7D.2 Particle in a 1D Box — Detailed Lecture Notes (English)

§1 Learning Objectives and Context

In 7D.1 we solved the free particle — no potential at all, continuous energy, plane-wave momentum eigenstates. This section adds the simplest possible confinement: two infinitely high walls with free space in between. Although extremely idealised, this model delivers one of quantum mechanics' most important messages: confinement forces the energy to be quantised.

Learning objectives:

State the infinite-well potential and boundary conditions, and justify them physically.
Derive $\psi_n(x)=\sqrt{2/L}\sin(n\pi x/L)$ and $E_n=n^2h^2/(8mL^2)$ from scratch.
Define and explain quantisation, zero-point energy, nodes, and orthogonality.
Compute expectation values $\langle x\rangle,\langle x^2\rangle,\langle p\rangle,\langle p^2\rangle$ and verify the Heisenberg uncertainty relation.
Use the model to estimate the optical spectra of real systems (e.g., conjugated polyenes).

Why it matters: this is the first model where you encounter a discrete energy spectrum. Every later discussion of molecular spectra, Zeeman splitting, tunnelling, or band structure returns here for intuition. The deeper lesson is that "boundary conditions → quantisation" is the universal logic of every bound system.

§2 Why Study the Particle in a Box?

2.1 Why this model is central

Solvable: one of the very few problems with a closed-form analytic solution.
Representative: it exhibits nearly every key feature of quantum mechanics — quantisation, zero-point energy, nodes, probability distributions, orthogonality, uncertainty.
Constructive: real potentials are often approximated as piecewise wells; 2D and 3D boxes and the tight-binding model build on this foundation.

2.2 Free particle vs box: side by side

	Free particle (7D.1)	Particle in a box (7D.2)
Potential	$V(x)=0$ everywhere	$V=0$ inside / $V=\infty$ outside
Boundary conditions	none	$\psi(0)=\psi(L)=0$
Wavefunction	$Ae^{ikx}+Be^{-ikx}$	$\sqrt{2/L}\sin(n\pi x/L)$
$k$	any positive real number	$k_n=n\pi/L$ (discrete)
Energy	continuous $E=\hbar^2k^2/2m$	discrete $E_n=n^2h^2/8mL^2$
Lowest energy	$E=0$ (rest allowed)	$E_1>0$ (zero-point energy)

Key comparison: the only new requirement is that $\psi$ must vanish on the walls. From that single constraint, the whole ladder of quantised energies emerges automatically.

§3 Model: The Infinite Square Well

3.1 Potential

$$V(x)=\begin{cases}0,&0

Diagram: two infinitely tall walls at $x=0$ and $x=L$, flat bottom inside.

3.2 Physical picture

The "walls" represent an impenetrable barrier — an idealisation of a very strong confinement.
Inside, no force acts: $F=-dV/dx=0$, so the particle moves freely.
Ignored: finite wall thickness/height, tunnelling beyond the walls, internal potential variations.

Although "$V=\infty$" sounds extreme, it is a useful approximation for electrons in strongly confined semiconductor nanostructures and metallic nanowires.

§4 Outside the Box: Why ψ Must Vanish

For $x\le 0$ or $x\ge L$, substituting $V=\infty$ into the stationary Schrödinger equation forces

$\psi(x)=0$ for $x\le 0$ or $x\ge L$.

Physically: infinitely high walls make the particle's probability of being outside strictly zero.

4.1 Continuity requirement

Wavefunctions must be continuous (otherwise the momentum operator $\hat p=-i\hbar\,d/dx$ would be ill-defined). Matching the interior solution to the exterior $\psi=0$ gives the boundary conditions:

$\psi(0)=0,\ \psi(L)=0.$

Note: the first derivative is not required to be continuous across the infinite wall — this is the singular limit of an otherwise continuous-derivative problem (see the finite-well case in §16).

