University Physics V · The Schrödinger Equation · 4.7
Boundary Conditions & the Discrete Spectrum
Anyone can solve −(ħ²/2m)ψ″ + Vψ = Eψ. The harder skill is knowing why almost none of its solutions are states. Here is where discreteness actually enters, how to find the levels by shooting in from both ends and hunting a sign change, and which energies the argument cannot reach.
Build the model
Connect the measurement to the mechanism.
A second-order linear ODE has a two-dimensional solution space at every energy, real or complex, so −(ħ²/2m)ψ″ + Vψ = Eψ rejects no value of E. What rejects it is membership of the Hilbert space. Fix E below the asymptotic value V∞, and far from the well the equation reads ψ″ = κ²ψ with one decaying and one growing branch at each end; demanding decay as x → −∞ picks out a single ray ψL, and demanding it as x → +∞ picks out ψR.
Two lines through the origin of a two-dimensional space generically miss each other, and their failure to coincide is measured by the Wronskian W(E) = ψL ψR′ − ψL′ ψR, which is independent of x and analytic in E. Eigenvalues are the zeros of W, and the zeros of an entire function that is not identically zero are isolated — that, and not any picture of waves fitting between walls, is why the bound spectrum is discrete. The criterion is energetic rather than geometric: the argument needs both asymptotic regions classically forbidden, so it runs only for min V < E < V∞.
Above V∞ both branches oscillate, nothing is rejected, every E survives, and the spectrum is continuous. The cost is that eigenvalues are now roots of a transcendental function; only in the hard-wall limit, where the decay length 1/κ collapses to zero and the condition degenerates to sin(kL) = 0, do they come out in closed form.
- Simple definition
- Quantisation is the statement that only isolated energies admit a solution of Hψ = Eψ that is square-integrable and satisfies the matching conditions defining the domain of Ĥ; every other energy still solves the differential equation, just not inside the Hilbert space.
- Example
- For an electron in an infinite well of width L = 0.50 nm, sin(kL) = 0 selects kₙ = nπ/L, so E₁ = 1.50 eV and E₂ = 6.02 eV. At E = 3.00 eV the ODE still has a two-dimensional solution space; nothing in it vanishes at both walls.
Fixes the shape but not the spectrum: ψ bends toward the axis where E > V, away from it where E < V, at every E.
m in kg, V and E in J, ħ = 1.055 × 10⁻³⁴ J s; the ratio has units m⁻²
An electron 1.00 eV below V∞ has κ = 5.12 × 10⁹ m⁻¹, a decay length of 0.195 nm.
V∞ = lim V(x) as |x| → ∞; κ in m⁻¹, real only while E < V∞
E is an eigenvalue exactly when W(E) = 0, and an entire function has isolated zeros — the discreteness in one line.
ψL decays as x → −∞, ψR as x → +∞; W is independent of x and entire in E
The one case where the matching condition is algebraic, because the exterior amplitude is exactly zero.
n = 1, 2, 3, …; L in m, E in J. Electron with L = 1.00 nm: E₁ = 0.376 eV
Replaces sin(kL) = 0 by transcendental roots: the levels are no longer spaced as n².
k = √(2m(E + V₀))/ħ and κ = √(−2mE)/ħ in m⁻¹, with k² + κ² = 2mV₀/ħ²
Counts the levels before any root is found, and shows N ≥ 1 always — a 1D well never fails to bind.
z₀ dimensionless; a in m, V₀ in J. Electron, a = 0.50 nm, V₀ = 5.0 eV: z₀ = 5.73
The differential equation rejects no energy
Write the eigenvalue problem in the position basis: ψ″(x) = [2m(V(x) − E)/ħ²] ψ(x). It is a second-order linear ODE, so for every complex E its solution set is a two-dimensional vector space — fix ψ(x₀) and ψ′(x₀) at one point and everything else follows. Nothing in that statement singles out any E. A free particle makes it sharp: for every E > 0 both e(ikx) and e(−ikx) solve it, and for every E < 0 both e(κx) and e(−κx) do. Quantisation is therefore never a property of the differential equation. It is a property of the operator, and an operator is a formula plus a domain: Ĥ acts on the square-integrable functions obeying the stated matching conditions, and E is an eigenvalue only if the two-dimensional solution space meets that domain in something other than the zero vector.
Decay at each end picks one ray out of two
Let V(x) → V∞ at both ends and fix E < V∞. Far out the equation becomes ψ″ = κ²ψ with κ = √(2m(V∞ − E))/ħ, whose branches are e(+κx) and e(−κx). Square-integrability as x → −∞ admits only e(+κx), fixing ψ up to scale: call that one-dimensional set of solutions ψL. The same demand as x → +∞ leaves ψR. Now count. The solution space is two-dimensional and we have chosen two lines through its origin; generically they are different lines, and the state fails to exist not because the ODE broke down but because the solution decaying on the left grows on the right. Their coincidence is measured by the Wronskian W(E) = ψL ψR′ − ψL′ ψR, constant in x because the equation carries no first-derivative term, and analytic in E. E is an eigenvalue exactly when W(E) = 0, and an entire function that is not identically zero has isolated zeros. That is the proof of discreteness, and it never mentions a wall.
