Classically forbidden regions and exponential decay
Where the energy runs out, ψ decays instead of stopping.
Every problem so far kept the particle where classical mechanics allowed it. This one does not. Read the sign of ψ″/ψ to see the exponential coming, measure its reach with a single length, 1/κ, and settle the question every class asks next: whether a particle found there carries negative kinetic energy.
Build the model
Where the energy runs out, ψ does not stop: it bends away from the axis and decays.
1The big idea
The time-independent Schrödinger equation is local: ψ″ = (2m/ħ²)(V − E)ψ fixes the curvature of ψ at each point from V − E there. Where E > V the curvature opposes ψ, the curve keeps bending back through zero, and ψ oscillates. Where V₀ > E the curvature has ψ's own sign, the curve bends away from the axis and can never turn over, so the only solutions are real exponentials e(±κx) with κ = √(2m(V₀ − E))/ħ. Classical mechanics stops the particle dead at the turning point; quantum mechanics only changes the shape of ψ there, from wave to decay, and the decaying branch leaves |ψ|² nonzero over a reach δ = 1/κ. For an electron 1 eV short of the wall that reach is 0.195 nm, the size of an atom, so the tails of neighbouring atoms overlap and matter; for a 1 g marble 1 mm short of the top of a rise it is 7.5 × 10⁻³¹ m, which is why no one has seen a marble do it. The picture has two edges. As E rises towards V₀ the reach diverges and the tail stops being short compared with anything. And it cannot be read as a particle sitting inside the wall with negative kinetic energy: a detection that places it in the tail supplies, on average, more energy than the deficit it was meant to reveal.
2In one sentence
A classically forbidden region is any stretch where the potential energy V exceeds the particle's energy E; where V is flat there, ψ″ = κ²ψ, so ψ is built from real exponentials and its decaying part falls by a factor e over the penetration depth 1/κ = ħ/√(2m(V − E)).
Example. An electron 1.00 eV short of the top of a wall has κ = √(2 × 9.11 × 10⁻³¹ kg × 1.60 × 10⁻¹⁹ J)/(1.055 × 10⁻³⁴ J s) = 5.12 × 10⁹ m⁻¹: ψ falls by a factor e every 0.195 nm, and |ψ|² by e² ≈ 7.4.
3Read the sign of ψ″/ψ before solving anything
Rearrange the time-independent Schrödinger equation into ψ″ = (2m/ħ²)(V − E)ψ and read it as a statement about shape. Where E > V the factor is negative, so ψ″ has the opposite sign to ψ: wherever ψ is positive it curves down, wherever it is negative it curves up, and the curve keeps bending back through zero. That is oscillation, with local wavenumber k = √(2m(E − V))/ħ; an electron with 1.0 eV of kinetic energy has k = 5.12 nm⁻¹ and a wavelength of 1.23 nm. Where V > E the factor is positive, ψ″ shares the sign of ψ, and the curve bends away from the axis. It cannot turn over and come back, so there is no oscillation left to have. At the classical turning point V = E, and ψ″ = 0: the place where a classical particle would stop and reverse is, for ψ, only an inflection point.
4In a flat forbidden region, two real exponentials
Take V = V₀ > E over a stretch of x. The equation becomes ψ″ = κ²ψ with κ = √(2m(V₀ − E))/ħ, and the functions whose second derivative is κ² times themselves are e(−κx) and e(κx), so ψ = C e(−κx) + D e(κx). Substitute to check: the second derivative of e(−κx) is κ² e(−κx). The oscillating trial cos(κx) fails, because its second derivative is −κ² cos(κx), the wrong sign. Check the dimensions: 2m(V₀ − E) is kg × J = kg² m² s⁻², its root is a momentum in kg m s⁻¹, and dividing by ħ in J s = kg m² s⁻¹ leaves m⁻¹, exactly what the exponent κx needs to be a pure number. Which branch stays is not this equation's business: a forbidden region running out to infinity forces D = 0, as the next topic shows, while a barrier of finite width keeps both.
51/κ is how far the particle reaches
Keep the decaying branch, ψ(x) = ψ(0) e(−κx), with x measured from the wall. ψ falls by a factor e over δ = 1/κ, and |ψ|² = |ψ(0)|² e(−2κx) falls by e² ≈ 7.39 over the same distance. Integrate from d to infinity and P(x > d) = |ψ(0)|² e(−2κd)/(2κ): 13.5% of the tail lies deeper than δ, 1.8% deeper than 2δ and 0.25% deeper than 3δ. At d = 0 the whole tail holds |ψ(0)|²/(2κ), a pure number because both |ψ(0)|² and κ are in m⁻¹, and normalisation caps it below 1. For an electron κ = 5.12 nm⁻¹ × √((V₀ − E)/1 eV): a 1.00 eV deficit gives δ = 0.195 nm, comparable to an atom, and a 0.10 eV deficit stretches it to 0.617 nm. Because δ ∝ 1/√m, a proton with the same 1.00 eV deficit reaches only 4.56 pm, √1836 = 42.85 times less far.