§5 Inside the Box: the Schrödinger Equation and General Solution

5.1 The equation

$\displaystyle -\frac{\hbar^2}{2m}\psi''(x)=E\psi(x).$

This is exactly the free-particle equation from 7D.1. Its general solution:

$\psi(x)=A\sin(kx)+B\cos(kx),\quad k=\sqrt{2mE}/\hbar.$

5.2 Why sines and cosines here?

Although $e^{\pm ikx}$ and $\sin/\cos$ are mathematically equivalent, the $\sin/\cos$ form is far more convenient for boundary-value problems. The solutions will be standing waves, not travelling waves.

§6 How Boundary Conditions Select Allowed k

6.1 At x=0

$\psi(0)=A\sin(0)+B\cos(0)=B=0.$

So $B=0$ and $\psi(x)=A\sin(kx)$.

6.2 At x=L

$\psi(L)=A\sin(kL)=0.$

Discarding the trivial case $A=0$:

$\sin(kL)=0\ \Rightarrow\ kL=n\pi,\ n=1,2,3,\dots$

Quantisation condition: $$k_n=\frac{n\pi}{L},\quad n\in\mathbb{Z}^+.$$ $k$ is no longer arbitrary — only wavelengths that fit an integer number of half-waves across the box survive.

Analogy: a plucked guitar string or organ pipe supports only standing waves of wavelength $\lambda_n=2L/n$. The quantum box is the same mathematics in a different guise.

6.3 Why not n=0 or n<0?

$n=0\Rightarrow\psi\equiv 0$: not a physical state.
$n<0$: $\sin(-n\pi x/L)=-\sin(n\pi x/L)$, differing only by an overall sign from the $n>0$ state, i.e., the same physical state.

§7 Energy Quantisation: E_n = n²h²/8mL²

From $k=\sqrt{2mE}/\hbar$ and $k_n=n\pi/L$:

$\left(\dfrac{n\pi}{L}\right)^2=\dfrac{2mE_n}{\hbar^2}$
$E_n=\dfrac{n^2\pi^2\hbar^2}{2mL^2}$
With $\hbar=h/(2\pi)$:
$E_n=\dfrac{n^2h^2}{8mL^2}.$

Central result: $$\boxed{\ E_n=\dfrac{n^2h^2}{8mL^2},\quad n=1,2,3,\dots\ }$$

7.1 Spacing

$E_n\propto n^2$ — not equally spaced.
$E_{n+1}-E_n=(2n+1)\,h^2/(8mL^2)$ — grows with $n$.
First ratios: $E_1:E_2:E_3:E_4 = 1:4:9:16$.

7.2 Dependence on size and mass

$E_n\propto 1/(mL^2)$:

Smaller box ($L$ small) → levels spread apart → stronger quantum effects.
Lighter particle ($m$ small) → stronger quantum effects.

Hence electrons (light) confined to nanometre-scale structures show dramatic quantum behaviour, while a ping-pong ball on a table does not.

§8 Normalisation: Fixing N

Write $\psi_n(x)=N\sin(n\pi x/L)$. Require $\int|\psi_n|^2\,dx=1$:

$\int_0^L N^2\sin^2(n\pi x/L)\,dx=1$
Use $\sin^2\theta=(1-\cos 2\theta)/2$:
$\int_0^L\sin^2(n\pi x/L)\,dx=L/2-0=L/2.$
Therefore $N^2\cdot L/2=1\Rightarrow N=\sqrt{2/L}$.

Normalised eigenfunctions: $$\boxed{\ \psi_n(x)=\sqrt{\dfrac{2}{L}}\sin\!\left(\dfrac{n\pi x}{L}\right),\ 0\le x\le L,\ n=1,2,3,\dots\ }$$

Global phase: $N$ can be multiplied by any $e^{i\phi}$ and still be normalised. Global phases have no observable effect, so we conventionally take $N>0$ real.

§9 Shape of the Wavefunctions and Nodes

9.1 First few states

$n$	$\psi_n(x)$	Nodes (inside)	Antinodes	$E_n/E_1$
1	$\sqrt{2/L}\sin(\pi x/L)$	0	1	1
2	$\sqrt{2/L}\sin(2\pi x/L)$	1	2	4
3	$\sqrt{2/L}\sin(3\pi x/L)$	2	3	9
4	$\sqrt{2/L}\sin(4\pi x/L)$	3	4	16

Sketch: four sine curves stacked on top of each other, each with one more oscillation inside the box.

