Quantum tunneling — concept, transmission coefficient: Exercises

Meet the four symbols before the toolkit (every problem uses exactly these):

Constants used throughout (keep these on a sticky note):

Intuition Where does the prefactor

$G$ come from — and can it be trusted? $G = \frac{16 E ( V _{0} - E )}{V _{0}^{2}}$ is not a fudge factor. It falls out of matching the wavefunction and its slope at both walls ( $x = 0$ and $x = L$ ) — the Wavefunction and Boundary Conditions that glue Region I to Region II to Region III. Physically it measures the impedance mismatch at the two steps: how much amplitude bounces off the walls rather than entering/leaving. When $E$ is close to $V_{0}$ the wave enters easily and $G$ approaches its maximum of $4$ ; when $E ≪ V_{0}$ the walls reflect hard and $G$ shrinks toward $0$ . Domain of validity: the compact form $T \approx G e^{- 2 κ L}$ is the thick-barrier limit ( $κ L ≳ 2$ ), obtained by keeping only the growing part of $sinh (κ L)$ . Because $G$ can be as large as $4$ , blindly evaluating $G e^{- 2 κ L}$ for a thin barrier ( $κ L \to 0$ , where $e^{- 2 κ L} \to 1$ ) can give the nonsense $T > 1$ . That is a signal you have left the approximation's domain — fall back to the exact $sinh^{2}$ formula, which always obeys $T \leq 1$ . In all problems below $κ L$ is comfortably large, so the approximation is safe.

Here is the picture behind every problem — keep it open as you work. Region I (blue, $x < 0$ ) is the incoming + reflected wave; Region II (red, $0 < x < L$ ) is the exponential decay $e^{- κ x}$ inside the barrier of width $L$ (marked by the double-headed arrow along the bottom axis); Region III (green, $x > L$ ) is the small surviving transmitted wave whose squared amplitude is $T$ . The white dashed line marks the energy $E$ sitting below the yellow barrier top $V_{0}$ — that vertical gap, labelled in the figure, is $(V_{0} - E)$ , and the red curve's steepness is exactly $κ$ . Every problem below is a numerical question about this figure:

Figure — Quantum tunneling — concept, transmission coefficient

Figure 1 — Rectangular barrier of height $V_{0}$ and width $L$ with $E < V_{0}$ . Blue: incoming/reflected wave in Region I ( $x < 0$ ). Red: exponential decay $e^{- κ x}$ across the barrier, Region II ( $0 < x < L$ ). Green: transmitted wave in Region III ( $x > L$ ), whose squared amplitude gives $T$ . The bottom double arrow is the barrier width $L$ ; the vertical arrow inside the plateau is the energy deficit $V_{0} - E$ .

L1 — Recognition

Problem 1.1

Inside the barrier the wavefunction behaves as $e^{- κ x}$ (the red curve in Figure 1). In words, what does $κ$ (kappa) physically mean, and what three quantities in its formula make it bigger?

Recall Solution 1.1

$κ$ is the decay rate of the wavefunction inside the classically forbidden region — how steeply the red curve in Figure 1 falls. Over a distance $1/ κ$ the amplitude of $ψ$ drops by a factor $e \approx 2.718$ . Read the three formal factors straight off $κ = \frac{2 m ( V _{0} - E )}{ℏ}$ :

mass $m$ — heavier particle → larger $κ$ → faster decay → less tunneling.
energy deficit $(V_{0} - E)$ — a taller/steeper barrier above the particle's energy → larger $κ$ .
$1/ℏ$ — $κ$ is inversely proportional to Planck's constant $ℏ$ . A smaller $ℏ$ gives a larger $κ$ ; in the hypothetical classical limit $ℏ \to 0$ , $κ \to \infty$ and tunneling vanishes entirely. This third factor is why tunneling is a purely quantum effect.

A bigger $κ$ means harder tunneling, because $T \propto e^{- 2 κ L}$ shrinks.

Problem 1.2

In the formula $T \approx G e^{- 2 κ L}$ , why is the exponent $- 2 κ L$ and not $- κ L$ ?

Recall Solution 1.2

The amplitude of the wavefunction decays as $e^{- κ x}$ , so across the full width $L$ it falls to $e^{- κ L}$ . But transmission is a probability, and probability $\propto ∣ ψ ∣^{2}$ — you square the amplitude: $∣ e^{- κ L} ∣^{2} = e^{- 2 κ L} .$ The factor of 2 is literally the square in $∣ ψ ∣^{2}$ .

L2 — Application

Problem 2.1

An electron with $E = 2 eV$ hits a barrier of height $V_{0} = 5 eV$ and width $L = 0.4 nm$ . Compute $κ$ .

