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2.8

The one-dimensional finite square well

Lower the walls to a finite height and three new things happen: the bound states become finite in number, the wavefunction leaks outside, and a transcendental equation enters the picture.

Recommended first

After this section you should be able to

  • Solve region by region and derive the transcendental energy condition from the matching conditions
  • Count the bound states of a well of given depth graphically
  • Explain why an arbitrarily shallow symmetric well in 1D always has at least one bound state
V(x)={−V0,−a<x<a0,∣x∣>a(V0>0)V(x)=\begin{cases}-V_0,&-a<x<a\\ 0,&\lvert x\rvert>a\end{cases}\qquad (V_0>0)

We put the zero of energy outside the well, so bound states are those with −V0<E<0-V_0<E<0.

Solving region by region

Full solution: the even-parity bound statesadvanced~10 min

The potential is symmetric, V(−x)=V(x)V(-x)=V(x), so the eigenstates have definite parity (the reason is in section 4.1; we simply use it here). Start with even parity.

The regions. Write

k=2m(E+V0)ℏ (inside),κ=−2mEℏ (outside)(2.8.1)k=\frac{\sqrt{2m(E+V_0)}}{\hbar}\ (\text{inside}),\qquad \kappa=\frac{\sqrt{-2mE}}{\hbar}\ (\text{outside})\tag{2.8.1}

A bound state −V0<E<0-V_0<E<0 guarantees that kk and κ\kappa are both positive real.

  • Inside (∣x∣<a|x|<a): ψ′′=−k2ψ\psi''=-k^2\psi; the even solution is ψ=Dcos⁡kx\psi=D\cos kx
  • Right (x>ax>a): ψ′′=κ2ψ\psi''=\kappa^2\psi; normalisability keeps only ψ=Fe−κx\psi=F\ee^{-\kappa x}
  • The left side is fixed automatically by even parity

Matching conditions. VV is finite, so both ψ\psi and ψ′\psi' must be continuous at x=ax=a:

Dcos⁡ka=Fe−κa(2.8.2)D\cos ka=F\ee^{-\kappa a}\tag{2.8.2}−Dksin⁡ka=−κFe−κa(2.8.3)-Dk\sin ka=-\kappa F\ee^{-\kappa a}\tag{2.8.3}

Eliminate the constants. Divide the second by the first (a standard trick that kills both unknown coefficients at once):

ktan⁡ka=κ(2.8.4)k\tan ka=\kappa\tag{2.8.4}

That is the condition determining the energy levels.

Non-dimensionalise. Let

z≡ka,z0≡aℏ2mV0(2.8.5)z\equiv ka,\qquad z_0\equiv\frac{a}{\hbar}\sqrt{2mV_0}\tag{2.8.5}

z0z_0 is the only dimensionless parameter; it packages depth and width into a single number. From the definitions,

(ka)2+(κa)2=2mV0a2ℏ2=z02  ⟹  κa=z02−z2(2.8.6)(ka)^2+(\kappa a)^2=\frac{2mV_0a^2}{\hbar^2}=z_0^2 \;\Longrightarrow\;\kappa a=\sqrt{z_0^2-z^2}\tag{2.8.6}

Substituting gives the transcendental equation

tan⁡z=z02−z2z(2.8.7)\tan z=\frac{\sqrt{z_0^2-z^2}}{z}\tag{2.8.7}

Graphical method: counting the bound states

Treat each side as a function of zz and plot them together:

  • y1=tan⁡zy_1=\tan z: a family of branches diverging at z=π/2, 3π/2, …z=\pi/2,\ 3\pi/2,\ \dots
  • y2=z02−z2/zy_2=\sqrt{z_0^2-z^2}/z: falls monotonically from +∞+\infty as z→0z\to0 and vanishes at z=z0z=z_0

The number of intersections is the number of even-parity bound states.

The picture

y2y_2 only exists for 0<z<z00<z<z_0. The nn-th rising branch of tan⁡z\tan z starts at z=(n−1)πz=(n-1)\pi. So every time z0z_0 crosses an integer multiple of π\pi, one more even-parity bound state appears.

For odd parity (redo everything with ψ=Dsin⁡kx\psi=D\sin kx) the condition becomes −kcot⁡ka=κ-k\cot ka=\kappa, and a new bound state appears whenever z0z_0 crosses an odd multiple of π/2\pi/2.

Together, the total number of bound states is

N=⌈2z0π⌉(2.8.8)N=\left\lceil \frac{2z_0}{\pi}\right\rceil\tag{2.8.8}

The mathematics

Key limits

z0→∞z_0\to\infty (very deep and wide): the intersections approach the poles of tan⁡z\tan z, z→nπ/2z\to n\pi/2, so k→nπ/2ak\to n\pi/2a — exactly the infinite well of width 2a2a. The limit checks out.

z0→0z_0\to0 (very shallow and narrow): the y2y_2 curve collapses towards the origin, but tan⁡z\tan z also goes to 0 as z→0z\to0, and tan⁡z≈z\tan z\approx z rises more slowly than y2y_2 falls — so there is always at least one intersection.