9.2 Node theorem

Node theorem: in a 1D bound system, the $n$-th eigenstate has $n-1$ interior nodes (not counting the zeros at the walls).

Node positions: $x_k=kL/n,\ k=1,\dots,n-1$.

9.3 More nodes → higher energy

More nodes ⇒ more curvature ⇒ larger kinetic energy ⇒ higher energy. This rule of thumb — "number of nodes indexes the eigenvalues" — applies across quantum mechanics, including the $1s,2s,3s$ node count of atomic orbitals.

§10 Probability Density and Interpretation

$P_n(x)=|\psi_n(x)|^2=(2/L)\sin^2(n\pi x/L)$.

10.1 Features

At nodes: $P=0$ — the particle is never found there.
At antinodes: $P$ reaches the maximum $2/L$.
The average density is $1/L$ — the classical uniform value.

10.2 Born interpretation

$\displaystyle P(a

10.3 Quantum vs classical

Sketch: (classical) flat horizontal line; ($n=1$) single peak in the middle; ($n=2$) two peaks; ($n=3$) three peaks.

A classical particle bouncing between the walls has uniform density $1/L$. The $n=1$ quantum state instead has the particle most likely at the centre, almost never at the walls — completely counter-intuitive, but exactly what the mathematics says.

§11 Orthogonality and Completeness

11.1 Orthogonality

$$\int_0^L\psi_m^*(x)\psi_n(x)\,dx=\delta_{mn}.$$

11.2 Derivation

$\int_0^L\tfrac{2}{L}\sin(m\pi x/L)\sin(n\pi x/L)\,dx$
Use $\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]$:
$=\tfrac1L\int_0^L[\cos((m-n)\pi x/L)-\cos((m+n)\pi x/L)]\,dx$
For $m\ne n$: both integrals vanish → result = 0.
For $m=n$: first term = $L$, second = 0 → result = 1.

11.3 Completeness

The set $\{\psi_n\}$ forms a complete orthonormal basis on $[0,L]$: any reasonable function $f(x)$ with $f(0)=f(L)=0$ can be expanded as

$\displaystyle f(x)=\sum_{n=1}^\infty c_n\psi_n(x),\quad c_n=\int_0^L\psi_n^*(x)f(x)\,dx.$

This is the classical Fourier sine series. Fourier analysis is essentially the mathematical skeleton of the 1D box.

Application: an initial wave shape $\psi(x,0)$ can be decomposed into eigenstates that evolve independently as $e^{-iE_n t/\hbar}$ — the recipe for time-dependent wave packets.

§12 Expectation Values and the Uncertainty Principle

12.1 $\langle x\rangle$

$\langle x\rangle=\int_0^L x|\psi_n|^2\,dx=\tfrac{2}{L}\int_0^L x\sin^2(n\pi x/L)\,dx$
By mirror symmetry of $|\psi_n|^2$ about $x=L/2$: $\langle x\rangle=L/2.$

12.2 $\langle x^2\rangle$ and $\sigma_x$

$\langle x^2\rangle=\tfrac{L^2}{3}-\tfrac{L^2}{2n^2\pi^2},\quad \sigma_x^2=\tfrac{L^2}{12}\!\left(1-\tfrac{6}{n^2\pi^2}\right).$

12.3 $\langle p\rangle$ and $\langle p^2\rangle$

$\langle p\rangle=0,\quad \langle p^2\rangle=2mE_n=n^2\pi^2\hbar^2/L^2,\quad \sigma_p=n\pi\hbar/L.$

Physical meaning of $\langle p\rangle=0$: a standing wave is an equal superposition of right- and left-moving plane waves, so the mean momentum is zero.

12.4 Verifying the Heisenberg relation

$\sigma_x\sigma_p=\frac{\hbar}{2}\sqrt{\frac{n^2\pi^2}{3}-2}.$

For $n=1$: $\sigma_x\sigma_p\approx 0.568\,\hbar>\hbar/2$ ✓.
For $n\to\infty$: $\sigma_x\sigma_p\sim n\pi\hbar/(2\sqrt3)\gg\hbar/2$.

