Potential barrier and tunneling

Roll a ball at a hill it does not have enough energy to climb, and classical mechanics is unambiguous: the ball rolls back down. A quantum particle facing the same situation — a region where its potential energy $V_0$ exceeds its total energy $E$ — obeys a stranger rule. Its wavefunction does not stop dead at the wall; it leaks into the forbidden region and decays there, and if the wall is thin enough some of that leaked wave re-emerges as a genuine, freely moving particle on the far side. This is quantum tunneling, and it is not a rare curiosity — it underlies radioactive alpha decay, the scanning tunneling microscope, and the leakage currents that limit how thin a transistor's insulating layer can be. This article builds the order-of-magnitude tunneling law from the time-independent Schrödinger equation, applies it to an electron and a proton hitting the same barrier, and shows why tunneling is essentially an electron phenomenon.

The core insight

A quantum particle with $E<V_0$ still has a wavefunction that decays exponentially — not zero — inside a potential barrier, so the transmission probability falls off as $T\sim e^{-2\kappa L}$ with $\kappa=\sqrt{2m(V_0-E)}/\hbar$ , making tunneling significant only for thin barriers and light, fast particles.

The core idea

Classically forbidden does not mean wavefunction-forbidden. Take a particle of energy $E$ moving in one dimension and meeting a rectangular barrier of height $V_0$ and width $L$ , with $E<V_0$ . Outside the barrier the particle is free and $\phi(x)$ oscillates like $e^{\pm ikx}$ , $k=\sqrt{2mE}/\hbar$ . Inside the barrier the time-independent Schrödinger equation reads

-\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2} + V_0\phi = E\phi \quad\Longrightarrow\quad \frac{d^2\phi}{dx^2} = \kappa^2\phi,\qquad \kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}.

where: $m$ is the particle's mass, $E$ its total energy, $V_0$ the barrier height, $\hbar$ the reduced Planck constant, and $\kappa$ (units of inverse length) sets how fast the wavefunction decays inside the barrier.

Why the solution is a real exponential. Because $E<V_0$ , the classical kinetic energy $E-V_0$ is negative, so the equation flips sign compared with the free-particle case and the solutions become real exponentials, $\phi(x)\sim e^{\pm\kappa x}$ , instead of oscillations. A classical particle could never occupy this region at all — but $\phi$ and $d\phi/dx$ must stay continuous at each wall of the barrier, matching the decaying interior solution smoothly onto the oscillating exterior waves. That continuity requirement is what forces $\phi$ to be nonzero, and decaying rather than vanishing, all the way across the barrier.

The transmission law. For a barrier wide enough that the decay is substantial ( $\kappa L\gg1$ ), the wavefunction amplitude that survives to the far wall — and hence the probability of the particle being found beyond the barrier — is governed by how much $\phi$ has decayed:

T \sim e^{-2\kappa L}, \qquad \kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}.

where: $T$ is the tunneling probability (dimensionless, between $0$ and $1$ ), $L$ is the barrier width, and $\kappa$ is as defined above. This is the order-of-magnitude estimate expected at this level — not the exact matched-boundary-condition transmission coefficient, which needs more algebra than the syllabus asks for.

The step case, for contrast. A potential step (no second wall) shows the same wave-mechanical stubbornness in a different guise. For $E>V_0$ at a step, classical mechanics guarantees transmission, but quantum mechanics predicts partial reflection at the discontinuity in $V(x)$ , because $\phi$ and $\phi'$ must still match continuously across it. For $E<V_0$ at a step, $\phi$ decays exponentially into the forbidden region but, with no far wall to escape through, reflection is eventually total.

Figure

Wavefunction oscillating on both sides of a barrier and decaying exponentially inside it, with the transmitted amplitude on the right smaller than the incident amplitude on the left

Derivation ATunneling probability across a rectangular barrier

The goal is to justify the order-of-magnitude tunneling law $T\sim e^{-2\kappa L}$ starting only from the interior Schrödinger equation and the requirement that the wavefunction and its slope stay continuous at the two walls of the barrier, without carrying out the full multi-region algebra that fixes the exact prefactor in front of the exponential.

