Tunneling That ClicksEpisode 3 / Climb over, or slip through

Cool it down and at some temperature the reaction refuses to stop ── and the 2π that appears there is the Unruh 2π

Climb over, or slip through There are two ways over a wall: climb it thermally, or slip through it.
Both are exponentials, and both can be written in the language of imaginary time ── so they must swap somewhere.
$$T_0=\frac{\hbar\omega_b}{2\pi k_B}$$

Tools you'll need: \(e^{-2S_E/\hbar}\) from Episode 1, "Temperature That Clicks" Episodes 2, 6 and 7 The heart of this episode: two periods coinciding

In chemistry you learn that reaction rates rise steeply with temperature ── the Arrhenius law \(k\propto e^{-E_a/k_BT}\), the probability of climbing over a wall thermally. But cool the system down and below some temperature the law stops working. The reaction should stop, and it does not. The rate becomes independent of temperature and settles at a constant value all the way toward absolute zero. It has given up climbing over and switched to slipping through. The switch happens at \(T_0=\hbar\omega_b/2\pi k_B\) ── and this \(2\pi\) is the same \(2\pi\), arriving for the same reason, as in the Unruh temperature \(T=\hbar a/2\pi ck_B\) from the finale of the sister series "Temperature That Clicks." Heat and tunneling were in the same arena from the start.

01Two ways over a wall

Climb over thermallySlip through
Probability\(e^{-E_b/k_BT}\)\(e^{-2S_E/\hbar}\)
Temperature dependenceyes (cooling stops it)none (cooling changes nothing)
What mattersonly the wall's heightthe wall's height, width and the particle's mass
Proposed byArrhenius (1889)Gamow and others (1928)

Set them side by side and one thing is immediate ── one changes with temperature and the other does not. So cooling far enough must flip which one wins. At high temperature heat wins; at low temperature tunneling wins.

Experimentally this shows up as a kink in the Arrhenius plot. Plot \(\log(\text{rate})\) against \(1/T\) and the thermal regime is a straight line of slope \(-E_b/k_B\), while the tunneling regime is horizontal. The position of the kink is \(T_0\).

02Finding the crossover temperature

The crossover comes from setting the two exponents equal. For that we need the tunneling exponent \(2S_E/\hbar\). The standard move is to approximate the top of the wall as a parabola ── call the (imaginary) frequency at the summit \(\omega_b\).

The Euclidean action of a parabolic wall

Take the wall to be \(V(x)=E_b-\frac12 m\omega_b^2x^2\) and the particle's energy to be \(0\), and feed it into Episode 1's formula. The turning points are at \(x_0=\sqrt{2E_b/m}\,/\,\omega_b\), and

$$S_E=\int_{-x_0}^{x_0}\!\!\sqrt{2m\big(V(x)\big)}\,dx =\sqrt{2mE_b}\;x_0\int_{-1}^{1}\!\sqrt{1-u^2}\,du=\frac{\pi E_b}{\omega_b}$$

So the tunneling exponent is

$$\frac{2S_E}{\hbar}=\frac{2\pi E_b}{\hbar\omega_b}$$

Here is the satisfying part. Read that in the same form as \(E_b/k_BT\) and ──

The heart of this episode ── the crossover temperature
$$\frac{2S_E}{\hbar}=\frac{2\pi E_b}{\hbar\omega_b}\ \equiv\ \frac{E_b}{k_BT_0} \qquad\Longleftrightarrow\qquad \boxed{\ T_0=\frac{\hbar\omega_b}{2\pi k_B}\ }$$

In other words ── the probability of tunneling is exactly the probability of getting over thermally at temperature \(T_0\). So for \(T>T_0\) heat is faster, and for \(T

Put another way: however far you cool it, the system keeps behaving as though a heat bath at temperature \(T_0\) were present. Tunneling is also an apparent temperature that survives absolute zero.

03This 2π is the Unruh 2π

Why did a \(2\pi\) appear? "It came out of the parabolic integral" is true as far as it goes, but there is a deeper reason.

One story: two periods coinciding
The "period" in the imaginary-time directionWhat it fixes
having a temperature ("Temperature That Clicks" Ep. 6)\(\beta\hbar=\hbar/k_BT\)the Boltzmann factor
tunneling (this episode)\(2\pi/\omega_b\)one round trip across the inverted valley
an accelerating observer ("Temperature That Clicks" Ep. 7)\(2\pi c/a\)the Unruh temperature

