University Physics V · Finite Potential Wells and Quantum Tunneling · 6.2
The Finite Well, Split by Parity
The infinite box handed you a spectrum in closed form. Lower the walls to a finite height and that stops: what survives is a symmetry that halves the matching, one dimensionless radius carrying every parameter, and two transcendental conditions you have to bracket and root. Count the levels first — the count is free.
Build the model
Connect the measurement to the mechanism.
Sharp walls gave a spectrum in closed form; finite ones do not. Put V(x) = −V₀ on |x| ≤ a and 0 outside. Because V is even, the parity operator Π commutes with Ĥ, and since 1D bound states are non-degenerate, every eigenvector is itself even or odd — four unknown coefficients collapse to one interior function and one exterior tail, matched at the single wall x = +a.
Inside, ψ oscillates with k = √(2m(E + V₀))/ℏ; outside it decays with κ = √(−2mE)/ℏ, the growing exponential rejected by normalisability. Continuity of ψ′/ψ then gives one scalar equation per parity: k tan(ka) = κ for even states, −k cot(ka) = κ for odd. In the reduced variables z = ka and κa, the identity k² + κ² = 2mV₀/ℏ² becomes a circle of radius z₀ = a√(2mV₀)/ℏ, so the whole family of wells is one picture: a quarter circle crossing fixed tangent and cotangent branches.
That picture charges twice. The crossings are transcendental — there is no closed form for z, only a bracketed brentq on a branch — but the count comes free before any root is found, N = ⌊2z₀/π⌋ + 1, which is why a symmetric one-dimensional well always binds at least one even state, however shallow it is.
- Simple definition
- The bound levels of a symmetric finite well are the energies at which the interior logarithmic derivative matches the exterior decay — k tan(ka) = κ for even states, −k cot(ka) = κ for odd — with k and κ tied by k² + κ² = 2mV₀/ℏ².
- Example
- For an electron with half-width a = 2.00 Å and depth V₀ = 15.24 eV, z₀ = a√(2mV₀)/ℏ = 4.00, so ⌊8/π⌋ + 1 = 3 states exist; the even ground root is z₁ = 1.2524, giving E₁ = −0.902V₀ = −13.75 eV.
Each eigenvector then carries definite parity automatically, so you match at x = +a only and the other wall follows by reflection.
Π is Hermitian and unitary, so its eigenvalues are +1 and −1; in 1D bound states are non-degenerate.
The two wavenumbers are not independent: the depth ties them together, and that constraint is what makes a graphical solution possible.
k, κ in m⁻¹; −V₀ < E < 0; ℏ²/2mₑ = 3.81 eV Ų, so divide by m*/mₑ in a semiconductor.
Matching the logarithmic derivative deletes the unknown coefficients, leaving one scalar equation per parity whose roots are the levels.
Continuity of ψ and ψ′ at x = a, taken as the ratio ψ′/ψ so the amplitudes cancel.
Three physical inputs collapse to one number, so a 10 nm GaAs well and a 4 Å atomic well of equal z₀ are the same problem.
z = ka and κa = √(z₀² − z²) are pure numbers; the pair (z, κa) sits on a circle of radius z₀.
N ≥ 1 for every z₀ above zero: a symmetric 1D well always binds one even state however shallow, and the 3D well does not.
Branches begin at z = 0, π/2, π, 3π/2, …, so N rises by one each time z₀ passes a multiple of π/2.
The crossing's height on the circle is κa, so a root near the axis is barely bound and one near z = 0 is bound almost to the floor.
zₙ is the nth root, ordered even, odd, even, …; E is measured from the top of the well, where E = 0.
