Finite walls change two things at once: the particle can be found where it has no business being, and only some energies admit a solution that stays finite.

Every potential so far has had infinitely high walls. That was a convenience, and it cost us two things: a particle in an infinite box can never be found outside it, and its energies are always discrete.

Chapter 5 lowers the walls. The moment they are finite, both statements fail — and what replaces them is most of what makes quantum mechanics useful.

The potential

Phillips picks a deliberately simple field — an infinite wall on the left, an attractive well of depth $V_0$ and width $a$ , and nothing beyond:

V(x) = \begin{cases} \infty & \text{if } -\infty < x < 0\\ -V_0 & \text{if } 0 < x < a\\ 0 & \text{if } a < x < \infty\end{cases}\tag{5.1}

A state of definite energy has the usual form from §4.3, $\Psi(x,t) = \psi(x)e^{-iEt/\hbar}$ , with

-\frac{\hbar^2}{2m}\frac{\mathrm d^2\psi}{\mathrm dx^2} + V(x)\psi = E\psi\tag{5.4}

V(x)
energy level Eₙ
ψₙ(x)
classically forbidden (V > E)

width3.0depth V₀12.0

selected level: E₀ = -11.5791
nodes in ψ: 0
bound states: 5
E₁ − E₀: 1.2549

Make the well shallower or narrower and watch the top level rise to zero and vanish into the continuum. Note that ψ does not stop at the wall — it leaks into the shaded forbidden region, which is tunnelling in its simplest form.

The technique: solve, then join

$V(x)$ is constant in each of three regions, so Eq. (5.4) is easy in each one separately. The work is in joining the pieces.

How smooth must ψ be?

Because $V$ jumps, we cannot demand that $\psi$ be as smooth as we like. Phillips is precise about how much smoothness survives:

$\psi$ itself is continuous everywhere — otherwise $|\psi|^2$ would jump, and a probability density cannot change abruptly at a point.
$\mathrm d\psi/\mathrm dx$ is continuous at $x = a$ , where $V$ jumps by a finite amount.
$\mathrm d\psi/\mathrm dx$ is discontinuous at $x = 0$ , where $V$ jumps to infinity.

The book’s footnote gives the honest justification for the last one: treat the wall as a finite step $V_1$ , require both continuities, and then let $V_1 \to \infty$ . The kink is what survives that limit. An infinite potential is an idealization, and this is the price it charges.

Bound states

A bound state has $E$ somewhere between $-V_0$ and 0. Write

E = -\epsilon

where $\epsilon$ is the binding energy — a positive number, the energy you would have to supply to free the particle.

ϵ is the binding energy, and ε₀ turns up later in this chapter

The book writes the binding energy as $\epsilon$ . Two things to keep straight:

$\epsilon$ is positive while $E$ is negative — they differ by a sign, not just a symbol. Deep binding means large $\epsilon$ and very negative $E$ .
$\epsilon$ is not $\epsilon_0$ . The permittivity of free space appears in §5.2’s Coulomb barrier, and the two look nearly identical in running text.

This page writes the binding energy as $\epsilon$ to match the book, and always says “binding energy” alongside it on first use in any argument.

And a third collision, coming up in a moment: $\alpha$ . Below, $\alpha$ is the decay constant of a bound state’s tail, defined by $\epsilon = \hbar^2\alpha^2/2m$ — a quantity with units of $\mathrm{m^{-1}}$ . That is the fourth thing this book has called $\alpha$ : §1.4 used it for a microscope’s half-angle, §0.2 for the dimensionless fine-structure constant $1/137$ , and ch03’s problem 7 for yet another decay constant. §5.2b brings the fine-structure meaning back, in the same chapter as this one. Read $\alpha$ from its equation, never from memory — and note that §5.2 uses $\beta$ for exactly this decay-constant role inside a barrier, because the book keeps the well and barrier situations under separate letters.

step 1 of 4

Three regions, three solutions, and in each one the physics picks which half of the general solution to keep.