Recall Solution 2.1

Step 1 (WHAT): find $V_{0} - E = 5 - 2 = 3 eV = 3 \times 1.602 \times 1 0^{- 19} = 4.806 \times 1 0^{- 19}$ J. Step 2 (WHY): $κ$ needs the energy deficit $(V_{0} - E)$ , the depth into the forbidden region. Step 3 (compute): $κ = \frac{2 ( 9.11 \times 1 0 ^{- 31} ) ( 4.806 \times 1 0 ^{- 19} )}{1.055 \times 1 0 ^{- 34}}$ Numerator: $2 \cdot 9.11 \times 1 0^{- 31} \cdot 4.806 \times 1 0^{- 19} = 8.756 \times 1 0^{- 49} = 9.357 \times 1 0^{- 25}$ . $κ = \frac{9.357 \times 1 0 ^{- 25}}{1.055 \times 1 0 ^{- 34}} \approx 8.87 \times 1 0^{9} m^{- 1} .$

Problem 2.2

Using the $κ$ from Problem 2.1, compute the exponent $2 κ L$ and the prefactor $G$ , then estimate $T$ .

Recall Solution 2.2

Exponent: $2 κ L = 2 (8.87 \times 1 0^{9}) (0.4 \times 1 0^{- 9}) = 7.10$ (here $L = 0.4$ nm is the barrier width). Prefactor: $G = \frac{16 E ( V _{0} - E )}{V _{0}^{2}} = \frac{16 ( 2 ) ( 3 )}{5 ^{2}} = \frac{96}{25} = 3.84$ . This is below the cap of $4$ , as expected. Because $κ L = 3.55 ≫ 1$ , we are safely in the thick-barrier domain, so $G e^{- 2 κ L} ≪ 1$ and no $T > 1$ pathology arises. Transmission: $T \approx G e^{- 2 κ L} = 3.84 e^{- 7.10} = 3.84 (8.23 \times 1 0^{- 4}) \approx 3.16 \times 1 0^{- 3}$ . So $T \approx 3.2 \times 1 0^{- 3} \approx 0.32%$ — small but nonzero, the tunneling fingerprint.

L3 — Analysis

Problem 3.1

For the electron of Problem 2.1 ( $κ = 8.87 \times 1 0^{9} m^{- 1}$ , $G = 3.84$ ), the barrier width is increased from $L = 0.4$ nm to $L = 0.8$ nm. By what factor does $T$ drop?

Recall Solution 3.1

$G$ is unchanged (it does not depend on the width $L$ ). Only the exponential changes: $\frac{T ( 0.8 )}{T ( 0.4 )} = \frac{e ^{- 2 κ (0.8 nm)}}{e ^{- 2 κ (0.4 nm)}} = e^{- 2 κ (0.4 nm)} = e^{- 7.10} .$ $e^{- 7.10} \approx 8.23 \times 1 0^{- 4} \Rightarrow T falls by a factor of about 1215.$ Reading: doubling the width didn't halve $T$ — it cut it by ~1200×. This exponential sensitivity is exactly why the Scanning Tunneling Microscope can resolve single atoms: a sub-nanometre tip-height change swings the current by orders of magnitude.

Problem 3.2

Compare the exact and approximate formulas for the electron of Problem 2.1 (recall $κ L = 3.55$ ). Is the thick-barrier approximation trustworthy here?

Recall Solution 3.2

The approximation replaces $sinh (κ L)$ by $\frac{1}{2} e^{κ L}$ . Check the size of $κ L$ : $κ L = 8.87 \times 1 0^{9} \times 0.4 \times 1 0^{- 9} = 3.55.$ Since $κ L = 3.55 ≫ 1$ , we have $sinh (3.55) = 17.31$ while $\frac{1}{2} e^{3.55} = 17.34$ — they agree to $0.2%$ . Exact: $T = [1 + \frac{V _{0}^{2} sinh ^{2} ( κ L )}{4 E ( V _{0} - E )}]^{- 1} = [1 + \frac{25 ( 17.31 ) ^{2}}{4 ( 2 ) ( 3 )}]^{- 1} = [1 + 312.1]^{- 1} = 3.19 \times 1 0^{- 3} .$ Approx: $3.16 \times 1 0^{- 3}$ (Problem 2.2). Difference $< 1%$ — the approximation is excellent whenever $κ L ≳ 2$ . (If instead $κ L$ were tiny, the exact formula would keep $T \leq 1$ while the approximation could overshoot — see the toolkit note on $G$ .)

L4 — Synthesis

Problem 4.1

Keep everything from Problem 2.1 (same $E$ , $V_{0}$ , $L$ ) but replace the electron with a deuteron (mass $\approx 3670 m_{e}$ ). By what factor does $κ$ change, and roughly what is the new exponent $2 κ L$ ? Comment on $T$ .

Recall Solution 4.1

$κ \propto m$ , so $\frac{κ _{deut}}{κ _{e}} = 3670 = 60.6.$ New exponent: $2 κ L \to 60.6 \times 7.10 \approx 430$ . $T \sim e^{- 430} \approx 1 0^{- 187} .$ Reading: a deuteron is $\sim 3670 \times$ heavier, its $κ$ is $\sim 61 \times$ larger, and the transmission collapses to an unimaginably tiny number. This is why macroscopic and even nuclear-mass objects effectively never tunnel across ordinary barriers — connecting directly to Alpha Decay, where tunneling only survives because the barrier is thin and the alpha (though heavy) faces enormous collision rates.

