Tunneling That ClicksEpisode 1 / Tunneling is just walking, in imaginary time

Getting to the far side of a wall without going over it ── what is strange is not the phenomenon but the fact that we are watching in real time

Tunneling is just walking, in imaginary time Rotate time 90° onto the imaginary axis and the wall flips into a valley.
The particle merely walks across a slope ── and the fee is \(e^{-2S_E/\hbar}\).
Tunneling is not "unlikely"; it rides on the exponent. That is the whole point.

Tools you'll need: waves and exponentials, square roots, imaginary time from "Temperature That Clicks" Episode 6 The heart of this episode: P ≈ e^(−2S_E/ℏ)

A ball rolling up a slope cannot get over a hill taller than itself. Obviously. Yet an electron does ── or rather, it does not go over; it appears on the far side. This is tunneling, and in most introductions it is presented as "isn't that strange" and left there. But what is strange is not the phenomenon; it is the fact that we are watching in real time. Rotate time 90° onto the imaginary axis (the Wick rotation from Episode 6 of the sister series "Temperature That Clicks") and the wall flips into a valley. Now the particle is not jumping over anything ── it is simply walking across an ordinary slope. The "action" spent on that stroll, \(S_E\), rides straight onto the exponent of the probability. This series uses that single viewpoint to thread together the burning of the Sun, superconductivity, and the collapse of the vacuum.

01First, what happens if you look in real time

Fire a particle of energy \(E

Inside the wall, the wave stops oscillating and starts decaying

Outside the wall (\(E>V\)) the wave oscillates with wavenumber \(k=\sqrt{2mE}/\hbar\). Inside (\(Epurely imaginary, and

$$\psi(x)\propto e^{ikx}\ \xrightarrow{\ k\to i\kappa\ }\ e^{-\kappa x}, \qquad \kappa=\frac{\sqrt{2m(V-E)}}{\hbar}$$

Oscillation has turned into decay. So the wave dies away inside the wall but never reaches zero. If the wall has finite thickness \(a\), a little is left on the far side, and there it starts oscillating again. That is tunneling.
The transmission probability falls off roughly as the square of the amplitude:

$$P\ \approx\ e^{-2\kappa a}$$

Put numbers in for an electron. With \(V-E=1\) eV, \(\kappa\approx5.1\ \mathrm{nm^{-1}}\). A wall 1 nm thick gives \(e^{-10}\approx4.5\times10^{-5}\) ── one attempt in twenty thousand gets through. Make it 2 nm and \(e^{-20}\approx2\times10^{-9}\): four orders of magnitude gone at a stroke.

That feel ── "double the thickness, lose four orders" ── is the protagonist of this whole series. Tunneling rides on the exponent.

02Rotate to imaginary time, and the wall becomes a valley

The decisive step in that calculation was where "the square root went negative and \(k\) became imaginary." When an imaginary number appears, rotate time ── that is the move learned in Episode 6 of the sister series "Temperature That Clicks."

The heart of this episode ── the wall inverts

Conservation of energy in real time reads

$$\tfrac12 m\dot x^2 = E-V(x)$$

Inside the wall the right-hand side is negative, so there is no real-velocity solution. Now set \(t=-i\tau\) (imaginary time), so that \(\dot x^2\to-(dx/d\tau)^2\):

$$\tfrac12 m\left(\frac{dx}{d\tau}\right)^2 = \underbrace{V(x)-E}_{\textstyle \text{the sign has flipped!}}$$

The right-hand side is now positive. In other words ── in imaginary time, the potential turns upside down. The mountain you could not cross in real time is a valley in imaginary time, and the particle rolls across it perfectly ordinarily.
Tunneling is not "something impossible happening by chance"; it is ordinary motion, seen along a different time axis.

