Part 3 · One-dimensional problems

3.4 The potential barrier and tunnelling

Source: the barrier, its matching conditions, $t$ below and above $U_{0}$ , $T (E)$ and the tip over a surface from the 2024/25 board #8; this year’s course has not reached it yet

Objectives

Understand why a particle can get through a barrier it classically cannot cross, and why it can bounce off one it classically sails over.
Recipe: find the reflection and transmission amplitudes from the matching conditions.
Picture: a wave hitting a barrier, decaying through it and continuing smaller; $T (E)$ against the classical step.
Theorem: $T \approx \frac{16 E ( U _{0} - E )}{U _{0}^{2}} e^{- 2 κa}$ for a thick barrier.
Vocabulary: reflection and transmission amplitude and probability, tunnelling, transmission resonance, scanning tunnelling microscope.

$U (x) = ⎩ ⎨ ⎧ 0 U_{0} 0 x \leq 0 0 < x \leq a x > a$ A wall of height $U_{0}$ and thickness $a$ , free motion on both sides. The particle comes from the left with energy $E$ . This is the reverse of the finite well: there the particle sat in a dip; here it meets a bump.

Classically, $E = E_{k} + U$ is conserved. A ball rolling at a hill with $E$ below the top climbs until $E_{k} = 0$ , stops and rolls back: always reflected. With $E$ above the top it slows over the hill and passes: always transmitted. Quantum mechanics gives neither.

The solution in each region

The notation: $k_{0} = 2 m E /ℏ$ outside, $κ = 2 m (U_{0} - E) /ℏ$ inside. Take $E < U_{0}$ first (Definition 3.1.1): $I (x \leq 0) : II (0 < x \leq a) : III (x > a) : u = A e^{i k_{0} x} + B e^{- i k_{0} x}, u = C e^{- κ x} + D e^{κ x}, u = E^{'} e^{i k_{0} (x - a)} + F e^{- i k_{0} (x - a)} .$ In I, $e^{i k_{0} x}$ is the forward wave, momentum $+ ℏ k_{0}$ , and $e^{- i k_{0} x}$ the reflected one. In II both exponentials stay: the region is finite, so neither blows up. In III the exponents are written in $x - a$ so they equal 1 at the boundary, a convenience. (The board calls the transmitted amplitude $E$ ; here it is $E^{'}$ , to keep it apart from the energy.)

Nothing comes from the right, so $F = 0$ . Only ratios matter: divide by the incoming $A$ .

Definition 3.4.1Reflection and transmission2024/25 board #8

$r = \frac{B}{A}, R = ∣ r ∣^{2}, t = \frac{E ^{'}}{A}, T = ∣ t ∣^{2} .$ $r$ and $t$ are the reflection and transmission amplitudes; $R$ and $T$ the probabilities of being reflected and of getting through.

Why $∣ t ∣^{2}$ is a probability (beyond the boards): what counts is the flux, particles per second, proportional to speed times $∣ amplitude ∣^{2}$ . Regions I and III have the same $U = 0$ , so the same speed, which cancels. Particles are not lost: $R + T = 1$ .

Matching and solving

Four unknowns, $r$ , $t$ , $C$ , $D$ ; four equations, $u$ and $u^{'}$ continuous at $0$ and at $a$ . The board’s trick is to add or subtract pairs so that one unknown drops out each time.

Recipe 3.4.2Solving the barrier2024/25 board #8

At $x = 0$ . Value times $i k_{0}$ : $i k_{0} (1 + r) = i k_{0} (C + D)$ . Slope: $i k_{0} - i k_{0} r = - κ C + κ D$ . ==Add them: $r$ drops out==, $2 i k_{0} = D (κ + i k_{0}) - C (κ - i k_{0})$ .
At $x = a$ . Value times $κ$ : $κ (C e^{- κa} + D e^{κa}) = κ t$ . Slope: $- κ C e^{- κa} + κ D e^{κa} = i k_{0} t$ . Adding and subtracting: $D = \frac{t}{2 κ} (κ + i k_{0}) e^{- κa}, C = \frac{t}{2 κ} (κ - i k_{0}) e^{κa} .$ (The board boxes them without the exponentials; the next line uses them.)
Substitute into step 1 and solve for $t$ .