Parity is a reduction you are entitled to, not a guess
Write V(x) = −V₀ for |x| ≤ a and 0 outside. Then V(−x) = V(x), so the parity operator Π, defined by Πψ(x) = ψ(−x), commutes with Ĥ. Π is both Hermitian and unitary with Π² = 1, so its eigenvalues are +1 and −1, and Ĥ and Π can be diagonalised together. In one dimension you get something stronger than the usual permission to choose: bound states of a 1D Hamiltonian are non-degenerate, so if ψ is an eigenvector then Πψ is an eigenvector at the same E, hence Πψ = cψ with c² = 1. Every bound state is therefore even or odd with no choice in the matter, and the spectrum interleaves — even, odd, even, odd in ascending energy, the nth state carrying n − 1 nodes. What this buys is bookkeeping. A naive attack has four unknown coefficients and four matching equations at x = ±a. Fixing the parity first reduces the interior solution to cos kx or sin kx alone, kills the growing exponential outside by normalisability, and leaves a single condition at the single wall x = +a; the wall at −a is then satisfied automatically by reflection.
Two regions, two wavenumbers, one surviving equation
Inside the well E + V₀ is positive, so ψ″ = −k²ψ with k = √(2m(E + V₀))/ℏ and ψ oscillates. Outside, E is negative, so ψ″ = +κ²ψ with κ = √(−2mE)/ℏ; the growing branch exp(+κx) is not square-integrable beyond the wall, so only exp(−κx) survives — and it is that rejection, not the matching, that makes the spectrum discrete. Take the even sector: ψ = A cos(kx) for |x| ≤ a and ψ = C exp(−κ(x − a)) beyond it. Because V is finite everywhere, ψ and ψ′ are both continuous at x = a; the kink you tolerated at an infinite wall is gone. Dividing the two conditions removes A and C at a stroke: the interior logarithmic derivative at the wall is −k tan(ka), the exterior one is −κ, so k tan(ka) = κ. Repeat with ψ = A sin(kx) in the odd sector, where the interior value is +k cot(ka), and the condition is −k cot(ka) = κ. Both demand a positive κ, which is the reminder that a root of the squared equation with tan(ka) negative is an artefact, not a state.
One number, z₀, replaces the mass, the width and the depth
Set z = ka and κa for the two reduced wavenumbers. The definitions give k² + κ² = 2mV₀/ℏ², so z² + (κa)² = z₀² with z₀ = a√(2mV₀)/ℏ, and the conditions become z tan z = κa and −z cot z = κa. Every physical input — mass, half-width, depth — is now packed into the single radius z₀, and the bound-state problem is the intersection of a quarter circle with a fixed set of branches that never move. Put numbers on it with ℏ²/2mₑ = 3.81 eV Ų, so z₀² = V₀a²/(3.81 eV Ų) for an electron. An electron in a = 2.00 Å at V₀ = 15.24 eV gives z₀ = 4.00. A GaAs quantum well 10.0 nm wide with a 0.300 eV offset and m* = 0.067mₑ has ℏ²/2m* = 56.9 eV Ų = 0.569 eV nm², so z₀ = 5.00 nm × √(0.300/0.569 eV nm²) = 3.63. The two systems differ by a factor of 25 in size and 50 in depth, and in reduced form they are nearly the same problem: three bound states each.
Counting the levels before finding any of them
The branches never move, so the count is geometry alone. z tan z rises from 0 to +∞ on every interval (nπ, nπ + π/2), and −z cot z does the same on every (nπ + π/2, (n + 1)π). The quarter circle starts at (0, z₀) and falls monotonically to (z₀, 0), so it crosses exactly once every branch whose left endpoint lies below z₀, and misses the rest. Those endpoints are 0, π/2, π, 3π/2, …, hence N = ⌊2z₀/π⌋ + 1, with parities alternating from even. For z₀ = 4.00 that is ⌊2.546⌋ + 1 = 3: even, odd, even. Two consequences. Because the first branch starts at z = 0, N is at least 1 for every positive z₀ — a symmetric well in one dimension always binds an even state however shallow. The three-dimensional spherical well does not: its l = 0 radial function obeys u(0) = 0, which is the odd condition, whose first branch starts at π/2, so a 3D well of z₀ below π/2 binds nothing. And at each threshold the new state arrives with E approaching 0, so κ approaches 0 and 1/κ diverges: it is bound, but spread far outside the well.
