Remove the wall behind the barrier and the encounter acquires a second outcome. Everything on this page is the arithmetic of how the incoming intensity divides between the two.

§5.1 put an infinite wall behind the well, so everything that went in came back out. The encounter had exactly one possible outcome, and the only thing left for the potential to change was a phase.

Take the wall away and put a barrier in the middle instead. Now there are two outcomes — the particle comes back, or it carries on — and which one happens is not decided by the equations. This is the section the chapter exists for.

Physics you need first — what "intensity" means for a wave you cannot normalize

Every wave on this page runs to infinity, so $\int|\psi|^2\,\mathrm dx$ diverges and no amplitude here has an absolute meaning — exactly the problem §5.1b flagged and deferred.

The way out is that we never need an absolute amplitude. Write the wave approaching the barrier as $A_Ie^{ikx}$ ; the quantity $|A_I|^2$ is called its intensity, and for a plane wave of fixed $k$ it is proportional to the rate at which probability flows past a point. Since the transmitted and reflected waves have the same $k$ as the incident one, the constant of proportionality is identical for all three, and it cancels in a ratio. That is why Eq. (5.35) below defines $R$ and $T$ as ratios and never as absolute quantities, and it is the whole reason this section can dodge the normalization problem instead of solving it.

Two other things carried over: §4.3 for $\hat H\psi = E\psi$ and §4.5 for what a superposition over a continuum of energies means. And from §5.1a, the one technique this chapter has: solve in each region, then join the pieces smoothly.

5.2 Barrier Penetration

The potential, and what classical physics predicts

V(x) = \begin{cases} 0 & \text{if } -\infty < x < 0\\ V_B & \text{if } 0 < x < a\\ 0 & \text{if } a < x < +\infty.\end{cases}\tag{5.25}

A classical particle arriving from the left has one of two futures, and which one is fully determined by its energy: below $V_B$ it bounces, above $V_B$ it crosses. There is nothing to compute.

situation⇅	classical particle⇅	quantum particle⇅
$E > V_{B}$	transmitted, with certaintyⓘ	usually transmitted — but $R \neq = 0$ ⓘ
$E < V_{B}$	reflected, with certaintyⓘ	usually reflected — but $T \neq = 0$ ⓘ
inside the barrier, $E < V_{B}$	never thereⓘ	a real, decaying probability densityⓘ
which one actually happens	determined in advanceⓘ	undetermined until the particle is detectedⓘ

Here is the whole problem, solved exactly, before any of the algebra. Drag the energy down through the barrier top and watch the wave inside stop oscillating and start decaying.

incident wave (Re)
reflected wave (Re)
the decaying wave inside, and what escapes
total ψ — real part
total ψ — imaginary part

the named wavestotal Re ψ and Im ψ|ψ|²

Inside the hatched strip the wave does not oscillate — it decays, by e^−βx with β = 2.236, so it falls to 1.07e-1 of its value across the width. Whatever is left at x = a is what leaks out as a travelling wave on the right, and the amplitude there never reaches zero. On the left the incident and reflected waves interfere and the total ripples; the ripples do not go all the way down to zero, and the shortfall is precisely the probability that leaked away.

energy E1.50height V_B4.00width a1.00

E / V_B 0.375

β (decay) 2.2361

R 0.958009

T 0.041991

R + T = 1.000000000000 — Eq. (5.36), and it holds to the last digit the arithmetic carries, for every setting of every slider. The particle’s energy is below the barrier, so classically T would be exactly zero. It is 0.0420. The wide-barrier formula Eq. (5.40) predicts 0.0428 — a 2% error, because e^−2βa = 1.14e-2 really is small.

Why we do not need time-dependent quantum mechanics

The honest description of “a particle arrives, and may bounce or may cross” is a non-stationary state — a wave packet, built from a continuum of energies, exactly the machinery of §4.5:

\Psi(x,t) = \int c(E')\,\psi_{E'}(x)\,e^{-iE't/\hbar}\,\mathrm dE'

with $|c(E')|^2\,\mathrm dE'$ the probability that the particle’s energy lies between $E'$ and $E' + \mathrm dE'$ . If $|c(E')|^2$ is sharply peaked at $E$ , this is a localized lump of probability representing a particle of energy $E$ arriving at the barrier.