Theorem 3.4.3Transmission amplitude below the barrier2024/25 board #8

For $E < U_{0}$ , $t = \frac{2 iκ k _{0}}{( k _{0}^{2} - κ ^{2} ) sinh κa + 2 iκ k _{0} cosh κa} .$

Proof

Step 3 gives $2 i k_{0} = \frac{t}{2 κ} [(κ + i k_{0})^{2} e^{- κa} - (κ - i k_{0})^{2} e^{κa}]$ . With $(κ \pm i k_{0})^{2} = κ^{2} - k_{0}^{2} \pm 2 iκ k_{0}$ (expanded beyond the boards) the bracket is $(k_{0}^{2} - κ^{2}) (e^{κa} - e^{- κa}) + 2 iκ k_{0} (e^{κa} + e^{- κa}) .$ Use $e^{κa} \mp e^{- κa} = 2 sinh κa, 2 cosh κa$ and divide top and bottom by 2.

== $t \neq = 0$ ==: some of the wave gets through a barrier the particle classically cannot cross. This is tunnelling.

Key observation 3.4.4How much gets throughbeyond the boards

$T = \frac{1}{1 + \frac{U _{0}^{2} sinh ^{2} κa}{4 E ( U _{0} - E )}} κa ≫ 1 \frac{16 E ( U _{0} - E )}{U _{0}^{2}} e^{- 2 κa} .$ ==Exponential in the thickness and in $U_{0} - E$ ==: this sets everything about tunnelling.

Proofbeyond the boards

$∣ denominator ∣^{2} = (k_{0}^{2} - κ^{2})^{2} sinh^{2} + 4 κ^{2} k_{0}^{2} cosh^{2} = 4 κ^{2} k_{0}^{2} + (k_{0}^{2} + κ^{2})^{2} sinh^{2}$ , with $cosh^{2} = 1 + sinh^{2}$ . Then $k_{0}^{2} + κ^{2} = 2 m U_{0} / ℏ^{2}$ and $k_{0}^{2} κ^{2} = (2 m / ℏ^{2})^{2} E (U_{0} - E)$ . For a thick barrier, $sinh κa \approx \frac{1}{2} e^{κa}$ and the 1 is negligible.

Example 3.4.5An electron of 0.5 eV against 1 eVbeyond the boards

$κ = 0.5/0.0381 = 3.62$ nm⁻¹.

$a$	$κa$	$T$
0.5 nm	1.81	0.10
1.0 nm	3.62	0.0029
1.5 nm	5.43	0.000076

Each extra half nanometre costs a factor of about 35. A 1 g ball short of the top by 1 mJ, against a 1 mm wall, has $2 κa \approx 3 \times 1 0^{28}$ : no tunnelling at human scale.

Above the barrier

For $E > U_{0}$ , $κ$ is imaginary, $κ = i α$ with $α = 2 m (E - U_{0}) /ℏ$ , as over the finite well: inside the barrier the wave oscillates, with a longer wavelength. The same $t$ holds with $sinh i α a = i sin α a$ and $cosh i α a = cos α a$ : $t = \frac{2 α k _{0}}{2 α k _{0} cos α a - i ( k _{0}^{2} + α ^{2} ) sin α a}, T = \frac{1}{1 + \frac{U _{0}^{2} sin ^{2} α a}{4 E ( E - U _{0} )}} .$ The board writes this without renaming $κ$ , so its symbols cannot be read consistently; the version here substitutes $κ = i α$ throughout (beyond the boards).

== $T \neq = 1$ == in general (the board’s mark): some reflection even when the particle classically sails over.
$T = 1$ exactly when $α a = nπ$ : a whole number of half-wavelengths fits in the barrier, and the reflections from its two edges cancel. These transmission resonances sit at $E = U_{0} + \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m a ^{2}}$ , the infinite well of width $a$ on top of $U_{0}$ (beyond the boards).

