Skip to main content
GIOPHYSICS
Menu
← University Physics V

University Physics V · The Schrödinger Equation · 4.6

Admissibility & Matching Conditions

Solve the differential equation and you have too much — a solution at every energy. This lesson is the sorting step: which solutions count as states, what you may impose at a step, a wall or a delta spike, and why choosing those conditions is choosing the operator rather than tidying up after it.

Open the lab ↓Practice & worked examples ↓
01

Build the model

Connect the measurement to the mechanism.

Solving −(ħ²/2m)ψ″ + Vψ = Eψ is not the hard part. As a second-order linear ODE it has a two-dimensional solution space for every complex E you care to name, so the equation on its own predicts a continuum of everything and quantises nothing. The physics enters when you say which functions count as states — that is, when you fix the domain on which Ĥ is self-adjoint.

Three requirements do the work. Square-integrability discards the solutions that blow up at infinity. Continuity of ψ is forced by the equation itself, because a jump in ψ would put a δ′ into ψ″ that nothing on the other side can balance.

And the behaviour of ψ′ at a point is not assumed at all but read off by integrating the equation across that point: Δψ′ = (2m/ħ²)∫(V − E)ψ dx, which vanishes wherever V is bounded, does not vanish at an infinite wall, and equals exactly −(2mα/ħ²)ψ(0) at V = −αδ(x). The cost is that the domain is a choice, not a consequence. The same differential expression on [0, L] under Dirichlet, Neumann, Robin or periodic conditions is four different self-adjoint operators with four different spectra — and a numerical grid quietly picks one of them on your behalf.

Simple definition
An admissible eigenfunction is one that lies in the domain of Ĥ — square-integrable, continuous, and joined at every point by exactly the jump in ψ′ that integrating the Schrödinger equation across that point demands.
Example
For an electron in V = −αδ(x) with α = 7.62 eV Å, and ħ²/2m = 3.810 eV Ų, the demanded kink is 2mα/ħ² = 2.000 Å⁻¹, so e(−κ|x|) is admissible only at κ = 1.000 Å⁻¹ — one state, at E = −3.810 eV.
Matching read off the equationψ′(x₀⁺) − ψ′(x₀⁻) = (2m/ħ²) lim(ε→0) ∫[x₀−ε, x₀+ε] (V − E) ψ dx

Every rule below is this one limit evaluated for a different V. Nothing beyond it is postulated.

ψ′ carries the units of ψ per metre; V and E in J; 2m/ħ² in J⁻¹ m⁻².

Bounded V: nothing jumpsψ(x₀⁺) = ψ(x₀⁻) and ψ′(x₀⁺) = ψ′(x₀⁻) wherever |V| < ∞

Two equations per interface — enough to cancel the piecewise amplitudes and leave one condition on E.

Holds across a finite step of any height; ψ″ still jumps, by (2m/ħ²)ΔV⋅ψ(x₀).

Infinite wall: Dirichlet onlyψ(xw) = 0, with ψ ≡ 0 beyond itψ′(xw) unconstrained

One condition, not two. Adding ψ′(xw) = 0 forces ψ ≡ 0 by uniqueness for a second-order ODE.

xw in m. It is the V₀ → ∞ limit of κ = √(2m(V₀ − E))/ħ, where the depth 1/κ → 0.

Delta well: the kink conditionψ′(0⁺) − ψ′(0⁻) = −(2mα/ħ²) ψ(0), for V(x) = −α δ(x)

The minus sign is why a well binds: e(−κ|x|) delivers a negative kink, while +βδ(x) would demand a positive one.

α > 0 in J m, or eV Å. ψ stays continuous there, which is what makes ψ(0) well defined.

Its one bound stateκ = mα/ħ² · E₁ = −mα²/2ħ² · ψ₁ = √κ e(−κ|x|)

Matching gave one linear equation, so exactly one state for any α: double α and the binding quadruples, the tail halves.

κ in m⁻¹, ψ₁ in m(−1/2). For α = 7.62 eV Å: κ = 1.000 Å⁻¹ and E₁ = −3.810 eV.

Self-adjoint domains on [0, L]ψ′(0) = γ₀ ψ(0), ψ′(L) = γL ψ(L), with γ₀, γL real

Same differential expression, different Ĥ, different spectrum — which is why the domain is physics and not bookkeeping.

γ in m⁻¹. γ = 0 is Neumann, γ → ∞ is Dirichlet; periodic conditions form a separate family.