1REGION 1, x < 0. The potential is infinite, so the only finite solution is zero. The particle is never found to the left of the wall.
$ψ (x) = 0 (- \infty < x < 0)$
2REGION 2, 0 < x < a. Here V = −V₀ and the equation is the oscillatory one, with k₀ set by how far the energy sits above the well bottom.
$\frac{d ^{2} ψ}{d x ^{2}} = - k_{0}^{2} ψ, E = \frac{ℏ ^{2} k _{0}^{2}}{2 m} - V_{0}$
Equation (5.5). The general solution is ψ = C sin(k₀x + γ), and continuity with region 1 at x = 0 forces γ = 0 — giving Eq. (5.6), ψ = C sin k₀x.
3REGION 3, x > a. Here V = 0 and the energy is NEGATIVE, so the equation flips sign: exponential, not oscillatory. This is the whole difference between a bound state and an unbound one.
$\frac{d ^{2} ψ}{d x ^{2}} = α^{2} ψ, E = - \frac{ℏ ^{2} α ^{2}}{2 m}$
Equation (5.7). The general solution is Ae^{−αx} + A′e^{+αx}.
4The growing exponential would blow up at infinity and could never be normalized, so A′ = 0. What is left decays.
$ψ (x) = A e^{- α x}$
Equation (5.8) — and note it is NOT zero. The particle has a real chance of being found beyond the well.

The joining condition

Now match at $x = a$ . The two requirements below — continuity of $\psi$ , and continuity of its slope — are the matching conditions , and they are the entire technique of this chapter. Continuity of $\psi$ :

C\sin k_0a = A e^{-\alpha a}\tag{5.9}

and continuity of $\mathrm d\psi/\mathrm dx$ :

k_0C\cos k_0a = -\alpha A e^{-\alpha a}\tag{5.10}

Dividing the second by the first eliminates both unknown amplitudes at once:

k_0\cot k_0a = -\alpha\tag{5.11}

Equation (5.11) is one equation in two unknowns, but $k_0$ and $\alpha$ are not independent. From the two definitions of $E$ ,

\alpha^2 + k_0^2 = w^2, \quad\text{where}\quad V_0 = \frac{\hbar^2w^2}{2m}\tag{5.12}

with $w$ the well-depth parameter . So we have two simultaneous equations, and Phillips solves them graphically.

well depth w2.00 π/a→ V₀ = 4.00 ħ²π²/2ma²

2 bound states. Binding energies ϵ = α², in units of ħ²π²/2ma²:

n = 1: k₀ = 0.8587 π/a, α = 1.8063 π/a → ϵ₁ = 3.2626
n = 2: k₀ = 1.6820 π/a, α = 1.0821 π/a → ϵ₂ = 1.1710

Reading the counting rule off the picture

The circle’s radius is $w$ . Each cotangent branch begins on the axis at $k_0 = \pi/2a,\ 3\pi/2a,\ 5\pi/2a,\dots$ and rises steeply. So a branch is reachable only once the circle is big enough to touch it, giving:

well depth	bound states
$w < \pi/2a$	none
$\pi/2a < w < 3\pi/2a$	one
$3\pi/2a < w < 5\pi/2a$	two
and so on

An attractive well is not enough

The first row deserves attention. In this potential — a well with an infinite wall on one side — a sufficiently shallow well binds nothing at all. Make it as attractive as you like below $w = \pi/2a$ and the particle still escapes.

That is not obvious, and it is not universal. A symmetric well with no wall (problem 1) always has at least one bound state, however shallow. The difference is the wall: it forces $\psi(0) = 0$ , so the wave function must complete a quarter-wavelength inside the well before it can match onto a decaying exponential, and a shallow well is not wide enough in $k$ to allow it.

The general lesson is worth keeping: whether a potential binds anything is a question about the potential’s shape and dimensionality, not just its depth. In one dimension a symmetric well always binds; in three dimensions it does not, which is why problem 2’s spherical well has a minimum depth — and, historically, why the deuteron is just barely bound while the di-neutron is not bound at all.

The worked example

Take $w = 2\pi/a$ , so $V_0 = 2\hbar^2\pi^2/ma^2$ . Two intersections, hence two bound states, with binding energies

\epsilon_1 = 3.26\,\frac{\hbar^2\pi^2}{2ma^2} \quad\text{and}\quad \epsilon_2 = 1.17\,\frac{\hbar^2\pi^2}{2ma^2}

Leaking into the forbidden region

Here is the result that separates this chapter from chapter 4. Outside the well,

\psi(x) = Ae^{-\alpha x}

which is small but not zero. The particle can be found where a classical particle absolutely cannot be — in the classically forbidden region , where $E < V(x)$ and classical kinetic energy would have to be negative.

