Matching ψ and ψ′ at the walls of a finite well
One ratio per wall; two walls fix phase and energy.
Four matching equations in four amplitudes and an energy look like a swamp. The logarithmic derivative drains it: at each wall the tail's size cancels, one equation survives, and two walls fix the interior phase and E. Every level of a symmetric finite well is a root of one of the two conditions this lesson produces.
Build the model
Divide ψ′ by ψ and the tail's amplitude drops out: one equation per wall.
1The big idea
After normalisability has thrown away the growing exponentials (UQMI-07.03), a bound state of the finite well is a cosine-and-sine combination inside and one decaying exponential beyond each wall: four amplitudes and an energy. The matching rules, taken as proved in UQMI-02.08, supply two equations per wall, because a finite step keeps both ψ and ψ′ continuous. Solving that homogeneous 4 × 4 system head-on works but hides the physics. The logarithmic derivative ψ′/ψ is the better tool. It is continuous wherever ψ and ψ′ are, it ignores how large ψ is, and outside the well it is exactly −κ beyond x = a and +κ beyond x = −a, since a lone exponential has the same fractional slope everywhere. So each wall hands over one equation with no tail amplitude in it, and the two walls together fix the interior phase and E. For a symmetric well the phase can only be 0 or π/2, which yields k tan(ka) = κ and −k cot(ka) = κ; with k² + κ² = 2mV₀/ħ² each is one transcendental equation for E. The method leans on its premises and changes where one fails: above the rim each outer region carries two travelling waves in a free ratio, so the outside no longer dictates ψ′/ψ and every E is allowed; across a heterojunction the effective mass changes, ψ′/m rather than ψ′ is continuous, and the quantity to match becomes ψ′/(mψ); and a graded wall has no closed-form tail, so shooting (UQMI-07.09) takes over.
2In one sentence
Matching at a finite wall means requiring ψ and ψ′ to be continuous there; their ratio ψ′/ψ, the logarithmic derivative, then takes the same value on both sides, and since the decaying tails fix it at −κ on the right and +κ on the left, each wall gives one amplitude-free equation tying the interior solution to E.
Example. An electron in a well 1.50 eV deep with a = 0.40 nm has its ground state at E = 0.294 eV: there k = 2.78 nm⁻¹, and cos kx reaches x = a with ψ′/ψ = −k tan(ka) = −5.63 nm⁻¹, exactly the tail's −κ.
3Count the unknowns before matching anything
Take the rules from UQMI-02.08 without re-deriving them: ψ is continuous everywhere, and ψ′ is continuous wherever V is finite, which a step of height V₀ is. Normalisability has already cut each outer region to one decaying exponential, so the candidate is G e(κx) for x < −a, C cos kx + D sin kx for |x| < a and F e(−κx) for x > a, with k = √(2mE)/ħ and κ = √(2m(V₀ − E))/ħ. Two walls times two rules give four equations, all linear and homogeneous in G, C, D and F. Such a system has a solution other than all zeros only when its 4 × 4 determinant vanishes, and that determinant depends on E alone: that is the quantisation condition. Normalisation plays no part yet; it fixes the overall scale at the very end. So the books balance as four equations against three amplitude ratios and one energy. What remains is to do the elimination without expanding a 4 × 4 determinant by hand.
4Divide, and the tail's amplitude cancels
Define the logarithmic derivative L = ψ′/ψ = d(ln ψ)/dx, in m⁻¹. If ψ and ψ′ are each continuous at x = a, so is their ratio, and one ratio condition replaces two. The payoff is outside: for F e(−κx), L = −κ whatever F is, and for G e(κx) on the left, L = +κ whatever G is. Each wall therefore gives one equation with no exterior amplitude in it: the interior solution must arrive at x = a already falling at the fractional rate κ. The division is always legal. If ψ(a) were zero, continuity would force F = 0, then ψ′(a) = 0 too, and a second-order equation with ψ = ψ′ = 0 at one point has only the solution ψ ≡ 0. For an electron in a well with V₀ = 1.50 eV and a = 0.40 nm at a trial E = 0.20 eV, cos kx reaches the wall with L = −2.99 nm⁻¹ while the tail needs −κ = −5.84 nm⁻¹, so 0.20 eV is not an eigenvalue.
