Bound states and what normalisability demands of the tails
Decaying tails, demanded at infinity, select the bound energies.
In the infinite well the walls did the quantising. In the finite well they do not: every energy between 0 and V₀ solves the equation and passes smoothly through both walls. This lesson finds what selects the levels instead, one demand made at infinity, and shows how it turns a continuous dial of trial energies into a short, exact list.
Build the model
The tails, not the walls, decide which energies are allowed.
1The big idea
Take the finite well, V = 0 for |x| < a and V₀ beyond, at an energy 0 < E < V₀. The Schrödinger equation has solutions at every such E: oscillating inside, and in each outer region a mix of e(−κs) and e(κs), where s = |x| − a is the distance past the nearer wall. The equation prefers neither exponential; the Born rule does. A state needs ∫|ψ|² dx = 1, and since the equation is linear, any solution with a finite integral can be rescaled to that, while one with an infinite integral cannot be rescued by any constant. A bound state is therefore a square-integrable solution, and on each side that one demand deletes exactly one term, leaving tails that fall as e(−κ|x|). The deletion is not free. Six constants become four, continuity of ψ and ψ′ at the two walls imposes four homogeneous conditions on them, and a nonzero solution survives only where their determinant vanishes. Equivalently, carry the decaying left tail across the well: it reaches the right wall with a growing part D(E) e(κs), and only at the isolated zeros of D(E) is the result a state. The discreteness lives in the tails. The argument stops where its premise does: above V₀ nothing can be deleted and every energy is allowed; and since normalisability is a statement about |x| → ∞, a well whose surroundings eventually fall below E, as in an applied field, holds no strictly bound state, only levels that leak.
2In one sentence
A bound state is a solution of the time-independent Schrödinger equation whose ∫|ψ|² dx over all x is finite, so that it can be normalised; in a finite well this forces 0 < E < V₀ and tails that die away as e(−κ|x|).
Example. For an electron in a well 1.00 nm wide and 2.00 eV deep, only 0.229, 0.887 and 1.815 eV give a solution with no growing exponential on either side. At 0.300 eV the solution that decays on the left leaves the right wall with a growing term −0.621 times the left tail's wall value.
3A bound state is a solution whose probability adds up
The Born rule reads |ψ|² as a probability density, so a state must satisfy ∫|ψ|² dx = 1 over the whole line. The time-independent equation is linear, so any solution with a finite integral can be rescaled to meet that; a solution with an infinite integral cannot, whatever constant multiplies it. The real demand is therefore square-integrability, and a bound state is a solution that has it. In the finite well, V = 0 for |x| < a and V₀ beyond, the inside is never the problem, because a continuous ψ on a finite interval always has a finite integral. Everything turns on what ψ does as |x| → ∞. The classical picture agrees. An electron with E = 0.229 eV in a well 1.00 nm wide and 2.00 eV deep turns round at x = ±a and never leaves; the quantum ground state at that energy puts 2.6% of its probability beyond the walls, yet the total still comes to one, because both tails are finite.
4Each outer region keeps one exponential
Past x = a, with E < V₀, the equation reads ψ″ = κ²ψ with κ = √(2m(V₀ − E))/ħ. Write s = x − a; the general solution is C e(−κs) + D e(κs), and both terms satisfy the equation exactly. Only the integral tells them apart. Out to a distance S past the wall, with real C and D, ∫|ψ|² dx = C²(1 − e(−2κS))/(2κ) + 2CDS + D²(e(2κS) − 1)/(2κ). The first term settles at C²/(2κ); the second grows linearly and the third exponentially, so the integral stays finite only if D = 0 exactly. On the left, with s = −a − x, the same algebra deletes the coefficient G of the term that grows as x → −∞ and keeps F e(−κs). One requirement applied on both sides gives tails falling as e(−κ|x|), with one κ because both walls have height V₀. For the ground state above, V₀ − E = 1.77 eV and κ = 6.82 nm⁻¹, so a stray growing term would be multiplied by about 910 for every nanometre it travelled.