Step 1. Write the interior equation. For $0<x<L$ with $E<V_0$ , the time-independent Schrödinger equation is $-\frac{\hbar^2}{2m}\phi''=(E-V_0)\phi$ . Since $E-V_0<0$ , rearrange as $\phi''=\kappa^2\phi$ with $\kappa=\sqrt{2m(V_0-E)}/\hbar$ real and positive. The general interior solution is $\phi(x)=Ae^{-\kappa x}+Be^{\kappa x}$ , a mixture of a decaying and a growing real exponential — no oscillation is possible because the effective "kinetic energy" $E-V_0$ is negative here.

Step 2. Match at the entrance wall, $x=0$ . Outside, for $x<0$ , the particle is free and its wavefunction is a mix of an incident wave $e^{ikx}$ and a reflected wave, with $k=\sqrt{2mE}/\hbar$ . Continuity of $\phi$ and $\phi'$ at $x=0$ fixes $A$ and $B$ in terms of the incident amplitude; the key point is that neither $A$ nor $B$ can be zero in general, so $\phi$ is nonzero everywhere inside the barrier, not just near the edges.

Step 3. Track the amplitude to the exit wall, $x=L$ . The decaying piece $Ae^{-\kappa x}$ shrinks by a factor $e^{-\kappa L}$ across the barrier while the growing piece $Be^{\kappa x}$ grows by $e^{\kappa L}$ ; continuity again at $x=L$ hands this combination over to an outgoing free wave $e^{ikx}$ for $x>L$ . When the barrier is wide ( $\kappa L\gg1$ ), the dominant surviving contribution to the transmitted amplitude scales as $e^{-\kappa L}$ , because the growing solution's coefficient is itself suppressed by matching conditions that keep the total wave bounded.

Step 4. Convert amplitude to probability. Transmission probability is proportional to the square of the transmitted wave's amplitude relative to the incident one, so a transmitted amplitude scaling as $e^{-\kappa L}$ gives $T\sim\left(e^{-\kappa L}\right)^2=e^{-2\kappa L}$ . This is the working formula: $T$ depends exponentially on both the barrier width $L$ and on $\sqrt{V_0-E}$ through $\kappa$ , so a barrier twice as wide does not halve $T$ — it roughly squares the already-small value, since $e^{-2\kappa(2L)}=\left(e^{-2\kappa L}\right)^2$ .

Worked example

An electron of kinetic energy $E=1.0\ \text{eV}$ approaches a rectangular barrier of height $V_0=2.0\ \text{eV}$ and width $L=0.50\ \text{nm}$ . Find the tunneling probability, then compare it with a proton facing the same energy deficit and width.

Step 1. Convert the energy deficit to SI: $V_0-E=1.0\ \text{eV}=1.0\times1.6\times10^{-19}\ \text{J}=1.6\times10^{-19}\ \text{J}$ .

Step 2. Compute $\kappa$ for the electron, using $m_e=9.11\times10^{-31}\ \text{kg}$ and $\hbar=1.055\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ : the product $2m_e(V_0-E)=2\times9.11\times10^{-31}\times1.6\times10^{-19}=2.92\times10^{-49}$ , whose square root is $5.40\times10^{-25}$ . Dividing by $\hbar$ gives $\kappa=5.40\times10^{-25}/1.055\times10^{-34}=5.12\times10^{9}\ \text{m}^{-1}$ .

Step 3. Compute $2\kappa L=2\times5.12\times10^{9}\times0.50\times10^{-9}=5.12$ , so $T\sim e^{-5.12}\approx6.0\times10^{-3}$ — about $0.6\%$ of electrons tunnel through, small but very much observable at nanometre scales.

