University Physics — Year 2 · Bachelor Year 2

31Potential Wells, Barriers and Tunnelling

A ball in a bowl rolls back and forth with whatever energy it was given; an electron in an atom, a nucleon in a nucleus, an electron in a nanometre-sized crystal can only have certain energies, and the lowest of them is not zero. A ball cannot cross a hill higher than its energy allows; an $\alpha$ particle leaves a nucleus through a Coulomb wall it could never climb, the needle of a scanning tunnelling microscope draws a current across a gap of vacuum, and the nitrogen atom of ammonia swings through the plane of its three hydrogens twenty-four billion times a second. This chapter solves the Schrödinger equation in the simplest potentials — wells with infinite and finite walls, a step, a barrier, a double well — and finds in them the two great quantum facts that classical mechanics cannot produce: the quantisation of bound energies, and tunnelling through classically forbidden regions, with its exponential sensitivity to the width and height of the barrier, which is why it can serve as a ruler of picometres and why radioactive lifetimes span thirty orders of magnitude.

A scanning tunnelling microscope: a metal tip held a few tenths of a nanometre above a surface; the tunnelling current, which changes tenfold per tenth of a nanometre, maps the surface atom by atom.

31.1 The infinite well

Proposition 31.1 (Infinite square well)

A particle confined to $0 < x < L$ by impenetrable walls ( $V = 0$ inside, $V = \infty$ outside, so $\varphi = 0$ at the walls) has the stationary states and energies

\varphi_n(x) = \sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}, \qquad E_n = \frac{n^2h^2}{8mL^2} = n^2E_1, \qquad n = 1, 2, 3, \dots

The energies are quantised, grow as $n^2$ , and the lowest is not zero: the confinement energy $E_1 = h^2/8mL^2$ is the price of localisation ( $\Delta x \sim L$ forces $\Delta p \sim h/2L$ ). The $\varphi_n$ are orthogonal, $\int\varphi_n\varphi_m\dd x = \delta_{nm}$ , and any state of the well is a superposition of them.

Proof. Inside, $\varphi'' = -k^2\varphi$ with $k^2 = 2mE/\hbar^2$ : $\varphi = A\sin kx + B\cos kx$ ; $\varphi(0) = 0$ kills $B$ , $\varphi(L) = 0$ requires $kL = n\pi$ — the standing waves of a string, with $E = \hbar^2k^2/2m$ . Normalisation gives $A = \sqrt{2/L}$ ; orthogonality is that of the sines. ∎

The infinite well: the first three levels, E_n n2, with their wave functions (left) and probability densities (right) — n - 1 nodes, and a ground state that is spread over the well and has a non-zero energy. — The infinite well: the first three levels, $E_n \propto n^2$ , with their wave functions (left) and probability densities (right) — $n - 1$ nodes, and a ground state that is spread over the well and has a non-zero energy.

Example 31.2 (Orders of magnitude, and colours)

$E_1 = h^2/8mL^2$ : an electron in $1\,\mathrm{nm}$ , $0.38\,\mathrm{eV}$ ; in $0.1\,\mathrm{nm}$ (an atom), $38\,\mathrm{eV}$ — the scale of atomic energies; a nucleon in $5\,\mathrm{fm}$ , $8\,\mathrm{MeV}$ — the scale of nuclear energies; a marble in a box, $10^{-63}\,\mathrm{J}$ — no one will notice. A semiconductor nanocrystal a few nanometres across (a quantum dot) confines its electrons: the confinement energy adds to the crystal’s gap, so the smaller the dot, the bluer the light it emits — the same material glows red at $6\,\mathrm{nm}$ and green at $3\,\mathrm{nm}$ , tuned by size alone.

31.2 The finite well

Proposition 31.3 (Finite square well)

For $V = 0$ in $|x| < a$ and $V = V_0$ outside, a state with $0 < E < V_0$ oscillates inside ( $k = \sqrt{2mE}/\hbar$ ) and decays outside ( $\kappa = \sqrt{2m(V_0 - E)}/\hbar$ ): $\varphi \propto \eu^{-\kappa|x|}$ beyond the walls. Matching $\varphi$ and $\varphi'$ at $x = \pm a$ gives, for the even and odd states,

k\tan ka = \kappa \quad\text{(even)}, \qquad -k\cot ka = \kappa \quad\text{(odd)}, \qquad\text{with}\quad (ka)^2 + (\kappa a)^2 = \frac{2mV_0a^2}{\hbar^2} \equiv R^2 .

Graphically, the solutions are the intersections of the curves $\eta = \xi\tan\xi$ , $\eta = -\xi\cot\xi$ with the circle $\xi^2 + \eta^2 = R^2$ ( $\xi = ka$ , $\eta = \kappa a$ ): there is always at least one bound state (the even ground state), and $1 + \lfloor 2R/\pi\rfloor$ in all. The particle penetrates the forbidden region over the depth $1/\kappa$ , and the levels lie below those of the infinite well of the same width.

