The Free Particle & Constant Potentials

Before solving any real potential, learn to read the Schrödinger equation the way a musician reads a score. Wherever $V$ is constant the solutions are plane waves or exponentials; everything else is stitching those pieces together at the seams. This lesson builds the stitching kit — continuity, curvature, parity, nodes — and ends with the punchline of the whole course: why bound energies come in discrete steps. A cat at a closed door already suspects the answer.

Learning Objectives

After this lesson you will be able to:

1. Solve the time-independent Schrödinger equation in any region of constant potential, for $E > V$ and $E < V$.
2. Explain why $e^{ikx}$ is a momentum eigenfunction but not a physical state, and how wave packets resolve this.
3. Compute the penetration depth $1/\kappa$ into a classically forbidden region.
4. Apply the matching conditions for $\psi$ and $\psi'$, including the jump condition at a $\delta$-function potential.
5. Sketch energy eigenfunctions qualitatively using the curvature relation, parity, and the node theorem.
6. Explain the qualitative origin of energy quantization and the idea of the shooting method.

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Intuition

Squint at the time-independent Schrödinger equation and it says one thing: the curvature of $\psi$ is proportional to $(V - E)\,\psi$. Where the particle is classically allowed ($E > V$), $\psi$ curves back toward the axis and oscillates, like a plucked string. Where it is forbidden ($E < V$), $\psi$ curves away, growing or decaying exponentially — and a physical wavefunction had better pick "decays." A bound state must decay on both sides, and tuning the energy so the decaying tail on the left evolves into a decaying tail on the right is possible only at special, discrete energies. That is quantization, before we solve a single potential exactly.

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Theory

The time-independent Schrödinger equation with constant $V$

From P.4.2, stationary states $\Psi(x,t) = \psi(x)e^{-iEt/\hbar}$ obey the time-independent Schrödinger equation (TISE)

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\,\psi = E\,\psi \qquad\Longleftrightarrow\qquad \frac{d^2\psi}{dx^2} = \frac{2m}{\hbar^2}\bigl(V(x) - E\bigr)\,\psi .$$

If $V(x) = V$ is constant in a region, this is a constant-coefficient ODE with exactly two regimes:

$$\begin{aligned} E > V \ (\text{allowed}):&\quad \psi'' = -k^2\psi, &\psi &= A\,e^{ikx} + B\,e^{-ikx}, & k &\equiv \sqrt{2m(E - V)}/\hbar;\\ E < V \ (\text{forbidden}):&\quad \psi'' = +\kappa^2\psi, &\psi &= F\,e^{-\kappa x} + G\,e^{+\kappa x}, & \kappa &\equiv \sqrt{2m(V - E)}/\hbar. \end{aligned}$$

Everything in this course is these two lines, glued together at seams.

The free particle

Set $V = 0$ everywhere. For any $E > 0$ the solutions are plane waves $e^{\pm ikx}$ with $k = \sqrt{2mE}/\hbar$. Acting with $\hat p = -i\hbar\,\partial_x$ from P.4.3 gives $\hat p\,e^{ikx} = -i\hbar\frac{d}{dx}e^{ikx} = \hbar k\,e^{ikx}$: $e^{ikx}$ is a momentum eigenfunction with eigenvalue $p = \hbar k$ (and $e^{-ikx}$ with $-\hbar k$) — a right- and a left-mover of the same energy $E = \hbar^2k^2/2m$, de Broglie's $\lambda = h/p$ as an eigenvalue statement. But these solutions are not normalizable: $\int\lvert e^{ikx}\rvert^2dx = \int 1\,dx = \infty$, and no constant fixes that. The honest patch is Dirac-delta normalization: with $\psi_k(x) = e^{ikx}/\sqrt{2\pi}$,

$$\int_{-\infty}^{\infty}\psi_{k'}^*(x)\,\psi_k(x)\,dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(k-k')x}dx = \delta(k - k') ,$$

a delta function where an orthonormal basis would have a Kronecker delta. The $\psi_k$ are not states but a continuum basis. The physical resolution is the one built in P.4.3: a real free particle is a wave packet, $\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int\phi(k)\,e^{i(kx - \hbar k^2t/2m)}dk$, normalizable whenever $\phi(k)$ is, and moving at the group velocity $\hbar k_0/m = p/m$ (the phase velocity is half that — the packet, not the ripples, carries the particle).