Step 4. Scale up to a proton of the same energy deficit and width. Since $\kappa\propto\sqrt{m}$ and $m_p/m_e\approx1836$ , $\kappa_p=\kappa_e\sqrt{1836}\approx5.12\times10^{9}\times42.8\approx2.19\times10^{11}\ \text{m}^{-1}$ , giving $2\kappa_pL\approx219$ .

Step 5. Estimate $T_p\sim e^{-219}\approx10^{-95}$ — utterly negligible. The electron tunnels with probability roughly $6\times10^{-3}$ , while a proton under identical conditions tunnels with probability of order $10^{-95}$ : because $\kappa\propto\sqrt{m}$ , tunneling is dramatically suppressed for heavier particles, which is why tunneling phenomena in electronics and atomic physics are almost always electron effects.

Common traps

Using $\kappa=\sqrt{2mE}/\hbar$ instead of $\sqrt{2m(V_0-E)}/\hbar$ . Inside the barrier the relevant quantity is the energy deficit $V_0-E$ , not the particle's own energy $E$ .
Thinking the particle "loses energy" while tunneling. It does not — the particle that emerges beyond the barrier still has kinetic energy $E$ , unchanged. Tunneling is a probability-of-transmission effect, not an energy-loss effect.
Assuming "enough energy always means getting through." At a potential step with $E>V_0$ , classical mechanics guarantees transmission, but quantum mechanics still gives partial reflection because $\phi$ and $\phi'$ must match continuously across the discontinuity.
Treating $T$ as roughly linear in $L$ . Because $T$ is exponential in $L$ , doubling the barrier width does not halve $T$ — it approximately squares the already-small probability.

Ask the tutor: If you want to see the full matched-boundary-condition transmission coefficient behind this order-of-magnitude formula, or want a fresh barrier-tunneling number checked step by step, ask the AI tutor to walk through it with you.

Why does kappa depend on V0 minus E and not on V0 or E separately?

Inside the barrier the Schrodinger equation becomes $\phi''=\kappa^2\phi$ with $\kappa=\sqrt{2m(V_0-E)}/\hbar$ , and this combination arises directly from rearranging $-\frac{\hbar^2}{2m}\phi''=(E-V_0)\phi$ . The physically meaningful quantity is how far short of the barrier's height the particle's energy falls — the energy deficit — not either $E$ or $V_0$ on its own. A particle with $E=1.0\ \text{eV}$ facing $V_0=2.0\ \text{eV}$ behaves identically, in terms of $\kappa$ , to one with $E=5.0\ \text{eV}$ facing $V_0=6.0\ \text{eV}$ , because both have the same deficit of $1.0\ \text{eV}$ .

Does the particle exist inside the barrier the way it exists outside?

Its wavefunction is nonzero and real (a decaying exponential) inside the barrier, so there is a nonzero probability density there, but the particle cannot be found there classically because its kinetic energy would have to be negative. This is exactly the tension quantum mechanics resolves: $\phi$ can be nonzero in a classically forbidden region as long as it decays smoothly and matches continuity conditions at the boundaries. Nothing about this violates energy conservation — the particle that eventually emerges beyond the barrier still carries its original energy $E$ , unchanged.

Why is tunneling essentially never seen for protons or heavier particles at nanometre scales?

Because $\kappa=\sqrt{2m(V_0-E)}/\hbar$ grows with the square root of the mass, and $T\sim e^{-2\kappa L}$ is exponentially sensitive to $\kappa$ . Even though a proton is only about $1836$ times heavier than an electron, $\kappa$ scales as $\sqrt{1836}\approx42.8$ , which multiplies the exponent by that same factor. For the worked example's barrier this turns a modest exponent of $5.12$ for the electron into an exponent of about $219$ for the proton, crushing $T$ from roughly $0.6\%$ down to about $10^{-95}$ — effectively zero.

Potential barrier and tunneling

The core idea

Worked example

Common traps

Before this

Where this leads