As Episode 1 showed, tunneling is a round trip in imaginary time. The inverted valley (= the original wall) is parabolic near the summit, so the period of that round trip is the harmonic one, \(2\pi/\omega_b\).
Meanwhile, in a system at temperature \(T\) the imaginary-time direction is a circle of circumference \(\beta\hbar\). Whether the round trip fits on that circle decides the contest ── if \(\beta\hbar > 2\pi/\omega_b\) (cold, long circle) the round trip fits and tunneling happens; if \(\beta\hbar < 2\pi/\omega_b\) (hot, short circle) it does not, and the only way over is thermal. The boundary is

$$\frac{\hbar}{k_BT_0}=\frac{2\pi}{\omega_b}\qquad\Longrightarrow\qquad T_0=\frac{\hbar\omega_b}{2\pi k_B}$$
Identical down to the lettering In Episode 7 of "Temperature That Clicks," the imaginary time of an accelerating observer becomes an angle; an angle has period \(2\pi\); so the period is \(2\pi c/a\) and out comes \(T=\hbar a/2\pi ck_B\).
Here, the round trip at the summit is harmonic motion; so its period is \(2\pi/\omega_b\), and out comes \(T_0=\hbar\omega_b/2\pi k_B\).
Both are the same single line: "the period in the imaginary-time direction is \(2\pi/(\text{some angular frequency})\)." The acceleration \(a/c\) and the barrier frequency \(\omega_b\) are sitting in the same seat. Setting these two formulas side by side is the best view in the first half of this series.

04Try it ── watch the kink

The figure plots the two lines just derived. Horizontal axis: temperature (log). Vertical: reaction rate (log). The line falling to the left is thermal, the flat line is tunneling, and they meet exactly at \(T_0\).

The preset buttons switch to real systems. Check that the same logic covers six orders of magnitude in temperature, from chemical reactions through hydrogen in metals to superconducting circuits.

Figure: an Arrhenius plot (horizontal = log temperature, vertical = log rate). Blue = climbing over thermally (falling to the left), red = tunneling (flat). They cross at T₀ = ℏω_b/2πk_B. Below T₀ the rate no longer falls however far you cool
climbing over, e^(−E_b/k_BT) tunneling, e^(−2S_E/ℏ) what is actually observed (the larger of the two)

05In orders of magnitude ── where the kink sits

Systemtypical ω_bT₀Meaning
a chemical reaction moving a hydrogen10¹⁴ s⁻¹about 120 Ktunneling takes over above liquid nitrogen
a reaction moving a heavy atom (carbon etc.)10¹³ s⁻¹about 12 Kinvisible in an ordinary laboratory
hydrogen diffusing in a metal2×10¹³ s⁻¹about 25 Kdiffusion becomes temperature-independent at low T
a Josephson junction (macroscopic quantum tunneling)10¹¹ s⁻¹about 0.1 Kdilution-refrigerator territory. On to Episode 4

The lighter the particle, the larger \(\omega_b\) and the higher \(T_0\) (\(\omega_b\propto1/\sqrt m\)). So "tunneling matters" applies first to hydrogen, then deuterium, then… That mass dependence becomes the tool for detecting tunneling inside enzymes in bonus ⑤ (swap hydrogen for deuterium; if the rate drops by orders, it was tunneling).

A chemical reaction that proceeds at 11 K In 2011, Schreiner and colleagues reported that a molecule called methylhydroxycarbene spontaneously isomerises inside solid argon at 11 K. The barrier is 28 kJ/mol ── about 300 times the thermal energy at 11 K. Thermally it would not happen once in \(10^{120}\) seconds. It proceeds in hours.
They called this "tunneling control." A product that is favoured neither thermodynamically nor kinetically gets selected for a third reason: the wall is thin. In low-temperature chemistry this genuinely happens.

06The neighbourhood of the boundary is messier

To be honest: \(T_0\) is not a clean "phase transition." In reality both routes are open at once, and the fastest path is often "climb partway thermally, then tunnel the rest" (thermally assisted tunneling). The kink in the figure is, in practice, rounded rather than sharp.

Even so, \(T_0\) is a good landmark. Above or below it, the kind of behaviour changes. And that the boundary can be written in the tidy form \(\hbar\omega_b/2\pi k_B\) is itself evidence that heat and tunneling are written in the same language.

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

Established: the Arrhenius law \(k\propto e^{-E_a/k_BT}\); that Arrhenius plots bend and become tunneling-dominated at low temperature (observed in many systems); the Euclidean action of a parabolic barrier \(S_E=\pi E_b/\omega_b\); the crossover temperature \(T_0=\hbar\omega_b/2\pi k_B\) (Langer 1967, Affleck 1981, within the Callan–Coleman framework); that finite-temperature tunneling is decided by whether the bounce solution fits on the imaginary-time circle \(\beta\hbar\); the observation of macroscopic quantum tunneling in Josephson junctions (Voss–Webb 1981, Devoret–Martinis–Clarke 1985 and others); and the isomerisation of methylhydroxycarbene at 11 K with tunneling control (Schreiner et al. 2011, Science). All standard results.

