# Step, Well & Barrier
> Three rectangles, and in them most of quantum technology. The step teaches that particles reflect
> where no classical particle would; the well teaches that confinement quantizes energy; the barrier
> teaches that walls are negotiable. The last one runs your world: it is why the Sun burns, how a
> scanning tunneling microscope sees atoms, and what beats inside every superconducting qubit. All
> of it falls out of the matching toolkit from the previous lesson — no new physics, just seams.
## Learning Objectives
After this lesson you will be able to:
1. Solve the **potential step** for $E > V_0$ and $E < V_0$, computing **reflection** and **transmission** coefficients from **probability currents**.
2. Derive the full spectrum and eigenfunctions of the **infinite square well**, verify their **orthonormality**, and list the four ways **quantum confinement** departs from the classical box — discreteness, **zero-point energy**, non-uniform density, and **nodes**.
3. Reduce the **finite square well** to transcendental equations via **parity**, solve them graphically, and count bound states.
4. Derive the exact **tunneling** probability through a rectangular **barrier** and its opaque-limit form $T \sim e^{-2\kappa a}$.
5. Locate **transmission resonances** above the barrier and connect tunneling to **α-decay**, the **STM**, and **Josephson junctions**.
---
## Intuition
A classical ball rolling toward a cliff edge never bounces back; a ball rolling at a wall taller
than its energy never gets through. Quantum mechanics breaks both rules with a single mechanism: at
every discontinuity of $V$ the wave must be stitched together continuously, and stitching a wave
means partial reflection — just like light hitting glass. Confine the wave between two walls and the
stitching only works at discrete wavelengths: a guitar string, hence discrete energies. Make one
wall thin, and the evanescent tail of Lesson 1 pokes out the far side with small but nonzero
amplitude: tunneling. Reflection, quantization, tunneling — three faces of wave matching.
---
## Theory
### The potential step, $E > V_0$
Let $V = 0$ for $x < 0$ and $V = V_0 > 0$ for $x > 0$, with a particle incident from the left at
energy $E > V_0$. With $k_1 = \sqrt{2mE}/\hbar$ and $k_2 = \sqrt{2m(E - V_0)}/\hbar$,
$$
\psi(x) =
\begin{cases}
e^{ik_1x} + r\,e^{-ik_1x}, & x < 0,\\
t\,e^{ik_2x}, & x > 0,
\end{cases}
$$
(incident amplitude 1; no left-mover on the right — nothing returns from $+\infty$). Matching $\psi$
and $\psi'$ at $x = 0$:
$$
1 + r = t, \qquad ik_1(1 - r) = ik_2\,t
\quad\Longrightarrow\quad
r = \frac{k_1 - k_2}{k_1 + k_2}, \qquad t = \frac{2k_1}{k_1 + k_2}.
$$
Amplitudes are not yet probabilities. The bookkeeping is done by the **probability current** of
[P.4.1](../c04-schrodinger-equation/01-schrodinger-equation-born-rule.md),
$j = \frac{\hbar}{m}\mathrm{Im}(\psi^*\psi')$, which gives $j_{\mathrm{inc}} = \hbar k_1/m$,
$j_{\mathrm{refl}} = \frac{\hbar k_1}{m}\lvert r\rvert^2$, and
$j_{\mathrm{trans}} = \frac{\hbar k_2}{m}\lvert t\rvert^2$. Reflection and transmission are **flux
ratios**:
$$
R = \lvert r\rvert^2 = \left(\frac{k_1 - k_2}{k_1 + k_2}\right)^{\!2},
\qquad
T = \frac{k_2}{k_1}\lvert t\rvert^2 = \frac{4k_1k_2}{(k_1 + k_2)^2},
\qquad
R + T = \frac{(k_1 - k_2)^2 + 4k_1k_2}{(k_1 + k_2)^2} = 1 . \checkmark
$$
Note $R > 0$ even though $E > V_0$: the wave reflects off the *discontinuity* — something no
classical particle does.