Rooting it: bracket the branch, never chase the pole
Do not hand F(z) = z tan z − √(z₀² − z²) to a root finder over a wide interval. F jumps from +∞ to −∞ at every z = π/2 + nπ, and a bracketing method reads that jump as a sign change and returns the pole, to machine precision, as a spurious root. Fix it twice over. Bracket by branch: the nth even root lies in (nπ, min(nπ + π/2, z₀)) and the nth odd root in (nπ + π/2, min((n + 1)π, z₀)). And clear the denominator first — multiply by cos z or sin z to get f(z) = z sin z − y cos z for even states and h(z) = z cos z + y sin z for odd, with y = √(z₀² − z²). Both are bounded and analytic, and neither gains a spurious zero, since f = 0 together with cos z = 0 would also need z sin z = 0. For z₀ = 4.00, f(0) = −4.000 and f(1.5) = +1.234, so scipy.optimize.brentq(f, 0, 1.5) returns z₁ = 1.2524. Cross-check by diagonalising the discretised Hamiltonian on a grid several decay lengths wider than the well; if the lowest eigenvalue drifts as you widen the box, the box is doing the binding, not the well.
What the root buys, and what the sharp wall costs
Convert the root: Eₙ = −V₀(1 − zₙ²/z₀²), measured from the top of the well. At z₀ = 4.00 with V₀ = 15.24 eV, z₁ = 1.2524 gives E₁ = −13.75 eV, so the ground state sits 1.49 eV above the floor. Infinite walls of the same full width 4.00 Å would put it at π²ℏ²/(2mL²) = 2.35 eV, 57% higher, and κ says why: with κa = 3.799 the decay length is 1/κ = a/3.799 = 0.527 Å, so the state occupies about 4.00 + 2 × 0.527 = 5.05 Å rather than 4.00 Å. Finite walls lower every level because they let the state spread. Now the honesty. Vertical faces are a fiction: a real heterojunction is graded over two or three monolayers, and the effective mass differs across it. The correct interface condition is then not ψ′ continuous but BenDaniel-Duke — ψ and (1/m*)ψ′ continuous — which turns the even condition into k tan(ka) = (mw/mb)κ. With equal masses it collapses to what you derived; with a 30% mass step every root moves, and a subband spacing quoted to three figures without it is quoting the model, not the sample.
Change one variable at a time
Make the relationship visible.
Raise z₀ past π/2, π and 3π/2 and a new crossing appears at each threshold, alternating even, odd, even, odd. Then drag the cursor onto a crossing until the matching residual reads 0.000 — that z is a root, and its height on the circle is κa.
BOUND STATES ⌊2z₀/π⌋+13
EVEN z sin z − κa cos z0.891
ODD z cos z + κa sin z3.074
LEVEL AT CURSOR E/V₀-0.809
Live interpretationBOUND STATES ⌊2z₀/π⌋+1: 3. EVEN z sin z − κa cos z: 0.891. ODD z cos z + κa sin z: 3.074. LEVEL AT CURSOR E/V₀: −0.809
Catch the common trap
Explain before calculating.
A symmetric one-dimensional well has depth parameter z₀ = a√(2mV₀)/ℏ = 4.00. How many bound states does it hold, and in what order of parity?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is confined by a symmetric well of full width L = 2a = 4.00 Å and depth V₀ = 15.0 eV. Find the depth parameter z₀, the number of bound states and their parities, and the depth that would be needed to bind a fourth state at the same width.
- z₀² = 2mV₀a²/ℏ² = V₀a²/(ℏ²/2m). With ℏ²/2mₑ = 3.81 eV Ų and a = 2.00 Å: z₀² = 15.0 × 4.00/3.81 = 15.75, so z₀ = 3.97.
- Count before rooting: 2z₀/π = 7.94/3.1416 = 2.53, so N = ⌊2.53⌋ + 1 = 3.
- Parities alternate from even, and the branches open at z = 0, π/2 and π: z₁ in (0, π/2) even, z₂ in (π/2, π) odd, z₃ in (π, z₀) even. That third branch opened at π = 3.14, only 0.83 below z₀, so the top state is weakly bound.
