Quantum Tunnelling Through a Barrier
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable solving the time-independent Schrödinger equation for piecewise-constant potentials and applying boundary conditions at interfaces, as covered in the previous lecture on step potentials.
Why this matters
Tunnelling is the definitive quantum phenomenon that has no classical analogue.
In classical mechanics, a particle with energy \( E \) cannot penetrate a potential barrier of height \( V_0 \) if \( E < V_0 \); it is simply reflected. Quantum mechanically, the wavefunction does not vanish abruptly at the barrier but decays exponentially within it. If the barrier is sufficiently thin, a non-zero amplitude emerges on the other side, resulting in a finite probability of transmission. This effect is not a minor correction but a fundamental feature of nature, governing nuclear fusion in stars, radioactive decay, and the operation of scanning tunnelling microscopes.
The Physical Picture
Imagine a ball rolling towards a hill. If the ball does not have enough kinetic energy to reach the top, it rolls back down. There is no way for it to appear on the other side. In the quantum world, the particle is described by a wave. When this wave encounters a region where its energy is less than the potential energy, the wave does not stop; it penetrates into the forbidden region. Inside the barrier, the wavefunction is not oscillatory (like a travelling wave) but exponential (decaying). If the barrier is wide, the wave decays to zero before reaching the other side, and the particle is reflected. However, if the barrier is narrow, the wavefunction retains a small but non-zero value at the exit. This residual amplitude allows the particle to be detected on the far side. The probability of this happening depends exponentially on the width of the barrier and the square root of the energy deficit.
Wavefunction in the Barrier
Consider a rectangular barrier of height \( V_0 \) and width \( a \), where the particle energy \( E \) satisfies \( 0 < E < V_0 \). In the region inside the barrier (\( 0 < x < a \)), the time-independent Schrödinger equation takes the form:
Rearranging this yields:
Defining the decay constant \( \kappa \) as:
The general solution inside the barrier is a linear combination of growing and decaying exponentials:
Unlike the oscillatory solutions in free space, these are real exponentials. The term \( e^{\kappa x} \) grows with \( x \), while \( e^{-\kappa x} \) decays. For a finite barrier, both terms are present to satisfy the boundary conditions at both ends, but the physical interpretation is that the wavefunction penetrates into the classically forbidden region.
The Transmission Coefficient
The transmission coefficient \( T \) is defined as the ratio of the transmitted probability current to the incident probability current. For a rectangular barrier, the exact expression is derived by matching the wavefunction and its first derivative at \( x=0 \) and \( x=a \). The result is:
Here, \( \sinh(\kappa a) = \frac{e^{\kappa a} - e^{-\kappa a}}{2} \). This formula is exact for any barrier width \( a \). It shows that \( T \) is never zero for finite \( a \), confirming the possibility of tunnelling. The term \( \sinh^2(\kappa a) \) dominates the denominator, making \( T \) small when \( \kappa a \) is large.
The Thick Barrier Approximation
In many physical situations, the barrier is "thick," meaning \( \kappa a \gg 1 \). In this limit, \( e^{\kappa a} \gg e^{-\kappa a} \), so \( \sinh(\kappa a) \approx \frac{1}{2} e^{\kappa a} \). Substituting this into the exact expression:
The transmission coefficient simplifies to:
Since \( \kappa = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}} \), the exponent becomes:
This exponential dependence on the barrier width \( a \) and the mass \( m \) is the hallmark of tunnelling. A small increase in width or mass leads to a dramatic decrease in transmission probability. This approximation is widely used in nuclear physics and semiconductor device modelling.
The derivation
Tunnelling Through a Rectangular Barrier
Derives the transmission coefficient for a particle incident on a rectangular barrier, exhibiting quantum tunnelling and the thick-barrier exponential suppression.
A common misconception is that the particle "borrows" energy from the vacuum to cross the barrier. This is incorrect. The energy of the particle is conserved and remains \( E \) throughout the process. The wavefunction inside the barrier is not a travelling wave with a well-defined momentum \( p = \sqrt{2mE} \); rather, it is a standing wave composed of evanescent modes. The probability current is continuous across the barrier, but the local momentum is not defined in the classical sense within the forbidden region. C
Worked example
Problem. An electron with kinetic energy \( E = 2.0\ \mathrm{eV} \) encounters a rectangular potential barrier of height \( V_0 = 5.0\ \mathrm{eV} \) and width \( a = 1.0\ \mathrm{nm} \). Calculate the transmission coefficient \( T \) using the thick barrier approximation. Use \( m_e = 9.11 \times 10^{-31}\ \mathrm{kg} \), \( \hbar = 1.055 \times 10^{-34}\ \mathrm{J \cdot s} \), and \( 1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J} \).
- Convert energies to Joules: \( E = 2.0 \times 1.602 \times 10^{-19} = 3.204 \times 10^{-19}\ \mathrm{J} \), \( V_0 = 5.0 \times 1.602 \times 10^{-19} = 8.010 \times 10^{-19}\ \mathrm{J} \).
- Calculate the energy deficit: \( V_0 - E = 4.806 \times 10^{-19}\ \mathrm{J} \).
- Compute \( \kappa \): \( \kappa = \sqrt{\frac{2(9.11 \times 10^{-31})(4.806 \times 10^{-19})}{(1.055 \times 10^{-34})^2}} = \sqrt{\frac{8.753 \times 10^{-49}}{1.113 \times 10^{-68}}} = \sqrt{7.864 \times 10^{19}} \approx 8.87 \times 10^9\ \mathrm{m^{-1}} \).
- Calculate the exponent \( 2\kappa a \): \( 2(8.87 \times 10^9)(1.0 \times 10^{-9}) = 17.74 \).
- Calculate the prefactor: \( \frac{16 E (V_0 - E)}{V_0^2} = \frac{16(3.204 \times 10^{-19})(4.806 \times 10^{-19})}{(8.010 \times 10^{-19})^2} = \frac{2.467 \times 10^{-36}}{6.416 \times 10^{-37}} \approx 3.845 \).
- Compute \( T \): \( T \approx 3.845 \times e^{-17.74} \approx 3.84 \times 1.98 \times 10^{-8} \approx 7.6 \times 10^{-8} \).
Units check. The prefactor is dimensionless as it is a ratio of energies. The exponent \( \kappa a \) is dimensionless (\( \mathrm{m^{-1}} \cdot \mathrm{m} \)). The result is a dimensionless probability, as required.
Three quick tests
- What happens to the transmission coefficient \( T \) if the barrier width \( a \) is doubled in the thick barrier approximation?
Answer
The exponent \( -2\kappa a \) becomes \( -4\kappa a \). Since \( T \propto e^{-2\kappa a} \), doubling \( a \) squares the exponential factor, reducing \( T \) by a factor of \( e^{-2\kappa a} \), which is a dramatic decrease. - Why is the wavefunction inside the barrier exponential rather than oscillatory?
Answer
Because \( E < V_0 \), the kinetic energy term in the Schrödinger equation is negative, leading to a positive coefficient for the second derivative. This yields real exponential solutions rather than complex sinusoidal ones. - Does the transmission coefficient depend on the mass of the particle?
Answer
Yes. \( \kappa \propto \sqrt{m} \), so the exponent \( -2\kappa a \propto -\sqrt{m} \). Heavier particles have a smaller transmission probability for the same barrier parameters.
Problem set for PU-202 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.