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.

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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.

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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, $j = \frac{\hbar}{m}\mathrm{Im}(\psi^*\psi')$, 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) 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): 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).

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). 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) — 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).
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) 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). 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).

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.

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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.

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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.

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).

# --- 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

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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.

E2 (easy). For the infinite-well ground state, find the probability of being in the middle third, $L/3 \le x \le 2L/3$.

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.

E3 (medium). For the step with $E < V_0$, show the evanescent region carries zero probability current, and reconcile $T = 0$ with $\lvert\psi\rvert^2 \neq 0$ at $x > 0$.

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 $-2\theta$.

E4 (medium). An electron sits in a finite well of half-width $a = 0.3\ \mathrm{nm}$. For what depths $V_0$ does the well hold exactly two bound states?

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 $V_0 \in [\,0.423(\pi/2)^2,\ 0.423\,\pi^2\,) \approx [\,1.04,\ 4.18\,)\ \mathrm{eV}$.

E5 (hard). From the $E < V_0$ transmission formula, obtain the $E > V_0$ result by analytic continuation, derive the resonance condition, and find the first two resonance energies for $V_0 = 10\ \mathrm{eV}$, $a = 0.5\ \mathrm{nm}$ (electron).

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).

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Checkpoint

1. Why does a particle with $E > V_0$ still reflect off a step, and why is $T = \frac{k_2}{k_1}\lvert t\rvert^2$ rather than $\lvert t\rvert^2$?
2. Where, mechanically, does the infinite well's $E_n \propto n^2$ quantization come from, and in what four ways does the confined particle differ from a classical one in the same box?
3. How do you solve the finite well graphically, and what is the minimum number of bound states of a 1D well?
4. Write the opaque-barrier estimate of $T$ and name the two quantities it is exponentially sensitive to.
5. State the above-barrier resonance condition and one physical appearance each of tunneling and of resonance.

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).

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Further Reading

• [Gri] Griffiths & Schroeter, §2.2, §2.5–2.6 — infinite well, scattering, finite well; §9.2 for Gamow's α-decay via WKB.
• [ER] Eisberg & Resnick, Ch. 6 — steps, barriers, and the classic applications (α-decay, tunnel diodes).
• [Sha] Shankar, Ch. 5 — the same problems with the propagator point of view in reach.

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