Proof. The even solution is $A\cos kx$ inside and $B\eu^{-\kappa|x|}$ outside; the continuity of $\varphi'/\varphi$ at $x = a$ gives $-k\tan ka = -\kappa$ . Same for the odd one with $\sin$ . The circle is the definition of $k$ and $\kappa$ . Each branch of $\xi\tan\xi$ starting at $\xi = n\pi$ and of $-\xi\cot\xi$ starting at $(n + \tfrac12)\pi$ meets the circle once if it starts inside it: hence the count. ∎

Graphical solution of the finite well: the bound states are the intersections of the circle of radius R = a√2mV_0/ with the branches (even states) and - (odd). Here R = 4: three bound states; the dashed small circle (R = 1.2) still cuts the first branch — a well always binds at least one state. — Graphical solution of the finite well: the bound states are the intersections of the circle of radius $R = a\sqrt{2mV_0}/\hbar$ with the branches $\xi\tan\xi$ (even states) and $-\xi\cot\xi$ (odd). Here $R = 4$ : three bound states; the dashed small circle ( $R = 1.2$ ) still cuts the first branch — a well always binds at least one state.

Example 31.4 (An electron in a nanometre well)

$V_0 = 1\,\mathrm{eV}$ , width $2a = 1\,\mathrm{nm}$ : $R = a\sqrt{2mV_0}/\hbar = 0.5 \times 10^{-9} \times 5.1 \times 10^9 = 2.6$ , so $1 + \lfloor 2R/\pi\rfloor = 2$ bound states; the ground state lies near $0.26\,\mathrm{eV}$ instead of the infinite well’s $0.38\,\mathrm{eV}$ , and leaks $0.2\,\mathrm{nm}$ into the walls ( $1/\kappa = \hbar/\sqrt{2m(V_0 - E)}$ ). Such wells, grown as layers of semiconductors a few nanometres thick, are the heart of the diode lasers of Chapter 23.

31.3 Step and barrier: tunnelling

Proposition 31.5 (Potential step)

A particle of energy $E$ coming from $x < 0$ onto the step $V = V_0$ for $x > 0$ :

$E > V_0$ : $\varphi = \eu^{\iu k_1x} + r\eu^{-\iu k_1x}$ on the left, $t\eu^{\iu k_2x}$ on the right, with $k_{1,2} = \sqrt{2m(E - V_{0,\text{left/right}})}/ \hbar$ ; continuity of $\varphi$ and $\varphi'$ gives $r = (k_1 - k_2)/(k_1 + k_2)$ and the reflection probability (ratio of currents) $R = r^2$ , $T = 1 - R = 4k_1k_2/(k_1 + k_2)^2$ — a particle can be reflected by a step it has the energy to climb, exactly as a wave on a string by a change of impedance (Chapter 15);
$E < V_0$ : on the right $\varphi = t\eu^{-\kappa x}$ , $\kappa = \sqrt{2m(V_0 - E)}/\hbar$ — an evanescent wave; $|r| = 1$ (total reflection, with a phase shift), no current flows into the step, but the particle is found there with a probability decaying over $1/\kappa$ .

Proof. Write the continuity equations at $x = 0$ and solve; for the currents use $j = (|A|^2 - |B|^2)\hbar k/m$ (Chapter 30). For $E < V_0$ , $r = (k - \iu\kappa)/(k + \iu\kappa)$ , of modulus one. ∎

Theorem 31.6 (Tunnelling through a barrier)

A particle of energy $E < V_0$ meeting a rectangular barrier of height $V_0$ and width $a$ is transmitted with the probability

T = \Big[1 + \frac{V_0^2\sinh^2(\kappa a)}{4E(V_0 - E)}\Big]^{-1} \ \approx\ 16\,\frac{E}{V_0}\Big(1 - \frac{E}{V_0}\Big)\eu^{-2\kappa a} \quad (\kappa a \gg 1), \qquad \kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar} .

This is the tunnel effect: classically impossible, quantum-mechanically exponentially small — and exponentially sensitive to the width $a$ , the height $V_0 - E$ and the mass $m$ . For an electron $1\,\mathrm{eV}$ below the top, $\kappa = 5.1\,\mathrm{nm}^{-1}$ : $T \sim 10^{-2}$ for $a = 0.5\,\mathrm{nm}$ , $10^{-4}$ for $1\,\mathrm{nm}$ , $10^{-9}$ for $2\,\mathrm{nm}$ ; for a proton, $\kappa$ is $43$ times larger and nothing passes at these widths.

