Admissibility and matching conditions on ψ
ψ never jumps; ψ′ jumps only where V is unbounded
Every solution in this course is built in pieces and glued at the boundaries, and most wrong answers are glue errors: a slope matched where it should jump, or a wall given two conditions instead of one. Derive the rules once, from one integral across the step, and you will know what to match at any feature before writing any algebra.
Build the model
Integrate the equation across the step, and every matching rule falls out.
1The big idea
The time-independent Schrödinger equation is second order, so between features of V it has a two-parameter family of solutions, and a problem is solved only when the pieces are joined. The joining rules are not extra postulates; the equation supplies them. Integrate ψ″ = (2m/ħ²)(V − E)ψ across a sliver of width 2ε and the change in slope equals the area under (2m/ħ²)(V − E)ψ. Where V is finite that area vanishes with ε, so ψ′ is continuous. A spike λδ(x − a) leaves the area (2mλ/ħ²)ψ(a), a corner of set size. An infinite wall is the limit in which V diverges while ψ is squeezed to zero within a decay length 1/κ; the product keeps a finite area, and what survives is ψ = 0 at the wall with ψ′ free. ψ itself never jumps: a jump would put a δ′ into ψ″ that no potential term can balance, and would cost infinite kinetic energy. These rules are local. Square-integrability is the global one: a bound state needs ∫|ψ|² dx finite, which discards every growing exponential. The rules stop where their derivation does. A mass that changes across the boundary, as in a semiconductor heterojunction, makes ψ′/m the continuous quantity; a potential more singular than a delta needs its own treatment; and plane waves and step solutions, which never decay, are admissible only as bounded scattering solutions, not as states.
2In one sentence
An admissible wavefunction is continuous everywhere, has a continuous slope wherever V is finite, has its slope jump by (2mλ/ħ²)ψ(a) at a spike λδ(x − a), vanishes at an infinite wall, and, if it is a bound state, is square-integrable.
Example. In a 1.0 nm infinite well, ψ = √2 sin(πx/1.0 nm) nm(−1/2) is admissible: continuous, zero at both walls, with ∫|ψ|² dx = 1, while its slope of √2 π = 4.44 nm(−3/2) at x = 0 drops to zero outside, a jump only a wall permits.
3Integrate the equation across the point
Write the time-independent equation as ψ″ = (2m/ħ²)(V − E)ψ and integrate both sides from a − ε to a + ε. The left side integrates exactly, with no assumption about smoothness, so ψ′(a + ε) − ψ′(a − ε) = (2m/ħ²) ∫ (V − E)ψ dx over that interval. This identity is the whole lesson. The change in slope across any interval equals the area under (2m/ħ²)(V − E)ψ, so every matching rule is a statement about what that area does as ε → 0, and the answer depends only on how badly V misbehaves at a. That fixes the working order for every problem that follows: name what sits at each boundary, a finite step, a delta spike or an infinite wall, before writing a single matching equation. Dimensions check the identity for free: in one dimension ψ is in m(−1/2), ψ′ in m(−3/2), and (2m/ħ²)(V − E) in m⁻², so the integrand times dx is in m(−3/2) too.
4Why ψ itself can never jump
Suppose ψ jumped by D at x = a. Then ψ′ would contain the spike Dδ(x − a), and ψ″ its derivative Dδ′(x − a), a doublet that, for D > 0, is infinitely positive just before a and infinitely negative just after. The right-hand side (2m/ħ²)(V − E)ψ can supply at worst a single δ, and only when V has one; nothing in it has the shape of δ′, so the equation fails. Energy gives the same verdict. Smear the jump over a width w and the slope there is about D/w, so the kinetic energy (ħ²/2m)∫|ψ′|² dx picks up about (ħ²/2m)D²/w. For an electron with D = 1.0 nm(−1/2) spread over w = 0.001 nm that is 0.0381 eV nm² × 1.0 nm⁻¹ ÷ 0.001 nm = 38 eV, and it grows without limit as w → 0. So ψ is continuous at every point, walls and spikes included: the one rule with no exceptions.
