Continuity Conditions at Potential Steps

Derivation

Continuity Conditions at Potential Steps

D-145 Home PU-202 Threads waves · matter Depends on Separation of Variables and Stationary States

Statement

Across any point x = a where the potential V(x) has a finite jump discontinuity, a solution of the one-dimensional time-independent Schrödinger equation and its first derivative are both continuous: ψ(a⁺) = ψ(a⁻) and ψ′(a⁺) = ψ′(a⁻). These are not extra postulates but consequences of the equation itself, valid whenever V stays bounded.

Why it matters

Almost every solvable quantum problem — the potential step, the finite square well, the rectangular barrier, tunnelling — is built by solving the Schrödinger equation separately in regions of constant or smooth potential and then stitching the pieces together. The matching conditions are the glue: they convert a piecewise collection of general solutions into a single physical wavefunction and fix the otherwise free amplitude and phase constants.

They are also what quantises energy. In a bound problem the demand that ψ and ψ′ match at every interface, together with normalisability, can be satisfied only for a discrete set of energies. Understanding why the derivative is continuous (and when it is not) is therefore the difference between applying a recipe and knowing what the recipe encodes.

Assumptions

The potential V(x) is bounded near the interface (finite jump).If V is unbounded — an infinite wall or a Dirac delta — the integral driving the argument no longer vanishes and ψ′ may jump or ψ may be forced to zero. The wavefunction ψ(x) is itself bounded in a neighbourhood of the interface.An unbounded ψ would make the source integral diverge as the interval shrinks, breaking the estimate; physically we discard such non-normalisable local behaviour. The energy E is finite and the state is a genuine (weak) solution of the second-order equation on an interval containing a.If one only imposes the equation away from a and treats the interface as an independent boundary, the matching conditions become inputs rather than derived facts, and consistency with the equation is lost. The problem is non-relativistic (Schrödinger, second order in x).For the Dirac equation, first order in space, the correct junction condition is continuity of the spinor components alone; imposing derivative continuity would over-constrain the solution.

Derivation

−(ℏ²/2m) ψ″(x) + V(x) ψ(x) = E ψ(x)

The time-independent Schrödinger equation, taken as the prior result governing the stationary state on an interval containing the interface at x = a. A

ψ″(x) = (2m/ℏ²) [ V(x) − E ] ψ(x)

Pure algebraic rearrangement to isolate the second derivative. This exposes ψ″ as a bounded quantity times ψ whenever V is bounded. A

∫_a−ε^a+ε ψ″(x) dx = ψ′(a+ε) − ψ′(a−ε)

Integrate both sides over a symmetric window straddling the interface; the left side collapses by the fundamental theorem of calculus, since ψ′ is an antiderivative of ψ″. B

ψ′(a+ε) − ψ′(a−ε) = (2m/ℏ²) ∫_a−ε^a+ε [ V(x) − E ] ψ(x) dx

Equate to the integral of the right-hand side of step 2. The jump in the derivative is thus entirely controlled by the area under the source term. B

| ψ′(a+ε) − ψ′(a−ε) | ≤ (2m/ℏ²) ( V_max + |E| ) ψ_max · (2ε)

Bound the integral by the supremum of the integrand times the interval length 2ε. Legal because, by assumption, V is bounded (V_max < ∞) and ψ is bounded (ψ_max < ∞) on the window. C

lim_ε→0 [ ψ′(a+ε) − ψ′(a−ε) ] = 0 ⇒ ψ′(a⁺) = ψ′(a⁻)

The bound in step 5 is proportional to ε and vanishes as the window shrinks. A finite jump in V carries zero area in the limit, so the one-sided derivatives coincide: ψ′ is continuous. C

ψ(a+ε) − ψ(a−ε) = ∫_a−ε^a+ε ψ′(x) dx → 0 ⇒ ψ(a⁺) = ψ(a⁻)