Shooting: the divergent tail is the root finder
Turn the Wronskian into an algorithm. Pick a matching point xₘ inside the well, integrate ψL rightwards from deep in the left forbidden region starting at ψ = e(κx), integrate ψR leftwards from the right, and form the mismatch Δ(E) = ψL′/ψL − ψR′/ψR at xₘ — a log-derivative difference, which is W(E) divided by ψL ψR and so shares its zeros. Δ changes sign as E crosses a level, so brentq on a bracketing pair converges in a dozen evaluations. Two practical points. Integrate into the forbidden region, never out of it: outward integration lets the growing branch swamp the decaying one, and a start accurate to 10⁻¹² is dominated by round-off after twenty decay lengths. And count nodes as you sweep E upward — the nth bound state carries n − 1 interior nodes, so the node count names the root you have found and reveals any you skipped.
The window is energetic, not geometric
That argument needed both asymptotic regions classically forbidden, and this sets the window exactly. For E > V∞ the asymptotic solutions are e(±ikx): both bounded, neither square-integrable, neither rejected, so every such E survives. The spectrum there is continuous and its kets are δ-normalised rather than members of L². For E < min V the sign flips everywhere: ψ″/ψ = 2m(V − E)/ħ² > 0 at every x, so ψ is convex away from the axis throughout and can never bend back to zero at both ends — the ground state must sit above the bottom of the well. Only min V < E < V∞ admits discrete eigenvalues. Two consequences the wall picture hides. A finite well with no walls at all, open to infinity on both sides, still holds bound states, because E < V∞ = 0 makes the outside forbidden. And an infinite well has V∞ = ∞, which is why every one of its states is discrete and it has no scattering sector at all.
Counting the levels before solving for them
For a symmetric well of half-width a and depth V₀ the matching conditions split by parity: k tan(ka) = κ for even states, −k cot(ka) = κ for odd ones, both subject to k² + κ² = 2mV₀/ħ². Write z = ka and z₀ = (a/ħ)√(2mV₀); every root then satisfies z tan z = √(z₀² − z²) or −z cot z = √(z₀² − z²) with 0 < z < z₀. The tangent branches start at z = 0, π, 2π, … and the cotangent branches at π/2, 3π/2, …, so a new root appears each time z₀ clears a multiple of π/2 and the count is N = ⌈2z₀/π⌉. Depth and width enter only through z₀, so quadrupling V₀ while halving a leaves N unchanged. Because z₀ > 0 always clears the first branch, a one-dimensional well binds at least one state however shallow — a statement that fails in three dimensions, where a shallow well binds nothing. An electron in a well of half-width 0.50 nm and depth 5.0 eV has z₀ = 5.73, so N = ⌈3.65⌉ = 4.
The infinite well is the degenerate case
Now send V₀ → ∞ at fixed width. The decay constant κ = √(2m(V∞ − E))/ħ diverges, the penetration depth 1/κ collapses to zero, and k tan(ka) = κ degenerates: tan(ka) → ∞, so ka → (2n − 1)π/2, and combining both parities the condition on the full width L = 2a is simply sin(kL) = 0. It is the same matching condition with the exterior amplitude driven to zero, and it is the only case whose roots are available in closed form: kₙ = nπ/L and Eₙ = n²π²ħ²/(2mL²) = n²h²/(8mL²). For an electron with L = 1.00 nm, E₁ = 0.376 eV and the levels sit at 1, 4, 9, 16 times it. The closed form is bought with a lie about the wall: continuity of ψ′ is abandoned there because no finite potential is present to enforce it, and a deep but finite well leaks over a decay length 1/κ of 0.195 nm for an electron 1.0 eV below the top.
Change one variable at a time
Make the relationship visible.
Leave z₀ at 2.66 and drag the trial energy: at −E/V₀ = 0.820, and again at 0.325, the right-hand tail folds back onto the axis and B passes through zero. Those two energies are the entire discrete spectrum; anywhere between them the tail leaves the frame and the state is not in the Hilbert space.
WAVENUMBER z = ka1.682
DECAY w = κa2.060
MISMATCH B-1.020
BOUND STATES2
Live interpretationWAVENUMBER z = ka: 1.682. DECAY w = κa: 2.060. MISMATCH B: −1.020. BOUND STATES: 2
Catch the common trap
Explain before calculating.
An electron moves in one dimension with V = −5.0 eV for |x| < 0.50 nm and V → 0 as |x| → ∞. Which description of the spectrum of Ĥ is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is confined to an infinite square well of width L = 0.50 nm: V = 0 for 0 < x < L and V = ∞ outside. Show where the quantisation condition comes from, and evaluate E₁ and E₂.