5Six constants, four survivors, four conditions
Now count. Inside, ψ = A cos kx + B sin kx with k = √(2mE)/ħ, and each outer region brings two constants, so the three regions carry six. Normalisability removes G and D and leaves four: F, A, B and C. The walls impose four conditions, ψ and ψ′ continuous at x = −a and at x = +a, each linear and homogeneous in those constants. Written as M(E) acting on (F, A, B, C), a nonzero solution exists only where det M(E) = 0, one real equation for one unknown, E, whose roots are isolated. Homogeneity leaves the overall scale free, and that is the one constant normalisation fixes; it never touches E. Undo the deletion and the count collapses: six constants against four conditions leave a two-parameter family of solutions at every E, so the walls on their own quantise nothing. UQMI-02.09 ran this count on a single step; the finite well needs it at both ends. In the 2.00 eV well, |det M| = 23.3 nm⁻² at 0.300 eV and zero at 0.229 eV.
6Scan the energy and watch the growing term change sign
The determinant has a more physical face. Fix the left tail as F = 1 and carry it through the well: inside it becomes cos k(x + a) + (κ/k) sin k(x + a), and at x = a it arrives with value ψ(a) and slope ψ′(a). Split that into the two outer terms and the growing part has coefficient D = ½[ψ(a) + ψ′(a)/κ] = cos 2ka + ½(κ/k − k/κ) sin 2ka, a smooth function of E that vanishes exactly where det M does, since det M = ±2kκD. For the electron in the 1.00 nm, 2.00 eV well, D = +2.01 at 0.100 eV, +0.342 at 0.200 eV and −0.621 at 0.300 eV. Below 0.229 eV the right tail runs off towards +∞, above it towards −∞, and at 0.229 eV it does neither: that energy is a level. Scanning on, D returns to zero at 0.887 and 1.815 eV and nowhere else below V₀, and each time the runaway flips sign one more node enters ψ. A continuous dial of trial energies has yielded three states, all below the 0.376, 1.504 and 3.384 eV of an infinite well of the same width, a pattern UQMI-07.08 explains.
7Deleting the growing term by hand breaks the wall
At a trial energy that is not a level, why not simply drop D? Because the second-order equation has already decided. Once the left tail is chosen, ψ(a) and ψ′(a) are fixed, and they fix both outer coefficients: C = ½[ψ(a) − ψ′(a)/κ] and D = ½[ψ(a) + ψ′(a)/κ]. At 0.300 eV in the 2.00 eV well, ψ(a) = −0.160 and ψ′(a)/κ = −1.083, so C = 0.461 and D = −0.621. Keep only C e(−κs) and ψ jumps from −0.160 to 0.461 at the wall, a jump of exactly |D|. Keep the value instead, ψ(a) e(−κs), and the slope jumps by 2κ|D| = 8.30 nm⁻¹. Either way a matching rule of UQMI-02.08 is broken. The two demands, smooth matching and a decaying tail, can be met together only when D = 0, which is the same as ψ′(a)/ψ(a) = −κ: the inside solution must reach the wall with the slope-to-value ratio of a pure decaying exponential. That ratio test is the form UQMI-07.04 puts to work.
8Where the demand stops selecting energies
Normalisability selects energies only while there is a growing term to delete. At E = V₀ the outer solution is a straight line, bounded only if it is constant, and no nonzero constant is square-integrable, so no level sits at the rim. Above V₀ the outer solutions are cos qs and sin qs with q = √(2m(E − V₀))/ħ; both stay bounded, neither can be deleted, and six constants against four conditions leave solutions at every energy, the unbound continuum of UQMI-02.09. At the other extreme, V₀ → ∞ sends κ → ∞, the tails shrink to nothing, and ψ(±a) = 0 returns, as UQMI-06.01 found. The last limit is physical. Normalisability is a statement about |x| → ∞, so it depends on V far from the well. Tilt a quantum well with a field ℰ and V outside falls below E a distance (V₀ − E)/(eℰ) past the wall; the tail must tunnel through that stretch and then oscillates away. For V₀ − E = 0.20 eV the stretch is 200 nm at 10⁶ V m⁻¹, and the leak is immeasurably slow; at 10⁸ V m⁻¹ it is 2 nm, and the levels broaden into resonances.