By step 6 the derivative ψ′ is now bounded across the interface, so its integral over a window of width 2ε is bounded by ψ′_max·2ε and vanishes as ε→0. Hence ψ itself is continuous. C

Result

ψ(a⁺) = ψ(a⁻) and ψ′(a⁺) = ψ′(a⁻) (finite V)

Reading. Wherever the potential is finite, the wavefunction is smooth to first order: no jumps and no kinks. A finite step in V produces only a discontinuity in the second derivative, i.e. a change of curvature, while the value and slope pass through unbroken. Physically, continuity of ψ keeps the probability density single-valued, and continuity of ψ′ keeps the probability current j = (ℏ/m) Im(ψ* ψ′) conserved across the interface.

Units check. In step 2, ψ″ has units [ψ]·m⁻². The right side: (2m/ℏ²)(V−E)ψ has units kg·J⁻²s⁻² × J × [ψ] = kg·J⁻¹s⁻²[ψ]. With J = kg·m²·s⁻², this is kg·(kg·m²·s⁻²)⁻¹s⁻²[ψ] = m⁻²[ψ]. Both sides carry [ψ]·m⁻². ✓

Limiting cases

Smooth potential (V continuous): both conditions hold automatically at every point; ψ is in fact twice continuously differentiable everywhere.
Finite step (V jumps by a finite amount): ψ and ψ′ continuous; ψ″ jumps — the curvature changes sign or magnitude at the edge.
High-barrier limit (V₀ → ∞ with the wall kept at finite location): the interior solution is squeezed toward zero at the wall, and derivative continuity degenerates into the single condition ψ = 0 there.
Zero-width window (ε → 0 first, before any limit on V): recovers exactly the pointwise matching statement used in every piecewise problem.

Breaks when

Infinite potential wall. Where V → ∞ the estimate in step 5 diverges; the solution is pushed to ψ = 0 at the wall and its derivative need not be continuous there (a kink is allowed just outside an impenetrable boundary).
Dirac-delta potential V(x) = α δ(x−a). The source integral in step 4 keeps a finite value (2mα/ℏ²)ψ(a) as ε→0, so ψ stays continuous but ψ′ jumps by exactly that amount.
Relativistic (Dirac) dynamics. The equation is first order in x, so only the wavefunction components are continuous; imposing continuity of the derivative would over-determine the junction and is physically wrong.
Singular or unbounded ψ. If the assumed local solution blows up at the interface (a rejected, non-normalisable branch), the boundedness used in steps 5–7 fails and neither condition can be asserted.

Failure modes

Matching |ψ|² instead of ψ. Students sometimes equate probability densities across the boundary; the correct conditions are on the complex amplitude ψ and on ψ′, not on |ψ|².
Dropping the derivative condition at a finite step. Using only ψ-continuity leaves the amplitude ratios undetermined and typically gives the wrong reflection coefficient.
Imposing derivative continuity at an infinite wall. At a hard wall only ψ = 0 is legitimate; forcing ψ′ to match yields no non-trivial solution.
Forgetting the delta jump. For a delta potential, writing ψ′(a⁺) = ψ′(a⁻) deletes the very interaction and gives a free particle instead of a bound state.
Sign/branch errors in κ. In a forbidden region keeping the growing exponential e^+κx instead of the decaying one violates boundedness and corrupts the matching.
Matching in the wrong variable. With a position-dependent mass m(x) the correct current-conserving junction is on (1/m)ψ′, not ψ′ itself; using plain ψ′ breaks continuity of j.

Discussion

The deep content of this result is that the smoothness of a quantum state is dictated by the smoothness of the potential, and precisely two orders less. The Schrödinger equation is second order, so the highest derivative appearing, ψ″, mirrors whatever discontinuity V has: a finite jump in V means a finite jump in ψ″, which is a change in curvature but leaves ψ and ψ′ intact. Integrating the equation once transfers a discontinuity in V down to a discontinuity in the second derivative; the two integrations in steps 6 and 7 are exactly the two orders of smoothing.