> **Caution.** $R$ and $T$ are **flux** probabilities, not "pieces of the particle." Every detection
> (Born rule, [P.4.1](../c04-schrodinger-equation/01-schrodinger-equation-born-rule.md)) finds one
> whole particle on one side; $R$ and $T$ are the odds. And the classic trap: $T \neq \lvert t\rvert^2$
> when $k_2 \neq k_1$ — the transmitted wave moves at a different speed, and it is *flux*
> $\frac{\hbar k}{m}\lvert\text{amp}\rvert^2$, not amplitude modulus, that is conserved.
### The step with $E < V_0$: total reflection with a fingerprint
For $E < V_0$, set $k_2 \to i\kappa$ with $\kappa = \sqrt{2m(V_0 - E)}/\hbar$; the right side
becomes evanescent, $t\,e^{-\kappa x}$. The same algebra gives
$$
r = \frac{k_1 - i\kappa}{k_1 + i\kappa} = e^{-2i\theta}, \qquad \theta = \arctan\frac{\kappa}{k_1},
\qquad \lvert r\rvert = 1 .
$$
Total reflection: a real exponential carries zero current, so $T = 0$ (E3). But the particle *does*
penetrate to depth $\sim 1/\kappa$, and the reflected wave picks up the **phase shift** $-2\theta$ —
as if it bounced off a point slightly *inside* the wall. That phase is the fingerprint of the
intrusion, and it becomes measurable the moment the wall is made thin.
### The infinite square well
Let $V = 0$ for $0 \le x \le L$, $V = \infty$ outside: $\psi$ must vanish at the walls ($\psi'$ may
kink there — Lesson 1). Inside, $\psi = A\sin kx + B\cos kx$ with $k = \sqrt{2mE}/\hbar$. Then
$\psi(0) = 0$ kills $B$, and $\psi(L) = 0$ forces $kL = n\pi$, $n = 1, 2, 3, \dots$ — exactly the
standing waves of a clamped string
([P.1.1](../c01-theoretical-mechanics/01-waves-and-the-wave-equation.md)): quantization is "a whole
number of half-wavelengths must fit." With $A$ fixed by
$\int_0^L\lvert A\rvert^2\sin^2(n\pi x/L)\,dx = \lvert A\rvert^2L/2 = 1$:
$$
\boxed{\ E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}, \qquad \psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\frac{n\pi x}{L}\ }
$$
The set is **orthonormal** — using $2\sin a\sin b = \cos(a - b) - \cos(a + b)$,
$$
\int_0^L\psi_m\psi_n\,dx = \frac1L\int_0^L\Bigl[\cos\frac{(m - n)\pi x}{L} - \cos\frac{(m + n)\pi x}{L}\Bigr]dx = \delta_{mn},
$$
since both cosines integrate to zero unless $m = n$, where the first contributes $L$. Checklist from
Lesson 1: $\psi_n$ has $n - 1$ nodes, and the nonzero $E_1$ is confinement's price — $\Delta x \sim L$
costs $\Delta p \sim \pi\hbar/L$ ([P.4.3](../c04-schrodinger-equation/03-matter-waves-uncertainty-momentum.md)).
We restrict $n$ to the **positive** integers, for two reasons worth stating explicitly. Since
$\sin(-a) = -\sin(a)$, the wavefunction for $-n$ is the one for $+n$ times the constant $-1$ — the
same physical state, since states are defined only up to a complex factor
([P.4.1](../c04-schrodinger-equation/01-schrodinger-equation-born-rule.md)). And $n = 0$ gives
$\psi \equiv 0$ everywhere, which is not a state at all: it normalizes to zero, not one.
### Quantum confinement: what the box actually changed
Put the result beside the classical particle rattling between two walls — same box, same mass — and
four differences stand out. They are not quirks of this potential; they recur in every bound problem
in this course, so they are worth naming here.
1. **The energy is discrete.** Classically *any* $E > 0$ is allowed; quantum mechanically only
$E_n \propto n^2$. Boundary conditions have done to the matter wave exactly what pinned ends do to
a guitar string ([P.1.1](../c01-theoretical-mechanics/01-waves-and-the-wave-equation.md)) — the
only new ingredient is de Broglie's $\lambda = h/p$ converting "allowed wavelength" into "allowed
energy."