That integral is not just a formality — it is a picture, and it is the picture that makes “the outcome is uncertain” mean something concrete. Below, it is evaluated directly: each $\psi_{E'}$ is one of the stationary states this page derives, so the animation introduces no new physics whatsoever.

time0.0

left of the barrier 100.0 %

inside 0.00 %

past the barrier 0.0 %

This is the book’s own formula on p. 95 — Ψ(x,t) = ∫ c(E′) ψ_E′(x) e^−iE′t/ħ dE′ — evaluated as a sum over 64 energies with a Gaussian c(E′) centred on the energy you chose. Every ψ_E′ in it is one of the stationary states the rest of this widget draws, so nothing new has been assumed: the moving picture and the standing one are the same solution, read two ways. Press play. The packet has a definite arrival time only because it does not have a definite energy.

energy E1.50height V_B4.00width a1.00

E / V_B 0.375

β (decay) 2.2361

R 0.958009

T 0.041991

Why we may throw all of that away and still get the right answer

The packet needs a spread of energies to be localized at all — that is $\delta t\,\Delta E \approx \hbar$ from §4.5. But if that spread is narrow compared with the features of $V(x)$ — here, compared with $V_B$ itself — then every energy in the packet sees very nearly the same barrier and gets very nearly the same $R$ and $T$ .

So the probabilities can be computed once, for a single sharp energy, using a stationary state. Time dependence is needed to understand the encounter and completely unnecessary to calculate it. The animation above shows the price of the shortcut: its packet is deliberately broad, and the transmitted fraction it measures differs from the single-energy $T$ by a few percent because $T$ varies sharply across the band.

So we take a stationary state,

\Psi(x,t) = \psi_E(x)\,e^{-iEt/\hbar}\tag{5.26}

whose spatial part obeys the energy eigenvalue equation:

-\frac{\hbar^2}{2m}\frac{\mathrm d^2\psi_E}{\mathrm dx^2} + V(x)\psi_E = E\psi_E\tag{5.27}

Three regions, three solutions

$V(x)$ is constant in each region, so Eq. (5.27) is the free-particle equation three times over with three different constants.

Left of the barrier ( $x < 0$ ), $V = 0$ :

\frac{\mathrm d^2\psi_E}{\mathrm dx^2} = -k^2\psi_E,\qquad\text{where}\quad E = \frac{\hbar^2k^2}{2m}\tag{5.28}

\psi_E(x) = A_Ie^{+ikx} + A_Re^{-ikx}\tag{5.29}

an incident wave of intensity $|A_I|^2$ travelling right, plus a reflected wave of intensity $|A_R|^2$ travelling left.

Inside the barrier, the form depends on which side of $V_B$ the energy sits. Above it, $(0 < x < a)$ is classically allowed:

\frac{\mathrm d^2\psi_E}{\mathrm dx^2} = -k_B^2\psi_E,\qquad\text{where}\quad E = \frac{\hbar^2k_B^2}{2m} + V_B\tag{5.30}

\psi_E(x) = Ae^{+ik_Bx} + A'e^{-ik_Bx}\tag{5.31}

Below it, $(0 < x < a)$ is a classically forbidden region and the sign on the right-hand side flips:

\frac{\mathrm d^2\psi_E}{\mathrm dx^2} = +\beta^2\psi_E,\qquad\text{where}\quad E = -\frac{\hbar^2\beta^2}{2m} + V_B\tag{5.32}

\psi_E(x) = Be^{-\beta x} + B'e^{+\beta x}\tag{5.33}

Right of the barrier ( $x > a$ ), $V = 0$ again, and nothing arrives from $+\infty$ , so only the rightward wave survives:

\psi_E(x) = A_Te^{+ikx}\tag{5.34}

symbol: $k = 2 m E /ℏ$
is: wave number OUTSIDE the barrier. It is the same on both sides, because the potential is zero on both sides — so the transmitted wave has the same wavelength as the incident one.
units: $m^{- 1}$
type: real scalar

Engineer's bridge — the barrier is a waveguide below cutoff

The substitution $q = i\beta$ is not a trick invented for quantum mechanics. You have already met it.