Key equations
What each relation says, and the units it is written in.
Bound means square-integrable
∫ |ψ(x)|² dx over −∞ < x < ∞ is finite
Linearity lets a finite integral be rescaled to 1. An infinite one cannot be, so that solution is not a state at any normalisation.
ψ in m(−1/2), so the integral is a pure number. Inside |x| ≤ a it is always finite; only the tails can make it infinite.
General solution past the right wall, E below V₀
ψ = C e(−κs) + D e(κs), s = x − a, κ = √(2m(V₀ − E))/ħ
Both terms solve the equation exactly at every E. Only the integral tells them apart, and it rejects D e(κs) unless D = 0.
κ in m⁻¹, V₀ and E in J, m in kg; C and D in m(−1/2). For an electron ħ²/2m = 0.0381 eV nm².
What normalisability leaves
ψ = F e(−κs) for x ≤ −a and ψ = C e(−κs) for x ≥ a, with s = |x| − a
One demand removes opposite exponentials on the two sides; each tail then holds |C|²/(2κ) or |F|²/(2κ) of probability.
F = ψ(−a) and C = ψ(a), in m(−1/2); one κ serves both sides because both walls have height V₀.
Growing coefficient at the right wall
D = ½[ψ(a) + ψ′(a)/κ]
D = 0 is the same as ψ′(a)/ψ(a) = −κ: the inside solution must reach the wall with the decaying tail's own slope-to-value ratio.
ψ(a) in m(−1/2) and ψ′(a) in m(−3/2) are the value and slope arriving from inside; D in m(−1/2).
The growing coefficient as a function of energy
D(E) = cos 2ka + ½(κ/k − k/κ) sin 2ka, with k = √(2mE)/ħ
A smooth function of E whose isolated zeros are the levels: 0.229, 0.887 and 1.815 eV for an electron in a 1.00 nm, 2.00 eV well.
For a left tail F = 1, so D is a pure number; k and κ in m⁻¹, a the half-width in m.
Constants against conditions
det M(E) = ±2kκ D(E)
Six constants less two growing ones leave four, against four homogeneous conditions: a nonzero ψ needs det M = 0, exactly where D = 0.
M is the 4 × 4 matrix of the matching equations for (F, A, B, C), rows as they come; the sign depends on row order. Units m⁻².
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At V₀ = 2.0 eV and E = 0.200 eV, D = 0.342 and the right tail shoots upward. Raise E slowly: near 0.114 V₀ the tail hugs the dashed curve for about 0.4 nm past the wall before peeling away, and just above that energy it peels downward. Find the next flips near 0.44 V₀ and 0.91 V₀, then lower V₀ and watch the flips thin out to one.
Live results
- TRIAL ENERGY E
- 0.200 eV
- DECAY CONSTANT κ
- 6.87 nm⁻¹
- GROWING COEFFICIENT D (F = 1)
- 0.342
- WELL STRENGTH z₀
- 3.62
Catch the common trap
The mistake this lesson is built to prevent.
Students often treat “discard the growing exponential” as a free choice made once at the start, like a sign convention, and then expect a finite-well state at every energy between 0 and V₀, or they find that the growing term is small at some energy and call that state approximately bound. Both readings get the logic backwards. Once the left tail is chosen, the second-order equation fixes ψ everywhere else, including its value and slope at x = a, and those fix D = ½[ψ(a) + ψ′(a)/κ]. Deleting D by hand then makes ψ jump at the wall by |D|, which continuity forbids. And a small D is not nearly zero in any useful sense: its term grows as e(κs), overtakes the decaying one at s = ln|C/D|/(2κ), and makes ∫|ψ|² dx infinite however small D is. Normalisability is a condition on E, met exactly at isolated energies, not a trimming step applied to every solution.