And since $\epsilon = \hbar^2\alpha^2/2m$ , a lower binding energy means a smaller $\alpha$ and a longer tail. Weakly bound particles are the leakiest.

Put a number on it — how far out, and how likely?

For the $w = 2\pi/a$ well, the $1/e$ penetration depth $1/\alpha$ is

ground state ( $\epsilon = 3.26$ ): $1/\alpha = 0.176\,a$
first excited state ( $\epsilon = 1.17$ ): $1/\alpha = 0.294\,a$

and integrating $|\psi|^2$ beyond $x = a$ gives the actual probability of finding the particle outside the well:

ground state: 2.8%
first excited state: 16.1%

A sixth of the time, the weakly bound particle is somewhere it has no classical right to be. Both numbers are computed in the block below rather than estimated.

This is not a curiosity. It is the same effect that lets an electron cross the gap in a scanning tunnelling microscope and lets two protons fuse in the sun — §5.2 turns the leak into a full crossing.

Engineer's bridge — evanescent waves, and a transcendental root-find

The physics half. $\psi \propto e^{-\alpha x}$ in the forbidden region is an evanescent wave, and you have met it: total internal reflection in a fibre leaves an evanescent field in the cladding; a waveguide below cut-off carries an exponentially decaying field rather than a propagating one; the skin depth in a conductor is the same mathematics. In every case a wave equation has its sign flipped by the medium and oscillation becomes decay. The quantum version differs only in what is waving.

The numerical half. Equation (5.11) with (5.12) is a transcendental system with no closed-form solution — and the graphical method is bisection carried out by eye. What the picture shows, a root-finder would report as “one sign change per branch”. The site’s own qm.ts solves the same well a completely different way, by discretizing $\hat H$ into a matrix and calling an eigensolver, and the widget at the top of this page uses exactly that. Two independent methods, the same energies — which is the only kind of agreement worth trusting.

Where it breaks. The matrix method needs a finite box, so it silently imposes a wall far out to the right that the real problem does not have. For bound states that is harmless, because $\psi$ has decayed to nothing long before it. For the unbound states of §5.1b it is not harmless at all: those extend to infinity, and a box turns their continuum into a dense set of discrete levels that are an artefact of the box size.

Reproduce it in NumPy

The counting rule, the worked example's two binding energies against the book's 3.26 and 1.17, and the leakage numbers the book does not give. Everything in units of π/a, where the branch structure is clearest.

import numpy as np

def bound(W, N=2000000):
    """Solve  alpha = -k0 cot(k0 a)  together with  alpha^2 + k0^2 = w^2.
    Everything in units of pi/a, so W = wa/pi and the branches sit on
    kappa in (n-1/2, n)."""
    out = []
    for n in range(1, 8):
        lo, hi = n - 0.5, n - 1e-12
        if W <= lo: break                       # circle too small to reach this branch
        f = lambda k: (-k/np.tan(np.pi*k))**2 + k*k - W*W
        a, b = lo + 1e-9, min(hi, W - 1e-12)
        if b <= a or f(a)*f(b) > 0: continue
        for _ in range(200):
            m = 0.5*(a+b)
            a, b = (a, m) if f(m)*f(a) <= 0 else (m, b)
        k = 0.5*(a+b)
        out.append((k, -k/np.tan(np.pi*k)))
    return out

print("how many bound states does a well of depth w support?")
print("=" * 62)
print("  the book: none below pi/2a, one up to 3pi/2a, two up to 5pi/2a, ...")
for W in (0.30, 0.49, 0.51, 1.00, 1.49, 1.51, 2.00, 2.51, 3.00):
    print(f"    w = {W:.2f} pi/a  ->  {len(bound(W))} bound state(s)")

print("\nthe worked example: w = 2 pi/a, i.e. V0 = 2 hbar^2 pi^2 / ma^2")
for i, (k, A) in enumerate(bound(2.0), start=1):
    print(f"  n={i}:  k0 = {k:.4f} pi/a   alpha = {A:.4f} pi/a   "
          f"epsilon_{i} = alpha^2 = {A*A:.4f}  (x hbar^2 pi^2 / 2ma^2)")
print("  the book states epsilon_1 = 3.26 and epsilon_2 = 1.17")