Evanescence. In a rectangular waveguide, the propagation constant is $\gamma = \sqrt{(\pi/w)^2 - \omega^2\mu\epsilon}$ . Above the cutoff frequency it is imaginary and the mode propagates; below cutoff it is real and the field decays as $e^{-\gamma z}$ without ever oscillating. Compare Eq. (5.32): $V_B$ plays the role of cutoff, $E$ the role of $\omega$ , and $\beta$ is $\gamma$ . A below-cutoff waveguide does not reflect perfectly either — make it short enough and power gets through. Microwave engineers call that a below-cutoff attenuator and calibrate it in dB per unit length, which is Eq. (5.40) written logarithmically.

Frustrated total internal reflection is the same equation again in optics: light totally reflected inside a prism has an evanescent field outside it, and bringing a second prism within a wavelength lets light cross the gap. That is optical tunnelling, and the transmitted intensity falls off as $e^{-2\beta a}$ with the gap.

Filter language. $\psi(x)$ here plays the role a signal plays in a transfer function: $A_T/A_I$ is a transmission coefficient with a magnitude and a phase, exactly $S_{21}$ . Eq. (5.36) is then the statement that the network is lossless — $|S_{11}|^2 + |S_{21}|^2 = 1$ — which is the same conservation statement one layer up.

Where it breaks, and it matters. In the waveguide, $e^{-\gamma z}$ is the attenuation of a field carrying energy, and the energy that does not get through is genuinely reflected as a wave. Here $|\psi|^2$ is a probability density, and no fraction of a particle crosses the gap. $T = 10^{-4}$ does not mean each electron arrives with $10^{-4}$ of its energy; it means one electron in ten thousand arrives whole, with all of its energy, and the rest arrive nowhere. The wave picture gets the number right and the ontology wrong, and that difference is the entire content of the Born interpretation.

Reflection and transmission probabilities

Join the pieces at $x = 0$ and $x = a$ and the amplitudes are fixed relative to $A_I$ . Their intensity ratios,

R = \frac{|A_R|^2}{|A_I|^2}\qquad\text{and}\qquad T = \frac{|A_T|^2}{|A_I|^2}\tag{5.35}

are the reflection probability and the transmission probability , and since those are the only two things that can happen,

R + T = 1\tag{5.36}

Why R + T = 1 needs proving at all

It looks like bookkeeping. It is not — it is a claim that nothing is lost, and $|\psi|^2$ alone cannot establish it, because $|\psi|^2$ is a density, not a flow. What has to balance is probability current: the rate at which probability arrives from the left must equal the rate at which it leaves, one way or the other.

The reason Eq. (5.35) can compare bare amplitudes and get away with it is that $j \propto k|A|^2$ for a plane wave, and all three waves share the same $k$ — the potential is zero on both sides of the barrier. Had the potential been different on the two sides, the $k$ ‘s would differ and $T$ would carry a ratio of wave numbers. That is a genuinely common mistake and this potential happens to hide it.

Problem 5 proves Eq. (5.36) from the probability current of §3.5 — and note that the site’s version of that current is the corrected one; the book prints its two terms the wrong way round.

Tunnelling through wide barriers

Now take $E < V_B$ and actually do it. The eigenfunction is Eq. (5.29), Eq. (5.33) and Eq. (5.34) stitched together:

\psi_E(x) = \begin{cases} A_Ie^{+ikx} + A_Re^{-ikx} & \text{if } -\infty < x < 0\\ Be^{-\beta x} + B'e^{+\beta x} & \text{if } 0 < x < a\\ A_Te^{+ikx} & \text{if } a < x < \infty,\end{cases}\tag{5.37}

with the five constants fixed by requiring $\psi_E$ and $\mathrm d\psi_E/\mathrm dx$ to be continuous at both edges. Four equations, five unknowns — and the missing fifth is the overall scale, which cancels out of every ratio in Eq. (5.35).

step 1 of 11

Two conditions at each of two interfaces, four equations, five unknowns — and one overall scale that never mattered. Every step is algebra except step 8, which is the single approximation.