5Finite V: the slope is continuous, the curvature is not
If V is bounded near a, the integrand is bounded, and the integral is at most 2ε times a finite number: it vanishes as ε → 0. So ψ′ is continuous at every finite step, however tall. What does jump is ψ″ = (2m/ħ²)(V − E)ψ, by (2m/ħ²)V₀ψ(a) at a step of height V₀. For an electron at a 3.0 eV step, (2m/ħ²)V₀ = 3.0 eV ÷ 0.0381 eV nm² = 79 nm⁻², so ψ bends at a new rate beyond the step but has no corner there. This is the quantum form of the classical picture. A classical particle crossing a step has its speed changed abruptly, and the local wavenumber √(2m(E − V))/ħ jumps just as abruptly; the wave follows by changing its wavelength, or turning into an exponential, while its slope runs straight through. Two conditions at each finite step are what make piecewise-constant problems solvable. Dividing one by the other, ψ′/ψ is continuous too, which the finite-well lesson UQMI-07.04 uses at each wall.
6A delta spike: a corner of prescribed size
Let V contain λδ(x − a), with λ in J m, negative for a well and positive for a barrier. Now the area does not vanish: the delta picks out λψ(a), while E and the finite parts of V still contribute nothing. So ψ′(a⁺) − ψ′(a⁻) = (2mλ/ħ²)ψ(a). ψ stays continuous, so the graph has a corner whose size is fixed by the spike and by ψ at the spike. For an electron, λ = −0.40 eV nm drops the slope by 10.5 nm⁻¹ times ψ(a), a peaked cusp. Because the jump is a real multiple of ψ(a), the current j = (ħ/m) Im(ψ* ψ′) gains (ħ/m) Im((2mλ/ħ²)|ψ(a)|²) = 0 across the spike: it bends ψ without creating probability. The lone well shows the rule at work. e(−κ|x|) is continuous and decays both ways; its corner −2κψ(0) must equal (2mλ/ħ²)ψ(0), so κ = −mλ/ħ² = 5.25 nm⁻¹ and E = −ħ²κ²/2m = −1.05 eV. As λ → 0, κ → 0 and the state spreads out and unbinds.
7An infinite wall: infinity times zero leaves a finite kick
Treat the wall as a step of height V₀ at x = 0 and let V₀ grow. For finite V₀ > E both ψ and ψ′ are continuous, and the admissible solution beyond the step is ψ(0)e(−κx) with κ = √(2m(V₀ − E))/ħ, so at the step ψ′/ψ = −κ. The slope inside is set by the oscillation there and stays finite, so ψ(0) = −ψ′(0)/κ is driven to zero. For an electron with V₀ − E = 3, 48 and 498 eV the tail length 1/κ is 0.113, 0.028 and 0.0087 nm. In the limit the tail has no width, and ψ′ falls from its inside value to zero across nothing. The integral says why that jump survives: over the tail, (2m/ħ²)∫(V₀ − E)ψ dx = κ²ψ(0)/κ = −ψ′(0) for every V₀, an infinite V times a vanishing ψ with a finite product. So a wall imposes one condition, ψ = 0, and leaves ψ′ free. Classically the particle bounces with p → −p; here the incident and reflected waves cancel at the wall, making the node.
8Square-integrability: the one global condition
Every rule so far is local and says nothing about infinity. A bound state must also satisfy ∫|ψ|² dx < ∞, and that is a separate demand. In any region running out to x → +∞ with V > E, the general solution is Ae(κx) + Be(−κx), and normalisability forces A = 0; at x → −∞ it is B that must vanish instead. Each discarded exponential is one more condition on the same unknown amplitudes, and together with the matching rules these conditions allow solutions only at special energies, which the next lesson counts. Solutions that oscillate out to infinity fail the test outright: a plane wave e(ikx) has |ψ|² = 1 everywhere, and the incoming side of every step solution never decays, so ∫|ψ|² dx diverges. Such solutions are kept as bounded but unnormalisable idealisations, made physical by building packets from them: from plane waves in Unit 5, from step solutions in Unit 8. A solution that grows without bound is never kept at all.