- Outside the walls κ = √(2m(V − E))/ħ → ∞, so the penetration depth 1/κ collapses to zero. The domain of Ĥ is the twice-differentiable square-integrable functions with ψ(0) = ψ(L) = 0. Continuity of ψ′ is given up at each wall, because no finite potential sits there to enforce it.
- Inside, ψ″ = −k²ψ with k = √(2mE)/ħ, so the general solution is ψ = A sin kx + B cos kx — two free constants, for every E > 0. The differential equation has quantised nothing yet.
- The left condition ψ(0) = 0 kills the cosine, so B = 0. The right condition needs A sin kL = 0, and A = 0 returns the zero vector, which is not a state. So sin kL = 0, giving kₙ = nπ/L with n = 1, 2, 3, … — n = 0 gives ψ ≡ 0, and −n repeats the same ray.
- Eₙ = ħ²kₙ²/(2m) = n²h²/(8mL²). With L = 0.50 nm, h²/(8mₑL²) = (6.626 × 10⁻³⁴)²/(8 × 9.109 × 10⁻³¹ × 2.5 × 10⁻¹⁹) = 2.410 × 10⁻¹⁹ J = 1.504 eV.
Answerkₙ = nπ/L, so E₁ = 1.50 eV and E₂ = 6.02 eV. The second boundary condition, not the differential equation, is what cut a continuum of k down to a countable set.
MediumAn electron sits in a finite well of half-width a = 0.30 nm and depth V₀ = 3.0 eV, with V → 0 far away. How many bound states does it hold? Then say what kind of solution, if any, exists at E = +1.0 eV and at E = −4.0 eV.
- Depth parameter: z₀ = (a/ħ)√(2mV₀). Here √(2mₑ × 3.0 eV) = √(1.822 × 10⁻³⁰ × 4.806 × 10⁻¹⁹) = 9.358 × 10⁻²⁵ kg m s⁻¹, and a/ħ = 3.0 × 10⁻¹⁰ ÷ 1.0546 × 10⁻³⁴ = 2.845 × 10²⁴, so z₀ = 2.662.
- A new root appears every π/2 in z₀, so N = ⌈2z₀/π⌉ = ⌈1.695⌉ = 2: one even ground state with no node and one odd state with a single node. N ≥ 1 for any V₀ > 0, so this well could not have failed to bind.
- E = +1.0 eV is above V∞ = 0. Both asymptotic branches are e(±ikx), bounded but not square-integrable, so neither is rejected and every such E is allowed: this energy belongs to the continuous spectrum, and its solution is a δ-normalised scattering state, not a bound state.
- E = −4.0 eV is below min V = −3.0 eV. Then 2m(V − E)/ħ² > 0 at every x, so ψ″ and ψ share a sign everywhere and ψ bends away from the axis throughout. A function convex away from the axis at every point cannot decay at both ends, so there is no solution of any kind.
AnswerTwo bound states. E = +1.0 eV lies in the continuous spectrum as a scattering state; E = −4.0 eV is below the well floor and admits no solution at all.
HardFor that same well — a = 0.30 nm, V₀ = 3.0 eV, z₀ = 2.662 — the even-parity matching condition at x = a is k tan(ka) = κ. Write it in the dimensionless variable z = ka, bracket the root, and get the ground-state energy.
- Put z = ka and w = κa. Then z² + w² = 2mV₀a²/ħ² = z₀² = 7.087, so w = √(z₀² − z²) and the even condition becomes z tan z = √(z₀² − z²), with exactly one root in (0, π/2).
- Bracket it. At z = 1.10: LHS = 1.10 tan 1.10 = 2.161 against RHS = √(7.087 − 1.210) = 2.424, so LHS − RHS < 0. At z = 1.20: LHS = 3.087 against RHS = √(7.087 − 1.440) = 2.376, so LHS − RHS > 0. The root is caught between.
- Bisect. z = 1.15 gives LHS = 2.570 > RHS = 2.401. z = 1.13 gives LHS = 2.395 < RHS = 2.410. z = 1.14 gives LHS = 2.480 > RHS = 2.406. So z₁ = 1.132 to four figures.
- Convert. ħ²/(2mₑₐ²) = (1.0546 × 10⁻³⁴)² ÷ (2 × 9.109 × 10⁻³¹ × 9.0 × 10⁻²⁰) = 6.783 × 10⁻²⁰ J = 0.4233 eV, and E + V₀ = z₁² × 0.4233 eV = 1.2808 × 0.4233 = 0.542 eV.
- So E₁ = 0.542 − 3.0 = −2.458 eV: 0.542 eV above the floor and 2.458 eV below the continuum edge. The odd condition −z cot z = √(z₀² − z²) gives z₂ = 2.181 and E₂ = −0.986 eV, and z₀ = 2.662 < π forbids a third root, so the count N = 2 checks.
Answerz₁ = 1.132 and E₁ = −2.458 eV, with E₂ = −0.986 eV above it. Two levels are the entire discrete spectrum of this well; every E > 0 belongs to its continuum.