- A fourth state needs the next branch, which opens at z₀ = 3π/2 = 4.712. At fixed a, V₀ = 3.81 z₀²/a² = 3.81 × 22.21/4.00 = 21.2 eV.
Answerz₀ = 3.97, so N = ⌊2z₀/π⌋ + 1 = 3 bound states, even, odd, even. A fourth needs z₀ ≥ 3π/2, that is V₀ ≥ 21.2 eV at the same width.
MediumThe same electron now sits in a well of half-width a = 2.00 Å and depth V₀ = 15.24 eV, chosen so that z₀ is exactly 4.00. Bracket and root the even ground state, quote E₁ in eV, and compare it with the infinite well of the same full width.
- Even states obey z tan z = κa with κa = √(z₀² − z²). Do not root that directly — it has a pole at π/2. Multiply through by cos z instead: f(z) = z sin z − √(16 − z²) cos z, bounded and analytic across the whole first branch.
- Bracket on (0, π/2): f(0) = −√16 = −4.000, and f(1.5) = 1.5 × 0.9975 − √13.75 × 0.07074 = 1.4962 − 0.2623 = +1.234. One sign change, so exactly one root.
- Narrow it: f(1.20) = 1.1184 − 3.8158 × 0.36236 = −0.264 and f(1.30) = 1.2526 − 3.7829 × 0.26750 = +0.241. brentq on [1.20, 1.30] returns z₁ = 1.2524, so κa = √(16 − 1.5685) = 3.7989.
- E₁ = −V₀(1 − z₁²/z₀²) = −15.24 × (1 − 1.5685/16) = −15.24 × 0.9020 = −13.75 eV, which is 1.49 eV above the well floor.
- Infinite walls of the same width L = 4.00 Å give π²ℏ²/(2mL²) = 9.8696 × 3.81/16.0 = 2.35 eV above the floor, 57% higher. The difference is leakage: 1/κ = a/3.7989 = 0.527 Å out of each face.
Answerz₁ = 1.2524 and E₁ = −13.75 eV, which is 1.49 eV above the well floor against 2.35 eV for infinite walls of the same width.
HardA 10.0 nm GaAs well between AlGaAs barriers has a conduction-band offset V₀ = 0.300 eV and effective mass m* = 0.067mₑ on both sides. Count the bound subbands, find their energies and the spacing E₂ − E₁, and give the width at which a fourth subband appears.
- Rescale: ℏ²/2m* = 3.81/0.067 = 56.9 eV Ų = 0.569 eV nm². With a = 5.00 nm, ℏ²/(2m*a²) = 0.569/25.0 = 0.02276 eV, so z₀² = 0.300/0.02276 = 13.18 and z₀ = 3.63.
- N = ⌊2 × 3.63/π⌋ + 1 = ⌊2.31⌋ + 1 = 3, in the order even, odd, even.
- Bracket and root each branch with the pole-free forms: even on (0, π/2) gives z₁ = 1.2264, odd on (π/2, π) gives z₂ = 2.4144, even on (π, z₀) gives z₃ = 3.455.
- Eₙ = −V₀(1 − zₙ²/z₀²): E₁ = −0.2658 eV, E₂ = −0.1674 eV, E₃ = −0.0285 eV. So E₂ − E₁ = 98.4 meV, a 12.6 μm intersubband transition, while E₃ lies only 28.5 meV below the barrier top — barely above kT = 26 meV at room temperature.
- For a fourth subband, z₀ = a√(V₀/(ℏ²/2m*)) = a√(0.300/0.569) = 0.726a with a in nm, so z₀ ≥ 3π/2 = 4.712 needs a ≥ 6.49 nm, that is L ≥ 13.0 nm.
- Caveat: this took ψ′ continuous. Across a real heterojunction the condition is BenDaniel-Duke, ψ and (1/m*)ψ′ continuous; with equal masses it reduces to what was used here, and with unequal ones every root shifts.
Answerz₀ = 3.63 gives three subbands, even, odd, even, at −0.266, −0.167 and −0.0285 eV; E₂ − E₁ = 98.4 meV (12.6 μm), and a fourth appears at L ≈ 13.0 nm.