01

The ODE is not yet the physics

Read −(ħ²/2m)ψ″ + Vψ = Eψ as an ODE and it is entirely undiscriminating. For any V and any E, real or complex, it has a two-dimensional space of solutions: hand it ψ(x₀) and ψ′(x₀) and integration supplies the rest. The equation alone therefore permits every energy, which is to say it selects none. Quantisation is a boundary-value statement, not a differential one. Put an infinite wall at x = 0 and a 10.0 eV well over 0 < x < 1.50 Å, and you can integrate outward at E = −2.4 eV, at −2.540 eV and at −2.6 eV alike. All three solve the ODE exactly. Only the middle one decays as x → ∞; the other two grow exponentially in the classically forbidden region and are not in L², so they are not states. Naming the admissible set — the domain of Ĥ — is precisely what turns a differential equation into a spectrum.

02

Integrate across the point and the rule falls out

Every matching rule in this topic comes from one manoeuvre. Write the equation as ψ″ = (2m/ħ²)(V − E)ψ and integrate it from x₀ − ε to x₀ + ε, giving ψ′(x₀+ε) − ψ′(x₀−ε) = (2m/ħ²)∫(V − E)ψ dx. Then ask what that integral does as ε → 0. If |V| ≤ Vₘₐₓ on the interval and ψ is continuous there, the integrand is bounded by (Vₘₐₓ + |E|)|ψ|ₘₐₓ and the whole integral is at most 2ε times that, so it vanishes: ψ′ is continuous, and stays continuous across a step of any finite height. If V = −αδ(x₀), the integral picks out −αψ(x₀) exactly, however small ε becomes, so Δψ′ = −(2mα/ħ²)ψ(x₀). If V is infinite over a region, the integral does not tend to zero at all, ψ′ is unconstrained, and the wall constrains ψ instead. Integrate a second time and the same bound gives Δψ = 0 in all three cases: ψ itself never jumps.

03

Match the logarithmic derivative, not the constants

Solving piecewise leaves an unknown amplitude on each side, and the instinct is to solve for them. You rarely need to. Divide the ψ′ condition by the ψ condition and the amplitudes cancel, leaving the logarithmic derivative L = ψ′/ψ. Across a bounded interface L is continuous; across V = −αδ(x) it jumps by exactly −2mα/ħ², because ψ is continuous there. For the wall-plus-well below, k cos(ka)/sin(ka) on the inside must equal −κ on the outside — one real equation in one unknown E, with no amplitudes anywhere. This is also what a shooting code compares: integrate in from the left, integrate in from the right, and match ψ′/ψ at a meeting point, since that ratio is invariant under the arbitrary scaling of each piece. In practice one matches the Wronskian ψL ψ′R − ψ′L ψR instead, so that a node sitting at the match point does not send the ratio to infinity.

04

The delta well end to end, and why a barrier binds nothing

Take V = −αδ(x) with α > 0. Away from the origin the equation is ψ″ = κ²ψ with κ = √(−2mE)/ħ, so square-integrability forces ψ = A e(−κ|x|): continuous at 0, with ψ′(0⁺) = −κA and ψ′(0⁻) = +κA, a delivered kink of −2κA. The demanded kink is −(2mα/ħ²)A. Matching gives κ = mα/ħ², hence E = −ħ²κ²/2m = −mα²/2ħ², and normalising ∫|A|²e(−2κ|x|)dx = |A|²/κ = 1 fixes A = √κ. For an electron with α = 7.62 eV Å this is κ = 1.000 Å⁻¹ and E = −3.810 eV. Two features deserve names. There is exactly one bound state for every α, because matching produced a single linear equation rather than a transcendental one. And a repulsive δ, V = +βδ(x), demands Δψ′ = +(2mβ/ħ²)ψ(0) > 0, which no decaying e(−κ|x|) can supply — the sign of the jump condition is the entire reason a well binds and a barrier does not.

05

An infinite wall is one condition, not two

An infinite wall is the V₀ → ∞ limit of a step, and taking that limit carefully shows what survives. Beyond a step of height V₀ the decaying solution is ψ ∝ e(−κx) with κ = √(2m(V₀ − E))/ħ, so for an electron κ ≈ √(V₀ / 3.810 eV Ų): 5.12 Å⁻¹ at V₀ = 100 eV and 51.2 Å⁻¹ at 10 keV. The penetration depth 1/κ collapses from 0.195 Å to 0.0195 Å, and in the limit ψ is pinned to zero at the wall. What does not survive is any statement about ψ′. The logarithmic derivative just beyond the wall is −κ, which diverges, so the matching condition degenerates from “ψ′/ψ is continuous” to “ψ = 0”, leaving ψ′ free to jump. That is the whole content of an infinite wall. Imposing ψ(xw) = 0 and ψ′(xw) = 0 together on a second-order ODE forces ψ ≡ 0 by uniqueness: the extra condition does not select a better state, it abolishes every state.

