The requirements on ψ(x, t), and the reason for each
Each condition on ψ is inherited, not assumed.
Students are usually handed four conditions on ψ as if they were house rules. They are not. Each one is forced either by reading |ψ|² as a probability density or by the Schrödinger equation, and knowing which tells you exactly where a rule may bend: at an infinite wall, at a delta well, and for a scattering state that never normalises.
Build the model
Each rule on ψ is inherited from the Born rule or the Schrödinger equation, not assumed.
1The big idea
Two inputs generate every requirement on ψ. The Born rule reads |ψ|² as a probability density, and the time-independent Schrödinger equation, derived in Unit 2 and used here as ψ″ = (2m/ħ²)(V − E)ψ, fixes how ψ may bend. The Born rule supplies two conditions. ψ must be single-valued, because one point carries one probability; and since states superpose, it is ψ itself, not only |ψ|², that must come back to its value, or an interference term would hand one point two densities. And ∫|ψ|² dx must be finite and non-zero, so that one constant can scale the total to 1. The equation supplies the other two. Where V and ψ are finite the right-hand side is finite, so ψ″ is finite and ψ′ cannot jump. A jump in ψ itself would put a delta function into ψ′ and its derivative δ′ into ψ″, which nothing of the form Vψ can match, and it would cost infinite kinetic energy besides. So ψ is continuous everywhere, an energy eigenstate's ψ′ is continuous wherever V is finite, and nothing stronger is promised: at a finite step ψ″ jumps. The rules bend exactly where their inputs fail. An infinite wall or a delta well is not a finite V, so ψ′ may jump there, by an amount the potential sets; and a scattering state such as e(ikx) has |ψ|² = 1 everywhere, so it is not square-integrable and stands in for a normalisable wave packet.
2In one sentence
An admissible wavefunction is single-valued, continuous and square-integrable, and an energy eigenstate also has ψ′ continuous wherever V is finite; the first and third conditions come from reading |ψ|² as a probability density, the second and fourth from the Schrödinger equation.
Example. In an infinite well of width a = 1.00 nm, ψ₁ = √(2/a) sin(πx/a) is continuous and zero at x = 0, but its slope jumps from 0 outside to π√(2/a)/a = 4.44 nm(−3/2) just inside, which is legal only because V is infinite at the wall.
3Single-valued: one point carries one probability
The Born rule makes |ψ(x, t)|² a density, and a density assigns one number to each point, so a ψ with two values at one x is out. The sharper form of the rule appears on a ring, where the angles φ and φ + 2π label the same place. The state e(iφ/2)/√(2π) looks harmless, since its |ψ|² = 1/(2π) everywhere. But states superpose. Add it to the uniform state 1/√(2π) and normalise, and the density becomes (1 + cos(φ/2))/(2π): 1/π = 0.318 rad⁻¹ at φ = 0 and zero at φ = 2π, which is the same point. So the condition is on ψ itself, not only on |ψ|²: ψ(φ + 2π) = ψ(φ). For e(imφ) that forces m to be an integer, and that integer is where Lz = mħ will come from when angular momentum arrives. On a line the condition is usually automatic, which is why it is so easily forgotten.
4Square-integrable: a question about tails and spikes
Normalising needs ∫|ψ|² dx to be finite and non-zero; the constant then does the rest. The integral can fail in only two places, far out in the tails or at a point where ψ blows up, and a power law settles both. If |ψ| falls as |x|(−p), then |ψ|² falls as |x|(−2p), and the integral out to infinity converges only for 2p > 1. So a 1/x tail normalises: ∫ dx/x² from a to L is 1/a − 1/L, which tends to 1/a. A 1/√x tail does not: ∫ dx/x from a to L is ln(L/a), only 20.7 for a = 1 nm and L = 1 m, but unbounded as L grows. Exponential and Gaussian tails, the ones bound states actually have, beat every power. Near a single point the inequality turns round: |x|(−p) is square-integrable there when p < ½, so a ψ that behaves as |x|(−1/4) near x = 0 and decays in its tails normalises although it is infinite at x = 0. Square-integrability alone does not exclude that spike; the next requirement does.
