Exactly Solvable Systems I — Boxes, Steps & Barriers
The particle in a box is the first system anyone solves, and the one most students learn to compute without understanding. This part treats it properly — where the quantisation actually comes from, why the zero-point energy cannot be removed, what the nodes mean — and then extends it to the cases the exam prefers: degeneracy in three dimensions, and tunnelling, which explains everything from ammonia inversion to why some kinetic isotope effects exceed their classical ceiling. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.
The 9 sections in Part 3
- 1The infinite square well: solving it from the boundary conditions Free below
- 2Normalisation, nodes, orthogonality and zero-point energy
- 3Expectation values and the uncertainty product
- 4Boxes in two and three dimensions: separation of variables
- 5Degeneracy, symmetry, and the free-electron model of conjugated molecules
- 6The free particle, the continuum, and the finite well
- 7The potential step
- 8The rectangular barrier and the tunnelling probability
- 9What tunnelling explains: isotope effects, STM, α-decay and NH3 inversion
The infinite square well: solving it from the boundary conditions
Free extractSection C.1 of Part 3, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.
A particle trapped between two walls. Nothing else. Watch the integer appear out of pure trigonometry.
Here is the whole problem. A particle of mass m is confined to move along a line, between x = 0 and x = L. Inside that region nothing acts on it — the potential energy is zero, so it is a free particle. Outside, the potential energy is infinite, which is a mathematician's way of saying that the walls are absolutely rigid: no amount of energy will get the particle through them.
Start outside, because that part takes one line. If V is infinite there, then for the total energy E to be finite the wavefunction must be zero there. (Look at the Schrödinger equation: the term Vψ is infinite unless ψ = 0.) So ψ(x) = 0 for all x ≤ 0 and all x ≥ L. The particle is never outside. That is the whole content of the infinite wall, and it is what makes this problem solvable.
Now go inside. There V = 0 and the time-independent Schrödinger equation is just the kinetic-energy operator acting on ψ:
Rearrange it into the standard form. Write k2 = 2mE/ℏ2, so that k has units of 1/length. Then
This is the most familiar differential equation in all of physics: it says that the second derivative of a function is proportional to minus the function itself. Its solutions are sines and cosines, and the general solution is a combination of both:
Step 1 — the wall at x = 0 kills the cosine
A wavefunction has to be continuous. It is zero just to the left of x = 0, so it must be zero at x = 0 as well — otherwise it would jump, and a jump in ψ would make d2ψ/dx2 catastrophically undefined. Impose ψ(0) = 0:
Because cos 0 = 1 and sin 0 = 0, the boundary condition at the left wall does nothing except delete the cosine. We are left with ψ(x) = A sin(kx). No quantisation yet: k can still be anything, and therefore so can E.
Step 2 — the wall at x = L is the one that quantises
Now impose the same condition at the other wall, ψ(L) = 0:
There are two ways to satisfy this. The first is A = 0, which makes ψ identically zero everywhere — the particle does not exist. That is not a wavefunction; it is the absence of one, and we discard it. (It is called the trivial solution, and rejecting it is worth a mark in a written answer.) The second is
There is the integer. It arrived because the sine function is zero only at multiples of π, so a sine that is pinned to zero at both ends of an interval of length L must fit a whole number of half-wavelengths into that interval. Nothing was assumed. The quantisation is a consequence of confinement plus continuity, and it would have appeared even if nobody had ever heard of Planck.
Step 3 — read off the energies
We had k = √(2mE)/ℏ, and now k = nπ/L. Set them equal and solve for E:
Substituting ℏ = h/2π puts it in the form everyone quotes:
Three features of that formula deserve to be said aloud, because they are what examiners actually test.
(i) The lowest energy is not zero. n starts at 1, not 0, because n = 0 would give ψ = 0 everywhere. So the particle can never be at rest; it retains E1 = h2/8mL2 however cold you make it. That is the zero-point energy, and we return to it in C.2.