Continuity of ψ′ is really a conservation law in disguise. The probability current j = (ℏ/m) Im(ψ* ψ′) must be the same on both sides of a static interface in a stationary state, otherwise probability would accumulate or drain at a point of measure zero. Continuity of both ψ and ψ′ guarantees continuity of j. This is why the correct junction condition generalises, for a spatially varying effective mass, to continuity of (1/m)ψ′ — the combination that keeps the current single-valued.

The cleanest way to see the whole hierarchy is distributional. Treat the Schrödinger equation as an equation between distributions on an interval. If V is an ordinary bounded function, the product V ψ is locally integrable, so ψ″ is locally integrable and hence ψ belongs to the Sobolev space H² ⊂ C¹: a member of H² automatically has a continuous representative for both ψ and ψ′. Push a delta into V and the equation forces a delta into ψ″, whose antiderivative is a step — precisely the jump in ψ′. The junction conditions are thus not boundary lore bolted onto the theory but the pointwise face of the regularity theorem ψ ∈ H².

Common misconceptions. A finite potential step does not make the wavefunction “bend sharply” with a kink; kinks (slope discontinuities) require an infinite or delta-like potential. And continuity of ψ′ is not an independent axiom of quantum mechanics — it is a theorem that follows from the equation, which is exactly why it fails in the singular cases where the derivation’s assumptions collapse.

Worked examples

Step down, over-barrier. Electron incident from the left on V = 0 (x<0), V = V₀ = 1.00 eV (x>0), energy E = 2.00 eV. Find the reflection coefficient by matching.

Set up piecewise solutions. Left: ψ_L = e^ikx + r e^−ikx. Right: ψ_R = t e^iqx, propagating since E > V₀. B

ψ continuous at 0: 1 + r = t | ψ′ continuous: ik(1 − r) = iq t

Apply the two matching conditions at the interface. Solving: r = (k − q)/(k + q), t = 2k/(k + q). A

k = √(2mE)/ℏ, q = √(2m(E−V₀))/ℏ

Insert numbers: m = 9.109×10⁻³¹ kg, ℏ = 1.0546×10⁻³⁴ J·s, E = 3.204×10⁻¹⁹ J, E−V₀ = 1.602×10⁻¹⁹ J. Gives k = 7.24×10⁹ m⁻¹, q = 5.12×10⁹ m⁻¹. B

r = (7.24 − 5.12)/(7.24 + 5.12) = 0.172, R = r² = 0.0294

Reflection probability. Cross-check with transmission T = (q/k)|t|² = (5.12/7.24)(1.172)² = 0.971. B

R ≈ 0.029, T ≈ 0.971, R + T = 1.000

Reading. Even with more than enough energy to pass, the electron reflects with ~3% probability purely because the potential changes abruptly — a wave phenomenon. Current conservation R+T=1 confirms the matching was done consistently.

Units check. k, q in m⁻¹; r, t, R, T dimensionless. ✓

Step up, under-barrier (total reflection). Same electron, now E = 1.00 eV incident on a step to V₀ = 2.00 eV. Find r and the penetration depth.

Right region is classically forbidden, so use a decaying solution ψ_R = t e^−κx with κ = √(2m(V₀−E))/ℏ; the growing branch is rejected by boundedness. B

1 + r = t, ik(1 − r) = −κ t ⇒ r = (ik + κ)/(ik − κ)

Same two matching conditions, with iq replaced by −κ. Symbolic result before numbers. A

k = √(2m·1.602×10⁻¹⁹)/ℏ = 5.12×10⁹ m⁻¹, κ = √(2m·1.602×10⁻¹⁹)/ℏ = 5.12×10⁹ m⁻¹

Here V₀−E = E = 1.00 eV, so κ = k numerically. B

r = (i + 1)/(i − 1) = −i, |r|² = 1; penetration depth 1/κ = 1.95×10⁻¹⁰ m

Total reflection with a −90° phase shift; the evanescent tail decays over ~0.20 nm into the barrier. B

r = −i, R = 1, 1/κ ≈ 0.20 nm

Reading. Below the step the electron is fully reflected, yet ψ is non-zero just inside the barrier: matching forces a continuous, exponentially decaying tail. The value and slope join smoothly even though no current flows to the right.

