University Physics IV · Quantum Potentials · 7.5

The Finite Square Well

Let the walls come down to a finite height and the tidy n² ladder of the box goes with them. What replaces it: even and odd interiors matched onto decaying tails, a transcendental condition solved by graph or by root-finder, and a level count fixed by a single dimensionless depth.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Give the walls a finite height V₀ and two things change at once. Outside the well the Schrödinger equation still has solutions, real exponentials e^(−κ|x|) that decay rather than vanish, so ψ leaks into the classically forbidden region and the condition ψ(±a) = 0 is gone. In its place stands continuity of ψ and ψ′ at each wall, which is the same as demanding that the logarithmic derivative ψ′/ψ match there.

Because the well is symmetric, parity commutes with H and every bound state can be taken even or odd, so one wall does the work of two. What comes out is not a formula for E but a transcendental equation, z tan z = √(z₀² − z²) for even states and −z cot z = √(z₀² − z²) for odd, whose roots you find by graph or by bisection. That is the price: no closed form, and no infinite ladder either.

Everything is set by one dimensionless depth z₀ = (a/ħ)√(2mV₀), the radius of a circle in the (z, κa) plane, and each time it grows past another π/2 the circle catches one more branch and the well gains one more level, alternating even, odd, even. What you buy is a model that is true of quantum wells, molecular potentials and nuclei: levels pushed below their infinite-well partners, a finite number of them, and a tail that can carry most of the probability.

Simple definition: A finite square well is a region of depth V₀ and width 2a in which a particle is bound only at those energies where the oscillating interior solution joins smoothly onto a decaying exterior tail, a matching condition with no closed-form solution.
Example: An electron in a 1.0 nm well 5.0 eV deep has z₀ = 5.73, so it holds floor(2z₀/π) + 1 = 4 bound states, at 0.27, 1.08, 2.38 and 4.06 eV above the floor. The ground state sits 28% below the infinite well's 0.376 eV.

Interior and exterior wavenumbersk = √(2m(E + V₀))/ħκ = √(−2mE)/ħk² + κ² = 2mV₀/ħ²

k sets how fast ψ oscillates inside, κ how fast it dies outside, and their squares always add to the same constant.

V = −V₀ for |x| ≤ a and 0 outside, so a bound state has −V₀ < E < 0 and both roots are real. k and κ are in m⁻¹.

Even-parity matching conditionψ = A cos kx insidek tan(ka) = κ

Fixes the ground state and every second level above it, carrying 0, 2, 4 nodes.

Continuity of ψ′/ψ at x = a. The amplitudes cancel, so no normalisation is needed to find an energy.

Odd-parity matching conditionψ = A sin kx inside−k cot(ka) = κ

Supplies the second, fourth and sixth levels, with 1, 3, 5 nodes, and the deuteron's only state.

Same wall, same rule. The odd family needs ka past π/2, so a well with z₀ under π/2 has none at all.

Dimensionless well strengthz = kaz₀² = 2mV₀a²/ħ²z² + (κa)² = z₀²

Collapses m, V₀ and a into one parameter, and turns both matching conditions into curves cut by a circle of radius z₀.

a is the half-width and z₀ is a pure number. For an electron, z₀ = 2.562 (L/nm) √(V₀/eV) with L = 2a.

How many bound statesN = floor(2z₀/π) + 1

A finite well always holds at least one state and never infinitely many. z₀ = 5.73 gives four, not the three a naive count returns.

Parity alternates upward from an even ground state; level n appears once V₀ passes the infinite well's Eₙ₋₁.

Shallow limit and the tailE_b → 2mV₀²a²/ħ² as z₀ → 01/κ = ħ/√(2mE_b)

Binding dies as the square of V₀a, so a shallow one-dimensional well still binds, but its tail is far wider than the well.

E_b is the binding energy below the rim, in J or eV; 1/κ is the e-folding length of the outside tail, in metres.

Parity does half of the work

A symmetric well has V(−x) = V(x), so the parity operator commutes with H and every bound state can be chosen even or odd. That is a theorem rather than a convenience: in one dimension bound states are non-degenerate, so there is no third option and no mixed case to worry about. The saving is real. Without parity you carry four constants, two inside and one in each tail, and four matching equations. With it, the interior is either A cos kx or A sin kx, the left tail is the mirror of the right, and a single wall at x = +a carries the whole condition. Even states have 0, 2, 4 nodes and odd states have 1, 3, 5, so the ground state is always the nodeless even one. Write that down before any root-finding starts and you already know the parity of every level you are about to find.

Match the logarithmic derivative, not the amplitude

At a finite step both ψ and ψ′ are continuous; only an infinite jump in V can break ψ′. Take an even state: ψ = A cos kx for |x| ≤ a, and ψ = B e^(−κx) for x beyond a. Continuity of ψ gives A cos ka = B e^(−κa), and continuity of ψ′ gives −Ak sin ka = −κB e^(−κa). Divide the second by the first and both amplitudes cancel, leaving k tan ka = κ. That division is the whole trick, and it is why the quantisation condition is a statement about the logarithmic derivative ψ′/ψ: you never touch normalisation to find an energy, and you normalise afterwards only if you want a probability. Repeat the argument with the sine and you get −k cot ka = κ. Each condition still contains two unknowns, k and κ, but they are not independent.

