How Science Works | How Quantum Tunnelling Lets Particles Cross a Barrier

Imagine a graph with position along the horizontal axis and potential energy up the vertical axis. Potential energy is zero to the left, rises to a constant positive height over a finite width, and returns to zero on the right. This is an ideal rectangular barrier.

Draw a horizontal total-energy line below the top of the rectangle. In classical mechanics, kinetic energy would be total energy minus potential energy. A classical particle cannot have the negative kinetic energy that this subtraction would require inside the rectangle. It turns back rather than crossing.

The graph is not a picture of a wall’s physical height. Its vertical coordinate is energy. An electrical potential, molecular interaction or another appropriate physical system can produce an energy barrier. The horizontal coordinate must also be defined: it could describe a particle’s position or, in a more complicated model, a collective configuration.

Quantum mechanics replaces the assumed classical path with a state governed by a wave equation. Calling the region classically forbidden identifies a failure of the classical trajectory description there. It does not mean quantum theory has suspended its own laws.

Nor does every particle facing a barrier tunnel. The predicted probability can be close to zero, appreciable or, under different conditions, very large. A useful explanation must connect the outcome to measurable properties of a specified system, rather than use tunnelling as a general word for surprising motion.

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For our model, assume one spatial dimension, a constant particle mass, a static barrier and no absorption or energy exchange with the surroundings. On the left, the stationary scattering state contains an incoming wave and a reflected wave. On the right, it contains an outgoing transmitted wave.

Inside the barrier, where the energy is below its height, the spatial solutions involve exponentials rather than freely travelling sinusoidal waves. Both growing and decaying exponential terms can be needed over the finite interval. Their coefficients are fixed by matching the wavefunction and its derivative at the two boundaries.

A growing term within a finite interval does not mean an unphysical explosion at infinite distance. The interval ends at the second boundary. Treating the interior as only a decaying exponential can be a helpful sketch in some limits, but it is not the complete finite-barrier solution.

Transmission probability is defined through the ratio of transmitted to incident probability current. Because our left and right regions have the same potential and mass, this becomes the squared magnitude of the transmitted-to-incident amplitude ratio. Different potentials on the two sides require an additional velocity factor.

For the conservative model, reflection probability plus transmission probability equals one. An individual detection gives a whole-particle outcome. The probabilities describe the distribution across repeated, equally prepared trials; they do not say that each detected particle arrives as a fixed fraction of itself.

The boundary-matching calculation is developed in MIT’s quantum transport lecture notes. Our numerical comparisons below are invented teaching cases for that ideal model.

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Let the barrier height be V₀, its full width be a, the particle energy be E and its mass be m. We restrict the example to 0 < E < V₀. Define κ = √[2m(V₀ − E)]/ℏ, where ℏ is the reduced Planck constant.

The exact transmission probability is T = 1 / {1 + [V₀² / (4E(V₀ − E))] sinh²(κa)}. The function sinh is the hyperbolic sine: sinh(x) = (exp(x) − exp(−x))/2. A scientific calculator can evaluate it directly.

There are two useful checks before calculating. First, κ has units of inverse length, so κa is dimensionless. Second, the expression gives a probability between zero and one. A result such as 1.4 would indicate a calculation error or use of an inappropriate approximation.

Choose the particularly simple case E = V₀/2. The factor multiplying sinh² becomes one, and T simplifies to 1/cosh²(κa), also written sech²(κa). The function cosh(x) = (exp(x) + exp(−x))/2.

For κa = 1, the transmission probability is approximately 0.420. For κa = 2, it is 0.0707. For κa = 3, it is 0.00987. These correspond to about 42.0%, 7.07% and 0.987% of identical incident preparations in the ideal ensemble.

The dimensionless numbers do not specify a particular laboratory material. Many combinations of mass, energy difference and width could produce the same κa. That is why the example is useful for understanding sensitivity without pretending to predict an actual device from missing material information.

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Hold energy, barrier height and mass fixed. Then κ stays fixed, so doubling the barrier width doubles κa. In our half-height example, changing κa from one to two reduces transmission from about 42.0% to 7.07%. Doubling width does much more than halve the probability.

Now return to the original width and consider a hypothetical particle with four times the mass, while keeping the same energy and potential-energy profile. Because κ depends on the square root of mass, κ again doubles. The probability changes by the same amount as in the width-doubling comparison.