5Continuous: a jump in ψ costs infinite kinetic energy
Continuity does not come from the Born rule. A density with a step, such as a uniform distribution on an interval, is a perfectly good probability distribution. It comes from the Schrödinger equation. Suppose ψ jumps by J at x₀. Then ψ′ contains J δ(x − x₀) and ψ″ contains J δ′(x − x₀), a spike that is huge and positive on one side and huge and negative on the other. The equation needs (2m/ħ²)(V − E)ψ to match it, but a finite V gives a finite right-hand side, and even a delta in V produces a δ multiplied by ψ(x₀), not a δ′; at a jump ψ(x₀) is not even defined. Energy gives the same verdict. Replace the jump by a straight ramp of width w: ψ′ = J/w across it, so the ramp alone contributes (ħ²/2m)J²/w to ⟨T⟩. For an electron with J = 1.0 nm(−1/2) that is 3.81 eV at w = 0.01 nm and 381 eV at w = 10⁻⁴ nm, growing without limit as the ramp sharpens.
6ψ′ continuous where V is finite: integrate across the point
Integrate ψ″ = (2m/ħ²)(V − E)ψ from x₀ − ε to x₀ + ε. The left side gives the change in slope across the interval. The right side is a bounded function integrated over a width 2ε, so it is at most 2ε times a finite number and vanishes as ε → 0. Hence ψ′(x₀⁺) = ψ′(x₀⁻) at every point where V is finite, including the edge of a finite step, where V itself jumps. Nothing stronger is guaranteed. Take an electron with E = 2.0 eV meeting a 4.0 eV step: k² = E/(ħ²/2m) and κ² = (V₀ − E)/(ħ²/2m) are both 52.5 nm⁻², so ψ″ = −52.5ψ(0) nm⁻² just before the step and +52.5ψ(0) nm⁻² just after. The curvature jumps by 105ψ(0) nm⁻², which is (2m/ħ²)V₀ψ(0), while ψ and ψ′ pass through unbroken. At a finite step, then, ψ is smooth to first order and no further, which is why matching at a step always means exactly two equations, one for ψ and one for ψ′.
7Where the rules bend: walls, delta wells and scattering states
Each relaxation sits where an input to the argument fails. At an infinite wall V is not finite, so the slope argument is void. ψ must vanish in the forbidden region, since any ψ there would make ⟨V⟩ infinite; continuity still holds, so ψ = 0 at the wall, and ψ′ may jump from its inside value to 0. A finite step shows the corner forming: for the 2.0 eV electron the tail beyond the step has length 1/κ = 0.138 nm at V₀ = 4 eV, 0.020 nm at 100 eV and 0.0020 nm at 10 keV, squeezing the slope change into ever less room. A delta well, V = −α δ(x − x₀), is infinite at one point and finite elsewhere, so ψ′ jumps there by −(2mα/ħ²)ψ(x₀) and nowhere else. A scattering state such as e(ikx) has |ψ|² = 1 everywhere, so square-integrability fails; it is kept as the limit of a normalisable wave packet and read as a flux, not a probability. Single-valuedness and continuity of ψ never relax.
Key equations
What each relation says, and the units it is written in.
Square-integrability, from the Born rule
∫ |ψ(x, t)|² dx over all x is finite and non-zero
Finite and non-zero is exactly what lets one constant scale the total probability to 1. ψ = 0 passes the first test and fails the second: it describes no particle.
|ψ|² in m⁻¹ in one dimension, so ψ carries m(−1/2)
Power-law tail test
|ψ| ∝ |x|(−p) as |x| → ∞ is square-integrable only if p > ½
A 1/x tail normalises and a 1/√x tail does not. Near a finite point the inequality turns round: |x|(−p) is square-integrable there when p < ½.
p is dimensionless; at p = ½ the integral out to |x| = L grows as ln L, slowly but without bound
Curvature form of the Schrödinger equation
ψ″(x) = (2m/ħ²)(V(x) − E) ψ(x)
Read it as a curvature budget: finite V and finite ψ give a finite ψ″, and every smoothness rule on ψ follows from that one fact.
m in kg, V and E in J, ħ in J s; the factor 2m(V − E)/ħ² carries m⁻²
Slope change across a point x₀
ψ′(x₀ + ε) − ψ′(x₀ − ε) = (2m/ħ²) ∫ (V − E) ψ dx, from x₀ − ε to x₀ + ε
A bounded integrand over a vanishing interval integrates to zero, so ψ′ is continuous. Only an infinite wall or a delta in V can keep the right side non-zero.
ε in m; ψ′ in m(−3/2); the right side vanishes as ε → 0 whenever V stays finite, even across a finite step
Kinetic energy as a slope integral
⟨T⟩ = (ħ²/2m) ∫ |ψ′|² dx
A jump J smoothed into a ramp of width w adds J²/w to the integral, which diverges as w → 0: a discontinuous ψ has infinite kinetic energy.
