Quantum Tunneling Probability

Q: Does tunneling mean the particle borrows energy?

No. In this picture the particle keeps its energy $E$. Tunneling is not a way of climbing over the barrier. The wave has a small chance of reaching the far side.

Q: Why does a thicker barrier block so much more?

The chance depends on $e^{-2\kappa L}$, so every extra bit of width multiplies the chance by the same factor. In the example, $2\kappa L$ grows by about 1.7 for each extra 0.1 nm, which cuts the chance to about 0.18 of its value.

Q: Is the formula exact?

No. It is a thick-barrier estimate, and it drops a prefactor that a fuller calculation contains. For the example here, the estimate and the full calculation differ by less than 0.2 percent, but for other energies the difference can be larger.

In one sentence: A quantum particle can sometimes pass through a barrier that is too high for it to climb over, and this formula gives a rough estimate of how often.

$T \approx e^{-2\kappa L}$

Read it out loud: “T is approximately equal to e to the power of minus 2 kappa L. T is the chance that the particle gets through the barrier, a number from 0 to 1. e is Euler's number, about 2.718. Kappa (KAP-uh) is how fast the wave fades inside the barrier. L is the width of the barrier. Minus 2 kappa L is the exponent: two times kappa times L, with a minus sign in front.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Powers of ten, like $3.0 \times 10^{-10}$ m: The example uses numbers such as a barrier width of $3.0 \times 10^{-10}$ m and an electron mass of $9.1093837 \times 10^{-31}$ kg. Each power of ten says how far to move the decimal point. The exponent $-10$ moves it ten places to the left, so $3.0 \times 10^{-10}$ m is 0.0000000003 m, which is 0.3 nanometers. To multiply powers of ten, add the exponents: $10^{-10} \times 10^{10} = 10^{0} = 1$ .
Euler’s number $e$ and the exponential $e^{x}$: The letter $e$ here stands for Euler’s number, about 2.71828, a fixed number from mathematics. It appears whenever something shrinks or grows in proportion to its own size. A negative power such as $e^{-5}$ means one divided by $e^{5}$ , which is a very small number, about 0.0067. The more negative the exponent, the smaller the result. This letter is not the electron’s charge.
Probability as a chance from 0 to 1: A probability is a chance written as a number from 0 to 1. Zero means the event never happens, and 1 means it always happens. A percent is a part out of one hundred, so a probability of 0.0058 is 0.58 percent. That means about 58 times in 10,000 tries. The probabilities of all possible outcomes add up to 1.
The Greek letter $\kappa$ (kappa): The Greek letter $\kappa$ is read KAP-uh. In this formula it is the decay constant, which says how fast the wave dies away as it enters the barrier. Its unit is 1/m, one per meter. So $\kappa L$ has no unit, because one per meter times meters cancels. A large $\kappa$ means the wave fades quickly, so the particle has a small chance of getting through.
Potential energy and the barrier height $V_0$: Potential energy is energy stored because of position, like a ball resting at the top of a hill. A barrier is a region where the potential energy is high. Its height is written $V_0$ , the potential energy at the top of the barrier, measured in joules or electron volts. A particle with energy $E$ below $V_0$ does not have enough energy to climb over the top in ordinary physics.
Electron volts and the joule: An electron volt (eV) is a small unit of energy, the energy an electron gains when pushed through one volt. It equals $1.602176634 \times 10^{-19}$ J. The joule (J) is much larger: it is roughly the energy needed to lift a 100 gram apple about one meter. Atomic energies are written in electron volts, because joules would give very long decimal numbers.