1Write down the general solution in each region separately. Each is just the free-particle equation with a different constant, so we already know both solutions in each.
$ψ_{E} (x) = ⎩ ⎨ ⎧ A_{I} e^{+ ik x} + A_{R} e^{- ik x} B e^{- β x} + B^{'} e^{+ β x} A_{T} e^{+ ik x} x < 0 0 < x < a a < x < \infty$
Equation (5.37). Five unknowns. Note what is NOT there: no e^{-ikx} term on the right, because nothing is coming back from x = +∞ — that is a physical choice, and it is the only place the physics of "incident from the left" enters.
2ψ must be continuous at x = 0. A kink is allowed; a jump is not, because a jump in ψ makes dψ/dx a delta function and d²ψ/dx² its derivative, which no finite V(x) in Eq. (5.27) could balance.
$A_{I} + A_{R} = B + B^{'}$
3dψ/dx must also be continuous at x = 0, and for the same reason one derivative up: V(x) has a finite step here, not an infinite one.
$ik A_{I} - ik A_{R} = - β B + β B^{'}$
4Add ik × (the first) to (the second) to eliminate A_R, since A_R is not what we are after.
$2 ik A_{I} = - (β - ik) B + (β + ik) B^{'}$
Equation (5.38). The incident amplitude is now written purely in terms of what happens inside the barrier.
5Now do the same at the far edge x = a. Continuity of ψ first.
$B e^{- β a} + B^{'} e^{+ β a} = A_{T} e^{ik a}$
6And continuity of dψ/dx at x = a.
$- β B e^{- β a} + β B^{'} e^{+ β a} = ik A_{T} e^{ik a}$
7Solve those two together for A_T and for B′ — both in terms of the single constant B.
$A_{T} e^{ik a} = \frac{2 β}{( β - ik )} B e^{- β a} and B^{'} = B e^{- 2 β a} \frac{( β + ik )}{( β - ik )}$
Equation (5.39). Everything is now expressed through B, and T = |A_T|²/|A_I|² follows from Eqs. (5.38) and (5.39) with no further physics — only algebra.
8Make the ONE approximation of this derivation: assume the barrier is wide, exp(−2βa) ≪ 1. Equation (5.39) then says B′ is smaller than B by exactly that factor, so drop it from Eq. (5.38).
$2 ik A_{I} \approx - (β - ik) B$
Physically: B′e^{+βx} is the wave growing toward x = a, and it is fed only by whatever reflects back off the far edge. Across a wide barrier almost nothing gets that far, so almost nothing comes back.
9Divide the two relations to eliminate B, which was never observable anyway.
$A_{T} e^{ik a} \approx - \frac{4 ik β e ^{- β a}}{( β - ik ) ^{2}} A_{I}$
10Take the squared modulus, using |β − ik|² = β² + k². The phases cancel, which is why a transmission PROBABILITY exists at all.
$T \approx [\frac{16 k ^{2} β ^{2}}{( β ^{2} + k ^{2} ) ^{2}}] e^{- 2 β a}$
11Finally substitute k = √(2mE)/ħ and β = √(2m(V_B − E))/ħ, so that β² + k² = 2mV_B/ħ² and the whole prefactor collapses onto energies.
$T \approx [\frac{16 E ( V _{B} - E )}{V _{B}^{2}}] e^{- 2 β a}$
Equation (5.40). The prefactor is a number of order 1 — it is at most 4, at E = V_B/2. The exponential is everything.

The exact eigenfunction for E = 1.50, V_B = 4.00, a = 1.00. Smooth everywhere — that is what the four conditions bought.

E / V_B 0.375

β (decay) 2.2361

R 0.958009

T 0.041991

The result:

T \simeq \left[\frac{16E(V_B - E)}{V_B^2}\right]e^{-2\beta a}\tag{5.40}

symbol: $e^{- 2 β a}$
is: the entire story. It ranges over every order of magnitude there is. Double a and you SQUARE this factor; this is what makes tunnelling either routine or unimaginably rare with nothing in between.
units: dimensionless
type: dimensionless, spans 10⁰ to 10⁻⁴⁰ and beyond

What this really says — tunnelling is not borrowing and not luck; the wave is simply there

Tunnelling is not the particle “borrowing energy” and it is not the particle “getting lucky”. Nothing here is stochastic: Eq. (5.27) is a linear differential equation, and its solution simply does not vanish inside the barrier. The wave decays there rather than oscillating — but a decaying exponential is never zero, so there is always something left at $x = a$ , and whatever is left has no choice but to continue as a travelling wave.