print("\nhow far does a bound particle leak into the forbidden region?")
print("  psi ~ exp(-alpha x) outside, so the 1/e penetration depth is 1/alpha:")
sol = bound(2.0)
for i, (k, A) in enumerate(sol, start=1):
    print(f"  n={i}:  epsilon = {A*A:5.2f}   alpha = {A*np.pi:6.3f}/a   "
          f"1/alpha = {1/(A*np.pi):.4f} a")
print("  the WEAKLY bound state leaks further -- smaller epsilon means smaller alpha")

print("\nand the probability of actually finding it out there")
for i, (k, A) in enumerate(sol, start=1):
    kk, aa = k*np.pi, A*np.pi
    xi = np.linspace(0, 1, 400001)                    # inside, in units of a
    xo = np.linspace(1, 60, 400001)                   # outside
    Ai = np.sin(kk)/np.exp(-aa)                       # continuity at x = a, with C = 1
    Ii = np.trapezoid(np.sin(kk*xi)**2, xi)
    Io = np.trapezoid((Ai*np.exp(-aa*xo))**2, xo)
    print(f"  n={i}:  P(x > a) = {Io/(Ii+Io):.4f}")
print("  a classical particle has exactly zero chance of being there")

printshow many bound states does a well of depth w support? ============================================================== the book: none below pi/2a, one up to 3pi/2a, two up to 5pi/2a, ... w = 0.30 pi/a -> 0 bound state(s) w = 0.49 pi/a -> 0 bound state(s) w = 0.51 pi/a -> 1 bound state(s) w = 1.00 pi/a -> 1 bound state(s) w = 1.49 pi/a -> 1 bound state(s) w = 1.51 pi/a -> 2 bound state(s) w = 2.00 pi/a -> 2 bound state(s) w = 2.51 pi/a -> 3 bound state(s) w = 3.00 pi/a -> 3 bound state(s) the worked example: w = 2 pi/a, i.e. V0 = 2 hbar^2 pi^2 / ma^2 n=1: k0 = 0.8587 pi/a alpha = 1.8063 pi/a epsilon_1 = alpha^2 = 3.2626 (x hbar^2 pi^2 / 2ma^2) n=2: k0 = 1.6820 pi/a alpha = 1.0821 pi/a epsilon_2 = alpha^2 = 1.1710 (x hbar^2 pi^2 / 2ma^2) the book states epsilon_1 = 3.26 and epsilon_2 = 1.17 how far does a bound particle leak into the forbidden region? psi ~ exp(-alpha x) outside, so the 1/e penetration depth is 1/alpha: n=1: epsilon = 3.26 alpha = 5.675/a 1/alpha = 0.1762 a n=2: epsilon = 1.17 alpha = 3.400/a 1/alpha = 0.2942 a the WEAKLY bound state leaks further -- smaller epsilon means smaller alpha and the probability of actually finding it out there n=1: P(x > a) = 0.0276 n=2: P(x > a) = 0.1608 a classical particle has exactly zero chance of being there

Where this leaves us

A finite well gives discrete bound states — but only finitely many, and only if it is deep enough. Their wave functions extend beyond the classical turning point, further the more weakly they are bound.

§5.1b asks what happens above $E = 0$ , where the particle is not trapped. The same joining condition applies — and turns out to impose nothing at all on the energy. What it fixes instead is a phase, and that phase turns out to be measurable.

Check yourself

0 / 7 answered

1.Why is $d ψ / d x$ continuous at $x = a$ but discontinuous at $x = 0$ ?
2.Dividing Eq. (5.10) by Eq. (5.9) gives $k_{0} cot k_{0} a = - α$ . What does that division accomplish?
3.In the graphical solution, what does each intersection represent?
4.A well with $w < π /2 a$ binds nothing at all. Is that a general feature of attractive wells?
5.For the $w = 2 π / a$ well, the ground state leaks 2.8% outside while the first excited state leaks 16.1%. Why is the *weakly* bound state leakier?
6. $ψ \propto e^{- α x}$ in the forbidden region. Where have you met that mathematics before?
7.The widget solves this well by discretizing $\hat{H}$ into a matrix, while the figure solves it graphically. Where does the matrix method quietly mislead?