Part 3 · One-dimensional problems

3.1 One-dimensional problems

Source: Centini’s whiteboard of 9 Oct, p. 10–11; the common method from the 2024/25 boards #6 and #20

Objectives

Understand what a one-dimensional problem is, and why it shows the key effects of confinement.
Recipe: the method for a potential made of flat pieces.
Picture: a solution that oscillates where $E > U$ and bends away from the axis where $E < U$ ; only special energies make it behave at both ends.
Vocabulary: bound and unbound state, classically forbidden region, matching conditions.

A particle moving along a line in a given potential energy $U (x)$ . Its states of definite energy solve the time-independent Schrödinger equation in one dimension, $- \frac{ℏ ^{2}}{2 m} \frac{d ^{2} u _{E}}{d x ^{2}} + U (x) u_{E} = E u_{E},$ with $∣ u_{E} (x) ∣^{2}$ the probability density of finding the particle at $x$ . Real nanostructures are three-dimensional, but a layer, a wire or a barrier is often thin in one direction only, and there the physics is one-dimensional. The 1D problems show every key effect with the least algebra: quantisation, zero-point energy, penetration into forbidden regions, tunnelling.

The common method

Rearranged, the equation says how $u$ curves: $u^{''} = - \frac{2 m}{ℏ ^{2}} (E - U) u .$ Where $U$ is constant this has three kinds of solution.

Definition 3.1.1Allowed and forbidden regions2024/25 board #6

In a region of constant $U$ :

$E > U$ : oscillating, $e^{\pm ik x}$ or $sin k x$ , $cos k x$ , with $k = 2 m (E - U) /ℏ$ ;
$E < U$ : exponential, $e^{\pm κ x}$ , with $κ = 2 m (U - E) /ℏ$ . Classically the particle could never be there: a classically forbidden region;
$U = \infty$ : $u = 0$ .

Recipe 3.1.2Solving a piecewise-constant potential2024/25 board #6

Split the line into regions where $U$ is constant.
Solve in each region, as above.
Keep it normalisable: drop any exponential that blows up at $\pm \infty$ ; for scattering, drop any wave coming in from where nothing is sent.
Match at each boundary: $u$ continuous, and $u^{'}$ continuous where $U$ is finite.
Read off the result. If $E$ is below $U$ on both sides, a bound state, the conditions can be met only for special $E$ : the energies are quantised. If not, an unbound state, every $E$ is allowed and the matching gives reflected and transmitted amplitudes.

Why $u^{'}$ is continuous: integrate the equation across the boundary. A finite jump in $U$ gives $u^{''}$ a finite jump, which integrates to no jump in $u^{'}$ . At an infinite wall that argument fails, and only $u = 0$ is left (beyond the boards).

Key observation 3.1.3Bending back, bending awaybeyond the boards

Where $E > U$ , $u^{''}$ has the opposite sign to $u$ : the curve always bends back towards the axis and oscillates, more tightly the larger the kinetic energy $E - U$ . Where $E < U$ , $u^{''}$ has the sign of $u$ : the curve bends away and dies off or blows up. ==A bound state is an energy at which the curve, started small on one side, ends small on the other.==

Loading the demo…

Figure 3.1.1 The infinite well of width 1 nm, solved by shooting: pick an energy, start with

u = 0

at the left wall, and let the equation draw the curve. For most energies

u

misses zero at the right wall. Drag the

energy

up until it hits:

E_{1} = 0.376

eV. Keep going for

E_{2}

and

E_{3}

. Each extra level fits one more half-wave.

The harmonic oscillator’s $U \propto x^{2}$ is not piecewise flat and is solved differently (§3.5 The harmonic oscillator), but the picture of bending back and bending away holds for it too.

The problems

The 2024/25 recap lists them, each $\hat{K} = \overset{p}{^}^{2} /2 m$ plus a different $U$ :

problem	$U (x)$	what it shows
free particle	$0$	any energy; plane waves
infinite well	$0$ inside, $\infty$ outside	quantised energies, zero-point energy
finite well	$0$ inside, $U_{0}$ outside	tails into the walls, finitely many bound states
barrier	$U_{0}$ on a stretch	tunnelling
harmonic oscillator	$\frac{1}{2} m ω^{2} x^{2}$	evenly spaced levels, ladder operators

Exercises

Exercise 3.1.1beyond the boards

For an electron with $E = 1$ eV, find $k$ in a region with $U = 0$ , and $κ$ in a region with $U = 2$ eV. Over what distance does the exponential fall by $e$ ?

Solution

With $ℏ^{2} /2 m = 0.0381$ eV nm²: $k = 1/0.0381 = 5.12$ nm⁻¹ ( $λ = 1.23$ nm), and $κ = (2 - 1) /0.0381 = 5.12$ nm⁻¹ too, so the decay length is $1/ κ = 0.20$ nm. Electrons leak a fraction of a nanometre into a forbidden region: the start of tunnelling.

Exercise 3.1.2beyond the boards

Why can the solution in a forbidden region not be $e^{+ κ x}$ for $x \to + \infty$ ? And why is the forbidden region not forbidden for the wavefunction?

Solution

$e^{κ x}$ grows without bound, so $\int ∣ u ∣^{2}$ diverges and the state cannot be normalised. In a forbidden region of finite width, or near its edge, $e^{- κ x}$ is non-zero: the particle can be found where it could never be classically. The classical argument needs a definite position and a definite kinetic energy at once, and $\overset{x}{^}$ and $\overset{p}{^}_{x}^{2} /2 m$ do not commute: a state of sharp $E$ has no “kinetic energy at the point $x$ ” to be negative.