The reason the effect is invisible in everyday life is $\beta$ , not principle. $\beta = \sqrt{2m(V_B - E)}/\hbar$ grows with mass, so a heavier particle tunnels exponentially less. A tennis ball against a wall has $2\beta a$ of order $10^{34}$ , so its $T$ is $e^{-10^{34}}$ — not small, but small in a way that has no physical meaning at all. An electron across a nanometre has $2\beta a \approx 10$ , and $e^{-10}$ is one in twenty thousand. That is a number an instrument can work with, which is why §5.2b can build a microscope out of it.

Put a number on it — an electron meeting a 5 eV barrier

An electron with $E = 1\ \mathrm{eV}$ meets a barrier $V_B = 5\ \mathrm{eV}$ high and $a = 0.5\ \mathrm{nm}$ wide.

Step 1 — the penetration parameter . $V_B - E = 4\ \mathrm{eV} = 6.409\times10^{-19}\ \mathrm{J}$ , so

\beta = \frac{\sqrt{2m_e(V_B - E)}}{\hbar} = \frac{\sqrt{2(9.109\times10^{-31})(6.409\times10^{-19})}}{1.055\times10^{-34}} = 1.025\times10^{10}\ \mathrm{m^{-1}}

That is a decay length $1/\beta = 0.098\ \mathrm{nm}$ — about one atomic diameter. Memorize this number: for an electron and a barrier a few eV high, the wave function dies out over roughly one atom.

Step 2 — the exponent. $2\beta a = 2(1.025\times10^{10})(0.5\times10^{-9}) = 10.25$ , so $e^{-2\beta a} = 3.55\times10^{-5}$ .

Step 3 — the prefactor. $16E(V_B-E)/V_B^2 = 16(1)(4)/25 = 2.56$ .

Step 4 — multiply. $T \simeq 2.56 \times 3.55\times10^{-5} = 9.1\times10^{-5}$ .

About one electron in eleven thousand crosses. The exact answer, with no approximation at all, is $9.0845\times10^{-5}$ — the wide-barrier formula is good to two parts in a hundred thousand here, because $e^{-2\beta a}$ really is small.

Note where the sensitivity lives. Widen the gap to $0.6\ \mathrm{nm}$ — a change of one fifth of an atomic diameter — and $T$ falls to $1.17\times10^{-5}$ , a factor of 7.8. Nothing in the prefactor can do that: over the whole range of energies it spans a factor of 5, while over these widths the exponential spans a factor of $7.5\times10^{16}$ .

T exact
Eq. (5.40) — a straight line, by construction

E / V_B0.25height V_B4.00

β 2.4495

T at a = 1 2.22e-2

T at a = 2 1.67e-4

The dashed line is Eq. (5.40). On these axes it is exactly straight — that is what an exponential is — with slope −2β/ln 10 = −2.13 decades per unit of width. The solid curve is the exact answer, and the two are indistinguishable everywhere except the first fraction of a decay length on the left, which is exactly the region where e^−2βa ≪ 1 is false.

The number to remember is the one that does not depend on any slider: widening the barrier by one decay length 1/β always costs a factor of e² ≈ 7.39, which is 0.869 of a decade. Doubling the width from 1 to 2 here takes T from 2.22e-2 to 1.67e-4 — a factor of 133. Nothing else in the formula behaves like this: the prefactor 16E(V_B − E)/V_B² can never leave the range 0 to 4, however hard you push the sliders.

Reproduce it in NumPy

The exact coefficients come from joining ψ and dψ/dx at both edges — the same four conditions the StepThrough applies by hand — with q = k_B above the barrier and q = iβ below it, so one line of code covers both cases.