⟨T⟩ in J; valid for a continuous, normalised ψ that vanishes at ±∞
Corner at a delta well
ψ′(x₀⁺) − ψ′(x₀⁻) = −(2mα/ħ²) ψ(x₀), for V = −α δ(x − x₀)
The size of a corner measures the delta that permits it. For e(−κ|x|) the slope drops by 2κψ(0), which fixes α = ħ²κ/m.
α in J m is the well's strength; δ(x − x₀) carries m⁻¹
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At E = 2 eV, drag V₀ from 4 to 100 eV: the tail shrinks from 0.14 to 0.02 nm and ψ(0) sinks towards zero, yet the tangent still touches both pieces. Only in the limit does the tail vanish, leaving the dashed curve's corner.
Live results
- ψ(0)/A = √(E/V₀)
- 0.707
- SHARED SLOPE ψ′(0)/A
- -5.12 nm⁻¹
- TAIL LENGTH 1/κ
- 0.138 nm
- ψ″ JUMP AT x = 0, PER A
- 74 nm⁻²
Catch the common trap
The mistake this lesson is built to prevent.
The commonest slip is to treat the four conditions as one package that holds everywhere. At an infinite wall a student reasons that ψ′ must be continuous and ψ′ is zero outside, so ψ′ = 0 at the wall. Together with ψ = 0 there, that is two conditions at one point on a second-order equation, and the only solution meeting both is ψ = 0 everywhere: no particle at all. The slope rule was derived for finite V, and at the wall V is not finite, so only ψ = 0 survives there. The mirror-image slip is to justify continuity of ψ by saying probability cannot jump. A probability density can have a step; uniform on an interval is a perfectly good distribution. ψ is continuous because the Schrödinger equation makes a jump cost infinite kinetic energy, not because of the Born rule.
Test it: what changes?
- ψ(0)/A = √(E/V₀) rises
- Tail length 1/κ grows
- Curvature jump at the step grows
- ψ and ψ′ stay continuous
- Tail length 1/κ shrinks towards zero
- ψ(0)/A = √(E/V₀) sinks towards zero
- ψ′ stays continuous while V₀ is finite
- Infinite limit: ψ = 0 at wall, ψ′ may jump
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
A trial state is ψ(x) = A/√(1 + x²/a²) with a = 0.40 nm. Show that it is square-integrable, find A, and find the probability that the particle is within a of the origin. Then explain why the similar-looking ψ = B/(1 + x²/a²)¹⁄⁴ cannot be normalised with any B.
View solution →
- Tail test first. At large |x|, |ψ| ≈ Aa/|x|, a power law with p = 1 > ½, so ∫|ψ|² dx converges. The function is smooth and single-valued, so the tail was the only thing that could fail.
- Normalise: |ψ|² = A²/(1 + x²/a²), and ∫ dx/(1 + x²/a²) over all x is a[arctan(x/a)] from −∞ to ∞ = πa. So A²πa = 1 and A = 1/√(πa) = 1/√(1.257 nm) = 0.892 nm(−1/2). Dimensions: |ψ|² comes out in nm⁻¹, as a one-dimensional density must.
- Probability: P(|x| < a) = A²a[arctan(1) − arctan(−1)] = (1/π)(π/2) = 0.500. It cannot depend on a, because a is the only length in ψ.
- Second candidate: |ψ|² = B²/√(1 + x²/a²) falls as B²a/|x|, the borderline p = ½. Out to ±L the integral is 2aB² asinh(L/a) ≈ 2aB² ln(2L/a): 44.7aB² at L = 1 m and 58.5aB² at L = 1 km. It never stops growing, so no finite B makes the total 1.
Answer. A = 0.892 nm(−1/2) and P(|x| < a) = 0.500. The second candidate sits at the borderline p = ½: its integral grows as ln L without limit, so it is not square-integrable.
MediumExample 2
An electron is held between infinite walls at x = 0 and x = L = 1.00 nm. A flat trial state, constant inside, would jump to zero at each wall. Replace each jump by a straight ramp of width w: ψ = Cx/w for 0 ≤ x ≤ w, ψ = C on the plateau, ψ = C(L − x)/w for L − w ≤ x ≤ L. Normalise, then find ⟨T⟩ = (ħ²/2m)∫|ψ′|² dx for w = 0.10 nm and w = 0.010 nm. Use ħ²/2mₑ = 0.0381 eV nm².
View solution →
- Audit first. The ramps make ψ continuous and put ψ = 0 at each wall, as an infinite wall requires, and ψ′ may jump there from 0 outside to C/w inside because V is not finite at a wall. But ψ′ also jumps, between ±C/w and 0, at x = w and x = L − w, where V = 0 is finite, so this ψ is not an energy eigenstate. A trial state need not be one: it must be single-valued, continuous and square-integrable with a finite ⟨T⟩, and a kink keeps ⟨T⟩ finite where a jump would not.