Points to note: (1) \(T_0\) is a crossover, not a sharp transition. Near it, thermally assisted tunneling dominates and the kink is rounded (for some barrier shapes it can be sharper, "first-order-like"; classifying this is itself a research topic). (2) \(S_E=\pi E_b/\omega_b\) is a parabolic approximation and deviates for real barrier shapes; the low-temperature magnitude in particular is sensitive to shape. (3) Real systems have dissipation (coupling to the environment), which lowers \(T_0\) and suppresses tunneling (Caldeira–Leggett theory). The figure idealises this away. (4) The "rate" in the figure is a relative value with the prefactor normalised to 1, not an absolute one. (5) The \(\omega_b\) and \(T_0\) in the table are order-of-magnitude guides and move by factors of a few from system to system.

Exercises (solvable with this episode's ideas)
  1. Why does an Arrhenius plot go flat at low temperature?
    See the answer
    Because the tunneling probability \(e^{-2S_E/\hbar}\) contains no temperature. The thermal probability falls as you cool; tunneling does not. So below some temperature only tunneling is left and the rate becomes constant.
  2. What would you change to raise \(T_0\)?
    See the answer
    \(T_0=\hbar\omega_b/2\pi k_B\), so increase \(\omega_b\). Since \(\omega_b\propto1/\sqrt m\), lighter particles have higher \(T_0\) ── hydrogen bends at around 100 K, heavy atoms at around 10 K. Note that the barrier height \(E_b\) does not affect \(T_0\) at all (it affects the absolute rate).
  3. What do this episode's \(2\pi\) and the Unruh temperature's \(2\pi\) have in common?
    See the answer
    Both come from the same single line: "the period in the imaginary-time direction equals \(2\pi/(\text{an angular frequency})\)." For Unruh, one turn of an angle is \(2\pi\); here, one round trip of harmonic motion in the inverted valley is \(2\pi/\omega_b\). The acceleration \(a/c\) and \(\omega_b\) occupy the same seat.
  4. What does "tunneling is an apparent temperature that survives absolute zero" mean?
    See the answer
    Because the tunneling probability \(e^{-2S_E/\hbar}\) equals exactly the probability of getting over thermally at temperature \(T_0\), \(e^{-E_b/k_BT_0}\). However far you cool, the system keeps reacting as though a heat bath at \(T_0\) were present. Once the real temperature drops below \(T_0\), the system's effective temperature stops there.

Episode 3 summaryHeat and tunneling were in the same arena

There are two ways over a wall. Climbing over, \(e^{-E_b/k_BT}\), depends on temperature; slipping through, \(e^{-2S_E/\hbar}\), does not. So cooling must eventually flip the winner, and the Arrhenius plot bends.

For a parabolic wall, \(2S_E/\hbar=2\pi E_b/\hbar\omega_b\). Read that as \(E_b/k_BT_0\) and you get \(T_0=\hbar\omega_b/2\pi k_B\) ── which also means tunneling is an apparent temperature \(T_0\) that survives absolute zero.

And this \(2\pi\) is the Unruh \(2\pi\). Both come from the line "the period in the imaginary-time direction is \(2\pi/(\text{an angular frequency})\)" ── temperature being the imaginary-time circle \(\beta\hbar\), tunneling the round trip \(2\pi/\omega_b\) across the valley, and the crossover being where the two periods coincide. About 120 K for hydrogen, about 0.1 K for a Josephson junction: systems six orders apart, lined up by one formula.

This document is Episode 3 of the "Tunneling That Clicks" series, a reading piece for physics-loving high-schoolers and undergraduates. The Arrhenius law, the low-temperature bend of Arrhenius plots into tunneling dominance, the Euclidean action of a parabolic barrier \(S_E=\pi E_b/\omega_b\), the crossover temperature \(T_0=\hbar\omega_b/2\pi k_B\) (Langer 1967, Affleck 1981, within the Callan–Coleman framework), the criterion of whether the bounce fits on the imaginary-time circle \(\beta\hbar\), the observation of macroscopic quantum tunneling in Josephson junctions (Voss–Webb 1981, Devoret–Martinis–Clarke 1985 and others), and the isomerisation of methylhydroxycarbene at 11 K (Schreiner et al. 2011) are all established results. That \(T_0\) is a crossover rather than a sharp transition with thermally assisted tunneling dominating nearby, that \(S_E=\pi E_b/\omega_b\) is a parabolic approximation, that dissipation (Caldeira–Leggett) lowers \(T_0\) in real systems, that the rate in the figure is a normalised relative value, and that the tabulated values are order-of-magnitude guides ── all spelled out in the body's "honest line." ── To print, use your browser's "Print" and "Save as PDF" (in the print version the sliders and answers are frozen and hidden). Adjacent episodes: Episode 2, The Sun burns because it tunnels / Episode 4, When the phase lines up, tunneling becomes a current / Contents / sister series Temperature That Clicks (Episodes 6 and 7, and Bonus ①).

Print / make a PDF: ⌘+P (Ctrl+P on Windows). On screen, moving the barrier height and summit frequency shifts the position of the kink T₀. The preset buttons travel six orders of magnitude in temperature, from chemical reactions to Josephson junctions. "See the answer" opens each solution.