Proof. $\varphi = \eu^{\iu kx} + r\eu^{-\iu kx}$ for $x < 0$ , $A\eu^{\kappa x} + B\eu^{-\kappa x}$ inside, $t\eu^{\iu kx}$ for $x > a$ ; four continuity conditions (of $\varphi$ and $\varphi'$ at $0$ and $a$ ) for four unknowns; eliminating $A$ , $B$ , $r$ gives $1/|t|^2 = 1 + (k^2 + \kappa^2)^2\sinh^2(\kappa a)/4k^2\kappa^2$ , which is the formula with $k^2 = 2mE/\hbar^2$ , $\kappa^2 = 2m(V_0 - E)/\hbar^2$ . For $\kappa a \gg 1$ , $\sinh\kappa a \approx \eu^{\kappa a}/2$ . For a barrier of arbitrary shape the exponent becomes $2\int\kappa(x)\dd x$ across the forbidden region (the Gamow factor, admitted). ∎

Tunnelling through a barrier: the wave oscillates on the left, decays exponentially inside the classically forbidden region, and emerges on the right with a reduced amplitude — the same wavelength, a small probability.

Example 31.7 (The scanning tunnelling microscope)

A sharp metal tip is brought within $d \approx 0.5\,\mathrm{nm}$ of a conducting surface and a small voltage applied: electrons tunnel across the vacuum gap, whose barrier height is the work function $\Phi \approx 4.5\,\mathrm{eV}$ , so $\kappa = \sqrt{2m\Phi}/\hbar = 1.1 \times 10^{10}\,\mathrm{m}^{-1}$ and the current $I \propto \eu^{-2\kappa d}$ changes by a factor $\eu^{2.2} \approx 10$ for every $0.1\,\mathrm{nm}$ . A feedback loop moves the tip to keep $I$ constant, and the tip’s height, recorded as it scans, draws the surface with a vertical resolution of a picometre — and laterally atom by atom, because the last atom of the tip, being nearest, carries most of the current (Binnig and Rohrer, 1981).

Example 31.8 (Alpha decay)

An $\alpha$ particle of $5\,\mathrm{MeV}$ inside a heavy nucleus faces the Coulomb barrier of the remaining charge $Ze$ , tens of MeV high at the nuclear surface and extending to the radius $b = 2Ze^2/4\pi\varepsilon_0E \approx 50\,\mathrm{fm}$ where $V = E$ . The Gamow factor $2\int\kappa\,\dd r$ is of order $80$ : $T \sim \eu^{-80} \sim 10^{-35}$ . The particle hits the wall some $10^{21}$ times per second, so it escapes at a rate $\sim 10^{-14} \,\mathrm{s}^{-1}$ : a lifetime of a million years. Because the exponent varies as $Z/\sqrt E$ , a $2\,\mathrm{MeV}$ change of $E$ shifts the lifetime by twenty orders of magnitude — the Geiger–Nuttall law, from microseconds to the age of the universe (Gamow, 1928: the first application of tunnelling).

Atomic resolution on a gold surface, imaged by a scanning tunnelling microscope: the rows of atoms of the reconstructed (100) face, a few tenths of a nanometre apart.

31.4 The double well

Proposition 31.9 (Tunnel splitting)

Two identical wells separated by a barrier: if the barrier were impenetrable each well would have the same ground level $E_0$ , twice. Tunnelling couples them and the two lowest states of the double well are the symmetric and antisymmetric combinations, $\varphi_\pm \approx (\varphi_{\text{L}} \pm \varphi_{\text{R}})/\sqrt2$ , with energies $E_0 \mp \Delta/2$ : the level is split by an amount $\Delta \propto \eu^{-\kappa a}$ (the tunnelling amplitude, not its square). A particle placed in the left well, $\psi = (\varphi_+ + \varphi_-)/\sqrt2$ , is not stationary: it oscillates between the wells at the frequency $\nu = \Delta/h$ — tunnelling oscillations, the two-level beats of Chapter 30.

Proof. The symmetric combination has no node under the barrier and a lower curvature, hence a lower energy; the antisymmetric one has a node and lies higher; the splitting is proportional to the overlap of the localised functions under the barrier, $\eu^{-\kappa a}$ . The beating is the two-level superposition already computed. ∎

The double well: the degenerate level of the two isolated wells splits into a symmetric state (lower) and an antisymmetric one (higher) by - a; a particle started on one side tunnels back and forth at the frequency /h. — The double well: the degenerate level of the two isolated wells splits into a symmetric state (lower) and an antisymmetric one (higher) by $\Delta \propto \eu^{-\kappa a}$ ; a particle started on one side tunnels back and forth at the frequency $\Delta/h$ .