5Two walls, two equations: the phase and the energy
Write the interior as A cos(kx − φ), so that tan φ = D/C carries the one amplitude ratio left inside. Its logarithmic derivative is −k tan(kx − φ), and matching at the two walls gives k tan(ka − φ) = κ at x = +a and k tan(ka + φ) = κ at x = −a. Two equations, two unknowns: φ and E. Set them equal. tan(ka − φ) = tan(ka + φ), and since the tangent repeats only every π, ka + φ = ka − φ + nπ, so φ = 0 or φ = π/2, up to an overall sign of ψ. The matching conditions by themselves make the interior a pure cosine or a pure sine; the structural reason, symmetry of V plus non-degeneracy, belongs to UQMI-07.05. The only symmetry this step used was that both walls share one κ. Give the walls different heights and the two wall equations, now carrying different values of κ, still fix φ and E, but φ lands at some other angle and the state leans toward the lower wall.
6Reading the even and odd conditions
φ = 0 turns the right-wall equation into k tan(ka) = κ, the even condition; φ = π/2 turns the interior into sin kx and the equation into −k cot(ka) = κ, the odd one. Each carries its own checks. Dimensions: k and κ are both in m⁻¹, and ka is a pure number, as a tangent's argument must be. Sign: κ > 0, so even states need tan(ka) > 0 and odd states cot(ka) < 0. Limiting case: as V₀ → ∞, κ → ∞ drives ka to π/2 for the lowest even state and to π for the lowest odd one, the first two levels of an infinite well of width 2a. For the 1.50 eV, 0.40 nm electron well the even condition holds at E = 0.294 eV, both sides 5.63 nm⁻¹, and the odd one at 1.079 eV, both sides 3.32 nm⁻¹. Classical correspondence: a classical particle never leaves |x| < a, and the tail length 1/κ = 0.18 nm of the ground state shrinks to zero as ħ → 0 at fixed E and V₀.
7Why no algebra will isolate E
k and κ are not independent: k² + κ² = 2mV₀/ħ² whatever E is. Substituting κ = √(2mV₀/ħ² − k²) makes each condition a single equation in E, and multiplying by a puts it in dimensionless form: z tan z = √(z₀² − z²) for even states and −z cot z = √(z₀² − z²) for odd, with z = ka and z₀ = (a/ħ)√(2mV₀). The mass, width and depth of the well have collapsed into one number, z₀ = 2.51 for the example well. E now sits inside a tangent and a square root at once, and no finite chain of algebraic steps and inverse functions frees it: that is what transcendental means. The levels come from a graph and a bracketed solver, which is the work of UQMI-07.06. The infinite well never faced this, because there ψ(a) = 0 gives cos ka = 0 directly, an equation whose roots are written down by inspection.
8Hunting E with the mismatch, then restoring the amplitudes
The logarithmic derivative also gives a direct search in E. Define the mismatch M(E) = Lᵢₙ(a⁻) − Lₒᵤₜ(a⁺), which for even states is κ − k tan(ka). Raising E makes the interior curve faster, so Lᵢₙ falls, while κ shrinks, so −κ rises: M decreases steadily, and a change of sign from + to − brackets an eigenvalue. Beware the poles. Where ψ(a) passes through zero, Lᵢₙ jumps from −∞ to +∞, a sign change from − to + with no root in it; the product form ψ′(a) + κψ(a) = 0 says the same thing and has no poles. Once E is found, undo the division. Continuity of ψ gives F = ψᵢₙ(a) e(κa), and continuity of ψ′ then holds with no further choice, because that is what the choice of E bought. Normalise last, and only then ask for probabilities, such as the fraction of the particle found outside the walls.
Key equations
What each relation says, and the units it is written in.
Logarithmic derivative
L(x) = ψ′(x)/ψ(x), with L(a⁻) = L(a⁺)
Scaling ψ by any constant leaves L unchanged, so a wall's two matching equations collapse into one that no amplitude enters.