What each symbol means

Symbol	Meaning
$T$	The transmission probability, the chance that a particle gets through the barrier. It is a number from 0 to 1 and has no unit. A value of 0.0058 means about 0.58 percent of the particles get through, on average.
$e$	Euler’s number, about 2.71828, a fixed number from mathematics and the base of natural logarithms. It has no unit. This letter is not the elementary charge of the electron, although that charge is also written $e$ in other places.
$\kappa$	The decay constant inside the barrier, in 1/m (per meter). It equals $\kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar}$ . A large $\kappa$ means the wave dies out quickly as it enters the barrier, so the chance of getting through is small.
$L$	The width of the barrier, the distance the particle would have to cross, in meters (m). In the example, $L = 0.30$ nm $= 3.0 \times 10^{-10}$ m. Doubling the width squares the chance, which makes it much smaller.
$m$	The mass of the particle, in kilograms (kg). For an electron, $m = 9.1093837 \times 10^{-31}$ kg. A heavier particle has a larger $\kappa$ , so it tunnels less often.
$E$	The energy of the particle, in joules (J), or in electron volts when that is more convenient. The formula assumes $E < V_0$ , so the particle does not have enough energy to climb over the barrier. In the example, $E = 0.20$ eV.
$V_0$	The height of the barrier, the potential energy at its top, in joules (J). In the example, $V_0 = 3.0$ eV, which is $4.8 \times 10^{-19}$ J, because one electron volt is $1.602 \times 10^{-19}$ J.
$\hbar$	The reduced Planck constant, $1.054571817 \times 10^{-34}$ J s, which is Planck’s constant divided by $2\pi$ . It sets how strongly quantum effects show up. A smaller $\hbar$ would make $\kappa$ larger.

The formula piece by piece

$V_0 - E$: Start with the missing energy. $V_0 - E$ is the barrier height minus the particle’s energy, which is how much energy the particle lacks to climb over the top. In the example, $3.0$ eV $- 0.20$ eV $= 2.8$ eV. A bigger gap makes the wave fade faster.
$\kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar}$: Next, $\kappa$ is found from the missing energy, the mass and $\hbar$ . The square root of $2m(V_0 - E)$ grows with both the mass and the missing energy, and dividing by $\hbar$ turns it into a fade rate in 1/m. Heavy particles and tall barriers therefore tunnel less.
$2\kappa L$: Multiply the fade rate by the width and by 2. The product $\kappa L$ counts how many fade lengths fit across the barrier. The factor 2 comes from the probability, which is the wave squared. In the example, $2\kappa L \approx 5.14$ .
$e^{-2\kappa L}$: Use $e$ raised to the power of minus $2\kappa L$ . The minus sign means the chance shrinks exponentially as the barrier gets wider or the fade rate gets bigger. In the example this gives a number of about 0.0058.
$T \approx e^{-2\kappa L}$: The whole formula in one sentence: the chance of getting through the barrier is about $e$ raised to the power of minus two times the fade rate times the width.

What the formula is doing, and why

In everyday life, a ball that lacks the energy to roll over a hill stays on its side. In quantum mechanics, the wave of a particle does not stop sharply at the edge of a barrier. It fades inside the barrier, and a small part of it can reach the far side. What goes in is the energy, the barrier height and width, and the mass. What comes out is the chance $T$ that the particle gets through.

Why the missing energy and the mass? The energy $E$ is the particle’s energy and $V_0$ is the barrier height, so $V_0 - E$ is the energy the particle is missing. The fade rate $\kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar}$ grows with the square root of the mass and with the missing energy. Heavy particles and tall barriers make the wave fade faster, so they tunnel less.

Why the exponential? The chance falls exponentially with the width $L$ . Each extra bit of width multiplies $T$ by the same fraction, rather than subtracting the same amount. In this estimate, doubling the width squares $T$ , which makes the chance drop very fast. The shape comes from the wave fading as $e^{-\kappa x}$ inside the barrier, and the probability is the wave squared, which doubles the exponent.

Does the particle climb over? No. Tunneling is not a way of climbing over the barrier. The particle keeps its energy $E$ and does not borrow energy. The result is a chance, and most attempts fail, so the particle is usually reflected. Quantum mechanics says the particle has a small chance of appearing on the far side, and no sudden jump of energy is needed.

How good is the estimate? This is a thick-barrier estimate. The fuller thick-barrier result carries an extra prefactor, $\frac{16E}{V_0}\left(1 - \frac{E}{V_0}\right)$ , which the simple formula drops. In the example below this prefactor is about 0.996, so it hardly changes the answer. It is not always close to 1. It equals 4 when $E$ is half of $V_0$ , and it is much less than 1 when $E$ is far below $V_0$ .

An everyday picture

Think of the sound of a neighbor’s music through a wall. A thin wall lets through plenty of bass, while a thick wall removes much more. Each extra bit of thickness takes away the same fraction of what is left, though a little always gets through. Tunneling works the same way: the wave fades exponentially inside the barrier, so a thicker barrier lets far less through, but never exactly nothing.

Worked example: An electron meets a 3.0 eV barrier

An electron with energy $E = 0.20$ eV meets a barrier of height $V_0 = 3.0$ eV and width $L = 0.30$ nm $= 3.0 \times 10^{-10}$ m. The electron mass is $m = 9.1093837 \times 10^{-31}$ kg. We estimate the chance $T$ that the electron tunnels through, and then how the chance changes if the barrier is twice as wide.