Key equations
What each relation says, and the units it is written in.
Slope change across any interval
ψ′(a + ε) − ψ′(a − ε) = (2m/ħ²) ∫ (V − E) ψ dx, taken over a − ε < x < a + ε
Exact for every ε. Each matching rule is this integral's limit as ε → 0, and which rule applies depends only on how V behaves at a.
m in kg, V and E in J, ħ in J s, ψ in m(−1/2) in one dimension; both sides are in m(−3/2).
ψ never jumps
ψ(a⁺) = ψ(a⁻) at every point a
A jump would make ψ″ contain δ′, which no (V − E)ψ term can match, and would make ∫(ħ²/2m)|ψ′|² dx, the kinetic energy, infinite.
Holds at finite steps, delta spikes and infinite walls alike; a and x in m.
Slope continuous where V is finite
ψ′(a⁺) = ψ′(a⁻) whenever V is bounded near a
Two conditions at every finite step. What jumps instead is the curvature ψ″, by (2m/ħ²)V₀ψ(a): the wave changes wavelength without a corner.
ψ′ in m(−3/2). A finite step of any height V₀ in J qualifies; V₀ = 10³ eV is still finite.
Jump at a delta spike λδ(x − a)
ψ′(a⁺) − ψ′(a⁻) = (2mλ/ħ²) ψ(a)
The corner scales with ψ(a), so a spike at a node of ψ is invisible, and the jump is a real multiple of ψ, so the current j stays continuous.
λ in J m (eV nm in practice), negative for a well; 2mλ/ħ² in m⁻¹. For an electron ħ²/2m = 0.0381 eV nm².
Infinite wall
ψ = 0 at the wall, with ψ′ left free
One condition per wall, not two: (2m/ħ²)∫(V₀ − E)ψ dx over the tail equals −ψ′ at the wall for every V₀, so the slope drops to zero across nothing.
Limit of a step V₀ → ∞ where ψ′/ψ = −κ, κ = √(2m(V₀ − E))/ħ in m⁻¹, and the tail length 1/κ → 0.
Square-integrability of a bound state
∫ |ψ|² dx < ∞ over the whole line
The one global condition. Plane waves and step solutions fail it: they are kept as bounded scattering solutions, never as states.
|ψ|² in m⁻¹, so the integral is a pure number. Where V > E out to x → ±∞, keep only e(∓κx).
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. With λ = 0, raise V₀ from 2 to 100 eV: the join stays smooth while ψ(0) and the tail shrink toward a node, the hard-wall limit. Then drag λ: a corner opens at x = 0, the slope dropping across a well and rising across a barrier, and it fades as V₀ grows, since the jump is proportional to ψ(0).
Live results
- DECAY LENGTH 1/κ
- 0.113 nm
- ψ(0) AT THE STEP
- 0.500 A
- SLOPE JUMP Δψ′
- 0.00 A nm⁻¹
- EQUIVALENT HARD WALL AT
- 0.102 nm
Catch the common trap
The mistake this lesson is built to prevent.
The commonest slip is to treat “ψ and ψ′ are continuous” as one law to impose at every boundary. Continuity of ψ′ is not a postulate; it is what the integrated equation gives when V is bounded, and it fails exactly where V is not. Impose it at an infinite wall and the problem collapses: ψ(0) = 0 and ψ′(0) = 0 together are initial data for a second-order linear equation, and the uniqueness theorem then allows only ψ ≡ 0, which is why a student who insists on both finds no states in the infinite well at all. At a delta the same habit erases the physics: forcing ψ′ continuous sets λψ(a) = 0, so the spike seems to do nothing. Ask first what V does at the point and let the integral decide: bounded V, match ψ and ψ′; a spike, match ψ and jump ψ′ by (2mλ/ħ²)ψ(a); an infinite wall, impose ψ = 0 alone.