Example 31.10 (Ammonia)

In NH $_3$ the nitrogen atom sits on one side of the plane of the three hydrogens or on the other — a double well with a barrier of about $0.25\,\mathrm{eV}$ . The tunnel splitting of the ground level is $\Delta = 1 \times 10^{-4}\,\mathrm{eV}$ : $\nu = \Delta/h = 24\,\mathrm{GHz}$ , the inversion frequency, the transition of the first maser and of the first atomic clock (1949). Replace hydrogen by deuterium and the heavier molecule tunnels less: $1.6\,\mathrm{GHz}$ . In heavier pyramids (PH $_3$ , AsH $_3$ ) the splitting collapses to kilohertz and below: the molecule stays on its side for years — which is why left- and right-handed molecules exist at all, and why sugar does not racemise on the shelf.

Charles Townes with the first maser (1954): a beam of ammonia molecules, sorted by an electric field into the upper inversion state, amplified the 24\, GHz transition in a cavity — stimulated emission’s first device, six years before the laser. — Charles Townes with the first maser (1954): a beam of ammonia molecules, sorted by an electric field into the upper inversion state, amplified the $24\,\mathrm{GHz}$ transition in a cavity — stimulated emission’s first device, six years before the laser.

Method 31.11 (Wells and barriers)

(1) Write $\varphi$ region by region: $\eu^{\pm\iu kx}$ where $E > V$ , $\eu^{\pm\kappa x}$ where $E < V$ , with $k, \kappa = \sqrt{2m|E - V|}/\hbar$ . (2) Match $\varphi$ and $\varphi'$ at each boundary (at an infinite wall, $\varphi = 0$ ). (3) Bound states: the matching has solutions only for discrete $E$ — count them graphically. (4) Scattering: currents $(|A|^2 - |B|^2)\hbar k/m$ give $R$ and $T$ . (5) Tunnelling: $T \sim \eu^{-2\kappa a}$ ; order of magnitude first, prefactor after. (6) Double well: splitting $\propto\eu^{-\kappa a}$ , beats at $\Delta/h$ .

31.5 Exercises

Exercise 31.1 ★

Infinite well: electron in $1\,\mathrm{nm}$ ( $E_1$ , $E_2$ , $E_3$ , wavelength of the $2 \to 1$ photon); electron in $0.1\,\mathrm{nm}$ ; nucleon in $5\,\mathrm{fm}$ ; a $1\,\mathrm{g}$ marble in $10\,\mathrm{cm}$ ( $E_1$ , and the quantum number for a speed of $1\,\mathrm{cm}/\mathrm{s}$ ).

Solution

Solution of Exercise 31.1.

$1\,\mathrm{nm}$ : $0.38\,$ , $1.50\,$ , $3.38\,\mathrm{eV}$ ; $2 \to 1$ : $1.13\,\mathrm{eV}$ , $1.1\,\text{µ}\mathrm{m}$ . $0.1\,\mathrm{nm}$ : $38\,\mathrm{eV}$ . Nucleon: $8\,\mathrm{MeV}$ . Marble: $E_1 = 5.5 \times 10^{-63}\,\mathrm{J}$ ; $n = 2mvL/h = 3 \times 10^{27}$ .

Exercise 31.2 ★

Quantum dots: a nanocrystal of gap $1.74\,\mathrm{eV}$ confines an electron–hole pair whose confinement energy is $h^2/8m^*L^2$ with $m^* = 0.1\,m_{\text{e}}$ . Emission wavelength for $L = 6$ , $4$ , $3$ , $2.5\,\mathrm{nm}$ ; which colours?

Solution

Solution of Exercise 31.2.

Confinement $3.76\,\mathrm{eV}\,\mathrm{nm}^{2}/L^2$ : $0.10$ , $0.24$ , $0.42$ , $0.60\,\mathrm{eV}$ ; total $1.84$ , $1.98$ , $2.16$ , $2.34\,\mathrm{eV}$ : $673\,$ , $628\,$ , $574\,$ , $530\,\mathrm{nm}$ — red, orange, yellow-green, green.

Exercise 31.3 ★

Tunnelling: electron with $V_0 - E = 4.5\,\mathrm{eV}$ : $\kappa$ ; $\eu^{-2\kappa a}$ for $a = 0.3$ , $0.5$ , $1$ , $2\,\mathrm{nm}$ ; factor per $0.1\,\mathrm{nm}$ ; same for a proton at $a = 0.3\,\mathrm{nm}$ .

Solution

Solution of Exercise 31.3.

$\kappa = 1.09 \times 10^{10}\,\mathrm{m}^{-1}$ ; $\eu^{-2\kappa a}$ : $1.5 \times 10^{-3}$ , $1.8 \times 10^{-5}$ , $3 \times 10^{-10}$ , $10^{-19}$ ; $\times 8.8$ per $0.1\,\mathrm{nm}$ ; proton: $\kappa$ is $43$ times larger, $\eu^{-280}$ at $0.3\,\mathrm{nm}$ .

