentum always exceeds $\Delta p$ , the average kinetic energy always exceeds $(\Delta p)^2/2m$ , which exceeds $\hbar^2/8ma^2$ .

That bound is loose by a factor of 40 — and it is still the right argument

$\hbar^2/8ma^2$ is $0.125\,\hbar^2/ma^2$ . The true answer, $E_1 = \pi^2\hbar^2/2ma^2$ , is $4.93\,\hbar^2/ma^2$ — about 39.5 times larger.

That is not a failure. The uncertainty argument is not trying to compute $E_1$ ; it is proving that $E_1$ cannot be zero, which is exactly the classical claim it needs to refute. A bound that is loose by 40× still does that job perfectly.

Worth internalising as a habit: an order-of-magnitude argument that gets within a factor of 40 of a hard number is usually telling you the mechanism is right. Reach for the exact calculation when you need the number, and for the bound when you need to know whether something is possible at all.

The shapes are the string’s. The spatial shape of a wave function with energy $E_n$ is identical to the normal mode of a string with angular frequency $\omega_n$ , and the number of nodes grows with $n$ .

Normalizing

The wave function is not observable, but observable things are built from it — and the first step is always normalization:

\int_0^a|\Psi(x,t)|^2\,\mathrm dx = 1

As the book notes, this was already done in §3.4, giving $N = \sqrt{2/a}$ . From there you can compute position and momentum probability densities exactly as §3.4 did in Fig. 3.3.

A three-dimensional box

Confinement in three dimensions changes nothing structural. A state of definite energy still has the form

\Psi(x,y,z,t) = \psi(x,y,z)\,e^{-iEt/\hbar}\tag{4.40}

with

\left[-\frac{\hbar^2}{2m}\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) + V(x,y,z)\right]\psi = E\psi\tag{4.41}

and the box is now

V(x,y,z) = \begin{cases} 0 & \text{if } 0 < x < a,\ 0 < y < b,\ 0 < z < c\\ \infty & \text{elsewhere}\end{cases}\tag{4.42}

Separating again — now three times, one per axis — gives eigenfunctions labelled by three quantum numbers:

\psi_{n_x,n_y,n_z}(x,y,z) = N\sin\left(\frac{n_x\pi x}{a}\right)\sin\left(\frac{n_y\pi y}{b}\right)\sin\left(\frac{n_z\pi z}{c}\right)\tag{4.43}

E_{n_x,n_y,n_z} = \frac{\hbar^2\pi^2}{2m}\left[\frac{n_x^2}{a^2} + \frac{n_y^2}{b^2} + \frac{n_z^2}{c^2}\right]\tag{4.44}

Degeneracy

Equation (4.44) shows the levels depend on the box’s dimensions — and that some levels coincide when the box has particular dimensions. For a cubical box with $a = b = c$ , the states $E_{1,1,2}$ , $E_{2,1,1}$ and $E_{1,2,1}$ all have the same energy. When several distinct states share an energy, the level is degenerate .

Phillips says degeneracies “arise because the interactions which confine electrons in atoms, nucleons in nuclei and quarks in hadrons have specific symmetry properties”, and that observed degeneracies can be used to deduce those symmetries. That claim only becomes convincing if you can break the symmetry yourself:

b / a1.00c / a1.00

a = b = c. The levels sit at 3, 6, 9, 11, 12 … and three of the first five are threefold degenerate: (1,1,2), (2,1,1) and (1,2,1) are genuinely different states — different wave functions, different shapes — that happen to share an energy. Nothing forces that except the cube's symmetry: permuting the axes maps each state onto another and cannot change the energy.

Fig. 4.3's top level is printed as 16; it should be 12

The book’s figure labels its highest level $E_{2,2,2} = 16\,\hbar^2\pi^2/2ma^2$ . Equation (4.44) with $n_x = n_y = n_z = 2$ and $a = b = c$ gives $4 + 4 + 4 = \mathbf{12}$ .

The stronger check is that 16 is not a sum of three positive squares at all. The Compute block below searches every triple with $n_x, n_y, n_z \le 19$ and finds none, so no state of a cubical box has that energy — it is not a different state mislabelled, it is an impossible number.

The likely origin is Fig. 4.2 on page 66, where $E_4 = 16$ is perfectly correct for the one-dimensional box. A 16 migrated from one figure to the other.