Test it: what changes?
- Join stays smooth, with no corner
- Decay length 1/κ shrinks
- ψ(0) falls toward a node
- Equivalent hard wall moves toward the step
- A corner opens in ψ at x = 0
- Slope drops across a well, rises across a barrier
- ψ itself stays continuous
- Probability current unchanged across the spike
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
A computed stationary state of an electron has ψ(0) = 1.20 nm(−1/2), slope +2.0 nm(−3/2) just left of x = 0 and −3.0 nm(−3/2) just right of it. V is finite everywhere except possibly at x = 0. What sits at the origin? If it is a delta spike, find its strength λ and say whether it is a well or a barrier.
View solution →
- Name the feature. ψ(0) = 1.20 nm(−1/2) is not zero, so this is no infinite wall, and ψ is continuous as it must be. The slope changes, and a finite V cannot change a slope, so V contains a spike λδ(x) at the origin.
- Measure the jump, right minus left: Δψ′ = ψ′(0⁺) − ψ′(0⁻) = −3.0 − (+2.0) = −5.0 nm(−3/2). Reversing the order flips the sign of λ.
- Invert the jump rule Δψ′ = (2mλ/ħ²)ψ(0): λ = (ħ²/2m) Δψ′/ψ(0) = 0.0381 eV nm² × (−5.0 ÷ 1.20) nm⁻¹ = 0.0381 × (−4.17) eV nm = −0.159 eV nm.
- Check the units: eV nm² × nm⁻¹ = eV nm, an energy times a length, as it must be since δ(x) carries m⁻¹. Check the sign: λ < 0 is a well, and a well turns a positive ψ into a peaked cusp, which is what a slope of +2.0 followed by −3.0 describes.
- Check the current: the jump is a real multiple of ψ(0), so (ħ/m) Im(ψ* Δψ′) = 0 and the probability current is the same on both sides, as a real V requires.
Answer. A delta well at x = 0 of strength λ = −0.159 eV nm.
MediumExample 2
A trial wavefunction for an electron is the tent ψ(x) = N(1 − |x|/a) for |x| ≤ a and ψ = 0 for |x| > a, with a = 0.20 nm. Normalise it. Then read the matching rules backwards: find the potential for which this ψ is a stationary state with E = 0, and confirm E = 0 from ⟨T⟩ + ⟨V⟩.
View solution →
- Normalise: ∫ N²(1 − |x|/a)² dx over −a < x < a is 2N²a/3 = 1, so N = √(3/(2a)) = √(3 ÷ 0.40 nm) = 2.74 nm(−1/2). ψ is continuous everywhere, zero at ±a from both sides, and square-integrable, so it passes the admissibility tests that do not involve V.
- Between the corners ψ″ = 0, so (2m/ħ²)(V − E)ψ = 0 with ψ ≠ 0: V = E = 0 on 0 < |x| < a.
- At x = ±a the slope jumps between ∓N/a and 0 while ψ = 0. A finite V allows no jump, and a spike would give (2mλ/ħ²) × 0 = 0. Only an infinite wall lets ψ′ jump at a node, so V = ∞ for |x| ≥ a, consistent with ψ = 0 there.
- At x = 0, Δψ′ = (−N/a) − (+N/a) = −2N/a with ψ(0) = N, so 2mλ/ħ² = −2/a and λ = −ħ²/(ma) = −2 × 0.0381 eV nm² ÷ 0.20 nm = −0.381 eV nm, a delta well. With no interior node, this is the ground state.