06

Same expression, different operator

Ĥ is not the formula; it is the formula together with a domain, and the choice changes the answer. On [0, L] the expression −(ħ²/2m)d²/dx² is self-adjoint under Dirichlet ψ(0) = ψ(L) = 0, under Neumann ψ′(0) = ψ′(L) = 0, under the Robin family ψ′(0) = γ₀ψ(0) and ψ′(L) = γLψ(L) with γ real, and under periodic conditions — and these are different operators, not different notations. For an electron on L = 10.0 Å, Dirichlet gives Eₙ = 0.376 n² eV, a ground state at 0.376 eV. Neumann admits the constant function and puts its ground state at exactly 0. Mixing them, ψ(0) = 0 with ψ′(L) = 0, gives kL = (n − ½)π and a ground state at 0.0940 eV. Three spectra, one differential expression. The moral is practical as well as formal: a finite-difference Hamiltonian on a box silently imposes Dirichlet at the outermost grid points, so it answers the boxed question — and the levels nearest the top of a well are the ones that notice.

02

Change one variable at a time

Make the relationship visible.

Interactive model
1.6 Å⁻¹
0.40 Å⁻¹

Set the well strength, then drag κ until the filled and open markers stack up — that happens at κ = mα/ħ², exactly half the strength, and the trial-energy readout is then the true bound state. Halve the strength and the binding falls by four, not two, because E goes as κ².

Interactive physics modelThe trial function ψ(x) = e^(−κ|x|) drawn across an attractive δ well at x = 0; the two short tangents at the cusp show the kink it delivers. Below is a number line for Δψ′(0): the filled marker is what the trial delivers, −2κ = −0.80 Å⁻¹, and the open marker what the delta demands, −2mα/ħ² = −1.60 Å⁻¹. They coincide only at the bound-state energy.ψ(x) = e(−κ|x|), κ = 0.40 Å⁻¹V(x) = −α δ(x), 2mα/ħ² = 1.60 Å⁻¹E = −0.61 eVx from −4 Å to +4 ÅΔψ′(0) / Å⁻¹−3.50● delivered −2κ ○ demanded −2mα/ħ²

WELL DEPTH α6.10 eV Å

TRIAL E = −ħ²κ²/2m-0.61 eV

KINK DELIVERED −2κ-0.80 Å⁻¹

KINK DEMANDED −2mα/ħ²-1.60 Å⁻¹

Live interpretationWELL DEPTH α: 6.10 eV Å. TRIAL E = −ħ²κ²/2m: −0.61 eV. KINK DELIVERED −2κ: −0.80 Å⁻¹. KINK DEMANDED −2mα/ħ²: −1.60 Å⁻¹

03

Catch the common trap

Explain before calculating.

Ĥ = −(ħ²/2m) d²/dx² + V has an infinite wall at x = 0 and an attractive δ well V = −α δ(x − d) at some d > 0. Which package of conditions is the domain that selects the bound-state energies?

Choose an answer to test the model.

04

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron is bound by V(x) = −α δ(x) with α = 7.62 eV Å. Take ħ²/2m = 3.810 eV Ų. Write down the admissible bound state, find κ and E, and give the probability of finding the electron within 1.00 Å of the origin.
  1. Away from x = 0 the equation is −(ħ²/2m)ψ″ = Eψ with E < 0, so ψ″ = κ²ψ with κ = √(−2mE)/ħ. Square-integrability kills the growing exponential on each side, leaving ψ(x) = A e(−κ|x|), continuous at the origin as any δ potential requires.
  2. The kink the trial function delivers: ψ′(0⁺) = −κA and ψ′(0⁻) = +κA, so ψ′(0⁺) − ψ′(0⁻) = −2κA.
  3. The kink the potential demands: integrating the equation across x = 0 gives ψ′(0⁺) − ψ′(0⁻) = −(2mα/ħ²)ψ(0) = −(7.62 eV Å / 3.810 eV Ų)A = −2.000 A Å⁻¹.
  4. Match them: 2κ = 2.000 Å⁻¹, so κ = 1.000 Å⁻¹ and E = −(ħ²/2m)κ² = −3.810 eV. One linear equation, one root — the delta well holds exactly one bound state whatever α is.
  5. Normalise: ∫|A|² e(−2κ|x|) dx = |A|²/κ = 1, so A = √κ = 1.000 Å(−1/2). Then P(|x| < 1.00 Å) = ∫ from −1 to 1 of κe(−2κ|x|) dx = 1 − e(−2κ⋅1.00 Å) = 1 − e(−2) = 0.865.