The level ordering is $3, 6, 9, 11, 12, 14, \dots$ , and the widget above shows 12. Everything the book says about Fig. 4.3 — that the first, second and third excited states are degenerate — is correct.

Engineer's bridge — degeneracy is a repeated eigenvalue, and symmetry is why

In matrix terms a degenerate level is a repeated eigenvalue, and the states sharing it span an eigenspace of dimension > 1. Two consequences you already know:

The eigenvectors are not unique. Any basis of that subspace will do — which is exactly Phillips’ remark in §4.5 that eigenfunctions belonging to a degenerate eigenvalue “are not uniquely determined”, and that you can use that freedom to make them orthogonal (Gram–Schmidt).
Repeated eigenvalues are numerically delicate. Perturb the matrix slightly and they split; eigenvector routines return an arbitrary basis of the subspace, so two runs can disagree while both being right.

The physics content is that repeated eigenvalues here are not accidental. They come from the operator commuting with a symmetry — for the cube, permuting the axes. That is the same statement as A and P sharing an eigenbasis when AP = PA.

Where it breaks. In linear algebra a repeated eigenvalue is usually an inconvenience — a sign your problem is ill-conditioned. In physics it is information: an observed degeneracy is evidence of a symmetry you may not have known about, and a level that splits when you switch on a field tells you you have broken one. That is how the Zeeman and Stark effects are read, and it is the argument chapters 9 and 11 run for real atoms.

Put a number on it — why atoms are eV and nuclei are MeV

Equation (4.38) has $E_n \propto 1/ma^2$ , and that single factor spans all of physics’ energy scales.

An electron in an atom-sized box, $a = a_0 = 0.529$ Å:

E_2 - E_1 = \frac{3\pi^2\hbar^2}{2m_ea_0^2} \approx 403\ \mathrm{eV}

the right order for atomic transitions — X-ray to ultraviolet.

A neutron in a nucleus-sized box, $a = 1$ fm:

E_2 - E_1 \approx 614\ \mathrm{MeV}

a factor of $1.5\times10^6$ larger. Two contributions: the neutron is ~1839 times heavier (lowering the levels), but the box is 529 000 times smaller, and that enters squared.

This is what Phillips means by “manifestations of confinement”. Nobody had to put different physics into the nucleus to get MeV instead of eV. The same formula, with $a$ changed from ångströms to femtometres, spans six orders of magnitude.

Reproduce it in NumPy

Eq. (4.38)'s consequences at both scales, the full degeneracy structure of the cubical box, and an exhaustive search showing why Fig. 4.3's printed 16 cannot be an energy of any state.

import numpy as np
from itertools import product

hbar, me, eV = 1.054571817e-34, 9.1093837015e-31, 1.602176634e-19
MeV, u = 1e6*eV, 1.66053906660e-27

print("1-D box: E_n = n^2 pi^2 hbar^2 / 2ma^2, in units of the ground state")
print("=" * 64)
for n in (1, 2, 3, 4):
    print(f"  n={n}:  E_n/E_1 = {n**2:2d}      gap to next / E_n = "
          f"{((n+1)**2-n**2)/n**2:6.3f}   2/n = {2/n:6.3f}")
print("  the fractional gap -> 2/n, so discreteness fades at high n")

print("\nzero-point energy vs the uncertainty-principle bound")
print("  E_1        = pi^2 hbar^2 / 2ma^2 = 4.9348 hbar^2/ma^2")
print("  the bound  = hbar^2 / 8ma^2      = 0.1250 hbar^2/ma^2")
print(f"  ratio      = {np.pi**2/2/0.125:.1f}x -- the bound is loose, but it is a BOUND:")
print("  it proves E_1 cannot be zero, which is all it is asked to do")

print("\nconfinement sets the scale: the same formula, atomic and nuclear")
a = 0.529177e-10                                   # Bohr radius
dE = (2**2 - 1**2)*np.pi**2*hbar**2/(2*me*a**2)
print(f"  electron, a = a_0 = {a*1e10:.4f} A :  E_2 - E_1 = {dE/eV:8.2f} eV")
a = 1e-15                                          # 1 fm
mn = 1.00866491595*u
dE = (2**2 - 1**2)*np.pi**2*hbar**2/(2*mn*a**2)
print(f"  neutron,  a = 1 fm           :  E_2 - E_1 = {dE/MeV:8.2f} MeV")
print("  atomic spacings in eV, nuclear in MeV -- six orders of magnitude, one formula")