Caution. $e^{ikx}$ is a useful fiction: infinitely delocalized, infinite norm. Every statement of the form "a free particle with momentum $p$" is secretly a statement about a narrow wave packet centered on $p$ — narrow enough in $k$ that the fiction is harmless, wide enough in $x$ to exist.

Allowed vs forbidden regions

For slowly varying $V(x)$ the constant-$V$ picture holds locally:

• Allowed ($E > V(x)$): $\psi$ oscillates with local de Broglie wavelength $\lambda(x) = h/p(x)$, where $p(x) = \sqrt{2m(E - V(x))}$.
• Forbidden ($E < V(x)$): real exponentials $e^{\pm\kappa x}$; a physical solution keeps only the decaying piece, dying off over the penetration depth $1/\kappa$.

Numbers. An electron $1\ \mathrm{eV}$ below a barrier top, using $\hbar c = 197.3\ \mathrm{eV\,nm}$ and $m_ec^2 = 511\ \mathrm{keV}$:

$$\kappa = \frac{\sqrt{2m_ec^2(V - E)}}{\hbar c} = \frac{\sqrt{2 \times 511000 \times 1}}{197.3}\ \mathrm{nm^{-1}} \approx 5.12\ \mathrm{nm^{-1}}, \qquad \frac{1}{\kappa} \approx 0.195\ \mathrm{nm}$$

— about two atomic radii. Quantum mechanics leaks into walls, but only by ångströms: why tunneling is huge for electrons and invisible for cats.

The matching toolkit

Real potentials are piecewise; the rules for sewing solutions across a seam come from the TISE:

1. $\psi$ is continuous everywhere. A jump in $\psi$ would put a $\delta'$ into $\psi''$, which nothing in the TISE could balance.
2. $\psi'$ is continuous wherever $V$ is finite. Integrate the TISE across a seam at $x_0$:$$\psi'(x_0 + \epsilon) - \psi'(x_0 - \epsilon) = \frac{2m}{\hbar^2}\int_{x_0-\epsilon}^{x_0+\epsilon}\bigl(V(x) - E\bigr)\psi\,dx \xrightarrow{\ \epsilon \to 0\ } 0$$whenever the integrand is bounded.
3. Jump condition at a $\delta$-spike. If $V(x) = \pm\alpha\,\delta(x - x_0)$, the integral above survives, since $\int\delta(x - x_0)\psi\,dx = \psi(x_0)$:$$\boxed{\ \psi'(x_0^+) - \psi'(x_0^-) = \pm\frac{2m\alpha}{\hbar^2}\,\psi(x_0)\ }$$— the derivative kinks in proportion to $\psi$ at the spike. (At a truly infinite wall, $\psi'$ may jump and instead $\psi = 0$ at the wall.)

This toolkit is the workhorse of Lesson 2: every step, well, and barrier problem is "general solution per region, then match."

Reading the curvature: sketching rules

Rewrite the TISE as a statement about relative curvature:

$$\frac{\psi''}{\psi} = \frac{2m}{\hbar^2}\bigl(V(x) - E\bigr).$$

In an allowed region $\psi''/\psi < 0$: $\psi$ always curves toward the axis — that is what oscillation is — and the more negative $V - E$, the tighter the curvature (shorter $\lambda(x)$). In a forbidden region $\psi''/\psi > 0$: curvature away, exponential growth or decay. At a turning point ($E = V$): $\psi'' = 0$, an inflection. One more rule finishes the kit: where the particle is fast (large $p(x)$) it spends little time, so the amplitude is small and the wavelength short; slow regions get tall, lazy waves. With these you can sketch any 1D eigenfunction before computing anything.

Parity

Let $V(-x) = V(x)$ and let $\psi(x)$ solve the TISE with energy $E$. Substituting $x \to -x$ (which leaves $d^2/dx^2$ invariant) shows $\psi(-x)$ is also a solution with the same $E$. Then $\psi_\pm(x) \equiv \psi(x) \pm \psi(-x)$ are solutions with energy $E$, one even, one odd, and at least one is nonzero: eigenfunctions of a symmetric potential can always be chosen even or odd. For 1D bound states they must be: bound levels are non-degenerate (proved in E5), so $\psi(-x) = c\,\psi(x)$ with $c^2 = 1$, i.e. $c = \pm1$. Parity halves every symmetric-well problem — the finite well of Lesson 2 leans on it hard.