2. **There is a floor.** The particle cannot have $E = 0$; the minimum is the **zero-point energy**
$E_1 = (\hbar^2/2m)(\pi/L)^2$. A classical particle may sit motionless at the bottom of the box; a
quantum one may not, because $\psi = 0$ is not a state and any nonzero $\psi$ vanishing at both
walls must curve, and curvature costs kinetic energy. The same floor reappears as $\tfrac12\hbar\omega$
for the oscillator ([Lesson 3](03-harmonic-oscillator.md)).
3. **The particle is not spread uniformly** — and how it is spread depends on which state it is in.
The classical particle, moving at constant speed, is equally likely anywhere: a flat density $1/L$.
The quantum densities $|\psi_n|^2 = (2/L)\sin^2(n\pi x/L)$ are anything but flat. In the ground
state the particle is found preferentially near the centre and almost never within a fraction of
$L/n$ of the walls (E2 makes this quantitative: $0.609$ of the probability in the middle third,
against the classical $1/3$).
4. **Higher states have more nodes** — and there are interior points where the particle is *never*
found. $\psi_n$ has exactly $n - 1$ interior zeros, one more for each step up the ladder, and at
each the density is strictly zero. That "one more node per level" pattern is not special to the
box; it holds for essentially every 1D bound-state spectrum (the **node theorem** of
[Lesson 1](01-free-particle-constant-potentials.md)) and is the fastest way to sanity-check a
numerically computed eigenfunction.
Measured from the well's centre the solutions also alternate in **parity** — $n$ odd gives an even
function, $n$ even an odd one — the definite-parity property that the symmetric finite well below will
exploit to halve the algebra.
*How big are these effects?* Everything scales as $1/L^2$, so confinement is negligible until the box
is atom-sized. Squeeze an electron into $L = 5\ \text{Å} = 0.5\ \mathrm{nm}$:
$$
E_1 = \frac{\hbar^2}{2m_e}\Big(\frac{\pi}{0.5\ \mathrm{nm}}\Big)^2
= 0.0381 \times 39.5\ \mathrm{eV} \approx 1.5\ \mathrm{eV},
\qquad E_2 - E_1 = 3E_1 \approx 4.5\ \mathrm{eV}.$$
A few electron-volts — precisely the characteristic spacing between major energy levels *in an atom*.
The crudest imaginable model of "an electron confined to atomic dimensions" lands on the right energy
scale, which is why the particle in a box, for all its rectangular honesty, is the first model anyone
reaches for.
### The finite square well
Now the realistic version: $V = -V_0$ for $-a \le x \le a$, zero outside; bound states have
$-V_0 < E < 0$. Let $l = \sqrt{2m(E + V_0)}/\hbar$ (inside) and $\kappa = \sqrt{-2mE}/\hbar$
(outside). The potential is symmetric, so eigenfunctions are even or odd
([Lesson 1](01-free-particle-constant-potentials.md)). For the **even** states,
$$
\psi(x) = \begin{cases} F\,e^{-\kappa\lvert x\rvert}, & \lvert x\rvert > a,\\ C\cos lx, & \lvert x\rvert \le a. \end{cases}
$$
Matching $\psi$ and $\psi'$ at $x = a$ and *dividing* the two equations (amplitudes cancel — parity's
gift) gives $\kappa = l\tan la$. In the Griffiths variables $z \equiv la$ and
$z_0 \equiv \frac{a}{\hbar}\sqrt{2mV_0}$, we have $\kappa a = \sqrt{z_0^2 - z^2}$, so
$$
\boxed{\ \tan z = \sqrt{(z_0/z)^2 - 1}\ }\ \ (\text{even}),
\qquad
-\cot z = \sqrt{(z_0/z)^2 - 1}\ \ (\text{odd, same steps with } \sin lx).
$$
Transcendental — no closed form, but the **graphical solution** is transparent: plot both sides for
$z \in (0, z_0)$; every intersection is a bound state. Two limits are worth owning:
- **Deep well** ($z_0 \gg 1$): intersections sit just below $z_n \approx n\pi/2$, i.e.