print("\n3-D cubical box: E/(hbar^2 pi^2/2ma^2) = nx^2+ny^2+nz^2")
lv = {}
for s in product(range(1, 5), repeat=3):
    lv.setdefault(sum(n*n for n in s), []).append(s)
for E in sorted(lv)[:6]:
    st = lv[E]
    print(f"  E = {E:2d}   degeneracy {len(st)}   {'  '.join('(%d,%d,%d)' % s for s in sorted(st))}")

print("\nis the book's Fig. 4.3 top level of 16 possible?")
hits = [s for s in product(range(1, 20), repeat=3) if sum(n*n for n in s) == 16]
print(f"  triples (nx,ny,nz) with nx^2+ny^2+nz^2 = 16, all n >= 1:  {hits if hits else 'NONE'}")
print(f"  E(2,2,2) = 2^2+2^2+2^2 = {2**2+2**2+2**2}")
print("  so the printed 16 is a misprint; the level is 12")

prints1-D box: E_n = n^2 pi^2 hbar^2 / 2ma^2, in units of the ground state ================================================================ n=1: E_n/E_1 = 1 gap to next / E_n = 3.000 2/n = 2.000 n=2: E_n/E_1 = 4 gap to next / E_n = 1.250 2/n = 1.000 n=3: E_n/E_1 = 9 gap to next / E_n = 0.778 2/n = 0.667 n=4: E_n/E_1 = 16 gap to next / E_n = 0.562 2/n = 0.500 the fractional gap -> 2/n, so discreteness fades at high n zero-point energy vs the uncertainty-principle bound E_1 = pi^2 hbar^2 / 2ma^2 = 4.9348 hbar^2/ma^2 the bound = hbar^2 / 8ma^2 = 0.1250 hbar^2/ma^2 ratio = 39.5x -- the bound is loose, but it is a BOUND: it proves E_1 cannot be zero, which is all it is asked to do confinement sets the scale: the same formula, atomic and nuclear electron, a = a_0 = 0.5292 A : E_2 - E_1 = 402.85 eV neutron, a = 1 fm : E_2 - E_1 = 613.53 MeV atomic spacings in eV, nuclear in MeV -- six orders of magnitude, one formula 3-D cubical box: E/(hbar^2 pi^2/2ma^2) = nx^2+ny^2+nz^2 E = 3 degeneracy 1 (1,1,1) E = 6 degeneracy 3 (1,1,2) (1,2,1) (2,1,1) E = 9 degeneracy 3 (1,2,2) (2,1,2) (2,2,1) E = 11 degeneracy 3 (1,1,3) (1,3,1) (3,1,1) E = 12 degeneracy 1 (2,2,2) E = 14 degeneracy 6 (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) (3,2,1) is the book's Fig. 4.3 top level of 16 possible? triples (nx,ny,nz) with nx^2+ny^2+nz^2 = 16, all n >= 1: NONE E(2,2,2) = 2^2+2^2+2^2 = 12 so the printed 16 is a misprint; the level is 12

Where this leaves us

Confinement quantizes energy, the mechanism is a boundary condition, and in three dimensions symmetry makes levels coincide. Every state so far has had a definite energy.

§4.5 asks the obvious next question: what describes a particle whose energy is uncertain? The answer reuses the superposition of Eq. (4.16) — the string again — and turns the expansion coefficients into probabilities.

Check yourself

0 / 7 answered

1.What is remarkable about the energy eigenvalue problem for a particle in a box?
2.Why is the ground-state energy not zero?
3.The uncertainty bound gives $ℏ^{2} /8 m a^{2}$ , but the true $E_{1}$ is about 40 times larger. Does that discredit the argument?
4.Why did nobody notice energy quantization until the twentieth century?
5.In a cubical box, $E_{1, 1, 2}$ , $E_{2, 1, 1}$ and $E_{1, 2, 1}$ coincide. Make the box non-cubical and they separate. What does that demonstrate?
6.Fig. 4.3 labels its top level $E_{2, 2, 2} = 16$ . What is wrong, and how can you be sure?
7.Atomic level spacings are electronvolts and nuclear ones megaelectronvolts. What accounts for the six orders of magnitude?