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beginner · Physics · Quantum Tunneling & Barriers

Tunneling Through a Barrier

Classically, a particle with energy EE cannot enter a region where the potential energy V0>EV_0 > E — it simply bounces back. Quantum mechanics is different: because the wavefunction must remain continuous and smooth, a non-zero amplitude penetrates into the barrier and emerges on the other side. This phenomenon is quantum tunneling.

Setting up the rectangular barrier

Consider a particle of mass mm and energy EE moving in one dimension. For 0≤x≤a0 \leq x \leq a a constant potential V0>EV_0 \gt E is present; outside the barrier V=0V = 0. The time-independent Schrödinger equation in each region reads

−ℏ22md2ψdx2+V ψ=E ψ.-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\,\psi = E\,\psi.

Region I (x<0x \lt 0): free propagation, so ψ=Aeikx+Be−ikx\psi = A e^{ikx} + B e^{-ikx} with k=2mE/ℏk = \sqrt{2mE}/\hbar.

Region II (0≤x≤a0 \leq x \leq a): the kinetic energy is negative, so instead of oscillating the wavefunction grows and decays exponentially. The solution is ψ=Ceκx+De−κx\psi = C e^{\kappa x} + D e^{-\kappa x} where

κ=2m(V0−E)ℏ.\kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar}.

Region III (x>ax \gt a): free propagation again, ψ=Feikx\psi = F e^{ikx}.

Matching ψ\psi and dψ/dxd\psi/dx at both boundaries (x=0x = 0 and x=ax = a) determines the coefficients. The transmission coefficient is T=∣F∣2/∣A∣2T = |F|^2 / |A|^2.

The thick-barrier approximation

Carrying out the full boundary-matching algebra yields an exact expression for TT. When κa≫1\kappa a \gg 1 (a barrier that is both wide and tall compared with the de Broglie wavelength inside), the growing exponential term is negligible and the expression simplifies to

T≈16EV0 ⁣(1−EV0)e−2κa.T \approx 16 \frac{E}{V_0}\!\left(1 - \frac{E}{V_0}\right) e^{-2\kappa a}.

The prefactor 16(E/V0)(1−E/V0)16(E/V_0)(1-E/V_0) is of order 1 for EE not too close to 00 or V0V_0, so the dominant factor is the exponential suppression e−2κae^{-2\kappa a}. The tunneling probability falls off exponentially with barrier width aa and with the square root of the energy deficit V0−EV_0 - E (because κ∝V0−E\kappa \propto \sqrt{V_0 - E}).

A minimal estimate — accurate to within a factor of a few for thick barriers — simply drops the prefactor:

T≈e−2κa.T \approx e^{-2\kappa a}.

A concrete example

Take an electron (m=9.109×10−31 kgm = 9.109 \times 10^{-31}\,\text{kg}) with energy 1 eV1\,\text{eV} below the barrier top, so V0−E=1 eV=1.602×10−19 JV_0 - E = 1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J}, and a barrier width a=0.5 nma = 0.5\,\text{nm}.

First, compute κ\kappa:

κ=2×9.109×10−31×1.602×10−191.0546×10−34≈5.12×109 m−1.\kappa = \frac{\sqrt{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19}}}{1.0546 \times 10^{-34}} \approx 5.12 \times 10^{9}\,\text{m}^{-1}.

Then the exponent is

2κa=2×5.12×109 m−1×0.5×10−9 m≈5.12,2\kappa a = 2 \times 5.12 \times 10^{9}\,\text{m}^{-1} \times 0.5 \times 10^{-9}\,\text{m} \approx 5.12,

and the thick-barrier estimate gives

T≈e−5.12≈0.006.T \approx e^{-5.12} \approx 0.006.

So roughly six electrons in a thousand pass through — a small but physically real probability that underlies technologies from alpha decay to modern semiconductor tunnel junctions.

Try it

This is a numerical exercise — return a number. Using the thick-barrier approximation T≈e−2κaT \approx e^{-2\kappa a}, compute the tunneling probability for an electron with an energy deficit of 1.0 eV1.0\,\text{eV} and a barrier width of 0.5 nm0.5\,\text{nm}.

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