Observation: the Heisenberg lower bound $\hbar/2$ is saturated only by Gaussian wave packets. Box eigenstates are sines, never Gaussian, so their product strictly exceeds $\hbar/2$.

§13 Zero-Point Energy

$E_1=\dfrac{h^2}{8mL^2}>0.$

13.1 Three explanations for $E_1\ne 0$

Mathematical: $E=0\Rightarrow k=0\Rightarrow\psi\equiv 0$, a non-physical state.
Boundary: at least a half-wavelength must fit inside the box, giving $\lambda_1=2L$ and minimum kinetic energy $h^2/(8mL^2)$.
Uncertainty: $\sigma_x\lesssim L\Rightarrow\sigma_p\gtrsim\hbar/L\Rightarrow E\gtrsim\hbar^2/(mL^2)$.

All three paths deliver the same conclusion: a confined particle can never be truly at rest.

13.2 Real-world consequences

Liquid helium remains liquid at 0 K because zero-point motion resists crystallisation.
H₂ vibrational zero-point energy ≈ 0.27 eV, generating observable isotope effects (H vs D reaction rates).
Vacuum electromagnetic modes have zero-point energy, producing the Casimir effect.

§14 Classical Correspondence

As $n\to\infty$, the rapidly oscillating probability density $(2/L)\sin^2(n\pi x/L)$ averages locally to $1/L$ — the classical uniform distribution.

Sketch: $n=1$ single hump → $n=10$ ten humps → $n=100$ fine oscillations blurring to a flat line.

Bohr correspondence principle: at large quantum numbers, quantum predictions must merge into classical results.

Relative level spacing $\Delta E_n/E_n=(2n+1)/n^2\to 0$: the ladder packs so tightly at high $n$ that it appears continuous.

§15 Application: π-Electrons in Conjugated Systems

15.1 Model

Along a linear conjugated polyene (e.g., a polyene or β-carotene), the π electrons move rather freely over the chain length $L$ and can be treated as particles in a 1D box.

15.2 Filling the levels

$N$ π electrons fill the lowest $N/2$ levels by the Pauli principle (two per level). HOMO = level $N/2$, LUMO = level $N/2+1$.

15.3 Absorption wavelength

$\Delta E=E_{N/2+1}-E_{N/2}=\tfrac{h^2(N+1)}{8m_eL^2},\quad\lambda=hc/\Delta E.$

15.4 β-carotene example

Chain length $L\approx 1.8$ nm (11 C=C double bonds, ~22 conjugated carbons);
$N=22$ π electrons;
Predicted $\Delta E\approx 3.0\times10^{-19}$ J, $\lambda\approx 660$ nm;
Experimental $\lambda_{max}\approx 450$ nm — the box model gives a reasonable order of magnitude.

The everyday observation that longer conjugated chains look redder (tomato vs carotene) has its quantum-mechanical origin precisely in $\Delta E\propto 1/L^2$.

§16 Preview of the Finite Well

If the walls are finite ($V=V_0<\infty$):

$\psi$ is non-zero outside, decaying as $e^{-\kappa x}$ — the particle has some probability of being found in the classically forbidden region.
$\psi$ and $d\psi/dx$ are both continuous at the wall (no infinite jumps).
Only finitely many bound levels exist; states with $E>V_0$ form a continuum.
Bound-state energies satisfy a transcendental equation, not the simple $n^2h^2/(8mL^2)$.

Connection to 7D.4: exponential decay into forbidden regions is the origin of tunnelling. The infinite well is the $V_0\to\infty$ limit.

§17 Worked Examples

Example 1 — Ground state of an electron in a 1 nm box

Compute $E_1$ for $m_e$ and $L=1.0$ nm.

$E_1=\frac{(6.626\times10^{-34})^2}{8(9.109\times10^{-31})(10^{-9})^2}$
$=6.03\times10^{-20}$ J $=0.376$ eV

Order of magnitude matches molecular electronic transitions — reasonable.