Answerκ = 1.000 Å⁻¹, E = −3.810 eV, ψ(x) = (1.000 Å(−1/2)) e(−|x| / 1.00 Å), and P(|x| < 1.00 Å) = 0.865.

MediumAn electron meets an infinite wall at x = 0 and a well of depth V₀ = 10.0 eV over 0 < x < a with a = 1.50 Å, with V = 0 beyond. Show that exactly one bound state exists, then find its energy. Use ħ²/2m = 3.810 eV Ų.
  1. Name the conditions before solving. At the wall, ψ(0) = 0 and ψ′(0) is free. At x = a the potential is bounded, so ψ and ψ′ are both continuous. As x → ∞, square-integrability forbids the growing exponential.
  2. Inside, k = √((E + V₀)/(ħ²/2m)) and ψ = A sin(kx) — the cosine is excluded by ψ(0) = 0. Outside, κ = √(−E/(ħ²/2m)) and ψ = B e(−κx). Note k² + κ² = V₀/(ħ²/2m) = 10.0/3.810 = 2.6247 Å⁻², a circle of radius R = 1.6201 Å⁻¹.
  3. Divide the ψ′ match at x = a by the ψ match: the amplitudes cancel and the log-derivative condition is k cot(ka) = −κ. A root needs ka > π/2, so it needs Ra = 1.6201 × 1.50 = 2.430 > 1.571 — true, so one state exists. A second would need Ra > 3π/2 = 4.712, so there is no second.
  4. Solve on the circle with u = ka and κa = √((Ra)² − u²), where (Ra)² = 5.906. The function u cot u + √(5.906 − u²) is +0.064 at u = 2.086 and −0.402 at u = 2.172; bisecting gives u = 2.0990.
  5. Unpack: k = 2.0990/1.50 = 1.3993 Å⁻¹, κ = √(5.906 − 4.406)/1.50 = 0.8165 Å⁻¹, and E = −3.810 × 0.8165² = −2.540 eV. Cross-check from inside: E + V₀ = 3.810 × 1.3993² = 7.460 eV, and −2.540 + 10.0 = 7.460 ✓.

AnswerExactly one bound state, at E = −2.540 eV: bound by 2.54 eV relative to the outside, sitting 7.46 eV above the well floor, with a tail decaying over 1/κ = 1.22 Å.

HardAdd an infinite wall at x = −a to the δ well V = −α δ(x) of the first example, keeping α = 7.62 eV Å so that κ₀ = mα/ħ² = 1.000 Å⁻¹ is the free-space answer. Find the transcendental condition on κ, the smallest a that still supports a bound state, and the energy at a = 2.00 Å.
  1. Domain first: ψ(−a) = 0 at the wall, ψ continuous at 0 with ψ′(0⁺) − ψ′(0⁻) = −2κ₀ψ(0), and ψ ∈ L² as x → ∞. So ψ = A sinh(κ(x + a)) on (−a, 0), the combination that vanishes at the wall, and ψ = B e(−κx) for x > 0, with E = −(ħ²/2m)κ².
  2. Continuity at x = 0 gives B = A sinh(κa). The jump condition gives −κB − κA cosh(κa) = −2κ₀ A sinh(κa).
  3. Substitute B and divide by A: κ[sinh(κa) + cosh(κa)] = 2κ₀ sinh(κa), that is κ e(κa) = κ₀(e(κa) − e(−κa)), which tidies to κ = κ₀(1 − e(−2κa)).
  4. Existence: the right-hand side leaves the origin with slope 2aκ₀ and is concave, so it meets the line κ away from zero only if 2aκ₀ > 1. A bound state therefore needs a > 1/(2κ₀) = ħ²/(2mα) = 0.500 Å. At a = 0.400 Å there is none — the wall has squeezed the state out of existence.
  5. At a = 2.00 Å iterate κ ← κ₀(1 − e(−4κ)) starting from κ = 1.000: 0.98168 → 0.98029 → 0.98018 → 0.98018 Å⁻¹.
  6. So E = −3.810 × 0.98018² = −3.660 eV, above the free-space −3.810 eV. Confining the tail costs 0.150 eV of binding, and that cost is set entirely by a boundary condition a distance a away from the well itself.

Answerκ = κ₀(1 − e(−2κa)); a bound state survives only for a > ħ²/(2mα) = 0.500 Å; at a = 2.00 Å, κ = 0.9802 Å⁻¹ and E = −3.660 eV.