- Energy check: ⟨T⟩ = (ħ²/2m)∫|ψ′|² dx = (ħ²/2m)(N²/a²)(2a) = 3(ħ²/2m)/a² = 3 × 0.0381 ÷ 0.040 eV = 2.86 eV, and ⟨V⟩ = λ|ψ(0)|² = −0.381 eV nm × 7.5 nm⁻¹ = −2.86 eV. They cancel, so E = 0.
- Limiting case: without the spike, this 0.40 nm well has ground state π² × 0.0381 eV nm² ÷ (0.40 nm)² = 2.35 eV; a spike of exactly ħ²/(ma) pulls it down to zero, and a weaker one leaves the ground state sin(k(a − |x|)), whose k → 0 limit is this tent.
Answer. N = 2.74 nm(−1/2). ψ is the E = 0 ground state of an infinite well with walls at x = ±0.20 nm, V = 0 inside, and a central spike λ = −ħ²/(ma) = −0.381 eV nm.
HardExample 3
An electron is held near a hard wall by a thin attractive layer: V = ∞ for x < 0 and V = −αδ(x − d) for x > 0, with α = 0.40 eV nm. On its own the spike binds one state at −1.05 eV. Find the equation that fixes κ = √(2m|E|)/ħ, the smallest d for which a bound state survives, and the energy of the state when d = 0.15 nm.
View solution →
- Admissibility first. For E = −ħ²κ²/2m < 0: ψ = 0 for x < 0 (the wall); on 0 < x < d, V = 0 and ψ = A sinh κx, since ψ(0) = 0 removes the cosh; for x > d, ψ = B e(−κx), since square-integrability removes e(+κx). Three rules are spent before any matching.
- At x = d, continuity gives B e(−κd) = A sinh κd, and the jump with λ = −α gives ψ′(d⁺) − ψ′(d⁻) = −κB e(−κd) − κA cosh κd = −(2mα/ħ²) A sinh κd.
- Substitute the first into the second and divide by −A sinh κd: κ(1 + coth κd) = 2κ₀, with κ₀ = mα/ħ² = 0.40 eV nm ÷ (2 × 0.0381 eV nm²) = 5.25 nm⁻¹, the lone spike's decay constant. Since 1 + coth u = 2/(1 − e(−2u)), this reads κ = κ₀(1 − e(−2κd)).
- Threshold. The right side starts at zero with slope 2κ₀d and bends over toward κ₀, so it meets the line κ at some κ > 0 only if 2κ₀d > 1. Hence dₘᵢₙ = 1/(2κ₀) = ħ²/(2mα) = 0.0381 eV nm² ÷ 0.40 eV nm = 0.095 nm; any closer and the wall squeezes the state out.
- At d = 0.15 nm, 2κ₀d = 1.57 > 1. Iterating κ ← κ₀(1 − e(−2κd)) from κ₀ gives 4.16, 3.74, 3.54, … and settles at κ = 3.30 nm⁻¹; check 5.25 × (1 − e(−0.990)) = 3.30. Then E = −0.0381 eV nm² × (3.30 nm⁻¹)² = −0.41 eV.
- Checks. As d → ∞, κ → κ₀ and E → −1.05 eV, the lone spike. As d → dₘᵢₙ, κ → 0 and the tail length 1/κ diverges, so the state stops being normalisable exactly at threshold. Units: mα/ħ² is kg × J m ÷ (J s)² = m⁻¹.
Answer. κ = κ₀(1 − e(−2κd)) with κ₀ = mα/ħ² = 5.25 nm⁻¹; a bound state needs d > ħ²/(2mα) = 0.095 nm; at d = 0.15 nm, κ = 3.30 nm⁻¹ and E = −0.41 eV, about 40% of the lone spike's binding.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Name the feature at each boundary before matching anything.
Where V is finite, match ψ and ψ′. At a spike λδ(x − a), match ψ and let ψ′ jump by (2mλ/ħ²)ψ(a). At an infinite wall, impose ψ = 0 alone. For a bound state, also discard every growing exponential. A mass that changes across the boundary makes ψ′/m continuous instead.