Exercise 31.4 ★

Step: electron of $2\,\mathrm{eV}$ onto a step of $1\,\mathrm{eV}$ : $k_1$ , $k_2$ , $R$ , $T$ . Same for a step down of $1\,\mathrm{eV}$ . Why is this impossible classically, and what is the analogue for light?

Solution

Solution of Exercise 31.4.

$k_1 = 7.3 \times 10^{9}\,\mathrm{m}^{-1}$ , $k_2 = 5.1 \times 10^{9}\,\mathrm{m}^{-1}$ ; $R = 0.03$ , $T = 0.97$ . Step down: $k_2 = 8.9 \times 10^{9}\,\mathrm{m}^{-1}$ , $R = 0.01$ . A classical particle never turns back when it has the energy; a light wave is partly reflected at any change of index.

Exercise 31.5 ★★

Infinite well in detail. (a) Derive the levels and normalise. (b) Show the orthogonality. (c) For $n = 1$ : $\langle x\rangle$ , $\Delta x = L\sqrt{1/12 - 1/2\pi^2}$ , $\langle p\rangle = 0$ , $\Delta p = \pi\hbar/L$ ; check Heisenberg. (d) Sketch $|\varphi_n|^2$ for large $n$ and compare with the classical probability of finding a bouncing particle.

Solution

Solution of Exercise 31.5.

(a), (b) Proposition 31.1. (c) $L/2$ ; $0.18L$ ; $0$ ; $\pi\hbar/L$ ; product $0.57\hbar > \hbar/2$ . (d) Rapid oscillations about the mean $1/L$ — the classical uniform density.

Exercise 31.6 ★★

Finite well. (a) Derive $k\tan ka = \kappa$ for the even states. (b) Show graphically there is always a bound state and count them for $R = 1$ , $4$ , $10$ . (c) Electron, $V_0 = 1\,\mathrm{eV}$ , $2a = 1\,\mathrm{nm}$ : $R$ , number of states, penetration depth of the ground state (take $E \approx 0.26\,\mathrm{eV}$ ). (d) What happens to the number of bound states as $V_0 \to \infty$ , and to their energies?

Solution

Solution of Exercise 31.6.

(a) Proposition 31.3. (b) $1$ , $3$ , $7$ . (c) $R = 2.6$ , two states; $\kappa = 4.4 \times 10^{9}\,\mathrm{m}^{-1}$ , depth $0.23\,\mathrm{nm}$ . (d) Infinitely many, tending to the infinite well’s from below.

Exercise 31.7 ★★

Step, $E < V_0$ . (a) Write $\varphi$ on both sides and find $r = (k - \iu\kappa)/(k + \iu\kappa)$ . (b) Show $|r| = 1$ and compute the phase of $r$ . (c) Show the current vanishes for $x > 0$ although $|\varphi|^2 \ne 0$ there. (d) Electron $1\,\mathrm{eV}$ below the top: penetration depth; compare with the evanescent wave of total internal reflection.

Solution

Solution of Exercise 31.7.

(a) $1 + r = t$ , $\iu k(1 - r) = -\kappa t$ . (b) Numerator and denominator are conjugates; phase $-2\arctan(\kappa/k)$ . (c) $\psi^*\psi'$ is real for a real exponential. (d) $0.2\,\mathrm{nm}$ ; like the evanescent wave of total reflection: present, carrying no flux, and able to feed a second medium — that is tunnelling.

Exercise 31.8 ★★

The barrier. (a) Write the four continuity conditions and derive $T$ . (b) Evaluate exactly and with the approximation for $E = V_0/2$ , $\kappa a = 3$ ; for $\kappa a = 1$ . (c) For $E > V_0$ show $T = [1 + V_0^2\sin^2(k_2a)/4E(E - V_0)]^{-1}$ and find the energies of perfect transmission. (d) Interpret them (think of the anti-reflection layer of Chapter 15).

Solution

Solution of Exercise 31.8.

(a) Theorem 31.6. (b) $\kappa a = 3$ : exact $1/(1 + \sinh^23) = 9.9 \times 10^{-3}$ , approximation $4\eu^{-6} = 9.9 \times 10^{-3}$ ; $\kappa a = 1$ : $0.42$ against $0.54$ . (c) $\kappa \to \iu k_2$ ; $T = 1$ for $k_2a = n\pi$ . (d) The reflections from the two faces cancel when the barrier is a whole number of half-wavelengths — resonant transmission.

