Quantum tunneling — concept, transmission coefficient: Question bank

Definition The map and every symbol used on this page (read first)

The whole problem is one particle of energy $E$ meeting a rectangular barrier: a region where the potential energy jumps up to height $V_{0}$ . Space splits into three pieces — look at the figure below:

Region I ( $x < 0$ ): flat ground, potential $V = 0$ . Here the particle is a free wave.
Region II ( $0 < x < L$ ): the barrier, potential $V = V_{0}$ , width $L$ . Here $E < V_{0}$ , so it is classically forbidden.
Region III ( $x > L$ ): flat ground again, $V = 0$ .

The symbols you will see below all come from solving Schrödinger's equation in those three pieces:

$k = \frac{2 m E}{ℏ}$ — the free-particle wavenumber: how fast the wave wiggles in the flat regions I and III (large $k$ = short wavelength). Here $m$ is the particle's mass and $ℏ$ is the reduced Planck constant.
$κ = \frac{2 m ( V _{0} - E )}{ℏ}$ — the decay constant (Greek "kappa"): how fast the wavefunction shrinks inside the barrier. Because $V_{0} > E$ the square root is real, so $κ$ is a real positive number and $e^{- κ x}$ is a genuine decay.
$A, B$ — amplitudes of the incoming and reflected waves in region I: $ψ_{I} = A e^{ik x} + B e^{- ik x}$ .
$C, D$ — amplitudes of the two barrier terms in region II: $ψ_{I I} = C e^{- κ x} + D e^{+ κ x}$ (a falling and a rising exponential).
$F$ — amplitude of the single transmitted wave in region III: $ψ_{I I I} = F e^{ik x}$ .
$T = \frac{∣ F ∣ ^{2}}{∣ A ∣ ^{2}}$ — the transmission coefficient: the fraction of incident particles that get through. $R$ is the reflected fraction, and $T + R = 1$ .
$T \approx G e^{- 2 κ L}$ with prefactor $G = \frac{16 E ( V _{0} - E )}{V _{0}^{2}}$ (order-1 number). Note: on this page $G$ always means this prefactor. When we deliberately write a wrong " $G e^{- ik x}$ " in a Spot-the-error item, that stray $G$ is the mistake being flagged.

Figure — Quantum tunneling — concept, transmission coefficient

The figure above is the mental picture behind every question below — keep it in view.

True or false — justify

A tunneling particle borrows energy to cross and pays it back after.

False — energy is conserved throughout; the particle's total energy

E

never changes, and it stays below the barrier height

V_{0}

the whole time. Nothing is borrowed; the wavefunction merely has nonzero amplitude in the forbidden region.

The particle's energy

E

is less than the potential everywhere along its journey.

False —

E < V_{0}

holds only inside the barrier (region II); in the flat regions I and III the potential is

0

, which is below

E

. The "forbidden" comparison

E < V_{0}

applies to the barrier alone.

Inside the barrier the particle is a decaying wave, so it slowly loses energy as it crosses.

False — the amplitude

∣ ψ ∣

decays with position

x

, not with time, and it encodes probability, not energy loss. A particle that does emerge carries full energy

E

If the barrier were infinitely wide,

T

would be exactly zero.

True —

T \approx G e^{- 2 κ L} \to 0

L \to \infty

, because the exponential decay

e^{- κ x}

ψ

leaves literally no amplitude reaching the far side.

Doubling the barrier width halves the transmission probability.

False — width sits in an exponent, so doubling

L

squares the tiny factor

e^{- 2 κ L}

. Concretely, for an electron with

E = 1 eV

V_{0} = 2 eV

: at

L = 0.5 nm

2 κ L \approx 5.12

; at

L = 1.0 nm

2 κ L \approx 10.24

. The ratio of

T

's is

e^{10.24 - 5.12} = e^{5.12} \approx 167

— about

170 \times

smaller, not

2 \times

For a barrier with

E > V_{0}

the transmission is always 100%.

False — even above the barrier there is partial reflection, and

T

oscillates with energy (Ramsauer–Townsend resonances) because

κ

becomes imaginary, turning

sinh

into

sin

A heavier particle at the same

E

and

V_{0}

tunnels less.

True —

κ = 2 m (V_{0} - E) /ℏ \propto m

, so more mass means faster decay of

ψ

and a smaller

e^{- 2 κ L}

; a proton tunnels vastly less than an electron (see de Broglie Wavelength for why heavy = short wavelength = "more classical").