Example 2 — Transition wavelength

Wavelength for $n=1\to n=2$ using the previous box.

$\Delta E=3E_1=1.81\times10^{-19}$ J = 1.13 eV
$\lambda=hc/\Delta E\approx 1.10\ \mu$m (near-IR)

Example 3 — Probability of finding the particle in $[0,L/4]$

For the $n=1$ state:

$P=\tfrac{2}{L}\int_0^{L/4}\sin^2(\pi x/L)\,dx=\tfrac{1}{L}\left[x-\tfrac{L}{2\pi}\sin(2\pi x/L)\right]_0^{L/4}$
$=\tfrac{1}{4}-\tfrac{1}{2\pi}\approx 0.091.$

Classical expectation is $0.25$; the quantum value is much lower because the $n=1$ wavefunction has little amplitude near the wall.

Example 4 — Zero-point energy from the uncertainty principle

Using $\sigma_x\sigma_p\ge\hbar/2$:

$\sigma_x\sim L$ ⇒ $\sigma_p\gtrsim\hbar/(2L)$
$E_{\min}\sim\langle p^2\rangle/(2m)\gtrsim\hbar^2/(8mL^2).$

The exact $E_1=\pi^2\hbar^2/(2mL^2)$ is 40× larger — same order of magnitude. A textbook example of "uncertainty-principle estimate".

Example 5 — β-carotene absorption

$L=1.8$ nm, 22 π electrons. HOMO-LUMO transition:

HOMO: $n=11$; LUMO: $n=12$
$\Delta E=(144-121)\,h^2/(8m_eL^2)=23\,h^2/(8m_eL^2)$
$\Delta E\approx 4.29\times10^{-19}$ J → $\lambda\approx 463$ nm (blue)

Experimentally $\lambda_{max}\approx 450$ nm — surprisingly accurate! This is why carrots appear orange (they absorb blue and reflect the complement).

Example 6 — Orthogonality check

Verify that $\psi_1$ and $\psi_3$ are orthogonal:

$\int_0^L\tfrac{2}{L}\sin(\pi x/L)\sin(3\pi x/L)\,dx=\tfrac{1}{L}\int_0^L[\cos(2\pi x/L)-\cos(4\pi x/L)]\,dx=0$ ✓

§18 Key Takeaways

Potential: $V=0$ inside, $V=\infty$ outside.
Boundary conditions: $\psi(0)=\psi(L)=0$.
Quantisation: $k_n=n\pi/L$, $n\in\mathbb Z^+$.
Wavefunctions: $\psi_n=\sqrt{2/L}\sin(n\pi x/L)$.
Energies: $E_n=n^2h^2/(8mL^2)\propto n^2$.
Zero-point energy: $E_1>0$; no bound state is ever at rest.
Node theorem: $n-1$ interior nodes for state $n$.
Orthonormality: $\int\psi_m\psi_n\,dx=\delta_{mn}$.
Completeness: $\{\psi_n\}$ is a Fourier sine basis.
Master principle: confinement + continuity = quantisation. This logic applies to every bound quantum system.

§19 Advanced Discussion Questions

Re-derive the wavefunctions if the box is centred: $[-L/2,L/2]$. How do parity and the sine/cosine mix change? Are the energies the same?
Prepare a state $\psi(x,0)=(\psi_1+\psi_2)/\sqrt 2$. Compute $\langle x\rangle(t)$ and explain the oscillation frequency.
Why is $\langle p\rangle=0$ but $\langle p^2\rangle>0$? Is this a contradiction? (Hint: think of the momentum distribution.)
Add a small perturbation $V'(x)=\epsilon x$ inside the box. Compute the first-order energy shift for $n=1$.
Split the box by inserting an infinite wall at $x=L/2$. How do the eigenenergies change? Which quantum numbers are allowed?
Expand the "uniform" initial state $\psi(x,0)=1/\sqrt L$ (non-eigenstate) in the $\{\psi_n\}$ basis. Which $c_n$ are nonzero?

Hints: (4) uses $\langle\psi_1|V'|\psi_1\rangle$; (6) previews time-dependent problems in 7F.