Loading the demo…

Figure 3.4.1 An electron arriving from the left at a

barrier

1 eV high. Top: the

real part

ψ

and

∣ ψ ∣^{2}

, with incident amplitude 1. Reflection and incoming wave interfere on the left, the wave decays through the barrier and continues, smaller and steady, on the right. Bottom:

T (E)

; drag the

marker

. Below 1 eV

T

is small but not zero; above, it climbs towards 1 and touches it at the resonances. Thicken the barrier and switch to the log scale: the straight line is

e^{- 2 κa}

Against the classical step. Classically $T$ is 0 below $U_{0}$ and jumps to 1 at $U_{0}$ . The quantum $T$ rises from 0 below, ==the quantum tunnelling effect==, and oscillates just under 1 above. At $E = U_{0}$ both formulas give $T = 1/ (1 + m U_{0} a^{2} /2 ℏ^{2})$ , so the curve is continuous (beyond the boards).

Tunnelling at the nanoscale

The board closes with two sketches and no text. One is a well with waves bouncing inside: with walls of finite thickness, each bounce has a chance $T$ of leaking out. The other is a sharp tip just above a surface, with a wave crossing the gap.

Example 3.4.6The scanning tunnelling microscopebeyond the boards

The tip over a surface is the scanning tunnelling microscope (Binnig and Rohrer, IBM, 1981; Nobel 1986). A metal tip is held a few tenths of a nm above a conducting sample, with a small voltage between them. The vacuum gap is a barrier about a work function high, $U_{0} - E \approx 4.5$ eV, and the gap $d$ wide.

The current goes as $T \propto e^{- 2 κ d}$ , with $κ = 4.5/0.0381 = 10.9$ nm⁻¹.
==Moving the tip 0.1 nm closer multiplies the current by $e^{2 \cdot 10.9 \cdot 0.1} \approx 9$ .==
Scanning with the current held constant, the tip traces the surface’s height to a fraction of an atom: atoms become visible.

Tunnelling also drives alpha decay, tunnel diodes, flash memory, which pushes charge through a thin oxide, and the leakage current through a transistor’s gate oxide once it is only a few nm thick (beyond the boards).

Exercises

The past sample sheets ask about tunnelling only through the finite well’s ground state (Exercise 3.3.2).

Exercise 3.4.1beyond the boards

An electron of 1 eV meets a barrier 2 eV high. How thick can the barrier be if at least 1% of the electrons must get through?

Solution

$κ = 5.12$ nm⁻¹; thick-barrier formula: $T \approx \frac{16 \cdot 1 \cdot 1}{4} e^{- 10.25 a} = 0.01$ gives $a = ln 400/10.25 = 0.58$ nm. Check with the full formula: $κa = 2.98$ , $sinh^{2} = 97$ , $T = 1/ (1 + 97) = 0.010$ .

Exercise 3.4.2beyond the boards

Show that at a resonance, $α a = nπ$ , nothing is reflected.

Solution

$sin α a = 0$ and $cos α a = \pm 1$ , so $t = \pm 1$ and $T = 1$ . $R + T = 1$ then needs $R = 0$ : the waves reflected at $x = 0$ and $x = a$ cancel exactly.

Exercise 3.4.3beyond the boards

A proton, with the electron’s numbers $E = 0.5$ eV, $U_{0} = 1$ eV, $a = 0.5$ nm. What is $T$ ?

Solution

$κ \propto m$ : $κ_{p} = 3.62 1836 = 155$ nm⁻¹, $2 κa = 155$ , $T \approx 4 e^{- 155} \approx 1 0^{- 67}$ . At these scales tunnelling is an effect of light particles.

Exercise 3.4.4beyond the boards

In the STM, the current at a gap of 0.5 nm is 1 nA. Estimate it at 0.6 nm and at 0.4 nm.

Solution

Each 0.1 nm changes it by $e^{2.18} = 8.9$ : about 0.11 nA at 0.6 nm, about 8.9 nA at 0.4 nm.