And computing the action of that stroll (the Euclidean action) gives:

The fee is the Euclidean action $$S_E=\int_{x_1}^{x_2}\sqrt{2m\big(V(x)-E\big)}\;dx \qquad\Longrightarrow\qquad \boxed{\ P\approx e^{-2S_E/\hbar}\ }$$

\(x_1,x_2\) are the two points where the wall meets the energy line (the turning points). For a square wall \(S_E=\hbar\kappa a\), which returns us to \(e^{-2\kappa a}\).
What makes this form powerful is that it works for a wall of any shape, in one line (the WKB approximation). The Sun in Episode 2 and the vacuum decay in Episode 7 are both applications of it.

The relation to "Temperature That Clicks," Episode 6 There we wrote that "temperature is the imaginary-time direction being a circle of circumference \(\hbar/k_BT\)." Here we write that "tunneling is one round trip in the imaginary-time direction."
Both are classical motion in imaginary time. The only difference is whether you go round a circle (heat) or back and forth across a valley (tunneling). Which is why, as Episode 3 will show, the two must swap places at some temperature ── they are in the same arena.

03Try it ── the wall, the wave, and the inverted valley

The figure below is an electron fired at a square wall (the exact solution, not an approximation). The upper panel is the real-time picture ── the wall, the energy line, and \(|\psi|\). You can see it thinning exponentially inside the wall. The dashed curve is the inverted potential as seen in imaginary time, where the wall is a valley.

The lower panel is the transmission probability on a logarithmic axis. Nudge a slider and the vertical axis jumps by orders of magnitude ── that is what "riding on the exponent" feels like. Check that adding just 0.1 nm of thickness changes the probability by a factor of ten.

Figure: exact solution for an electron hitting a 5 eV square wall. Above = the wall (shaded), the energy line, and |ψ| (decaying exponentially inside). The dashed line is the inverted potential in imaginary time ── the wall is a valley. Below = transmission probability on a log axis, which moves by orders with thickness and energy
|ψ| (exact) energy E inverted potential in imaginary time transmission probability

04In orders of magnitude ── why alpha decay spans 24 of them

The power of "riding on the exponent" first ran wild in alpha decay. In 1928 Gamow (and independently Gurney and Condon) explained it as tunneling ── the first application of quantum mechanics to the nucleus.

An alpha particle is trapped inside a nucleus and tunnels out through the Coulomb barrier. What is interesting is how wildly the half-lives scatter from nucleus to nucleus.

Nuclidealpha energyhalf-life
²¹²Po8.78 MeV0.3 microseconds
²²²Rn5.49 MeV3.8 days
²²⁶Ra4.87 MeV1600 years
²³⁸U4.27 MeV4.5 billion years

The energies differ by only a factor of two, yet the half-lives differ by 24 orders of magnitude. "That's just how it is" will not do. But with \(P\approx e^{-2S_E/\hbar}\) it is inevitable ── change the number on the exponent by a few tens, and the answer changes by a few tens of orders. The empirical relation (the Geiger–Nuttall rule) had been known since 1911; Gamow's theory explained what was inside the exponent.

How to read this series

From here on, the only thing we watch is what is riding on the exponent. The prefactor only has to be roughly right. Since a change of a few tens on the exponent moves the answer by a few tens of orders, a factor of a few out front is within the noise ── please carry that instinct with you.
It is also what decides the verdict in bonus episode ①, on cold fusion: why the argument "the metal lattice must help a bit" fails as a matter of orders of magnitude.

05While it is tunneling, where is the particle?

Let me answer in advance the question that always comes up. "During the time it is inside the wall, isn't the particle violating energy conservation?"

It is not. There is no such observation as "the particle inside the wall." That \(\psi\ne0\) inside the wall does not mean you can find the particle there (any attempt to look would inject energy through the measurement itself). What is observable is "it went in" and "it came out"; in between it is a matter of amplitudes.

Then how many seconds does tunneling take? ── this is, in fact, still an unresolved problem. Several definitions exist and they give different answers, and in some readings the effective speed appears to exceed light. It is interesting enough that bonus episode ④ is devoted entirely to it.

◇ ◇ ◇
The honest line ── how far is this established?

