Visual walkthrough — Quantum tunneling — concept, transmission coefficient

We are studying one specific scene: a particle of energy $E$ flying rightward, meeting a rectangular wall of height $V_0$ and width $L$. We take $E<V_0$ — the wall is taller than the particle's energy, so a classical ball would bounce. Let's watch what a wave does instead.

---

Step 0 — The vocabulary, anchored to a picture

Before any algebra, meet the four words we'll lean on. Each is just a plain idea with a picture.

• Wavefunction $\psi (x)$ — a number attached to every position $x$. Its square $|\psi {|}^2$ tells you how likely the particle is to be found there. Think of $\psi$ as the "fog density" from the parent's tennis-ball story.
• Energy $E$ — how much oomph the incoming particle carries. A fixed number for the whole trip.
• Barrier height $V_0$ — the "cost" of being inside the wall. Since $E<V_0$, the region $0<x<L$ is a place the particle "can't afford" classically.
• Barrier width $L$ — how thick the wall is, from $x=0$ to $x=L$.

---

Step 1 — The one equation that governs everything

Every quantum shape obeys the time-independent Schrödinger Equation:

$\underset{\text{how "stiff"}}{\underbrace{-\frac{{\hslash }^2}{2m}}}\underset{\text{curvature of }\psi }{\underbrace{\frac{d^2\psi }{d{x}^2}}}+\underset{\text{cost of being here}}{\underbrace{V(x)\psi }}=\underset{\text{total energy}}{\underbrace{E\psi }}$

WHAT each piece is doing:

• $\frac{d^2\psi }{d{x}^2}$ — the curvature (how sharply $\psi$ bends). This is the quantity we solve for.
• $\hslash$ — Planck's constant divided by $2\pi$; sets the quantum scale. $m$ — the particle's mass.
• $V(x)$ — the potential energy at position $x$ ($0$ outside the wall, $V_0$ inside).
• $E$ — the fixed total energy.

WHY this tool and not Newton's $F=ma$? Newton tracks a point. But a quantum particle is a spread-out wave, and we want its shape. The Schrödinger equation is the rule that shape must satisfy. Let's rearrange it to isolate the curvature, because curvature is what makes a wave either wiggle or decay:

$\frac{d^2\psi }{d{x}^2}=\frac{2m(V(x)-E)}{{\hslash }^2}\psi$

PICTURE: the sign of the factor $(V(x)-E)$ decides the whole story. Watch it below.

---

Step 2 — Region I: the incoming and reflected waves

Outside the wall on the left, $V=0$, so $V-E=-E$, a negative number. Write it as $-k^2$:

$\frac{d^2{\psi }_I}{d{x}^2}=-\frac{2mE}{{\hslash }^2}{\psi }_I\equiv -k^2{\psi }_I,k=\frac{\sqrt{2mE}}{\hslash }$

WHAT is $k$? It's the wavenumber — how many radians of wiggle per metre. Large $E$ → large $k$ → short wavelength (this is the de Broglie Wavelength $\lambda =2\pi /k$ in disguise).

WHY does a negative curvature give a wave? Because the function that curves back toward zero everywhere is a sine/cosine — i.e. $e^{\pm ikx}$. The solution:

${\psi }_I=\underset{\text{moving right (incoming)}}{\underbrace{A{e}^{ikx}}}+\underset{\text{moving left (reflected)}}{\underbrace{B{e}^{-ikx}}}$

• $A$ — amplitude of the incoming wave (what we fire in).
• $B$ — amplitude of the reflected wave (what bounces back).

PICTURE: an incoming ripple, part of which will bounce off the wall like light off glass — see Potential Barrier and Reflection.

---

Step 3 — Region II: inside the wall, the wave decays

Inside, $V=V_0>E$, so $V-E=+(V_0-E)$, a positive number. Write it as $+{\kappa }^2$:

$\frac{d^2{\psi }_{II}}{d{x}^2}=+\frac{2m(V_0-E)}{{\hslash }^2}{\psi }_{II}\equiv +{\kappa }^2{\psi }_{II},\boxed{\kappa =\frac{\sqrt{2m(V_0-E)}}{\hslash }}$

WHAT is $\kappa$ (kappa)? The decay constant — how fast $\psi$ shrinks per metre inside the wall. It is the single most important number on this page.

WHY a real exponential, not a wave? Positive curvature means the graph bends away from the axis. The functions that do that are $e^{+\kappa x}$ (grows) and $e^{-\kappa x}$ (shrinks):

${\psi }_{II}=\underset{\text{decaying}}{\underbrace{C{e}^{-\kappa x}}}+\underset{\text{growing}}{\underbrace{D{e}^{+\kappa x}}}$

Both are allowed for a finite wall (we can't yet drop $D$ because the wall ends at $x=L$ before the growth blows up).

PICTURE: the fog doesn't stop at the wall — it fades smoothly through it.

---

Step 4 — Region III: the survivor emerges

Past the wall, $V=0$ again — same equation as Region I, same $k$. But now only one kind of wave can exist:

${\psi }_{III}=\underset{\text{transmitted, moving right}}{\underbrace{F{e}^{ikx}}}$

WHY no leftward wave here? Nothing lies beyond $+\infty$ to reflect the wave back. So the coefficient of $e^{-ikx}$ is zero. Only the transmitted amplitude $F$ survives.

WHAT is $F$? The amplitude that made it through — smaller than $A$, but not zero. That "not zero" is tunneling.

PICTURE: the whole scene in one frame — big wave in, a faded exponential across, a small wave out.