$E_n + V_0 \approx \frac{n^2\pi^2\hbar^2}{2m(2a)^2}$ — the infinite well of width $2a$. ✓
- **Shallow well** ($z_0 \to 0$): the even branch always crosses once, however small $z_0$.
**A 1D well holds at least one bound state.** In general $N = \lceil 2z_0/\pi\rceil$, one new state
(alternating even/odd) each time $z_0$ passes a multiple of $\pi/2$.
### The rectangular barrier: tunneling
Flip the well: $V = V_0 > 0$ for $0 \le x \le a$, zero outside, incidence with $E < V_0$. Five
amplitudes (incident 1, $r$, interior $Ae^{\kappa x} + Be^{-\kappa x}$ — *both* kept, since the
region is finite — and $t$), four matching equations at the two seams. Eliminating $A, B, r$ yields
the exact transmission
$$
\boxed{\ T(E) = \left[1 + \frac{V_0^2\sinh^2(\kappa a)}{4E(V_0 - E)}\right]^{-1}}\ ,
\qquad \kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar}.
$$
$T$ is *never zero*: the evanescent tail always delivers amplitude to the far side. In the **opaque
limit** $\kappa a \gg 1$, $\sinh(\kappa a) \approx \tfrac12e^{\kappa a}$ gives
$$
T \approx 16\,\frac{E}{V_0}\Bigl(1 - \frac{E}{V_0}\Bigr)e^{-2\kappa a}:
$$
an $\mathcal{O}(1)$ prefactor times $e^{-2\kappa a}$, the tail's probability decay across the width.
Tunneling is *exponentially* sensitive to $a$ and to $\sqrt{V_0 - E}$. For $E > V_0$ the substitution
$\kappa \to ik_2$, $k_2 = \sqrt{2m(E - V_0)}/\hbar$, turns $\sinh^2(\kappa a)$ into $-\sin^2(k_2a)$
(E5), so
$$
T(E) = \left[1 + \frac{V_0^2\sin^2(k_2a)}{4E(E - V_0)}\right]^{-1},
$$
generally $< 1$ (reflection *above* the barrier!) except at the **resonances** $k_2a = n\pi$, where
$T = 1$ exactly: the doubly-reflected internal wave cancels the direct reflection — the physics of
anti-reflective lens coatings, and of the **Ramsauer–Townsend effect** (noble-gas atoms turning
nearly transparent to electrons of just the right sub-eV energy).
Both formulas describe a plane wave of a single energy. A real electron is a wave packet, and it
scatters *in time*. Below, the time-dependent Schrödinger equation runs on a grid for a packet 1.2 nm
wide. Once the reflected and transmitted pieces separate, their areas are $R$ and $T$. Compare $T$ with
the plane-wave $T(E_0)$ above, and with $T(E)$ averaged over the packet's spread of energies, which is
what the packet should actually reproduce:
```qm-demo barrier-tunneling
```
### Tunneling runs the world
- **α-decay (Gamow).** The α-particle rattles against its Coulomb barrier $\sim\!10^{21}$ times per
second, escaping with probability $\sim e^{-2\int\kappa\,dx}$ per attempt. Lifetimes exponential
in the barrier integral explain 20-plus orders of magnitude in half-lives.
- **Scanning tunneling microscope.** Electrons tunnel across the tip–surface vacuum gap; with
$1/\kappa \sim 0.1\ \mathrm{nm}$ the current changes tenfold per ångström — enough to image atoms.
- **Josephson junctions.** Cooper pairs tunnel coherently through a nanometer of oxide between two
superconductors: a lossless *nonlinear* circuit element, the anharmonic heart of the transmon and
of every superconducting qubit (Term 4.4) — a direct descendant of this rectangle.