L in m⁻¹ (nm⁻¹ in practice); ψ in m(−1/2), ψ′ in m(−3/2). ψ(±a) ≠ 0 for any bound state, so dividing is safe.
What the tails impose
L(x) = −κ sgn(x) for |x| > a
The tail fixes how fast ψ must be falling as it reaches the wall, never how large it is: F and G drop out.
κ = √(2m(V₀ − E))/ħ in m⁻¹, with m in kg, V₀ and E in J, ħ in J s. The left tail is e(+κx), hence +κ there.
Matching at both walls
k tan(ka ∓ φ) = κ at x = ±a, for ψ = A cos(kx − φ) inside
Two equations for two unknowns, φ and E. Since tan repeats every π, both hold only if φ = 0 or π/2.
k = √(2mE)/ħ in m⁻¹, a in m. φ is the interior phase, tan φ = D/C for C cos kx + D sin kx.
Even states, φ = 0
k tan(ka) = κ
As V₀ → ∞, κ → ∞ forces ka → π/2, the ground state of an infinite well of width 2a: the limiting-case check.
Both sides in m⁻¹, ka dimensionless. Needs tan(ka) > 0: ka in (0, π/2), (π, 3π/2), and so on.
Odd states, φ = π/2
−k cot(ka) = κ
The first odd branch cannot start before ka = π/2, so a shallow well has no odd state; as V₀ → ∞, ka → π.
Both sides in m⁻¹. Needs cot(ka) < 0: ka in (π/2, π), (3π/2, 2π), and so on.
The relation that leaves E the only unknown
k² + κ² = 2mV₀/ħ²
Substituting makes each condition one equation in E, such as z tan z = √(z₀² − z²): transcendental, so rooted numerically.
m⁻² on both sides, independent of E. With z = ka and z₀ = (a/ħ)√(2mV₀) it reads κa = √(z₀² − z²).
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Keep φ = 0° and slide E: both corners vanish between 0.29 and 0.30 eV, the even ground state. At φ = 90° the odd state appears near 1.08 eV. At any other φ, tuning E clears one wall but never both: the two walls fix the phase and the energy together.
Live results
- INTERIOR k
- 2.29 nm⁻¹
- TAIL κ
- 5.84 nm⁻¹
- MISMATCH AT x = −a
- -2.85 nm⁻¹
- MISMATCH AT x = +a
- 2.85 nm⁻¹
Catch the common trap
The mistake this lesson is built to prevent.
The common slip is to write ψ′/ψ = −κ at both walls, on the grounds that a tail always decays. On the left the admissible tail is G e(+κx): it decays as x → −∞, so it grows toward the wall, and its logarithmic derivative is +κ. The error does not produce nonsense, which is what makes it dangerous. With −κ at x = −a the wall equations become tan(ka − φ) = κ/k and tan(ka + φ) = −κ/k, which together force 2ka = nπ: the levels of an infinite well of width 2a, whatever V₀ is, attached to a lopsided interior with φ ≠ 0. A spectrum that ignores the depth of the well fails the limiting-case check in reverse, being right only at V₀ = ∞. Differentiate the tail you actually wrote, not the word decay.
Test it: what changes?
- Inside ψ′/ψ = −k tan(ka) falls between poles
- Tail's required −κ rises towards zero
- Corners at both walls vanish together at even levels
- Tuning E clears one wall, never both
- A corner persists at one wall at least
- No trial energy gives a bound state
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
A classmate's shooting code has converged on the ground state of an electron in a finite well with V₀ = 0.600 eV and half-width a = 0.50 nm, but the line giving E is lost. What survives is the interior solution at the right wall, scaled so that ψ(0) = 1: ψ(a⁻) = 0.5175 and ψ′(a⁻) = −1.757 nm⁻¹. Recover E from the wall alone, then check it two independent ways. Use ħ²/2m = 0.0381 eV nm².
View solution →
- Outside, the admissible tail is F e(−κx), whose logarithmic derivative is −κ whatever F is, and ψ′/ψ is continuous at a finite step. So κ = −ψ′(a⁻)/ψ(a⁻) = 1.757 ÷ 0.5175 = 3.395 nm⁻¹. Neither the unknown F nor the arbitrary scaling ψ(0) = 1 entered.