One dimensionless depth, and a circle

The well ties k and κ together: k² + κ² = 2mV₀/ħ² for every bound state, because E cancels between the two definitions. Multiply by a², write z = ka and z₀ = (a/ħ)√(2mV₀), and the pair becomes z² + (κa)² = z₀², while the matching conditions become z tan z = √(z₀² − z²) and −z cot z = √(z₀² − z²). Now everything sits on one graph: the tangent and cotangent branches climb away from the z axis, a quarter circle of radius z₀ sweeps down to meet them, and every intersection is one bound state. The point of the rewrite is that z₀ is the only parameter left. Two wells with the same z₀, whatever their separate mass, width and depth, have the same spectrum in units of ħ²/2ma². An electron in a 1.0 nm well 5.0 eV deep has z₀ = 5.73.

Counting the levels, and the off-by-one

The tangent branches begin at z = 0, π, 2π and the cotangent branches at π/2, 3π/2, so a new branch opens every π/2 and the circle catches one more each time z₀ grows past another half-multiple of π: N = floor(2z₀/π) + 1. Since the first branch starts at z = 0, N is never zero. Translate that threshold back into physics. Setting z₀ = (n − 1)π/2 gives V₀ = (n − 1)²π²ħ²/2mL² with L = 2a, which is exactly level n − 1 of the infinite well of the same width. So state n appears when V₀ passes Eₙ₋₁, not Eₙ, and the finite well always holds one more level than there are infinite-well levels below its rim. For the 5.0 eV, 1.0 nm well those are 0.376, 1.50 and 3.38 eV, three of them, and the finite well holds four: 0.27, 1.08, 2.38 and 4.06 eV above the floor.

The tail is not a correction

Outside the well |ψ|² falls as e^(−2κ|x|), with e-folding length 1/κ = ħ/√(2mE_b) set by how far the state sits below the rim, so the weaker the binding the wider the tail. Deeply bound states barely notice: the ground state of the 5.0 eV, 1.0 nm well has 1/κ = 0.090 nm and keeps 99% of its probability inside. The top state of the same well, bound by only 0.94 eV, has 1/κ = 0.20 nm and leaks 23% out. Push further and the tail takes over: a 0.20 eV well 0.40 nm wide binds by 0.033 eV, has 1/κ = 1.07 nm, and puts 70% of the electron outside the well that holds it. A useful summary is that the finite well behaves like an infinite well widened by one penetration depth at each wall. For that ground state π/k = 1.176 nm against L + 2/κ = 1.180 nm, agreement to 0.3%, though the rule degrades as the binding weakens.

One dimension always binds; three does not

Let the well go shallow. For small z₀ the even condition gives z ≈ z₀ and κa ≈ z₀², so the binding energy tends to E_b = ħ²κ²/2m → 2mV₀²a²/ħ², quadratic in the product V₀a rather than linear in the depth. It is tiny, but it is not zero: a symmetric one-dimensional well binds however shallow it is. Three dimensions behave differently. For an s state put u = rR(r); then u obeys the same one-dimensional equation with the extra condition u(0) = 0, which discards the entire even family. Only odd solutions survive, and those need z₀ past π/2, so a spherical well has a minimum depth, V₀ ≥ π²ħ²/8μR². The deuteron sits right at that edge: with R ≈ 2.0 fm the threshold is 25.6 MeV, the fitted depth is about 36.6 MeV, and z₀ = 1.88 against 1.571. The result is a nucleus bound by 2.22 MeV with no excited state and 64% of its wavefunction outside the range of the force.

Change one variable at a time

Make the relationship visible.

Interactive model

Well depth V₀5.0 eV

Full width L1.00 nm

Start at 5.0 eV and 1.0 nm, then drag the depth down and watch the circle shrink past each dashed line, losing a crossing each time. It never loses the last one: in one dimension a symmetric well always binds.

WELL STRENGTH z₀5.73

z₀ IN UNITS OF π/23.65

BOUND STATES4

NEXT LEVEL NEEDS V₀6.02 eV

Live interpretationWELL STRENGTH z₀: 5.73. z₀ IN UNITS OF π/2: 3.65. BOUND STATES: 4. NEXT LEVEL NEEDS V₀: 6.02 eV

Catch the common trap

Explain before calculating.

Misconception check

The picture to drop is an infinite well with slightly leaky walls. Students set k = nπ/L, fit whole half-wavelengths between the walls, then ask which of those levels fall out when the well is made shallow. Neither half survives. ψ does not vanish at ±a, so k is fixed by z tan z = √(z₀² − z²) and not by an integer; the interior wavelength is longer than the box value, so every level sits below its infinite-well partner, and the gap widens towards the rim: 0.27 eV against 0.376 eV for the ground state of a 5.0 eV, 1.0 nm well, and 4.06 eV against 6.02 eV for the fourth. The level count fails the same way. The rule is not the number of infinite-well levels below V₀ but one more than that, because state n appears when V₀ reaches Eₙ₋₁ rather than Eₙ. The naive count returns three where the well holds four.