Test it: what changes?
- D(E) passes through zero and changes sign
- Runaway tail flips to the opposite infinity
- ψ gains one more node
- Normalisable only exactly at the level
- κ → ∞, so the tails shrink to nothing
- Kept-tail integral 1/(2κ) → 0
- Infinite-well condition ψ(±a) = 0 returns
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is in a bound state of energy E = 0.45 eV in a symmetric finite well of depth V₀ = 1.20 eV. Find κ in the outer regions and write the tail each side must carry. Then compare the integrals of the kept and the discarded exponentials past the right wall, and find how far past the wall a growing admixture D = 10⁻⁶ C would take over.
View solution →
- κ = √(2m(V₀ − E))/ħ = √((V₀ − E)/(ħ²/2m)) = √(0.75 eV ÷ 0.0381 eV nm²) = 4.44 nm⁻¹. Units: eV over eV nm² is nm⁻², whose root is nm⁻¹.
- With s = |x| − a the distance past the nearer wall, e(κs) grows on both sides, so normalisability leaves ψ = C e(−κs) for x ≥ a and ψ = F e(−κs) for x ≤ −a: tails that fall as e(−κ|x|).
- Kept term: ∫ e(−2κs) ds from 0 to ∞ = 1/(2κ) = 0.113 nm, finite; times |C|² it is the probability beyond the right wall.
- Discarded term: ∫ e(2κs) ds from 0 to S = (e(2κS) − 1)/(2κ), which is 805 nm at S = 1.00 nm and 5.7 × 10⁶ nm at S = 2.00 nm, with no limit as S → ∞. No constant can bring it to 1.
- With D = 10⁻⁶ C the two terms are equal where e(2κs) = 10⁶, at s = ln(10⁶)/(2κ) = 13.8 ÷ 8.87 nm⁻¹ = 1.56 nm. Beyond that the growing term dominates and the integral diverges: a small D buys distance, not normalisability.
- Limiting case: as V₀ → ∞, κ → ∞ and the kept tail's 1/(2κ) → 0, the infinite-well result. A proton at the same energies would have κ larger by √1836 = 42.8, a tail 42.8 times shorter.
Answer. κ = 4.44 nm⁻¹; ψ = C e(−κs) on the right and F e(−κs) on the left, s = |x| − a. The kept tail integrates to 0.113 nm; the discarded one to 805 nm by 1.00 nm out, without limit. A 10⁻⁶ admixture takes over 1.56 nm past the wall.
MediumExample 2
An electron in a finite well of half-width a = 0.50 nm and depth V₀ = 2.00 eV has its lowest level at E₁ = 0.229 eV. Starting from the left tail e(−κs) with F = 1, show that the solution leaves the right wall with no growing term, then normalise it including both tails and find the probability of finding the electron outside the well.
View solution →
- k = √(E₁/0.0381 eV nm²) = 2.451 nm⁻¹ and κ = √((V₀ − E₁)/0.0381 eV nm²) = 6.818 nm⁻¹, so κ/k = 2.781 and 2ka = 2.451 rad, with cos 2ka = −0.771 and sin 2ka = 0.637.
- Carried across, ψ = cos k(x + a) + 2.781 sin k(x + a) inside. At x = a: ψ(a) = −0.771 + 2.781 × 0.637 = 1.000 and ψ′(a)/κ = cos 2ka − (k/κ) sin 2ka = −0.771 − 0.229 = −1.000. So D = ½(1.000 − 1.000) = 0 and C = 1.000: the right tail is e(−κs), the mirror image of the left one.
- Write the inside as √(1 + (κ/k)²) cos(k(x + a) − φ) with tan φ = κ/k: φ = 1.226 rad, which equals ka, so ψ = 2.956 cos kx inside, even about the centre although nothing assumed it.
- Normalisation integral: the tails give 2 × 1/(2κ) = 0.1467 nm; the inside gives 8.736 × [a + sin(2ka)/(2k)] = 8.736 × 0.6299 nm = 5.503 nm. The total is 5.649 nm, so N = 1/√(5.649 nm) = 0.421 nm(−1/2).