---
## Worked Examples
### Example 1 — An electron in a 1 nm box
For an electron in an infinite well of width $L = 1\ \mathrm{nm}$, using
$\hbar^2/2m_e = (\hbar c)^2/2m_ec^2 = 0.0381\ \mathrm{eV\,nm^2}$:
$$
E_n = n^2\,\pi^2 \times 0.0381\ \mathrm{eV} \approx 0.376\,n^2\ \mathrm{eV}
\quad\Longrightarrow\quad E_1 \approx 0.376,\ E_2 \approx 1.50,\ E_3 \approx 3.38\ \mathrm{eV}.
$$
The $2 \to 1$ transition emits $E_2 - E_1 \approx 1.13\ \mathrm{eV}$:
$\lambda = hc/\Delta E \approx 1240/1.13 \approx 1100\ \mathrm{nm}$, near infrared. Not a toy —
semiconductor quantum wells and dots are engineered this way, emission color set by box size.
### Example 2 — Tunneling through a real barrier
An electron with $E = 5\ \mathrm{eV}$ meets a barrier $V_0 = 10\ \mathrm{eV}$, $a = 0.5\ \mathrm{nm}$:
$\kappa = \sqrt{(V_0 - E)/0.0381} \approx 11.46\ \mathrm{nm^{-1}}$, so $\kappa a \approx 5.73$
(opaque). Exact:
$$
T = \left[1 + \frac{100\,\sinh^2(5.73)}{4 \times 5 \times 5}\right]^{-1} \approx 4.2\times10^{-5},
$$
and the opaque estimate $16 \times \tfrac12 \times \tfrac12\,e^{-11.46} \approx 4.2\times10^{-5}$
agrees to the digits shown. The STM punchline: widen the gap by $0.1\ \mathrm{nm}$ and $T$ drops by
$e^{-2\kappa(0.1\,\mathrm{nm})} = e^{-2.29} \approx 0.10$ — one atomic step, tenfold current change.
---
## Hands-on (Python)
Piecewise-constant potentials suit the **transfer matrix**: in each region
$\psi = Ae^{ikx} + Be^{-ikx}$ (complex $k$ handles forbidden regions automatically), and each seam is
a $2\times2$ matrix. Real units: eV and nm.
```python
import numpy as np
import matplotlib.pyplot as plt
HBAR2_2M = 0.0381 # hbar^2 / 2 m_e in eV nm^2
def transmission(E, Vs, xs):
"""Transfer-matrix T(E) for regions Vs (outer two = leads) with interfaces xs."""
ks = [np.sqrt(complex(E - V) / HBAR2_2M) for V in Vs] # complex k: evanescent OK
M = np.eye(2, dtype=complex)
for j, x in enumerate(xs): # match at each seam
ka, kb = ks[j], ks[j + 1]
M = M @ (0.5 * np.array(
[[(1 + kb/ka) * np.exp(1j*(kb - ka)*x), (1 - kb/ka) * np.exp(-1j*(kb + ka)*x)],
[(1 - kb/ka) * np.exp(1j*(kb + ka)*x), (1 + kb/ka) * np.exp(-1j*(kb - ka)*x)]]))
t = 1.0 / M[0, 0] # no left-mover in last region
return (ks[-1].real / ks[0].real) * abs(t)**2 # flux ratio, NOT |t|^2
def T_exact(E, V0=10.0, a=0.5):
"""Closed-form check for the rectangular barrier."""