import numpy as np

hbar, me, eV = 1.054571e-34, 9.10938e-31, 1.602177e-19

def coeffs(E, VB, a, m=1.0, hb=1.0):
    """Exact A_R, A_T with A_I = 1. q = k_B (E>VB) or i*beta (E<VB) — the
    complex square root covers both cases in one expression."""
    k = np.sqrt(2*m*E + 0j)/hb
    q = np.sqrt(2*m*(E - VB) + 0j)/hb
    D = (k + q)**2*np.exp(-1j*q*a) - (k - q)**2*np.exp(1j*q*a)
    return (k**2 - q**2)*(np.exp(-1j*q*a) - np.exp(1j*q*a))/D, 4*k*q*np.exp(-1j*k*a)/D

def approx(E, VB, a, m=1.0, hb=1.0):
    beta = np.sqrt(2*m*(VB - E))/hb
    return 16*E*(VB - E)/VB**2 * np.exp(-2*beta*a)

E, VB = 1*eV, 5*eV
beta = np.sqrt(2*me*(VB - E))/hbar
print("exact vs Eq. (5.40): an electron, VB = 5 eV, E = 1 eV")
print("=" * 70)
print(f"  beta        = {beta:.4e} /m     decay length 1/beta = {1e9/beta:.4f} nm")
print(f"  prefactor 16E(VB-E)/VB^2 = {16*E*(VB-E)/VB**2:.4f}  (bounded by 4, always)\n")
print(f"  {'a/nm':>7} {'2*beta*a':>11} {'T exact':>11} {'T Eq.(5.40)':>14} {'approx/exact':>14}")
worst, Ts = 0.0, []
for a_nm in (0.10, 0.30, 0.50, 0.60, 1.00, 2.00):
    a = a_nm*1e-9
    r, t = coeffs(E, VB, a, me, hbar)
    Te, Ta = abs(t)**2, approx(E, VB, a, me, hbar)
    worst = max(worst, abs(abs(r)**2 + Te - 1))
    Ts.append(Te)
    print(f"  {a_nm:7.2f} {2*beta*a:11.3f} {Te:11.3e} {Ta:14.3e} {Ta/Te:14.4f}")
print(f"  R + T = 1 to 1e-15 at every one of those widths." if worst < 1e-15 else f"  R+T off by {worst:.1e}")
print(f"  0.5 nm -> 0.6 nm, one fifth of an atom wider: T falls by {Ts[2]/Ts[3]:.1f}x")

print("\nthe prefactor cannot matter: 16 x (1-x) over the whole range")
xs = [0.05, 0.10, 0.25, 0.50, 0.75, 0.95]
pre = [16*x*(1 - x) for x in xs]
for i in range(3):
    print(f"  E/VB = {xs[i]:.2f} -> {pre[i]:5.3f}      E/VB = {xs[i+3]:.2f} -> {pre[i+3]:5.3f}")
print(f"  prefactor spans a factor of {max(pre)/min(pre):.1f}; over the same widths")
print(f"  the exponential spans a factor of {max(Ts)/min(Ts):.1e}.  That is the asymmetry.")

print("\nwhere Eq. (5.40) breaks: it can exceed 1")
r, t = coeffs(0.5, 1.0, 0.5)
print("  VB=1, a=0.5, E=0.5 (natural units, hbar=m=1)")
print(f"    exp(-2 beta a) = {np.exp(-2*np.sqrt(2*0.5)*0.5):.4f}   -- NOT small")
print(f"    T exact        = {abs(t)**2:.6f}")
print(f"    T Eq. (5.40)   = {approx(0.5, 1.0, 0.5):.6f}   <-- a probability greater than one")

print("\nresonances above the barrier: T = 1 exactly when kB*a = n*pi")
for n in (1, 2, 3):
    E2 = 1.0 + (n*np.pi)**2/2          # VB = 1, a = 1
    r2, t2 = coeffs(E2, 1.0, 1.0)
    print(f"  n={n}  E/VB = {E2:5.2f}  T = {abs(t2)**2:.12f}")