- Normalise: each ramp gives ∫(Cx/w)² dx = C²w/3 and the plateau gives C²(L − 2w), so C²(L − 4w/3) = 1.
- Slopes: ψ′ = ±C/w on the ramps and 0 on the plateau, so ∫|ψ′|² dx = 2(C²/w²)w = 2C²/w. Then ⟨T⟩ = (ħ²/2m)(2C²/w) = 2(ħ²/2m)/[w(L − 4w/3)]. Dimensions: eV nm² divided by nm² is an energy.
- w = 0.10 nm: w(L − 4w/3) = 0.10 × 0.8667 = 0.08667 nm², so ⟨T⟩ = 0.0762/0.08667 = 0.879 eV. w = 0.010 nm: 0.010 × 0.9867 = 0.009867 nm², so ⟨T⟩ = 7.72 eV.
- Limit: for small w, ⟨T⟩ ≈ 2(ħ²/2m)/(wL), growing as 1/w, so the flat state with genuine jumps has infinite kinetic energy. Check: both results lie above the true ground-state energy π²(ħ²/2m)/L² = 0.376 eV, as every state in this well must; the best ramp, w = 3L/8, gives 0.406 eV, only 8% above it.
Answer. ⟨T⟩ = 0.879 eV for w = 0.10 nm and 7.72 eV for w = 0.010 nm, rising as 1/w; the discontinuous limit costs infinite kinetic energy. The kinks at x = w and x = L − w stop this ψ from being an energy eigenstate, but they leave ⟨T⟩ finite.
HardExample 3
An electron's proposed bound state is ψ(x) = A e(−x/b₁) for x ≥ 0 and A e(x/b₂) for x ≤ 0, with b₁ = 0.10 nm and b₂ = 0.25 nm. Audit it against the four requirements, normalise it, find P(x > 0), and work out what V(x) must be for this to be an energy eigenstate. Take V = 0 far to the left and ħ²/2mₑ = 0.0381 eV nm².
View solution →
- Audit: single-valued; continuous, since both pieces equal A at x = 0; square-integrable, since both tails are exponential. But ψ′(0⁺) = −A/b₁ and ψ′(0⁻) = +A/b₂, so the slope drops by A(1/b₁ + 1/b₂) at x = 0. With finite V that is forbidden, so V must contain a delta at the origin.
- Normalise: ∫|ψ|² dx = A²(b₁/2 + b₂/2) = 1, so A = √(2/(b₁ + b₂)) = √(2/0.35 nm) = 2.39 nm(−1/2). Then P(x > 0) = A²b₁/2 = b₁/(b₁ + b₂) = 0.286.
- Energy from the left tail, where V = 0: ψ″ = ψ/b₂² = (2m/ħ²)(0 − E)ψ, so E = −(ħ²/2m)/b₂² = −0.0381/0.0625 = −0.610 eV.
- Right tail: ψ″ = ψ/b₁² = (2m/ħ²)(V₀ − E)ψ, so V₀ − E = 0.0381/0.0100 = 3.81 eV and V₀ = 3.20 eV. The two decay lengths differ, so V must also step up by 3.20 eV at x = 0.
- Corner: ψ′(0⁺) − ψ′(0⁻) = −A(1/b₁ + 1/b₂) must equal −(2mα/ħ²)A, so α = (ħ²/2m)(1/b₁ + 1/b₂) = 0.0381 × 14.0 = 0.533 eV nm. That is an attractive delta well sitting on top of the step.
- Checks: α in eV nm is an energy times a length, as a delta strength must be. The limit b₁ = b₂ = b gives V₀ = 0, α = ħ²/(mb) and E = −ħ²/(2mb²); that is the plain delta well. And E lies below V on both sides, so a classical electron could be in neither region: only the delta binds it.
Answer. A = 2.39 nm(−1/2), P(x > 0) = 0.286, E = −0.610 eV. V(x) = 0 for x < 0 and 3.20 eV for x > 0, plus an attractive delta −(0.533 eV nm) δ(x) at the step.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
ψ is always continuous; an eigenstate's ψ′ wherever V is finite.
Single-valuedness and square-integrability come from the Born rule; continuity of ψ and, for an energy eigenstate, of ψ′ come from the Schrödinger equation. At an infinite wall only the condition ψ = 0 survives; at a delta well V = −αδ(x − x₀), ψ′ jumps by −(2mα/ħ²)ψ(x₀). Scattering states are not square-integrable.