Part 3 · One-dimensional problems

3.3 The finite well

Source: the symmetric well and its circle from the 2024/25 board #6; the well on $[0,a]$ with a phase from the 2024/25 board #20; this year’s course has not reached it yet

The infinite well with walls that can be climbed: $$U(x)=\{\begin{array}{ll}0& -a\le x\le a\\ U_0& |x|>a\end{array}$$ A well of width $2a$ and depth $U_0$. Two kinds of state:

• $E<U_0$: classically the particle is trapped, bouncing between the walls. Quantum: bound states, discrete energies, the wavefunction leaking a little into the walls.
• $E>U_0$: classically it flies over the well and away. Quantum: unbound states, any energy, waves everywhere.

A real nanostructure is this kind of well: an electron in a thin GaAs layer between AlGaAs, or in a nanocrystal, is held by a finite step of 0.1–1 eV, not an infinite wall (beyond the boards).

Here $a$ is the half-width, as on board 6. The notation: $k_0=\sqrt{2mE}/\hslash$ inside, $\kappa =\sqrt{2m(U_0-E)}/\hslash$ in the walls (Definition 3.1.1).

Bound states

Inside, $u=A\cos k_{0}x$ or $B\sin k_{0}x$, as in the infinite well. Outside, with $E<U_0$, $u=C{e}^{\kappa x}+D{e}^{-\kappa x}$. (The board’s first $\kappa$ has $\hslash$ under the root, not ${\hslash }^2$; the later lines have it right.)

==A state must be normalisable==, so $u$ cannot blow up at $\pm \infty$: on the left keep $C{e}^{\kappa x}$, on the right keep $D{e}^{-\kappa x}$. Inside, the wave oscillates; outside it decays. ==The particle is found in the walls with non-zero probability==, falling off as $e^{-2\kappa |x|}$: a penetration depth of about $1/\kappa$ (beyond the boards). That is the seed of tunnelling.

The pieces join at $\pm a$ with $u$ and $u^{\prime }$ continuous (Recipe 3.1.2). By parity, as in the symmetric infinite well, each bound state is even or odd, so it is enough to match at $x=+a$.

Even states. Value and slope: $$B\cos k_{0}a=C{e}^{-\kappa a},-k_{0}B\sin k_{0}a=-\kappa C{e}^{-\kappa a}.$$ ==Divide the second by the first==: $B$, $C$ and $e^{-\kappa a}$ cancel, leaving $k_0\mathrm{tg}(k_{0}a)=\kappa$. The odd states give $k_0\mathrm{ctg}(k_{0}a)=-\kappa$ the same way (the board writes only the result).

These are transcendental equations: $E$ sits inside a tangent and under square roots and cannot be isolated. ==The bound energies are quantised because only some waves inside join smoothly onto a decaying tail.== And every possible energy sits on one circle of radius $R$, fixed by the well alone.

The graphical solution

The board draws only the circle; the rest is beyond the boards. A new branch, alternately even and odd, starts at every multiple of $\pi /2$, and the circle crosses each branch it reaches once.

Second route: the well on $[0,a]$

Board 20, the later board, redoes the bound states with the well from $0$ to $a$, this year’s geometry. Here ==$a$ is the full width==. Inside, the general solution is written as one sine with a phase, $A\sin (k_{0}x+\phi )$, which saves sorting even from odd. Outside, $C{e}^{\kappa x}$ on the left and $D{e}^{-\kappa x}$ on the right.

At $x=0$: $C=A\sin \phi$ and $\kappa C=A{k}_0\cos \phi$; dividing, $$\mathrm{tg}\phi =\frac{k_0}{\kappa }.$$ At $x=a$: $A\sin (k_{0}a+\phi )=D{e}^{-\kappa a}$ and $A{k}_0\cos (k_{0}a+\phi )=-\kappa D{e}^{-\kappa a}$; dividing and multiplying by $a$, with $X=k_{0}a$, $Y=\kappa a$, $$X\mathrm{ctg}(X+\phi )=-Y,X^2+Y^2=\frac{2m{U}_0{a}^2}{{\hslash }^2},$$ the same circle, with $a$ now the full width.

Read it as the infinite well’s $k_{0}a=n\pi$ minus a correction: the wave need not reach zero at the walls, it enters each with phase $\phi$, between $0$ and $\frac{\pi }{2}$. So $k_0$ is smaller than $n\pi /a$, the energy lower. As $U_0\to \infty$, $\phi \to 0$. Odd $n$ are board 6’s even states; check on the example, width 1 nm: $\pi -2\mathrm{arctg}(1.119/2.304)=2.238=2\times 1.119$ (beyond the boards).

Unbound states

Above the rim $U_0-E<0$, $\kappa$ is imaginary, $\kappa =\pm i\alpha$ with $\alpha =\sqrt{2m(E-U_0)}/\hslash$, and outside $$u(x)e^{-i\omega t}=(C{e}^{i\alpha x}+D{e}^{-i\alpha x})e^{-i\omega t},$$ a wave travelling right and one travelling left. Nothing blows up, nothing is discarded, and no condition is left to restrict $E$: ==every $E>U_0$ is allowed==, a continuous spectrum. (The board writes $e^{+i\omega t}$; the stationary-state factor is $e^{-i\omega t}$.)

The wavelength over the well. $E_k=E-U$: outside it is $E-U_0$, inside the full $E$. ==The particle moves faster over the well==, so $k_0>\alpha$ and the wavelength is shorter there. Over two wells the same rule holds point by point: the wave squeezes over each dip and stretches over the plateau between. The amplitude changes too, since the wave partly reflects at each step (§3.4 The potential barrier and tunnelling).

Exercises