printsexact vs Eq. (5.40): an electron, VB = 5 eV, E = 1 eV ====================================================================== beta = 1.0246e+10 /m decay length 1/beta = 0.0976 nm prefactor 16E(VB-E)/VB^2 = 2.5600 (bounded by 4, always) a/nm 2*beta*a T exact T Eq.(5.40) approx/exact 0.10 2.049 3.029e-01 3.298e-01 1.0887 0.30 6.148 5.467e-03 5.474e-03 1.0012 0.50 10.246 9.085e-05 9.085e-05 1.0000 0.60 12.296 1.170e-05 1.170e-05 1.0000 1.00 20.493 3.224e-09 3.224e-09 1.0000 2.00 40.985 4.060e-18 4.060e-18 1.0000 R + T = 1 to 1e-15 at every one of those widths. 0.5 nm -> 0.6 nm, one fifth of an atom wider: T falls by 7.8x the prefactor cannot matter: 16 x (1-x) over the whole range E/VB = 0.05 -> 0.760 E/VB = 0.50 -> 4.000 E/VB = 0.10 -> 1.440 E/VB = 0.75 -> 3.000 E/VB = 0.25 -> 3.000 E/VB = 0.95 -> 0.760 prefactor spans a factor of 5.3; over the same widths the exponential spans a factor of 7.5e+16. That is the asymmetry. where Eq. (5.40) breaks: it can exceed 1 VB=1, a=0.5, E=0.5 (natural units, hbar=m=1) exp(-2 beta a) = 0.3679 -- NOT small T exact = 0.786448 T Eq. (5.40) = 1.471518 <-- a probability greater than one resonances above the barrier: T = 1 exactly when kB*a = n*pi n=1 E/VB = 5.93 T = 1.000000000000 n=2 E/VB = 20.74 T = 1.000000000000 n=3 E/VB = 45.41 T = 1.000000000000

The approximation has a domain, and you can see its edge

Two more things become visible once $T(E)$ is plotted over the whole energy range, and neither is in the book’s text for this section.

T exact
T from Eq. (5.40)
R exact

log scale (the only way to see how small T gets)

Three things are worth waiting for on this plot. In the hatched region (E < V_B) T is not zero — that is tunnelling, and on the log scale you can watch it fall through ten orders of magnitude as the barrier widens. The dashed curve is Eq. (5.40): it tracks the exact answer beautifully deep in the hatched region and then fails near E = V_B, where it can even exceed 1, because e^−2βa → 1 there and the approximation’s one assumption has failed. And above the barrier, T returns to 1 at a discrete set of resonances, k_B a = nπ — a whole number of half wavelengths across the barrier, which is the same interference condition that makes an anti-reflection coating work.

energy E1.50height V_B4.00width a1.00

E / V_B 0.375

β (decay) 2.2361

R 0.958009

T 0.041991

The first is that $T$ is not zero anywhere — the hatched region is where classical physics forbids transmission entirely, and the curve simply passes through it. The second is that above the barrier, $R$ is not zero either, except at a discrete set of energies where $k_Ba = n\pi$ and the barrier goes completely transparent. That is the same interference condition that makes an anti-reflection coating work: the reflections from the two faces cancel when the round trip is a whole number of wavelengths. Problem 6 derives it.

Where this goes next

Equation (5.40) is a formula with two parameters, and the exponential means that plausible values of those parameters produce wildly different worlds. §5.2b takes it to the two places where the numbers land somewhere interesting: electrons across a vacuum gap, where $2\beta a \approx 10$ and a change of one hundredth of an atomic diameter is measurable — that is the scanning tunnelling microscope — and protons in the centre of the sun, where the barrier is not square, the exponent is about 22, and the answer is why stars burn for billions of years instead of exploding.

Check yourself

0 / 6 answered

Open the "R and T vs energy" view above and drag the energy up past the top of the barrier ( $E / V_{B} > 1$ ).
1.What does the transmission probability $T$ do once the particle has more energy than the barrier is high?
2.The page's electron facing a $4 eV$ energy deficit has a decay length $1/ β = 0.0976 nm$ . Replace it with a proton at the same deficit. What is the proton's decay length?
3.An electron beam is fired at a barrier with $T = 1 0^{- 4}$ , and a perfect detector waits on the far side. What does it record?
4.Why does §5.2 insist that the exponential in Eq. (5.40), not the prefactor $16 E (V_{B} - E) / V_{B}^{2}$ , is what matters?
Take $V_{B} = 1$ , $a = 0.5$ , $E = 0.5$ in natural units ( $ℏ = m = 1$ ). Eq. (5.40) returns $T = 1.47$ ; the exact answer is $0.786$ .
5.A probability of 1.47. What has gone wrong?
The log-scale figure above draws $T$ against barrier width as a straight line.
6.By how much does $T$ fall when the barrier is widened by exactly one decay length $1/ β$ ?