Exercise 31.9 ★★

Gamow. For the Coulomb barrier from the nuclear radius $R_{\text{n}}$ to $b = 2Ze^2/4\pi\varepsilon_0E$ , the exponent is $G = 2\int\kappa\,\dd r = (2\sqrt{2mE}/\hbar)\,b\,[\arccos\sqrt{R_{\text{n}}/b} - \sqrt{(R_{\text{n}}/b)(1 - R_{\text{n}}/b)}]$ . (a) $Z = 90$ , $E = 5\,\mathrm{MeV}$ , $R_{\text{n}} = 8\,\mathrm{fm}$ : $b$ , $G$ , $T$ . (b) Frequency of hits $v/2R_{\text{n}}$ and the lifetime. (c) Repeat for $E = 4$ and $6\,\mathrm{MeV}$ : ratio of lifetimes. (d) Why do nuclei emit $\alpha$ particles rather than protons or $^{12}$ C?

Solution

Solution of Exercise 31.9.

(a) $b = 52\,\mathrm{fm}$ ; $G = 82$ ; $T = \eu^{-82} \approx 2 \times 10^{-36}$ . (b) $v = 1.55 \times 10^{7}\,\mathrm{m}/\mathrm{s}$ , $10^{21}$ hits per second: rate $2 \times 10^{-15}\,\mathrm{s}^{-1}$ , lifetime $\sim 10^7$ years. (c) $G = 100$ and $69$ : $\eu^{18} \approx 10^8$ longer, $\eu^{-13} \approx 10^{-6}$ shorter. (d) The $\alpha$ is tightly bound and leaves with positive energy; a proton would not; a $^{12}$ C has three times the charge and a far larger exponent (cluster decay exists, at $10^{-10}$ ).

Exercise 31.10 ★★★

Ammonia. Splitting $\Delta = 1 \times 10^{-4}\,\mathrm{eV}$ . (a) Inversion frequency and the period of the tunnelling oscillation. (b) The N atom is prepared on one side: write $\psi(t)$ and the probability of finding it on the other side. (c) ND $_3$ tunnels at $1.6\,\mathrm{GHz}$ : deduce $\kappa a$ from the ratio of the splittings if $\Delta \propto \eu^{-\kappa a}$ and $\kappa \propto\sqrt m$ ( $m_{\text{D}} = 2m_{\text{H}}$ , barrier unchanged). (d) Why can a chiral molecule be left-handed for years?

Solution

Solution of Exercise 31.10.

(a) $24\,\mathrm{GHz}$ ; $41\,\mathrm{ps}$ . (b) $\psi = (\varphi_+\eu^{-\iu E_+t/\hbar} + \varphi_-\eu^{-\iu E_-t/\hbar})/\sqrt2$ ; $P = \sin^2(\pi\nu t)$ . (c) $\eu^{(\sqrt2 - 1) \kappa a} = 15$ : $\kappa a = 6.5$ . (d) Heavier groups and higher barriers make the tunnelling period longer than years.

Exercise 31.11 ★★★

The STM. $\Phi = 4.5\,\mathrm{eV}$ , $d = 0.5\,\mathrm{nm}$ , bias $0.1\,\mathrm{V}$ , $I = 1\,\mathrm{nA}$ . (a) $\kappa$ , the factor of $I$ per $0.1\,\mathrm{nm}$ , the height resolution if $I$ is measured to $1\%$ . (b) Electrons per second in the current. (c) The tip’s apex atom is $0.25\,\mathrm{nm}$ closer than its neighbours (laterally $0.25\,\mathrm{nm}$ away, so $d' = \sqrt{d^2 + 0.25^2}$ ): fraction of the current it carries. (d) A $1\,\mathrm{cm}$ steel frame expands by $\alpha L\Delta T$ with $\alpha = 1 \times 10^{-5}\,\mathrm{K}^{-1}$ : what temperature stability does a picometre need, and how do designs get round it?

Solution

Solution of Exercise 31.11.

(a) $1.09 \times 10^{10}\,\mathrm{m}^{-1}$ ; $\times 8.8$ ; $1\%$ of $I$ is $0.5\,\mathrm{pm}$ . (b) $6 \times 10^9$ . (c) Each neighbour at $d' = 0.56\,\mathrm{nm}$ carries $\eu^{-2\kappa \times 0.06\,\text{nm}} = 0.28$ of the apex’s current; the apex carries about half. (d) $10^{-5}\,\mathrm{K}$ — impossible; symmetric designs in which tip and sample expand together, fast scans, low-expansion materials, and the feedback.

Exercise 31.12 ★★★

A quantum-well laser. A $10\,\mathrm{nm}$ layer of GaAs (gap $1.42\,\mathrm{eV}$ , $m_{\text{e}}^* = 0.067\,m_{\text{e}}$ , $m_{\text{h}}^* = 0.45\,m_{\text{e}}$ ) between barriers $0.3\,\mathrm{eV}$ higher. (a) Ground confinement energies of electron and hole (infinite well). (b) Photon energy and wavelength of the laser. (c) Width for $780\,\mathrm{nm}$ . (d) With the finite barrier, how many bound electron states does the $10\,\mathrm{nm}$ well hold?