- Probability outside: 0.1467/5.649 = 0.0260, 1.30% beyond each wall, where a classical electron would never be found. Units: N² × nm is dimensionless, as a probability must be. The inside amplitude, 2.956N = 1.24 nm(−1/2), sits below the infinite-well 1/√a = 1.41 nm(−1/2); as V₀ → ∞ the tails vanish, sin 2ka → 0, and 1/√a returns.
Answer. D = 0 and C = F = 1 at 0.229 eV; the normalised state has N = 0.421 nm(−1/2), and the tails hold 2.60% of the probability, 1.30% beyond each wall.
HardExample 3
An electron moves in V = 1.00 eV for x < 0, V = 0 for 0 < x < L = 0.80 nm, and V = 0.40 eV beyond x = L. Count constants against conditions to say which energies can be bound. Then carry the normalisable left tail across the well, find the growing coefficient D on the right at 0.20 eV and 0.21 eV, and locate the lowest level to three significant figures.
View solution →
- Count for 0 < E < 0.40 eV: both outer regions are forbidden, normalisability deletes one of the six constants on each side, and four remain against four homogeneous matching conditions, so levels are isolated. The decay constants are κ₁ = √((1.00 eV − E)/(ħ²/2m)) on the left and κ₂ = √((0.40 eV − E)/(ħ²/2m)) on the right. E ≤ 0 gives none, since ψ would bend away from the axis everywhere.
- For 0.40 < E < 1.00 eV the right region oscillates and nothing there can be deleted: five constants against four conditions leave only the scale free, so a solution exists at every E and none is normalisable. Classically the electron escapes over the lower wall. Above 1.00 eV six constants face four conditions, and two solutions exist at every E.
- Carry the left tail e(κ₁x) across: inside, ψ = cos kx + (κ₁/k) sin kx, and at x = L, D = ½[ψ(L) + ψ′(L)/κ₂]. At 0.20 eV, k = κ₂ = 2.291 nm⁻¹, κ₁ = 4.582 nm⁻¹ and kL = 1.833 rad, so ψ(L) = −0.2591 + 2.000 × 0.9658 = 1.673, ψ′(L) = −2.291 × 0.9658 + 4.582 × (−0.2591) = −3.400 nm⁻¹, and D = ½(1.673 − 1.484) = +0.094.
- At 0.21 eV: k = 2.348 nm⁻¹, κ₁ = 4.554 nm⁻¹, κ₂ = 2.233 nm⁻¹ and kL = 1.878 rad, giving ψ(L) = 1.546 and ψ′(L) = −3.615 nm⁻¹, so D = ½(1.546 − 1.619) = −0.036. D has changed sign, so a level lies between. Interpolating, E₁ ≈ 0.20 eV + 0.01 eV × 0.094/(0.094 + 0.036) = 0.2072 eV, where |D| has dropped below 2 × 10⁻⁴.
- D keeps falling above E₁, to −1.22 at 0.30 eV and −2.16 at 0.35 eV, so there is no second level below 0.40 eV. Checks: 0.207 eV lies in the bound window and below the infinite-well value (ħ²/2m)π²/L² = 0.588 eV. The tails, 1/κ₁ = 0.219 nm into the high wall and 1/κ₂ = 0.445 nm into the low one, lean toward the side a more energetic electron would escape through.
Answer. Bound states only for 0 < E < 0.40 eV, and there is exactly one, at E₁ = 0.207 eV, 35% of the 0.588 eV infinite-well value. From 0.40 eV to 1.00 eV every energy is allowed but none is normalisable.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
A level is an energy where the growing coefficient vanishes.
Carry the decaying left tail across the well; at x = a the growing coefficient is D = ½[ψ(a) + ψ′(a)/κ]. Levels are the isolated zeros of D(E), equivalently ψ′(a)/ψ(a) = −κ. Deleting D by hand breaks continuity, and above V₀ nothing can be deleted, so every energy is allowed.