6Check the limits, and the classical one especially
Raise V₀ with E fixed and κ → ∞, δ → 0: the tail shrinks onto the wall and ψ there is squeezed towards zero, which is the boundary condition the infinite well imposed by decree, now recovered as a limit. For a single step the matching conditions give |ψ(0)|² = (E/V₀) × the peak density inside, which indeed vanishes as V₀ → ∞. Let E rise towards V₀ and κ → 0, δ → ∞; at E = V₀ exactly ψ″ = 0 and ψ is a straight line, the boundary case between bending back and bending away. Finally make the particle big. A 1.0 g marble rolling up a rise it falls 1.0 mm short of topping has V₀ − E = mgh = 9.81 × 10⁻⁶ J, so δ = ħ/√(2m(V₀ − E)) = 7.5 × 10⁻³¹ m, some 10¹⁵ times smaller than a proton. Quantum mechanics gives the marble a tail too; it simply puts it where no experiment can look, which is the classical correspondence.
7E < V₀ is not a negative kinetic energy
Classically, kinetic energy is E − V(x) at each point, and in the tail that is negative. Quantum mechanically E = T + V holds for operators acting on the whole state, not for numbers attached to a point. A measurement of kinetic energy returns p²/2m for some momentum p, which is never below zero. The expectation value is ⟨T⟩ = −(ħ²/2m)∫ψ*ψ″ dx, and one integration by parts, with ψ → 0 far away, turns it into (ħ²/2m)∫|ψ′|² dx, which cannot be negative. The local ratio −(ħ²/2m)ψ″/ψ does equal E − V₀ < 0 at every point of the tail, but that ratio is not an eigenvalue of anything and no instrument reads it. The apparent paradox needs the particle's position and its kinetic energy to be sharp at once, and position and momentum are exactly the pair the uncertainty principle refuses to sharpen together.
8Catching it in the tail costs more than the deficit
To claim the particle is in the forbidden region a detector must confine it to a window lying beyond the wall, and because the tail holds the particle only within about δ of the wall, the natural window has width δ. After such a detection ψ vanishes outside a width w, and no state confined to width w has an average kinetic energy ⟨T⟩ below the infinite-well ground state of that width, π²ħ²/(2mw²). With w = δ = 1/κ that is π²ħ²κ²/(2m) = π²(V₀ − E) ≈ 9.9(V₀ − E): for an electron 1.00 eV short of the top, ⟨T⟩ is at least 9.9 eV on top of V₀. This bounds the average; single readings of p²/2m still scatter, some below it. A wider window lowers the floor but never to zero, and inside it V = V₀, so after any such detection the mean energy exceeds V₀. The rougher uncertainty-principle estimate agrees in kind: Δp ≳ ħ/δ = ħκ gives (Δp)²/2m ≳ V₀ − E. Energy is conserved because the detector paid. The particle is never caught inside the wall with E < V₀; whatever is found there has had its mean energy lifted above V₀ by the finding.
Key equations
What each relation says, and the units it is written in.
Curvature from the energy gap
ψ″/ψ = 2m(V(x) − E)/ħ²
Its sign sorts the regions before any solving: negative bends ψ back towards the axis, positive bends it away, and zero marks an inflection at the turning point.
ψ″ = d²ψ/dx²; m in kg, V and E in J, ħ = 1.055 × 10⁻³⁴ J s, so ψ″/ψ is in m⁻²
Decay constant
κ = √(2m(V₀ − E))/ħ
Its reciprocal δ = 1/κ is the penetration depth. For an electron κ = 5.12 nm⁻¹ × √((V₀ − E)/1 eV); quadruple the deficit or the mass and δ halves.
κ in m⁻¹ with V₀ − E > 0 in J; the root √(2m(V₀ − E)) is a momentum in kg m s⁻¹
Solution in a flat forbidden region
ψ(x) = C e(−κx) + D e(κx)
Two real exponentials and no oscillating part. Which branch survives is decided by where the region ends, not by this equation.
x in m, measured from the wall into the region; C and D in m(−1/2) for a one-dimensional ψ
Probability beyond depth d
P(x > d) = |ψ(0)|² e(−2κd)/(2κ)
The 2 in the exponent is the square: |ψ|² decays twice as fast as ψ, so only e(−2) = 13.5% of the tail lies deeper than δ.
|ψ(0)|² in m⁻¹ at the wall, d in m, P dimensionless; d = 0 gives the whole tail
Average kinetic energy
⟨T⟩ = (ħ²/2m) ∫ |ψ′(x)|² dx ≥ 0
The integrand |ψ′|² is never negative, so the total cannot be. E − V₀ is the ratio −(ħ²/2m)ψ″/ψ at a point, not a kinetic energy anyone measures.