Solution

Solution of Exercise 31.12.

(a) $0.056\,$ and $0.008\,\mathrm{eV}$ . (b) $1.48\,\mathrm{eV}$ , $835\,\mathrm{nm}$ . (c) Confinement $0.17\,\mathrm{eV}$ : $L = 6.1\,\mathrm{nm}$ . (d) $R = 3.6$ : three.

31.6 Problem: The STM, the alpha particle and the ammonia maser

Problem 31.1

Weekend problem — three tunnels

Data: $\hbar = 1.05 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}$ , $h = 6.63 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}$ , $m_{\text{e}} = 9.11 \times 10^{-31}\,\mathrm{kg}$ , $m_\alpha = 6.64 \times 10^{-27}\,\mathrm{kg}$ , $e^2/4\pi\varepsilon_0 = 1.44\,\mathrm{MeV}\,\mathrm{fm}$ , $1\,\mathrm{eV}$ $= 1.60 \times 10^{-19}\,\mathrm{J}$ .

Part I — The scanning tunnelling microscope. Tip and sample are tungsten ( $\Phi = 4.5\,\mathrm{eV}$ ); gap $d$ ; bias $50\,\mathrm{mV}$ ; the current is $I = I_0\eu^{-2\kappa d}$ .

$\kappa$ for electrons at the Fermi level facing the vacuum barrier $\Phi$ .
Factor of change of $I$ for $\Delta d = 0.1\,\mathrm{nm}$ ; for $1\,\mathrm{pm}$ .
The feedback holds $I$ constant to $1\%$ : height noise.
$I = 1\,\mathrm{nA}$ : electrons per second; average time between two electrons, compared with the tunnelling “time” $\sim\hbar/\Phi$ .
The apex atom and its neighbours $0.25\,\mathrm{nm}$ away laterally, $d = 0.5\,\mathrm{nm}$ : share of the apex atom; why a single atom is enough to image atoms.
An adsorbed atom $0.1\,\mathrm{nm}$ high: change of $I$ in constant-height mode; why constant-current mode is preferred.
A $2\,\mathrm{cm}$ metal frame, $\alpha = 1 \times 10^{-5}\,\mathrm{K}^{-1}$ : temperature change that moves the tip by $1\,\mathrm{pm}$ ; how do instruments cope (symmetry, speed, low-expansion materials)?
With a $1\,\mathrm{V}$ bias the barrier is no longer rectangular: sketch it and say whether $I$ grows faster or slower than linearly with $V$ .
On an oxidised patch the work function is $3\,\mathrm{eV}$ instead of $4.5\,\mathrm{eV}$ : by how much does the tip retract in constant-current mode over a perfectly flat surface? What does the STM image, then?

Part II — Alpha decay of a nucleus $Z = 92$ , $A = 238$ . The $\alpha$ ( $E = 4.2\,\mathrm{MeV}$ ) moves in a nucleus of radius $R_{\text{n}} = 8\,\mathrm{fm}$ ; outside, the Coulomb energy is $V(r) = 2(Z - 2)e^2/4\pi\varepsilon_0r$ .

Height of the barrier at $R_{\text{n}}$ ; classical turning point $b$ .
Speed of the $\alpha$ inside and its frequency of hits on the wall.
Gamow exponent $G = (2\sqrt{2mE}/\hbar)\,b\,[\arccos\sqrt{R_{\text{n}}/b} - \sqrt{(R_{\text{n}}/b)(1 - R_{\text{n}}/b)}]$ and $T = \eu^{-G}$ .
Decay rate and half-life; compare with the measured $4.5 \times 10^9$ years.
The same with $E = 5.2\,\mathrm{MeV}$ (another isotope): half-life; the Geiger–Nuttall sensitivity.
Taking $R_{\text{n}} = 9\,\mathrm{fm}$ instead of $8\,\mathrm{fm}$ : effect on the half-life; comment on the precision of such estimates.
Why is the tunnelling of a proton (charge $1$ , mass $1/4$ ) not observed from the same nucleus, and that of a $^{12}$ C (charge $6$ ) either?
The $\alpha$ emerges with the full $4.2\,\mathrm{MeV}$ although it “passed under” a $30\,\mathrm{MeV}$ wall: is energy conserved?

Part III — The ammonia maser. The N atom tunnels through the H $_3$ plane; the ground level is split by $\Delta = 9.8 \times 10^{-5}\,\mathrm{eV}$ .