- Invert κ = √(2m(V₀ − E))/ħ: V₀ − E = (ħ²/2m)κ² = 0.0381 eV nm² × 11.53 nm⁻² = 0.439 eV, so E = 0.600 − 0.439 = 0.161 eV. Units: eV nm² × nm⁻² = eV.
- Check with the even condition. k = √(0.161 ÷ 0.0381) nm⁻¹ = 2.056 nm⁻¹, so ka = 1.028 and k tan(ka) = 2.056 × 1.657 = 3.41 nm⁻¹, equal to κ = 3.40 nm⁻¹ within the rounding of E.
- Check with the shape. An even interior is cos kx, so ψ(a)/ψ(0) must be cos(ka) = cos 1.028 = 0.517, which is the 0.5175 the code reports.
- Limiting case. The infinite well of the same width, 1.00 nm, has its ground state at 0.0381 × (π/1.00)² eV = 0.376 eV; the finite-well level lies below it, and its tail length 1/κ = 0.295 nm is a large fraction of a, as a level only 0.44 eV below the rim should have.
Answer. κ = 3.40 nm⁻¹ and E = 0.161 eV, confirmed by k tan(ka) = κ and by ψ(a)/ψ(0) = cos(ka) = 0.517.
MediumExample 2
An electron is bound in a well with V₀ = 1.20 eV and half-width a = 0.45 nm, for which z₀ = 2.53. Find its odd state from the mismatch M(E) = k cot(ka) + κ, bracketing it with trial energies 0.80 eV and 0.90 eV. Then recover the tail amplitude from continuity of ψ, normalise, and find the probability that the electron is outside the walls. Use ħ²/2m = 0.0381 eV nm².
View solution →
- An odd interior A sin kx has ψ′/ψ = k cot(kx), and the tail needs −κ at x = a, so the odd condition −k cot(ka) = κ is M(E) = k cot(ka) + κ = 0. Since z₀ = 2.53 exceeds π/2, the first odd branch, π/2 < ka < π, is reachable.
- At 0.80 eV: k = 4.582 nm⁻¹, ka = 2.062, k cot(ka) = −2.451 nm⁻¹ and κ = 3.240 nm⁻¹, so M = +0.789 nm⁻¹. At 0.90 eV: k = 4.860 nm⁻¹, ka = 2.187, k cot(ka) = −3.443 nm⁻¹ and κ = 2.806 nm⁻¹, so M = −0.637 nm⁻¹. ka stays inside (π/2, π), so no pole lies between, and the change from + to − brackets a root.
- Linear interpolation gives 0.80 + 0.10 × 0.789/1.426 = 0.855 eV, and refining by bisection settles E = 0.857 eV, where k = 4.743 nm⁻¹, ka = 2.134, κ = 3.000 nm⁻¹ and −k cot(ka) = 2.996 nm⁻¹, equal to three figures.
- Undo the division. Continuity of ψ at x = a gives F e(−κa) = A sin(ka), so F = A sin(ka) e(κa) = 0.845 × e(1.350) A = 0.845 × 3.858 A = 3.26 A. Continuity of ψ′ needs −κ A sin(ka) = kA cos(ka), which is the odd condition itself: it holds with no further choice.
- Normalise. Inside, the integral of A² sin²(kx) from −a to a is A²(a − sin(2ka)/(2k)) = 0.5452 A² nm; the two tails add 2 × A² sin²(ka)/(2κ) = 0.2382 A² nm. So A² = 1/(0.7834 nm) = 1.276 nm⁻¹, and the probability outside is 0.2382/0.7834 = 0.304.
- Checks. A² in nm⁻¹ gives ψ in nm(−1/2), right for one dimension. As V₀ → ∞ the odd condition forces ka → π, the n = 2 level of a 0.90 nm infinite well at 0.0381 × (π/0.45)² eV = 1.86 eV, far above 0.857 eV. Classically the electron never leaves |x| < a; here 30% of the probability sits in the forbidden regions, since E is only 0.34 eV below the rim.