Tunneling violates conservation of energy.

False — it violates only the classical rule that

E \geq V_{0}

is needed to pass; total energy is exactly conserved. What's "violated" is our classical intuition, not a conservation law.

The exact

sinh^{2}

formula and the

e^{- 2 κ L}

formula give the same answer for a thin, low barrier.

False — the approximation assumes

κ L ≫ 1

; for thin barriers

κ L

is small and you must use the exact form (see the next item for why

κ L ≫ 1

is needed).

Spot the error

"Inside the barrier

ψ = C e^{- iκ x} + D e^{+ iκ x}

, an oscillating wave."

Error: the

i

is wrong. Since

V_{0} > E

, Schrödinger's equation reads

ψ^{''} = + κ^{2} ψ

with real solutions

e^{\pm κ x}

(

C

falling,

D

rising) — pure decay/growth, no oscillation. See Potential Barrier and Reflection.

"In region III we write

ψ = F e^{ik x} + G e^{- ik x}

to be general."

Error: the second term is a leftward wave coming from

+ \infty

, but nothing is incident from the right — so its amplitude must be zero. (Also,

G

properly denotes the transmission prefactor, not a wave amplitude; the label collision is itself a red flag.) Only the transmitted wave

F e^{ik x}

survives.

T = ∣ F ∣^{2} /∣ A ∣^{2}

works because energy is the ratio of amplitudes."

Error:

T

is a ratio of probability fluxes (probability per second crossing a plane), not energies. Flux is speed

\times ∣ amplitude ∣^{2}

; here

A

is the incoming amplitude in region I and

F

the transmitted amplitude in region III. It collapses to

∣ F ∣^{2} /∣ A ∣^{2}

only because $k$ is identical in regions I and III, so the speeds cancel; with different potentials on the two sides you'd keep the velocity factors.

"We drop the

D e^{+ κ x}

growing term inside the barrier because it isn't normalizable."

Error: for a finite barrier the growing term is kept — the region II is finite (

0 < x < L

) so nothing blows up. You only drop

e^{+ κ x}

for a semi-infinite barrier. The four boundary conditions here fix both

C

and

D

"Since

T \approx G e^{- 2 κ L}

and

G

is order 1, we can ignore

G

entirely and

T = e^{- 2 κ L}

Error:

G = 16 E (V_{0} - E) / V_{0}^{2}

can be small when

E ≪ V_{0}

(energy near the bottom), so dropping it changes the number. The scaling lives in the exponent, but the prefactor still sets the actual value.

"The factor of 2 in

e^{- 2 κ L}

comes from there being two barrier walls."

Error: it comes from

∣ ψ ∣^{2}

. Amplitude decays as

e^{- κ L}

; probability is amplitude squared, giving

e^{- 2 κ L}

. Nothing to do with two walls.

"Tunneling needs the barrier to be higher than

E

, so a taller barrier means easier tunneling."

Error: reversed. Taller

V_{0}

increases

(V_{0} - E)

, hence

κ

, hence the decay rate — making tunneling harder, not easier.

"At

x = 0

we only need

ψ

itself to match; the slope can jump."

Error: both

ψ

and its slope

ψ^{'}

must be continuous at

x = 0

and at

x = L

. That gives four equations (

ψ_{I} = ψ_{I I}

ψ_{I}^{'} = ψ_{I I}^{'}

0

;

ψ_{I I} = ψ_{I I I}

ψ_{I I}^{'} = ψ_{I I I}^{'}

L

) — exactly enough to fix

B, C, D, F

relative to

A

. See Wavefunction and Boundary Conditions.

Why questions

Why is

ψ

exponential rather than oscillatory inside the barrier?

Because

V_{0} > E

flips the sign of the curvature term to

ψ^{''} = + κ^{2} ψ

; the only real solutions are

e^{\pm κ x}

. Oscillation (

e^{\pm ik x}

) requires the negative-curvature case

E > V

, which holds in regions I and III.

Why can

sinh (κ L)

be replaced by

\frac{1}{2} e^{κ L}

for a thick/high barrier?