A symmetric one-dimensional well of full width L = 1.0 nm and depth V₀ = 5.0 eV holds an electron, and its well-strength parameter is z₀ = (L/2)√(2mV₀)/ħ = 5.73. The infinite well of the same width has levels at 0.376, 1.50, 3.38 and 6.02 eV. How many bound states does the finite well have, and what is the parity of the highest one?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron is confined in a symmetric well of full width L = 1.0 nm and depth V₀ = 5.0 eV. Find the well-strength parameter z₀, the number of bound states, and the parity of each. Use ħc = 197.3 eV nm and m_ec² = 511 keV.

The half-width is a = L/2 = 0.500 nm, and z₀ = (a/ħ)√(2mV₀). Multiply inside and out by c so everything is an energy: √(2m_ec²V₀) = √(2 × 511 000 × 5.0) eV = 2261 eV.
z₀ = a √(2m_ec²V₀) / (ħc) = 0.500 nm × 2261 eV ÷ 197.3 eV nm = 5.73.
A new level enters at every half-multiple of π, so N = floor(2z₀/π) + 1 = floor(3.65) + 1 = 4.
Parity alternates upward from the nodeless even ground state: even, odd, even, odd. The fourth state is odd and carries three nodes.

Answerz₀ = 5.73, so the well holds four bound states, with parities even, odd, even, odd from the bottom.

MediumA symmetric well of full width L = 0.40 nm and depth V₀ = 0.20 eV holds an electron. Show that it has exactly one bound state, find its binding energy from the even condition z tan z = √(z₀² − z²), and compare the penetration depth 1/κ with the well itself.

z₀ = a√(2m_ec²V₀)/(ħc) = 0.200 nm × √(2 × 511 000 × 0.20) eV ÷ 197.3 eV nm = 0.200 × 452.1 ÷ 197.3 = 0.458, so z₀² = 0.2100. Then 2z₀/π = 0.29, giving N = 0 + 1 = 1, and the single state is even.
Bracket the root on 0 < z < π/2. At z = 0.40 the left side is 0.169 and the right side 0.224; at z = 0.43 they are 0.197 and 0.158. The root lies between, and bisection gives z = 0.4187.
κa = √(z₀² − z²) = √(0.2100 − 0.1753) = 0.1862, so the binding energy below the rim is E_b = (ħc)²(κa)²/(2m_ec²a²) = 38 938 × 0.03467 ÷ (2 × 511 000 × 0.0400) eV = 0.033 eV.
The energy above the floor is V₀ − E_b = 0.167 eV, and 1/κ = a/0.1862 = 0.200 ÷ 0.1862 = 1.07 nm, nearly three times the whole width of the well.
Integrating |ψ|² inside and outside with these values puts about 70% of the probability beyond the walls, so the state is held far more loosely than the word bound suggests.

AnswerOne even state, bound by 0.033 eV below the rim and so 0.167 eV above the floor, with 1/κ = 1.07 nm and roughly 70% of the probability outside the well.

HardThe deuteron is the only bound state of a neutron and a proton, bound by 2.22 MeV. Model the s-wave problem as a spherical well of radius R = 2.0 fm: u = rR(r) obeys the one-dimensional equation with u(0) = 0, so only odd solutions survive. Find the depth V₀ and show why there is no excited state. Take ħc = 197.3 MeV fm and μc² = 469.5 MeV.

u(0) = 0 kills the cosine, so the condition at r = R is the odd one, −z cot z = κR with z = kR.
κ = √(2μc²E_b)/(ħc) = √(2 × 469.5 × 2.225) MeV ÷ 197.3 MeV fm = 45.70 ÷ 197.3 = 0.2316 fm⁻¹, hence κR = 0.463 and 1/κ = 4.3 fm, already twice the well radius.
Solve −z cot z = 0.463 on π/2 < z < π by bisection: z = 1.820. The energy above the floor is (ħc)²z²/(2μc²R²) = 38 938 × 3.312 ÷ (2 × 469.5 × 4.0) = 34.3 MeV, so V₀ = 34.3 + 2.22 = 36.6 MeV.
z₀ = √(z² + (κR)²) = √(3.312 + 0.215) = 1.878. The next odd branch opens only at 3π/2 = 4.71, so no excited state exists; and had z₀ fallen below π/2 = 1.571, meaning V₀ below π²(ħc)²/8μc²R² = 25.6 MeV, the deuteron would not bind at all.
That threshold is the three-dimensional contrast in one line: the s-wave problem never gets the even family, so a spherical well has a minimum depth, while a one-dimensional symmetric well binds at any depth.

AnswerV₀ ≈ 36.6 MeV with z₀ = 1.88, only 20% above the π/2 threshold, so the deuteron is barely bound, has no excited state, and keeps most of its wavefunction outside the well.