⟨T⟩ in J; integral over all x, from −(ħ²/2m)∫ψ*ψ″ dx by parts with ψ → 0 at ±∞
Cost of catching it in the tail
⟨T⟩ ≥ π²ħ²/(2m w²)
The tail holds the particle within about δ of the wall, so take w = δ: ⟨T⟩ ≥ π²(V₀ − E) ≈ 9.9(V₀ − E). Any window beyond the wall leaves ⟨T⟩ > 0 where V = V₀, so the mean energy ends above V₀.
w in m is the width of the window a detection confines the particle to; the bound, in J, is that window's infinite-well ground-state energy
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Drag V₀ − E down to 0.1 eV and the tail stretches past 0.6 nm while its height at the wall climbs towards the peak. Then set the mass to 4 mₑ: the wavelength inside and δ both halve, the direction in which tails vanish for classical bodies.
Live results
- DECAY CONSTANT κ
- 5.12 nm⁻¹
- PENETRATION DEPTH δ = 1/κ
- 0.195 nm
- |ψ|² AT WALL ÷ PEAK
- 0.50
- |ψ|² AT 0.2 nm ÷ AT WALL
- 0.129
Catch the common trap
The mistake this lesson is built to prevent.
The tempting reading writes kinetic energy as E − V(x) point by point and concludes that a particle in the tail carries negative kinetic energy there. Kinetic energy cannot be assigned to a point, and not even to a region. Take a tail ψ = ψ(0)e(−κx) holding probability Pₜₐᵢₗ. The densities −(ħ²/2m)ψψ″ and (ħ²/2m)|ψ′|² both integrate over all x to the same ⟨T⟩, yet over the tail alone the first gives −(V₀ − E)Pₜₐᵢₗ and the second +(V₀ − E)Pₜₐᵢₗ, because they differ by a boundary term at the wall. A quantity whose share of a region flips sign with the choice between two equally valid formulas is not located there. Only the total is physical, and it is positive.
Test it: what changes?
- κ = √(2m(V₀ − E))/ħ falls
- δ = 1/κ grows, diverging as V₀ − E → 0
- At a single step, wall |ψ|²/peak = E/V₀ rises
- Detection floor π²(V₀ − E) on ⟨T⟩ falls
- κ grows as √m
- δ = 1/κ shrinks as 1/√m; 4m halves it
- Inside wavelength shrinks by the same factor
- Detection floor π²(V₀ − E) is unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron bound in a well has energy 2.5 eV below the potential V₀ outside the walls. Find the decay constant κ and the penetration depth beyond a wall, then the depth at which |ψ|² has fallen to 1.0% of its value at the wall.
View solution →
- Beyond the wall V₀ − E = 2.5 eV = 2.5 × 1.602 × 10⁻¹⁹ J = 4.005 × 10⁻¹⁹ J, so ψ″ = κ²ψ with κ = √(2m(V₀ − E))/ħ.
- 2m(V₀ − E) = 2 × 9.109 × 10⁻³¹ kg × 4.005 × 10⁻¹⁹ J = 7.296 × 10⁻⁴⁹ kg² m² s⁻², whose root is 8.542 × 10⁻²⁵ kg m s⁻¹, a momentum as it should be.
- κ = 8.542 × 10⁻²⁵ kg m s⁻¹ ÷ (1.0546 × 10⁻³⁴ J s) = 8.10 × 10⁹ m⁻¹, so δ = 1/κ = 1.23 × 10⁻¹⁰ m = 0.123 nm. The electron shortcut agrees: 5.12 nm⁻¹ × √2.5 = 8.10 nm⁻¹.
- |ψ|² goes as e(−2κx), so 1.0% at depth d means e(−2κd) = 0.010, or d = ln(100)/(2κ) = 4.605 ÷ (1.620 × 10¹⁰ m⁻¹) = 2.84 × 10⁻¹⁰ m.
- Sanity check: 0.284 nm is 2.3δ, and the 1% point always sits at ln(100)/2 = 2.3 penetration depths, whatever the particle. Forgetting to square and solving e(−κd) = 0.010 would give twice this, 0.57 nm.
Answer. κ = 8.10 × 10⁹ m⁻¹ (8.10 nm⁻¹), δ = 0.123 nm, and |ψ|² falls to 1.0% of its wall value 0.284 nm into the forbidden region.