Established: the exact transmission coefficient for a rectangular barrier; that the wavenumber becomes purely imaginary for \(E

Points to note: (1) WKB is an approximation. It is good when \(S_E/\hbar\gg1\) (tunneling sufficiently unlikely) and needs corrections near the turning points. The upper panel of the figure is exact; the button overlays WKB so the discrepancy is visible. (2) "Walking in imaginary time" is a restatement of the fact that the path integral's saddle point lies on the imaginary-time side, not a claim that the particle traverses a second, physically existing time (the same caveat as Episode 6 of "Temperature That Clicks"). (3) A particle inside the barrier cannot be directly observed, and \(\psi\ne0\) does not mean "it will be found there." (4) How long tunneling takes is an unresolved problem and is not treated here (bonus ④). (5) The figure idealises a 1D, stationary, spinless problem; in real solids and nuclei many-body effects matter.

Exercises (solvable with this episode's ideas)
  1. Double the wall's thickness. What happens to the transmission probability?
    See the answer
    Since \(P\approx e^{-2\kappa a}\), sending \(a\to2a\) gives \(P\to P^2\) ── not "half" but "squared." What was \(10^{-5}\) becomes \(10^{-10}\). That is what riding on the exponent means.
  2. Why does moving to imaginary time turn the wall into a valley?
    See the answer
    Setting \(t=-i\tau\) in \(\frac12m\dot x^2=E-V\) flips the sign of \(\dot x^2\), giving \(\frac12m(dx/d\tau)^2=V-E\). The sign on the right swaps ── i.e. the potential turns upside down. Only inside the wall does real motion become possible.
  3. Why do half-lives span 24 orders when alpha energies differ by only a factor of two?
    See the answer
    Because the half-life goes as \(e^{+2S_E/\hbar}\) and \(S_E/\hbar\) changes a great deal with energy. A change of a few tens on the exponent means a few tens of orders in the answer. It looks anomalous to linear intuition and is obvious in exponential terms.
  4. Fire a proton (about 1836 times heavier) at the same wall instead of an electron. Does it tunnel more easily?
    See the answer
    Orders of magnitude less easily. \(\kappa=\sqrt{2m(V-E)}/\hbar\propto\sqrt m\), so 1836 times the mass means about 43 times the \(\kappa\). With the exponent 43 times larger the probability is annihilated. Tunneling is a privilege of light particles ── a fact that is the key to both the muon (207 times the electron) in bonus ② and the isotope effect in bonus ⑤.

Episode 1 summaryWhat was strange was not the phenomenon but watching in real time

A particle without enough energy appears beyond the wall. Written in real time, the wavenumber inside the wall becomes purely imaginary and the wave stops oscillating, decaying as \(e^{-\kappa x}\) ── it dies away but never reaches zero, so a thin wall leaks a little to the far side.

Rotate time onto the imaginary axis and the strangeness disappears. Setting \(t=-i\tau\) flips the sign in the conservation of energy and the wall becomes a valley. The particle leaps over nothing; it walks across in imaginary time. The action of that stroll is \(S_E=\int\sqrt{2m(V-E)}\,dx\), and the probability is \(P\approx e^{-2S_E/\hbar}\).

What matters is that this quantity rides on the exponent. Alpha half-lives span 24 orders of magnitude for a factor-of-two change in energy. From here on this series watches only what is on the exponent, not the prefactor.

This document is Episode 1 of the "Tunneling That Clicks" series, a reading piece for physics-loving high-schoolers and undergraduates. The exact transmission coefficient for a rectangular barrier, exponential decay of the wavefunction for \(EEpisode 2, The Sun burns because it tunnels / Contents / sister series Temperature That Clicks (Episode 6) and Quantum That Clicks.

Print / make a PDF: ⌘+P (Ctrl+P on Windows). On screen, the energy and thickness sliders move the decay of the wave and the transmission probability. Check that adding just 0.1 nm of thickness makes the vertical axis jump by an order of magnitude. "See the answer" opens each solution.