if E < V0:
ka = np.sqrt((V0 - E) / HBAR2_2M) * a
return 1.0 / (1.0 + V0**2 * np.sinh(ka)**2 / (4*E*(V0 - E)))
k2a = np.sqrt((E - V0) / HBAR2_2M) * a
return 1.0 / (1.0 + V0**2 * np.sin(k2a)**2 / (4*E*(E - V0)))
V0, a = 10.0, 0.5
for E in [1.0, 5.0, 9.0]:
print(f"E={E:4.1f} eV TM: {transmission(E,[0,V0,0],[0,a]):.4e} exact: {T_exact(E):.4e}")
# E= 1.0 eV TM: 3.0443e-07 exact: 3.0443e-07
# E= 5.0 eV TM: 4.2354e-05 exact: 4.2354e-05
# E= 9.0 eV TM: 8.6068e-03 exact: 8.6068e-03 <- machine-precision agreement
E = np.linspace(0.05, 30.0, 2000)
plt.semilogy(E, [transmission(e, [0, V0, 0], [0, a]) for e in E])
plt.axvline(V0, color="gray", ls=":"); plt.xlabel("E (eV)"); plt.ylabel("T(E)")
plt.title("Rectangular barrier: V0 = 10 eV, a = 0.5 nm"); plt.show()
# Below V0: T climbs ~7 decades (log scale). Above V0: oscillations with perfect
# resonances T = 1 at E ~ 11.50 eV and 16.02 eV (k2*a = pi, 2*pi).
```
```python
# --- Finite well: graphical solution and bound-state count ---
from scipy.optimize import brentq
def bound_states(z0):
"""Roots of tan z = sqrt((z0/z)^2 - 1) (even) and -cot z = ... (odd) on (0, z0)."""
f_even = lambda z: np.tan(z) - np.sqrt((z0/z)**2 - 1)
f_odd = lambda z: 1.0/np.tan(z) + np.sqrt((z0/z)**2 - 1)
roots = []
for n in range(int(2*z0/np.pi) + 1): # scan tan/cot branches
for f, lo, hi in [(f_even, n*np.pi, n*np.pi + np.pi/2),
(f_odd, n*np.pi + np.pi/2, (n+1)*np.pi)]:
lo, hi = lo + 1e-9, min(hi - 1e-9, z0 - 1e-12)
if lo < hi and f(lo) * f(hi) < 0:
roots.append(brentq(f, lo, hi))
return sorted(roots)
V0, a_nm = 20.0, 0.3 # electron parameters
z0 = a_nm * np.sqrt(V0 / HBAR2_2M)
roots = bound_states(z0)
print(f"z0 = {z0:.3f} -> {len(roots)} bound states") # z0 = 6.873 -> 5 bound states
for z in roots:
print(f" z = {z:.4f} E = {HBAR2_2M*(z/a_nm)**2 - V0:+8.4f} eV")
# z = 1.3701 E = -19.2053 eV (even ground state, then alternating parity)
# z = 2.7327 E = -16.8387 eV
# z = 4.0774 E = -12.9622 eV
# z = 5.3834 E = -7.7315 eV
# z = 6.5776 E = -1.6844 eV
for z0 in [0.5, 1.0, 2.0, 4.0, 8.0]:
print(f"z0 = {z0:4.1f}: N = {len(bound_states(z0))}")
# N = 1, 1, 2, 3, 6 -- matching N = ceil(2 z0 / pi): even a whisker of a well binds
```
---
## Exercises
**E1 (easy).** A particle hits an upward step with $E = 2V_0$. Find $R$ and comment.
Solution
$k_2/k_1 = \sqrt{(E - V_0)/E} = 1/\sqrt2$, so
$R = \bigl(\frac{\sqrt2 - 1}{\sqrt2 + 1}\bigr)^2 \approx 0.029$: with *twice* the needed energy, the
particle still reflects $\approx 3\%$ of the time — pure wave behavior at the discontinuity. And
$T = 1 - R \approx 0.971 \neq \lvert t\rvert^2 \approx 1.37$: flux, not amplitude.
Solution
$P = \frac2L\int_{L/3}^{2L/3}\sin^2\frac{\pi x}{L}dx = \frac1L\int_{L/3}^{2L/3}\bigl(1 - \cos\frac{2\pi x}{L}\bigr)dx = \frac13 - \frac{1}{2\pi}\bigl[\sin\tfrac{4\pi}{3} - \sin\tfrac{2\pi}{3}\bigr] = \frac13 + \frac{\sqrt3}{2\pi} \approx 0.609$
— well above the classical $1/3$: the ground state
piles up in the middle.