Frequency and wavelength of the inversion transition.
Write the symmetric and antisymmetric states in terms of “N up” and “N down”; which is lower, and why?
A molecule prepared “N up”: $\psi(t)$ , the probability of “down” at time $t$ , the period.
Boltzmann ratio of the two levels at $300\,\mathrm{K}$ : can the thermal population give a maser? How is the inversion obtained (a state-selecting electric field)?
Photon energy in joules; number of molecules that must emit per second for an output of $1 \times 10^{-10}\,\mathrm{W}$ .
ND $_3$ tunnels at $1.6\,\mathrm{GHz}$ : deduce $\kappa a$ for NH $_3$ assuming $\Delta \propto \eu^{-\kappa a}$ with $\kappa \propto \sqrt{m}$ .
The maser was the first atomic clock: why is a tunnelling frequency a good clock and which effects shift it?
Summarise: quantisation from confinement, tunnelling from evanescence, and what fixes the exponents.

Solution

Solution of Problem 31.1.

1. $\kappa = \sqrt{2m\Phi}/\hbar = 1.09 \times 10^{10}\,\mathrm{m}^{-1}$ .

2. $\eu^{2.2} = 8.8$ ; $\eu^{0.022}$ : $2.2\%$ per picometre.

3. $0.5\,\mathrm{pm}$ .

4. $6 \times 10^9$ per second; $0.16\,\mathrm{ns}$ apart, against $\hbar/\Phi \sim 10^{-16}\,\mathrm{s}$ : one at a time.

5. About half; the rest decays so fast with distance that the image is that of one atom.

6. $\times 8.8$ ; constant current keeps the tip safe and turns the exponential into a linear height.

7. $\Delta T = 10^{-12}/(10^{-5} \times 0.02) = 5 \times 10^{-6}\,\mathrm{K}$ ; symmetric mounts, speed, low-expansion materials, and a feedback that follows the drift.

8. A trapezoid, lowered on the collecting side: the effective barrier thins and $I$ grows faster than linearly.

9. $\kappa' = 0.89 \times 10^{10}\,\mathrm{m}^{-1}$ ; at $0.5\,\mathrm{nm}$ the exponent drops from $10.9$ to $8.9$ , $I \times 7$ : the tip retracts by $\ln 7/2\kappa' \approx 0.1\,\mathrm{nm}$ — a bump that is electronic, not geometric.

10. $V(R_{\text{n}}) = 2 \times 90 \times 1.44/8 = 32\,\mathrm{MeV}$ ; $b = 62\,\mathrm{fm}$ .

11. $v = 1.4 \times 10^{7}\,\mathrm{m}/\mathrm{s}$ ; $9 \times 10^{20}$ hits per second.

12. $G = 96$ ; $T = \eu^{-96} \approx 10^{-42}$ .

13. Rate $\sim 10^{-21}\,\mathrm{s}^{-1}$ , half-life $\sim 10^{13}$ years — a thousand times the measured value: for an exponent of a hundred, a crude model does well to land within a few orders of magnitude.

14. $G = 79$ : $\eu^{17} \approx 10^7$ times shorter, about $10^6$ years — one extra MeV, seven orders of magnitude.

15. $G = 92$ : $\eu^{4.5} \approx 90$ times shorter — a fermi of radius is two orders of magnitude.

16. A proton is not pre-formed with positive energy (it is bound by some $7\,\mathrm{MeV}$ ); a $^{12}$ C has three times the charge and an exponent three times larger.

17. Yes: the $\alpha$ always has $4.2\,\mathrm{MeV}$ ; its wave function merely has an evanescent part under the barrier, where no measurement finds it with negative kinetic energy.

18. $23.7\,\mathrm{GHz}$ ; $1.27\,\mathrm{cm}$ .

19. $(|{\uparrow}\rangle \pm |{\downarrow}\rangle)/\sqrt2$ ; the symmetric one is lower — no node, less curvature.

20. $\psi = (\varphi_+\eu^{-\iu E_+t/\hbar} + \varphi_-\eu^{-\iu E_-t/\hbar})/\sqrt2$ ; $P_{\downarrow} = \sin^2(\pi\nu t)$ ; period $42\,\mathrm{ps}$ .

21. $\eu^{-\Delta/k_BT} = 0.996$ : almost equal, no inversion; an inhomogeneous electric field sorts the beam and sends the upper-state molecules into the cavity.

22. $1.6 \times 10^{-23}\,\mathrm{J}$ ; $6 \times 10^{12}$ molecules per second.

23. $\kappa a = 6.5$ .

24. It is fixed by the molecule alone, identical for every molecule, and insensitive to the outside; shifts come from the Doppler effect in the beam, stray electric fields, collisions and the cavity’s pulling.

25. Confinement quantises ( $E_n \propto n^2/L^2$ ); evanescence lets the wave through ( $T \sim \eu^{-2\kappa a}$ ); the exponents are set by $\sqrt{2m(V_0 - E)}/\hbar$ and the width — mass, height, distance.