This is a controlled mathematical comparison. Replacing one chemical isotope or material with another can alter additional properties. An actual mass-effect experiment must consider changes in vibrational energies, geometry and the environment before attributing every difference to the square-root factor alone.

Increasing the difference V₀ − E increases κ and strengthens attenuation, while the exact prefactor also changes. Barrier height and particle energy therefore belong in the full calculation. The formula should not be extended to E above V₀ by feeding a negative number into the same real-valued square root.

For sufficiently opaque rectangular barriers, an approximation is T ≈ [16E(V₀ − E)/V₀²] exp(−2κa). It captures the strong exponential dependence. At E = V₀/2 and κa = 3, it gives 0.00992, close to the exact 0.00987. At κa = 1, it gives 0.541 rather than 0.420 and is much less accurate.

The exponent contains two because probability involves a squared amplitude. Dropping the prefactor and treating exp(−2κa) as the exact answer creates another error. An approximation is valuable when its conditions are checked, not merely because it looks simpler.

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Suppose an imagined apparatus prepares 10,000 independent particles with the same narrow energy distribution and sends them toward a barrier described by our κa = 2, E = V₀/2 case. Ignore detector losses and interactions between the particles.

The expected transmitted count is 10,000 × 0.0706508, or about 707 particles. This is an average across repeated runs of the complete experiment. It is not a promise that the next run will contain exactly 707 detections.

For independent trials with constant probability p, the count’s standard deviation is √[Np(1 − p)]. Here it is about 25.6 particles. A difference of several detections between two runs would therefore be unsurprising even if the barrier had not changed at all.

Now imagine that only 90% of transmitted particles are detected. Under a simple independent-efficiency model, the expected recorded count becomes about 636. The physical transmission probability has not changed, but the observation has. Inferring transmission requires a detector-efficiency estimate and an incident-count estimate.

A broad energy spread introduces another issue: different particles have different transmission probabilities. We must average the transmission function over the incident distribution, rather than automatically insert its mean energy into the formula. Nonlinear functions generally do not preserve that substitution.

This thought experiment gives a practical evidence checklist. Ask how the incident state was prepared, what was counted, which particles were missed and whether independence is a reasonable assumption. A quantum prediction is tested through a measurement model as well as a wave equation.

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In our static, elastic model, a transmitted particle emerges with the same total energy it had initially. Since the potential energy is also the same on both sides, its outgoing kinetic energy is the same as its incoming kinetic energy.

The statement does not assign a classical kinetic energy to a hidden trajectory inside the forbidden region. That trajectory was precisely the assumption that failed. Combining a quantum transmission result with a classical path through every intermediate point produces the apparent paradox.

The familiar claim that a particle borrows enough energy to climb over the barrier, then pays it back, is misleading. Energy–time uncertainty is not permission to violate conservation for a short interval. The stationary barrier solution already predicts tunnelling without such a loan.

A time-dependent barrier, an applied electromagnetic field or interaction with the surroundings can exchange energy with the particle. Those are legitimate physical processes, but they require a different model with an explicit energy source or sink. They should not be silently added to explain the conservative example.

Likewise, transmission probability alone does not tell us a unique traversal time. Different timing measurements ask different operational questions. Nothing in the simple probability calculation licenses instantaneous messaging or faster-than-light communication. It predicts which outcomes occur and with what probability under its stated assumptions.

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Thermal activation occurs when energy exchange allows a system to reach configurations above a barrier. Tunnelling can connect states across a barrier without the corresponding classical over-barrier trajectory. Real systems can support both contributions, so their relative importance may change with temperature.

A temperature-independent escape rate at low temperature can be relevant evidence, but it is not sufficient by itself. Instrument noise, imperfect temperature measurement, unintended radiation or another temperature-independent process could also produce a plateau. Researchers need independent parameter measurements and quantitative predictions.

In a 1985 Josephson-junction experiment by Devoret, Martinis and Clarke, parameters were established using measurements in a thermal regime. At lower temperatures, the observed escape rate became temperature-independent and agreed with the macroscopic quantum-tunnelling prediction without adjustable parameters. The combined comparison was stronger than a plateau alone.

Notice that escape rate and transmission probability are different quantities. A rate has units of inverse time; T has no units. A system initially trapped in a metastable well requires a dynamical escape model. One cannot substitute our rectangular-barrier T directly for a measured number of escapes per second.