Solution
$\psi = t\,e^{-\kappa x}$ gives
$j = \frac{\hbar}{m}\mathrm{Im}(-\kappa\lvert t\rvert^2e^{-2\kappa x}) = 0$: the argument is real.
Nonzero $\lvert\psi\rvert^2$ with zero $j$ means a position measurement *can* find the particle
inside the wall, but there is no steady flow through it — water soaking a sponge with no river
beyond. Consistently $\lvert r\rvert = 1$: everything that flows in flows back, delayed by the phase
$-2\theta$.
Solution
$N = \lceil 2z_0/\pi\rceil = 2$ requires $\pi/2 \le z_0 < \pi$ (first odd state appears at
$z_0 = \pi/2$, second even at $\pi$). From $z_0^2 = 2mV_0a^2/\hbar^2$,
$V_0 = z_0^2\,\frac{\hbar^2/2m_e}{a^2} = z_0^2 \times \frac{0.0381}{0.09}\ \mathrm{eV} = 0.423\,z_0^2\ \mathrm{eV}$,
so $V_0 \in [\,0.423(\pi/2)^2,\ 0.423\,\pi^2\,) \approx [\,1.04,\ 4.18\,)\ \mathrm{eV}$.
Solution
For $E > V_0$, $\kappa = ik_2$ with $k_2 = \sqrt{2m(E - V_0)}/\hbar$ real. Since
$\sinh(ix) = i\sin x$, $\sinh^2(\kappa a) = -\sin^2(k_2a)$, and $V_0 - E = -(E - V_0)$: the two sign
flips cancel, giving $T = \bigl[1 + \frac{V_0^2\sin^2(k_2a)}{4E(E - V_0)}\bigr]^{-1}$. $T = 1$ iff
$\sin(k_2a) = 0$, i.e. $k_2a = n\pi$: the internal round trip is $2k_2a = 2\pi n$, so the
doubly-reflected wave cancels the directly reflected one. Energies:
$E_n = V_0 + \frac{\hbar^2}{2m}\bigl(\frac{n\pi}{a}\bigr)^2 = 10 + 0.0381(n\pi/0.5)^2\ \mathrm{eV}$,
so $E_1 \approx 11.50$ and $E_2 \approx 16.02\ \mathrm{eV}$ — the perfect-transmission spikes in the
Hands-on plot. This is the Ramsauer–Townsend mechanism (with the atom's attractive well playing the
rectangle).
At $E_1$ a plane wave passes untouched. A packet centred on $E_1$ carries energies on both sides of
the resonance, so part of it still reflects:
```qm-demo barrier-tunneling
{ "energy": 11.5, "height": 10, "width": 0.5 }
```
Answers
1. Matching a wave across a discontinuity always produces a reflected component (light at glass).
$T$ is a ratio of *fluxes* $\frac{\hbar k}{m}\lvert\text{amp}\rvert^2$, and the transmitted wave
has a different $k$.
2. $\psi$ must vanish at both walls, so $kL = n\pi$ — an integer number of half-wavelengths, the
clamped string. $E \propto k^2 \propto n^2$. Versus the classical particle: (i) only discrete
energies are allowed; (ii) the lowest is $E_1 > 0$, a zero-point energy — the particle cannot be
at rest; (iii) the density $|\psi_n|^2$ is non-uniform and state-dependent, not the classical flat
$1/L$; (iv) $\psi_n$ has $n-1$ interior nodes, points where the particle is never found.
3. Plot $\tan z$ (and $-\cot z$) against $\sqrt{(z_0/z)^2 - 1}$ on $(0, z_0)$; each crossing is a
bound state. Minimum one (the even ground state survives any $z_0 > 0$).
4. $T \approx 16\frac{E}{V_0}(1 - \frac{E}{V_0})e^{-2\kappa a}$: exponential in the width $a$ and in
$\sqrt{V_0 - E}$ (through $\kappa$).
5. $k_2a = n\pi \Rightarrow T = 1$. Tunneling: α-decay, STM, Josephson junctions. Resonance:
Ramsauer–Townsend transparency (optically: anti-reflection coatings).