(ii) The levels are not evenly spaced. The gap between successive levels is En+1 − En = (2n+1)E1, which grows as you climb. This is the opposite of the harmonic oscillator, whose levels are equally spaced, and mixing the two up is one of the most common errors in this topic.
(iii) Confinement costs energy. E ∝ 1/L2, so squeezing the box raises every level, and squeezing it a lot raises them a lot. This single fact explains why electrons in small nanocrystals absorb at shorter wavelengths than electrons in large ones — the basis of the quantum-dot colour tuning used in displays.
Getting a feel for the numbers
Formulae are easier to trust once you have put numbers in them. For an electron in a box of length L quoted in Ångström, the constant works out as
So an electron in a 1 Å box — roughly an atom — has E1 ≈ 38 eV, comparable with an ionisation energy. In a 10 Å = 1 nm box — roughly a long conjugated chain — E1 falls by a factor of a hundred to 0.3760 eV, and the first few transitions land in the visible and ultraviolet. That is exactly the regime of organic dyes, and it is why the box model is worth taking seriously for π systems at all.
Now do the same for something macroscopic. A 1-milligram bead confined to a one-centimetre track has E1 = 5.488 × 10−58 J. If the bead is ambling along at 1 mm s−1, its kinetic energy corresponds to a quantum number of about n = 3.018 × 1022. The spacing between adjacent levels at that n is roughly 2nE1 ≈ 3.313 × 10−35 J — unmeasurable by any conceivable instrument. The energy of the bead is quantised; it is simply that the steps are so fine that the staircase is indistinguishable from a ramp. Quantum mechanics does not stop applying to large objects. It stops being noticeable.
Why only ψ and not ψ′ is continuous at an infinite wall; the operator-theoretic reading of the boundary condition; and what is being idealised away.
The elementary derivation above is correct but glosses over three points that separate a safe answer from a good one. Each is a favourite of interviewers and of the harder Part-C questions.
(a) Continuity of ψ but not of ψ′
The standard requirement is that ψ and dψ/dx both be continuous everywhere. At an infinite potential step, the second requirement is abandoned. The reason is visible in the equation itself. Integrating the Schrödinger equation across a discontinuity at x = a from a−ε to a+ε gives
For a finite V the integral vanishes as ε → 0, so the derivative is continuous. For V = ∞ the right-hand side is an indeterminate ∞ × 0 and the argument fails: the derivative is permitted to jump. And it must, because ψ = A sin(nπx/L) has ψ′(0) = Anπ/L ≠ 0 while ψ′ = 0 just outside. The kink at the wall is real, it is a consequence of the idealisation, and it disappears the moment the wall is made finite — which is precisely what we shall do in C.6.
(b) The boundary condition as a statement about the Hamiltonian
A sharper way to say all of this: the boundary conditions are not extra information imposed on the solutions, they are part of the definition of the operator. The kinetic-energy operator −(ℏ2/2m)d2/dx2 is a symbol until you state the space of functions it acts on. Choosing the domain to be functions on [0, L] vanishing at both endpoints makes the operator self-adjoint, and self-adjointness is what guarantees real eigenvalues and a complete orthogonal set of eigenfunctions. Different boundary conditions define a different self-adjoint operator with a different spectrum: periodic conditions ψ(0) = ψ(L), ψ′(0) = ψ′(L) give the particle on a ring, with En = n2h2/2mL2, n = 0, ±1, ±2, …, doubly degenerate and including n = 0. Same differential equation, different physics, entirely because of the boundary condition.