MediumExample 2
A numerical solver returns a normalised bound state for an electron. Beyond the right wall (x = 0 there) it prints |ψ|² = 0.800 nm⁻¹ at x = 0, 0.362 nm⁻¹ at 0.10 nm and 0.164 nm⁻¹ at 0.20 nm. Confirm that the tail is one decaying exponential, extract κ and V₀ − E, and find the probability of finding the electron beyond the wall and beyond 0.30 nm.
View solution →
- Ratios over equal 0.10 nm steps: 0.362/0.800 = 0.453 and 0.164/0.362 = 0.453. A constant ratio per step is the signature of a single exponential e(−2κx), with no growing branch mixed in.
- Use the end points: 2κ × 0.20 nm = ln(0.800/0.164) = ln 4.878 = 1.585, so κ = 3.96 nm⁻¹ = 3.96 × 10⁹ m⁻¹ and δ = 0.252 nm.
- Invert κ = √(2m(V₀ − E))/ħ: V₀ − E = ħ²κ²/(2m) = (1.0546 × 10⁻³⁴ J s × 3.962 × 10⁹ m⁻¹)² ÷ (2 × 9.109 × 10⁻³¹ kg) = 9.58 × 10⁻²⁰ J = 0.598 eV.
- Whole tail: P(x > 0) = |ψ(0)|²/(2κ) = 0.800 nm⁻¹ ÷ 7.924 nm⁻¹ = 0.101. Units cancel to a pure number, and it is below 1 as a normalised state requires.
- Beyond 0.30 nm: P = 0.101 × e(−2 × 3.962 nm⁻¹ × 0.30 nm) = 0.101 × e(−2.377) = 0.101 × 0.0928 = 9.4 × 10⁻³.
Answer. The ratio is a constant 0.453 per 0.10 nm, so the tail is one exponential; κ = 3.96 nm⁻¹ and V₀ − E = 0.60 eV; P(beyond the wall) = 0.10 and P(beyond 0.30 nm) = 9.4 × 10⁻³.
HardExample 3
An electron is bound 0.40 eV below the rim of a well, and at the right wall its normalised |ψ(0)|² = 0.50 nm⁻¹. A detector registers the electron only in the window 0 < x < δ just beyond that wall, with δ = 1/κ. Find the probability that it fires, the least average kinetic energy ⟨T⟩ the electron can have once it has fired, and how the window and that floor change for a proton with the same 0.40 eV deficit.
View solution →
- 0.40 eV = 6.409 × 10⁻²⁰ J, so √(2m(V₀ − E)) = √(2 × 9.109 × 10⁻³¹ kg × 6.409 × 10⁻²⁰ J) = 3.417 × 10⁻²⁵ kg m s⁻¹. Then κ = 3.417 × 10⁻²⁵ ÷ (1.0546 × 10⁻³⁴ J s) = 3.24 × 10⁹ m⁻¹, and δ = 1/κ = 0.309 nm.
- Probability in the window: ∫₀δ |ψ(0)|² e(−2κx) dx = |ψ(0)|² (1 − e(−2))/(2κ) = 0.50 nm⁻¹ × 0.8647 ÷ 6.480 nm⁻¹ = 0.067.
- Once it fires, ψ vanishes outside a window of width w = δ. No state confined to width w has an average kinetic energy ⟨T⟩ below the infinite-well ground state π²ħ²/(2mw²), and with w = 1/κ that is π²ħ²κ²/(2m) = π²(V₀ − E).
- π² × 0.40 eV = 3.95 eV. The electron now sits where V = V₀ with ⟨T⟩ ≥ 3.95 eV, so its mean energy is at least V₀ + 3.95 eV, some 4.35 eV above where it started. The detector supplied the difference.
- Proton: δ ∝ 1/√m, so δₚ = 0.3086 nm ÷ √1836 = 0.3086 nm ÷ 42.85 = 7.20 pm. But π²(V₀ − E) contains no mass: the narrower window raises π²ħ²/(2mw²) by exactly the factor the larger mass lowers it, and the floor on ⟨T⟩ stays 3.95 eV.
Answer. The detector fires with probability 0.067; afterwards ⟨T⟩ ≥ 3.9 eV, so the electron's mean energy is at least 3.9 eV above V₀. For a proton the window shrinks to 7.2 pm, but the floor on ⟨T⟩ is the same 3.9 eV.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
ψ decays over δ = 1/κ; |ψ|² decays twice as fast.
Where V = V₀ > E is flat, ψ = C e(−κx) + D e(κx), κ = √(2m(V₀ − E))/ħ; a region running to infinity keeps only e(−κx). Probability beyond d is |ψ(0)|² e(−2κd)/(2κ). The reach diverges as E → V₀, yet a particle found there never has negative kinetic energy.
Next: Bound states and what normalisability demands of the tails →