Units check. κ in m⁻¹, so 1/κ in m; r dimensionless. ✓

Problems

Show explicitly that a finite step in V leaves ψ′ continuous but makes ψ″ discontinuous, and give the size of the jump in ψ″ at x=a.
Solution
From step 2, ψ″ = (2m/ℏ²)(V−E)ψ. Since ψ is continuous (step 7), the jump in ψ″ is (2m/ℏ²)·[V(a⁺)−V(a⁻)]·ψ(a) = (2m/ℏ²) ΔV ψ(a). It is finite and non-zero whenever ΔV≠0, while steps 6–7 already gave zero jump in ψ′ and ψ. Hence exactly the second derivative carries the discontinuity.
For the delta potential V(x) = −α δ(x) (α > 0), integrate the Schrödinger equation across x=0 and derive the derivative-jump condition. Hence find the single bound-state energy.
Solution
Integrating step 4 with the delta: ψ′(0⁺)−ψ′(0⁻) = (2m/ℏ²)∫(−α δ − E)ψ dx = −(2mα/ℏ²)ψ(0) (the −E term vanishes as ε→0). Take a bound state ψ = √κ e^−κ|x| with E = −ℏ²κ²/2m. Then ψ′(0⁺)−ψ′(0⁻) = −2κψ(0). Equating: −2κ = −2mα/ℏ² ⇒ κ = mα/ℏ², giving E = −mα²/2ℏ². There is exactly one bound state.
An electron of energy E = 5.0 eV meets a step down to V₀ = −3.0 eV (i.e. the potential drops by 8.0 eV). Compute R.
Solution
k = √(2mE)/ℏ with E = 8.01×10⁻¹⁹ J ⇒ k = 1.145×10¹⁰ m⁻¹. Right-region kinetic energy E−V₀ = 8.0 eV = 1.282×10⁻¹⁸ J ⇒ q = √(2m·1.282×10⁻¹⁸)/ℏ = 1.449×10¹⁰ m⁻¹. Then r = (k−q)/(k+q) = (1.145−1.449)/(1.145+1.449) = −0.117, R = r² = 0.014. About 1.4% reflects off the down-step.
At an interface the effective mass changes: m₁ for x<0 and m₂ for x>0, with the same potential. Argue which quantity must be continuous in place of ψ′, and write both matching conditions.
Solution
Probability current j = (ℏ/m)Im(ψ*ψ′) must be continuous for conservation. With m differing across the boundary, plain ψ′ continuity would break j. The current-conserving (BenDaniel–Duke) conditions are: ψ continuous, and (1/m)ψ′ continuous, i.e. ψ(0⁺)=ψ(0⁻) and (1/m₂)ψ′(0⁺) = (1/m₁)ψ′(0⁻). This follows by writing the kinetic term as −(ℏ²/2) d/dx[(1/m(x)) dψ/dx] and integrating across the interface.
A particle sits in an infinite square well of width L (V=0 inside, ∞ outside). Explain why derivative continuity is not imposed at the walls, and find the ground-state energy for an electron with L = 0.50 nm.
Solution
At an infinite wall the boundedness assumption (step 5) fails: V_max→∞, so the derivation does not yield ψ′ continuity. Only ψ continuity survives, which forces ψ = 0 at each wall; the slope is discontinuous there (a kink). Solutions ψ_n = √(2/L) sin(nπx/L), E_n = n²π²ℏ²/(2mL²). For n=1, L = 5.0×10⁻¹⁰ m: E₁ = π²(1.0546×10⁻³⁴)²/(2·9.109×10⁻³¹·(5.0×10⁻¹⁰)²) = 2.41×10⁻¹⁹ J = 1.5 eV.