Penetration depth

Outside the well ψ∝e−κ∣x∣\psi\propto\ee^{-\kappa|x|}, with characteristic decay length

δ=1κ=ℏ2m∣E∣(2.8.9)\delta=\frac{1}{\kappa}=\frac{\hbar}{\sqrt{2m\lvert E\rvert}}\tag{2.8.9}

The more weakly bound the state (the smaller ∣E∣|E|), the further it leaks. This δ\delta is the key quantity for understanding tunnelling: if a barrier’s thickness is comparable to δ\delta, the wavefunction can emerge on the far side — section 2.11 works that out quantitatively.

Worked example: a concrete welladvanced~6 min

Take an electron, a=0.5 nma=0.5\ \text{nm} (well width 1 nm), V0=5 eVV_0=5\ \text{eV}.

z0=a2mV0ℏ=aℏc2(mc2)V0=0.5197.32×511000×5(2.8.10)z_0=\frac{a\sqrt{2mV_0}}{\hbar}=\frac{a}{\hbar c}\sqrt{2(mc^2)V_0} =\frac{0.5}{197.3}\sqrt{2\times511000\times5}\tag{2.8.10}=0.5197.3×2260=5.73(2.8.11)=\frac{0.5}{197.3}\times2260=5.73\tag{2.8.11}

Number of bound states: N=⌈2×5.73/π⌉=⌈3.65⌉=4N=\lceil 2\times5.73/\pi\rceil=\lceil3.65\rceil=4.

Ground state. Solving tan⁡z=5.732−z2/z\tan z=\sqrt{5.73^2-z^2}/z on (0,π/2)(0,\pi/2) numerically gives z≈1.31z\approx1.31, so

E+V0=ℏ2z22ma2=(ℏc)2z22(mc2)a2=197.32×1.3122×511000×0.25 eV≈0.26 eV(2.8.12)E+V_0=\frac{\hbar^2z^2}{2ma^2}=\frac{(\hbar c)^2z^2}{2(mc^2)a^2} =\frac{197.3^2\times1.31^2}{2\times511000\times0.25}\ \text{eV}\approx0.26\ \text{eV}\tag{2.8.12}E≈−4.74 eV(2.8.13)E\approx-4.74\ \text{eV}\tag{2.8.13}

Compare with the infinite well of width 2a=12a=1 nm: E1∞=0.376E_1^{\infty}=0.376 eV (the numeric quiz question of the previous section), higher than the 0.26 eV here. Just as expected: the wavefunction leaks out, the effective width grows, and the energy drops.

Penetration depth: with ∣E∣=4.74|E|=4.74 eV,

δ=ℏc2(mc2)∣E∣=197.32×511000×4.74≈0.09 nm(2.8.14)\delta=\frac{\hbar c}{\sqrt{2(mc^2)\lvert E\rvert}}=\frac{197.3}{\sqrt{2\times511000\times4.74}}\approx0.09\ \text{nm}\tag{2.8.14}

About 0.9 Å, close to the size of an atom. So if two such wells are placed within 2 Å of each other their wavefunctions overlap appreciably — which is exactly where chemical bonding and band formation begin.

Key formulas

  • Even-parity level condition

    ktan⁡ka=κ,k=2m(E+V0)ℏ, κ=−2mEℏk\tan ka=\kappa,\qquad k=\frac{\sqrt{2m(E+V_0)}}{\hbar},\ \kappa=\frac{\sqrt{-2mE}}{\hbar}

    Odd parity gives −k cot ka = κ

  • Dimensionless form

    tan⁡z=z02−z2z,z0=aℏ2mV0\tan z=\frac{\sqrt{z_0^2-z^2}}{z},\qquad z_0=\frac{a}{\hbar}\sqrt{2mV_0}

    z₀ packages depth and width into the single parameter

  • Number of bound states

    N=⌈2z0π⌉N=\left\lceil\frac{2z_0}{\pi}\right\rceil

    A 1D symmetric well has at least one, however shallow

  • Penetration depth

    δ=1κ=ℏ2m∣E∣\delta=\frac{1}{\kappa}=\frac{\hbar}{\sqrt{2m\lvert E\rvert}}

    Weaker binding leaks further; the key length scale for tunnelling

Self-check3 questions

  1. 1.

    Compared with an infinite well of the same width, the ground-state energy of a finite well (measured from the bottom of the well) is:

  2. 2.

    In a symmetric one-dimensional finite well, if V₀ is made very small (a very shallow well), the number of bound states:

  3. 3.

    Why must both ψ and ψ′ be continuous at x = a here, when only ψ was required to be continuous in the infinite well?

What comes next

The next exactly solvable potential is a parabola. Its importance goes far beyond “one more exercise”: every potential looks parabolic near a minimum, which makes the harmonic oscillator the single most reused model in physics.

Section 16 of 106 · use ← → to turn the page