---

Step 5 — Stitching the regions: boundary conditions

We have three separate solutions with five unknowns $A,B,C,D,F$. Schrödinger's equation demands that at each join ($x=0$ and $x=L$) both $\psi$ and its slope ${\psi }^{\prime }$ match — no jumps, no kinks.

${\psi }_I(0)={\psi }_{II}(0),{\psi }_I^{\prime }(0)={\psi }_{II}^{\prime }(0),{\psi }_{II}(L)={\psi }_{III}(L),{\psi }_{II}^{\prime }(L)={\psi }_{III}^{\prime }(L)$

WHY must they match? A jump in $\psi$ would make $|\psi {|}^2$ (a probability) two-valued at one point — nonsense. A kink (jump in slope) would make the curvature infinite, blowing up Schrödinger's equation. So both are outlawed. This is exactly Wavefunction and Boundary Conditions.

WHAT we get: four equations for the ratios. Grinding the algebra (multiply, eliminate $C,D$, solve for $F/A$) gives the exact result below.

---

Step 6 — The thick-wall limit (the formula you actually use)

For a wall that's fat or tall, $\kappa L\gg 1$. Then $e^{-\kappa L}$ is tiny, so:

$\sinh (\kappa L)=\frac{1}{2}(e^{\kappa L}-e^{-\kappa L})\approx \frac{1}{2}{e}^{\kappa L}\Rightarrow {\sinh }^2(\kappa L)\approx \frac{1}{4}{e}^{2\kappa L}$

WHY drop the $-e^{-\kappa L}$ and the $1$? Because $e^{2\kappa L}$ is astronomically larger than $1$, the "$1+$" and the small term vanish beside it. Substituting:

$T\approx \frac{4E(V_0-E)}{V_0^2\cdot \frac{1}{4}{e}^{2\kappa L}}=\frac{16E(V_0-E)}{V_0^2}{e}^{-2\kappa L}$

---

Step 7 — Edge and degenerate cases (never skip these)

The formula must behave sensibly in every corner. Let's check all of them.

• Vanishing width, $L\to 0$: $\kappa L\to 0$, $\sinh (0)=0$, so $T\to [1+0{]}^{-1}=1$. No wall, everything passes. ✔
• Very thick/tall wall, $\kappa L\to \infty$: $e^{-2\kappa L}\to 0$, so $T\to 0$. Nothing gets through. ✔
• Heavy particle, $m\to \infty$: $\kappa \propto \sqrt{m}\to \infty$, so $T\to 0$. Bowling balls don't tunnel — this is why Alpha Decay (a heavy $\alpha$ particle) is slow and rare, and why macroscopic objects never tunnel. ✔
• Energy meets the top, $E\to V_0^{-}$: $\kappa \to 0$, and the exact formula's ratio $\frac{{\sinh }^2(\kappa L)}{4E(V_0-E)}$ stays finite (both numerator and denominator $\to 0$ together). $T$ is smooth here — no blow-up. ✔
• Above the barrier, $E>V_0$: now $V_0-E<0$, so $\kappa$ becomes imaginary, $\kappa =i{k}^{\prime }$. Then $\sinh (\kappa L)=\sinh (i{k}^{\prime }L)=i\sin (k^{\prime }L)$, turning the decay into an oscillation. $T$ now oscillates and can hit exactly 1 at resonances — the Ramsauer–Townsend effect. Do not plug $E>V_0$ into the decay formula blindly. ✔

---

The one-picture summary

Everything above, compressed into a single frame: wave in (amplitude $A$), exponential fade across the wall (rate $\kappa$), wave out (amplitude $F$), with the final formula annotated onto its own geometry.

Recall Feynman retelling — the whole walkthrough in plain words

A fuzzy wave-cloud rolls in from the left (that's $A$). It hits a wall taller than it can climb. Part bounces back (that's $B$). But the cloud can't just stop dead — the rules say it must fade smoothly, so inside the wall it shrinks exponentially at a rate $\kappa$ set by how heavy the particle is and how much taller the wall is than the particle's energy. If the wall is thin, the cloud is still faintly there when it reaches the far edge — and out crawls a small wave (that's $F$). The chance of getting across is $T=|F{|}^2/|A{|}^2$, and because the fade is exponential in the width, $T\approx G{e}^{-2\kappa L}$. Thin wall, light particle, low wall → more leaks through. That leak is tunneling — the engine behind the Scanning Tunneling Microscope and Alpha Decay.

Recall Active recall

Where does the sign flip in curvature come from? ::: From $V(x)-E$: negative outside ($E>V$) gives waves $e^{\pm ikx}$; positive inside ($V_0>E$) gives decay $e^{\pm \kappa x}$. Why keep both $e^{-\kappa x}$ and $e^{+\kappa x}$ inside the wall? ::: The wall is finite (ends at $L$), so the growing term doesn't blow up and can't be discarded. Why only $F{e}^{ikx}$ in Region III? ::: Nothing beyond $+\infty$ reflects the wave, so no leftward component exists. What forces $\psi$ and ${\psi }^{\prime }$ to match at the joins? ::: A jump makes probability two-valued; a kink makes curvature infinite — both break Schrödinger's equation. What does $L\to 0$ give and why? ::: $T\to 1$: $\sinh (0)=0$ removes the barrier term — no wall, full transmission. What happens when $E>V_0$? ::: $\kappa$ turns imaginary, $\sinh \to i\sin$, giving oscillating $T$ with resonances (over-barrier transmission), not simple decay.