Find the missing energy. The barrier is 3.0 eV and the electron has 0.20 eV, so the missing energy is $V_0 - E = 3.0 - 0.20 = 2.8$ eV. The formula needs joules, so convert with $1\ \text{eV} = 1.602 \times 10^{-19}$ J: $2.8 \times 1.602 \times 10^{-19} = 4.486 \times 10^{-19}$ J.
Compute the fade rate $\kappa$ . Multiply $2m(V_0 - E)$ : $2 \times 9.1093837 \times 10^{-31} \times 4.486 \times 10^{-19} \approx 8.173 \times 10^{-49}$ kg J. For the square root, write this as $81.73 \times 10^{-50}$ so the exponent is even, then $\sqrt{81.73 \times 10^{-50}} \approx 9.041 \times 10^{-25}$ . Divide by $\hbar$ : $\kappa \approx \frac{9.041 \times 10^{-25}}{1.0546 \times 10^{-34}} \approx 8.573 \times 10^{9}$ m $^{-1}$ .
Multiply the fade rate by the width. $\kappa L = (8.573 \times 10^{9}\ \text{m}^{-1})(3.0 \times 10^{-10}\ \text{m})$ . The front numbers give $8.573 \times 3.0 = 25.72$ , and the powers of ten give $10^{9} \times 10^{-10} = 10^{-1}$ . So $\kappa L \approx 2.572$ , with the meters canceling.
Double it to get the exponent. The exponent is $2\kappa L = 2 \times 2.572 = 5.144$ . The factor 2 is there because the chance is the wave squared, so the fade is counted twice.
Estimate the chance. $T \approx e^{-5.144}$ . A calculator gives about 0.0058, which is 0.58 percent. In words, roughly 1 particle in 170 gets through this barrier, which is small but not zero.
Check the dropped prefactor. The fuller result multiplies by $\frac{16E}{V_0}\left(1 - \frac{E}{V_0}\right) = 16 \times \frac{0.20}{3.0} \times \frac{2.8}{3.0} \approx 0.996$ . That changes the answer very little. A complete calculation for this rectangular barrier gives 0.584 percent, which agrees with the estimate to less than 0.2 percent.
Double the width. Doubling $L$ doubles the exponent, so the chance becomes $T$ squared. In this estimate the chance is about 171 times smaller: $e^{2\kappa L} = e^{5.144} \approx 171$ . Going from 0.30 nm to 0.60 nm makes tunneling far less likely.

Answer: The chance that the electron tunnels is about 0.58 percent, or roughly 1 in 170. That is small but not zero. Doubling the barrier width to 0.60 nm would make the chance about 171 times smaller.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Thinking a particle with less energy than the barrier can never get through. Quantum particles have a small chance of tunneling, and the formula estimates that chance.
Using the barrier height $V_0$ where the formula needs the missing energy $V_0 - E$ . Subtract first, then take the square root.
Leaving the energies in electron volts. The formula needs joules, so convert $V_0 - E$ with $1\ \text{eV} = 1.602 \times 10^{-19}$ J. Skipping this makes $\kappa$ about $2.5 \times 10^{9}$ times too large.
Treating the answer as exact. The formula drops a prefactor, so it estimates the exponential part only. For some energies it can be off by a factor of a few.

Where you will meet it

Scanning tunneling microscopes, which pass a tiny current across a gap of less than a nanometer. The current depends steeply on the gap, which lets the microscope image single atoms.
Alpha decay, where an alpha particle escapes a heavy nucleus by tunneling through the energy barrier that holds it inside.
Nuclear fusion in the Sun, where protons must tunnel through their electrical repulsion to get close enough to fuse.
Flash memory, where electrons tunnel through a thin insulating oxide layer to store or erase data.

Questions people ask

Does tunneling mean the particle borrows energy?

No. In this picture the particle keeps its energy $E$ . Tunneling is not a way of climbing over the barrier. The wave has a small chance of reaching the far side.

Why does a thicker barrier block so much more?

The chance depends on $e^{-2\kappa L}$ , so every extra bit of width multiplies the chance by the same factor. In the example, $2\kappa L$ grows by about 1.7 for each extra 0.1 nm, which cuts the chance to about 0.18 of its value.

Is the formula exact?