| Boundary condition | Allowed k | Energies | Degeneracy | Zero-point energy? |
|---|---|---|---|---|
| ψ(0) = ψ(L) = 0 (box) | nπ/L, n = 1,2,3… | n2h2/8mL2 | non-degenerate | Yes, h2/8mL2 |
| periodic, ψ(0) = ψ(L) (ring) | 2nπ/L, n = 0,±1,±2… | n2h2/2mL2 | 2 for n ≠ 0 | No — n = 0 is allowed |
| none (free particle) | any real k | ℏ2k2/2m ≥ 0 | 2 (±k) | No — continuum from zero |
(c) Why n = 0 and negative n are excluded, stated properly
n = 0 gives k = 0 and ψ ≡ 0, which cannot be normalised and describes no particle: it is not a state, so it is excluded. Negative n gives sin(−nπx/L) = −sin(nπx/L), which differs from the n-th solution only by an overall sign. Two wavefunctions differing by a global phase factor (here, by −1 = eiπ) represent the same physical state, because all observable quantities depend on ψ through |ψ|2 or through matrix elements ⟨ψ|Â|ψ⟩, and both are unchanged. So negative n is redundant rather than forbidden. That distinction — forbidden for n = 0, redundant for n < 0 — is exactly the kind of precision that separates full marks from most.
(d) What the model idealises away, and when that matters
It is worth being explicit about the physics that has been discarded, because the model is used far beyond the regime where it is literally true.
| Idealisation | Physical reality | Consequence of the idealisation |
|---|---|---|
| Walls of infinite height | Any real confining potential is finite — an ionisation limit, a work function, a band offset | The real system has only a finite number of bound states and ψ leaks into the wall (C.6); computed levels are always too high |
| Perfectly flat bottom | A conjugated chain has alternating single and double bonds, so V(x) is periodic, not flat | The real spectrum develops a gap at the zone boundary; this is why the free-electron model over-predicts λmax for polyenes (C.5) |
| One particle, no interactions | π electrons repel one another | The model gives orbital energies, not state energies; electron repulsion and exchange shift the observed transition |
| Sharp boundary at a defined L | Where does a molecule end? | L is a fitted parameter, not a measured one; results depend on the convention chosen, and any comparison must state it |
Solve the box from scratch, stating every step you are allowed to skip and every step you are not Medium
An electron is confined to a box of length 0.50 nm. Find the wavelength of the n = 1 → n = 2 transition. Easy
⚠ Common mistakes & exam traps
- Writing En = n2h2/8π2mL2. The π2 belongs with ℏ, not with h. Either En = n2h2/8mL2 or En = n2π2ℏ2/2mL2. Mixing them is the single commonest algebraic error in this topic; check by remembering ℏ = h/2π so π2ℏ2/2 = π2h2/8π2 = h2/8.
- Allowing n = 0. It is not a state: ψ would be identically zero. Note the contrast with the particle on a ring, where n = 0 is allowed because the constant function is a perfectly good normalisable solution there. If a question involves a ring, the zero-point energy is zero.
- Saying the walls are nodes. They are boundary conditions. The number of nodes is n − 1, counting only the interior zeros. Counting the walls gives n + 1 and wrecks every question about nodal structure.
- Treating the level spacing as constant. ΔE = (2n+1)h2/8mL2 grows with n. Only the harmonic oscillator has uniform spacing.
- Forgetting that A = √(2/L) is independent of n. Students often try to re-derive a different normalisation constant for each state. The integral of sin2 over a whole number of half-periods is always L/2.
- Using the mass of the box, the mass of the molecule, or the reduced mass. m is the mass of the confined particle. For π-electron problems it is me, always.
Show solution
For the ring the boundary condition is periodicity, not vanishing, and the solutions are ψ = (2π)−1/2einφ. n = 0 gives the constant function, which is perfectly normalisable, so the particle on a ring has no zero-point energy. The difference is entirely in the boundary condition: vanishing at two points forces a half-wave to fit, periodicity does not.
Show solution
Show solution
Show solution
Further reading. Levine Quantum Chemistry, Atkins & Friedman Molecular Quantum Mechanics, McQuarrie Quantum Chemistry, Szabo & Ostlund Modern Quantum Chemistry, Griffiths Introduction to Quantum Mechanics, and Pilar Elementary Quantum Chemistry.
Read the rest of Part 3
The remaining 8 sections of this part — Normalisation, nodes, orthogonality and zero-point energy, Expectation values and the uncertainty product, Boxes in two and three dimensions: separation of variables, Degeneracy, symmetry… — and all nine parts of Quantum Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.
See plans Open in the app