Answer. E = 0.857 eV with κ = 3.00 nm⁻¹; F = 3.26 A with A = 1.13 nm(−1/2); the electron is outside the walls with probability 0.304.
HardExample 3
A lopsided well holds an electron: V = 0.80 eV for x < 0, V = 0 for 0 < x < L with L = 0.60 nm, and V = 2.40 eV for x > L. Parity is gone, so use the logarithmic derivative at each wall to find the condition fixing the ground-state energy, solve it, and find the interior phase and the ratio ψ(0)/ψ(L) of the wavefunction at the two walls. Use ħ²/2m = 0.0381 eV nm².
View solution →
- Name the regions. For 0 < E < 0.80 eV: ψ = G e(κ₁x) for x < 0, A sin(kx + δ) for 0 < x < L, and F e(−κ₂(x − L)) for x > L, with k = √(2mE)/ħ, κ₁ = √(2m(0.80 eV − E))/ħ and κ₂ = √(2m(2.40 eV − E))/ħ. Once the amplitudes are divided out, the unknowns are the phase δ and E.
- Left wall: ψ′/ψ inside is k cot(kx + δ), which at x = 0⁺ must equal the tail's +κ₁, so δ = arctan(k/κ₁). Right wall: k cot(kL + δ) = −κ₂, so kL + δ = nπ − arctan(k/κ₂). Eliminating δ, the ground state (n = 1) obeys kL = π − arctan(k/κ₁) − arctan(k/κ₂): each wall gives back a phase arctan(k/κ) smaller than π/2.
- Solve. Φ(E) = kL + arctan(k/κ₁) + arctan(k/κ₂) − π is −0.438 at 0.30 eV and +0.008 at 0.40 eV, and it rises steadily with E, so the root lies just below 0.40 eV. Refining gives E = 0.398 eV, with k = 3.232 nm⁻¹, κ₁ = 3.248 nm⁻¹ and κ₂ = 7.249 nm⁻¹; then kL + 0.783 + 0.419 = 1.939 + 1.202 = 3.141, which is π.
- The interior phase, measured at the left wall, is δ = arctan(3.232/3.248) = 0.783 rad = 44.9°. To compare with a symmetric well, measure it from the centre instead: sin(kx + δ) = cos(k(x − L/2) − φ) with φ = π/2 − kL/2 − δ = 1.571 − 0.970 − 0.783 = −0.182 rad, about −10.4°. That is neither 0 nor ±π/2, so the interior is neither a pure cosine nor a pure sine about the centre. Its peak, where kx + δ = π/2, sits at x = (1.571 − 0.783)/3.232 = 0.244 nm, left of the centre at 0.300 nm: the state leans toward the lower wall.
- Wall values. Inside, ψ² + (ψ′/k)² = A² everywhere; at a wall ψ′ = ±κψ, so ψwall = Ak/√(k² + κ²) = Ak/√(2mVwall/ħ²), positive at both walls for this nodeless state. Hence ψ(0)/ψ(L) = √(2.40/0.80) = √3 = 1.73, set by the two rim heights alone and not by E or L, and the left tail is longer: 1/κ₁ = 0.308 nm against 1/κ₂ = 0.138 nm.
- Checks. Equal walls, κ₁ = κ₂ = κ with L = 2a, give ka = π/2 − arctan(k/κ), which is k tan(ka) = κ, the even condition. E lies between the same-width symmetric wells with both walls at 0.80 eV (0.326 eV) and both at 2.40 eV (0.506 eV), and below the 1.04 eV of the infinite well, the V → ∞ limit where kL = π.
Answer. kL = π − arctan(k/κ₁) − arctan(k/κ₂), giving E = 0.398 eV; the interior phase is δ = 0.783 rad (44.9°) at the left wall, or −0.182 rad measured from the centre, and ψ(0)/ψ(L) = √3 = 1.73.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Match ψ′/ψ: −κ at x = +a, +κ at x = −a.
The tail amplitude cancels from ψ′/ψ, so each wall gives one equation. Two walls fix the interior phase and E; in a symmetric well φ = 0 or π/2, giving k tan(ka) = κ and −k cot(ka) = κ. Where the effective mass changes, match ψ′/(mψ) instead.