By definition

sinh (κ L) = \frac{1}{2} (e^{κ L} - e^{- κ L})

. When

κ L ≫ 1

the shrinking term

e^{- κ L}

is negligible next to the growing

e^{κ L}

(e.g. at

κ L = 3

it is already ~

0.25%

of it), so

sinh (κ L) \approx \frac{1}{2} e^{κ L}

and hence

T \approx G e^{- 2 κ L}

. The relative error is roughly

e^{- 2 κ L}

, tiny for thick barriers.

Why can't the wavefunction just drop to zero at the wall instead of decaying?

Because Wavefunction and Boundary Conditions require both

ψ

and its slope

ψ^{'}

to be continuous everywhere; an abrupt drop to zero would break the continuity of the slope, which Schrödinger's equation forbids.

Why does STM (see Scanning Tunneling Microscope) achieve atomic resolution?

Because

T \propto e^{- 2 κ L}

is exponentially sensitive to the tip–surface gap

L

; a change of a fraction of an atom's diameter changes the tunneling current by a large factor, so height variations are hugely amplified.

Why do electrons tunnel but bowling balls never do?

κ \propto m

, so the enormous mass of a macroscopic object makes

2 κ L

astronomically large and

e^{- 2 κ L}

effectively zero. Light electrons keep

2 κ L

modest, so appreciable amplitude survives.

Why does alpha decay depend so violently on the nucleus (see Alpha Decay)?

The escaping alpha particle must tunnel through the Coulomb barrier, and

T \sim e^{- 2 κ L}

makes the escape probability — hence the half-life — swing over 20+ orders of magnitude for small changes in energy or barrier size.

Why is the transmitted particle's energy still exactly

E

Region III has

V = 0

just like region I, and the same wavenumber

k = 2 m E /ℏ

appears in both. Same

k

means same kinetic energy — no energy was added or removed.

Why does

T

never reach 1 for

E < V_{0}

A finite barrier always reflects some amplitude (

B \neq = 0

), so probability is split between transmission and reflection with

T + R = 1

and

T < 1

. Perfect transmission is impossible below the barrier.

Edge cases

What happens to

T

E \to V_{0}^{-}

(energy approaches barrier height)?

κ \to 0

, the decay vanishes, and using

sinh (κ L) \approx κ L

the exact formula gives

T \to [1 + \frac{m V _{0} L ^{2}}{2 ℏ ^{2}}]^{- 1}

— finite and often close to 1, not divergent. This is exactly why you must not use the

κ L ≫ 1

approximation here.

What does the exact formula do at

E \to 0^{+}

(very slow particle)?

The prefactor

G = 16 E (V_{0} - E) / V_{0}^{2} \to 0

and

sinh^{2} (κ L)

stays finite, so

T \to 0

. A nearly-still particle barely tunnels because it carries almost no incident flux.

What happens when

E > V_{0}

(above the barrier)?

(V_{0} - E)

goes negative, so

κ = 2 m (V_{0} - E) /ℏ

becomes imaginary; write

κ = i k^{'}

and

sinh (κ L) \to i sin (k^{'} L)

. Then

T = 1

whenever

sin (k^{'} L) = 0

(resonant full transmission) and dips between — the Ramsauer–Townsend effect.

What is

T

for a barrier of zero width,

L \to 0

sinh (0) = 0

so the exact formula gives

T = 1

. With no barrier to cross there is nothing to reflect from — total transmission, as expected.

What if the barrier is infinitely tall (

V_{0} \to \infty

) at fixed

E

and

L

κ \to \infty

, so

e^{- 2 κ L} \to 0

and

T \to 0

. This recovers the classical hard wall: an infinite barrier is impenetrable.

Can

ψ

be exactly zero somewhere inside a finite barrier?

Generally no for the ground-tunneling case —

ψ_{I I} = C e^{- κ x} + D e^{+ κ x}

is a sum of monotone exponentials with no oscillation, so it has no interior nodes; it decays smoothly across the barrier.

Recall One-line self-test before you leave

Cover everything: state (1) why $ψ$ decays not oscillates, (2) where the "2" in $e^{- 2 κ L}$ comes from, (3) what $T$ does when $E > V_{0}$ , and (4) whether the emerging particle's energy changed. Answers ::: (1) $V_{0} > E$ gives $ψ^{''} = + κ^{2} ψ$ , real exponentials; (2) probability is $∣ ψ ∣^{2}$ , squaring $e^{- κ L}$ ; (3) partial transmission with $T = 1$ resonances (imaginary $κ$ ); (4) unchanged, still $E$ .