Problem 4.2

Two barriers give the same transmission $T$ . Barrier A: $V_{0} - E = 3$ eV, width $L_{A} = 0.4$ nm. Barrier B: $V_{0} - E = 12$ eV. Ignoring the order-1 prefactor, find the width $L_{B}$ that reproduces barrier A's exponent.

Recall Solution 4.2

Equal $T$ (prefactor aside) means equal exponents: $κ_{A} L_{A} = κ_{B} L_{B}$ . Since $κ \propto V_{0} - E$ : $\frac{κ _{B}}{κ _{A}} = \frac{12}{3} = 4 = 2.$ Therefore $L_{B} = \frac{κ _{A}}{κ _{B}} L_{A} = \frac{1}{2} (0.4 nm) = 0.2 nm .$ Reading: a barrier 4× taller needs only half the width to leak the same amount — because the height enters under a square root while width enters linearly in the exponent. Height and width trade off through $V_{0} - E \cdot L = const$ .

L5 — Mastery

Problem 5.1 (design)

You are building a Scanning Tunneling Microscope tip. The tunnel gap acts as a barrier with $V_{0} - E = 4.0$ eV (a typical metal work function above the electron energy). You want the tunnel current to change by a factor of $e$ (about $2.718 \times$ ) when the tip moves. How large a vertical tip motion $Δ L$ gives one factor of $e$ in current? (Current $\propto T \propto e^{- 2 κ L}$ , where $L$ is the tip-to-surface gap width.)

Recall Solution 5.1

A factor of $e$ in current means the exponent changes by 1: $2 κ Δ L = 1 \Rightarrow Δ L = \frac{1}{2 κ} .$ Compute $κ$ with $V_{0} - E = 4.0 eV = 6.408 \times 1 0^{- 19}$ J: $κ = \frac{2 ( 9.11 \times 1 0 ^{- 31} ) ( 6.408 \times 1 0 ^{- 19} )}{1.055 \times 1 0 ^{- 34}} = \frac{1.0806 \times 1 0 ^{- 24}}{1.055 \times 1 0 ^{- 34}} \approx 1.024 \times 1 0^{10} m^{- 1} .$ $Δ L = \frac{1}{2 ( 1.024 \times 1 0 ^{10} )} = 4.88 \times 1 0^{- 11} m \approx 0.049 nm .$ Reading: moving the tip by only ~0.05 nm — a fraction of an atom's diameter — changes the current by a full factor of $e$ . That extreme sensitivity is the entire operating principle of the STM: it is a ruler that reads atoms because tunneling current is exponential in the gap width.

Problem 5.2 (reason backwards)

An experiment measures $T = 1.0 \times 1 0^{- 5}$ for electrons crossing a barrier of width $L = 1.0$ nm. Taking the prefactor $G \approx 4$ , estimate the energy deficit $V_{0} - E$ (in eV).

Recall Solution 5.2

Step 1 — isolate the exponent. From $T \approx G e^{- 2 κ L}$ : $e^{- 2 κ L} = \frac{T}{G} = \frac{1 0 ^{- 5}}{4} = 2.5 \times 1 0^{- 6} .$ $- 2 κ L = ln (2.5 \times 1 0^{- 6}) = - 12.90 \Rightarrow 2 κ L = 12.90.$ Step 2 — solve for $κ$ . With the width $L = 1.0 \times 1 0^{- 9}$ m: $κ = \frac{12.90}{2 ( 1.0 \times 1 0 ^{- 9} )} = 6.45 \times 1 0^{9} m^{- 1} .$ Step 3 — invert $κ$ for the energy deficit. From $κ = 2 m (V_{0} - E) /ℏ$ : $V_{0} - E = \frac{( ℏ κ ) ^{2}}{2 m} = \frac{( 1.055 \times 1 0 ^{- 34} \cdot 6.45 \times 1 0 ^{9} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )} .$ Numerator: $(6.805 \times 1 0^{- 25})^{2} = 4.63 \times 1 0^{- 49}$ . $V_{0} - E = \frac{4.63 \times 1 0 ^{- 49}}{1.822 \times 1 0 ^{- 30}} = 2.54 \times 1 0^{- 19} J = 1.59 eV .$ So the barrier sits about 1.6 eV above the electron energy — a plausible tunnel-junction value.

Recall Master checklist (fill from memory)

Units of $κ$ ? ::: inverse length, m $^{- 1}$
What does $L$ stand for? ::: the barrier width — the horizontal thickness of the forbidden region the wave must decay across
Convert energies to ___ before computing $κ$ ? ::: joules
$T$ scales how with $L$ ? ::: exponentially, as $e^{- 2 κ L}$
$κ$ scales how with mass? ::: as $m$
Height–width trade-off to keep $T$ fixed? ::: $V_{0} - E \cdot L = const$
One factor of $e$ in STM current needs $Δ L =$ ? ::: $1/ (2 κ)$
Where does the prefactor $G$ come from, and what caps it? ::: from boundary-condition matching at both walls; it is capped at $4$ (reached when $E \to V_{0}$ )

Related build-up notes: Wavefunction and Boundary Conditions, Potential Barrier and Reflection, Schrödinger Equation, de Broglie Wavelength.