For a paper-only investigation, imagine two explanations for a low-temperature plateau: tunnelling and detector background. Ask what each predicts when the barrier is changed in a calibrated way, and what happens in a control with no intended escape signal. A useful test separates predictions that previously looked alike.

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The word macroscopic can cause confusion. In an appropriate superconducting circuit, the relevant quantum coordinate can be a collective electrical variable, rather than the position of a single free particle. An effective energy landscape can contain a well and an escape barrier for that coordinate.

In their separate 1985 energy-level experiment, Martinis, Devoret and Clarke reported quantised levels associated with the phase difference across a current-biased Josephson junction. This helps clarify what the circuit’s quantum state describes.

The 2025 Nobel Prize in Physics recognised Clarke, Devoret and Martinis for macroscopic quantum tunnelling and energy quantisation in an electric circuit. The result concerns deliberately prepared circuit states under carefully controlled conditions. It does not show an ordinary room-temperature object casually passing through a solid wall.

Our one-particle rectangular formula is therefore an introduction, not a universal equation for every system called tunnelling. A collective coordinate has its own effective dynamics, potential landscape and coupling to the environment. Those ingredients must be established before making a quantitative prediction.

The same caution applies to atomic rearrangements. A 2025 study by Müller and colleagues combined spectroscopy and theoretical modelling to identify fluorine tunnelling in a confined polyfluoride complex. Temperature-dependent spectral structure supported the assignment, while the model identified a rotational barrier influenced by the surrounding neon matrix.

That study demonstrates why mass alone is not a complete verdict on whether tunnelling is observable. Barrier shape, confinement and the coordinate involved matter. Its specialist chemical preparation is unnecessary for this reader exercise; the transferable lesson is how measured spectral features are compared with a specific physical model.

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For a safe worksheet, draw three potential-energy graphs with identical height V₀ and energy E = V₀/2. Label their widths a, 2a and 3a. Choose the first width so that κa = 1. Under each graph, write the exact probabilities 0.420, 0.0707 and 0.00987.

On a separate sketch, show an incoming component, a reflected component, an attenuated interior state and a transmitted component. Keep this diagram distinct from the energy graph: wavefunction amplitude and potential energy have different units and should not share an unlabeled vertical axis.

Now make a second comparison with fixed width and masses m, 4m and 9m. With energy and barrier profile held fixed, the dimensionless barrier factors again become one, two and three. The mathematical probabilities match the width comparison, even though the changed physical parameter is different.

Next, predict counts for 1,000 ideal independent incident preparations. The means are about 420, 70.7 and 9.87. Fractional expected counts are allowed because they describe averages across repeated experiments, while any individual recorded count remains a whole number.

Finally, annotate every assumption that could fail in a real experiment: nonrectangular potential, varying effective mass, energy spread, time dependence, interactions, environmental coupling and imperfect detection. Choose one failure and describe what additional measurement would be needed. For example, an energy spread calls for characterising the incoming distribution.

A scanning tunnelling microscope uses related sensitivity to a barrier in a practical measurement. Its tip, feedback and interpretation of surface images are covered in the separate Scanning Tunnelling Microscopy guide. The worksheet here establishes the probability reasoning beneath that application.

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Does a transmitted particle need more energy than the barrier height? Not in the tunnelling case. Our stationary model allows transmission with E below V₀ while conserving total energy. The classical over-barrier requirement does not apply to the quantum state in the same way.

Why is barrier width important? The dimensionless factor κa measures how strongly the forbidden region attenuates the state. Increasing width can sharply reduce transmission even when height is unchanged. Width and height are independent inputs, not interchangeable descriptions.

Does a 7.07% probability mean every fourteenth particle passes? No. It predicts a statistical distribution across comparable preparations. Independent outcomes fluctuate, and no regular every-fourteenth schedule follows from that probability.

Would four times the mass always give the same change in a real chemical system? Only if the other assumptions in our comparison held. Chemical substitutions may alter the potential landscape and state energies as well as mass. A controlled model comparison is not automatically a controlled laboratory substitution.

What makes an experimental tunnelling claim persuasive? A defined quantum coordinate and barrier, measured preparation and detection conditions, predictions checked against independent parameters, and tests that constrain competing mechanisms. A surprising signal alone is not enough.

Quantum tunnelling becomes understandable when we replace the impossible classical path with a calculable quantum state. The important question is then how the state, barrier and measurement determine an outcome distribution. Continue through the Quantum Mechanics guide or return to the How Science Works master to connect this example with the wider study of models and evidence.

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