Nodes, and the origin of quantization

Node theorem. Ordering bound states by energy, $E_0 < E_1 < \cdots$, the $n$th excited state $\psi_n$ has exactly $n$ nodes (interior zeros). Motivation: raising $E$ tightens the curvature in the allowed region — one more half-oscillation fits before the tails must decay — and $\psi_n$ must be orthogonal to all lower states, forcing sign changes. (Rigor: Sturm oscillation theory.)

Why bound energies are discrete. Pick a well and a trial $E$; integrate the TISE from the far left, starting on the physical decaying exponential. Cross the well ($\psi$ oscillates), emerge on the right: generically $\psi = Fe^{-\kappa x} + Ge^{+\kappa x}$ with $G \neq 0$ — not normalizable. Only at special energies does $G(E) = 0$, letting $\psi$ decay at both infinities: those are the bound states. As $E$ sweeps through an eigenvalue the divergent tail flips sign — and hunting that flip numerically is the shooting method, previewed below and used in earnest in Lesson 3.

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Worked Examples

Example 1 — An electron's wavelength and its reach into a wall

An electron with kinetic energy $E = 10\ \mathrm{eV}$ in a region with $V = 0$ has

$$k = \frac{\sqrt{2m_ec^2E}}{\hbar c} = \frac{\sqrt{2\times511000\times10}}{197.3}\ \mathrm{nm^{-1}} \approx 16.2\ \mathrm{nm^{-1}}, \qquad \lambda = \frac{2\pi}{k} \approx 0.39\ \mathrm{nm},$$

comparable to atomic spacings — why electrons diffract off crystals (P.3.2). Meeting a region with $V = 11\ \mathrm{eV}$ (a $1\ \mathrm{eV}$ deficit), $\kappa \approx 5.12\ \mathrm{nm^{-1}}$: after $1\ \mathrm{nm}$ of wall the amplitude is down by $e^{-5.12} \approx 6\times10^{-3}$, the probability density by $e^{-10.2} \approx 4\times10^{-5}$.

Example 2 — The $\delta$-function well: the toolkit in action

Let $V(x) = -\alpha\,\delta(x)$, $\alpha > 0$, and seek a bound state ($E < 0$). For $x \neq 0$ the TISE gives $\psi'' = \kappa^2\psi$ with $\kappa = \sqrt{-2mE}/\hbar$, so the normalizable solution is $\psi = B\,e^{-\kappa\lvert x\rvert}$ — continuous at $0$ (rule 1) and even, as parity demands. The jump condition (rule 3, with $-\alpha$):

$$\psi'(0^+) - \psi'(0^-) = -2\kappa B \stackrel{!}{=} -\frac{2m\alpha}{\hbar^2}B \quad\Longrightarrow\quad \kappa = \frac{m\alpha}{\hbar^2} \quad\Longrightarrow\quad E = -\frac{\hbar^2\kappa^2}{2m} = -\frac{m\alpha^2}{2\hbar^2}.$$

One equation, one $\kappa$: exactly one bound state. Normalization gives $B = \sqrt{\kappa}$, from $\lvert B\rvert^2\!\int e^{-2\kappa\lvert x\rvert}dx = \lvert B\rvert^2/\kappa = 1$. The energy is pinned by matching alone — quantization with almost no algebra.

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Hands-on (Python)

Integrate the TISE for a smooth Gaussian well $V(x) = -V_0e^{-x^2/2\sigma^2}$ and watch the divergent tail flip sign as $E$ sweeps through an eigenvalue. Units: length in $\sigma$, energy in $\hbar^2/m\sigma^2$ — i.e. $\hbar = m = \sigma = 1$ in the code only.

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import solve_ivp

hbar = m = 1.0                            # code units (see note above)
V0 = 4.0
V = lambda x: -V0 * np.exp(-x**2 / 2.0)   # smooth Gaussian well

def rhs(x, y, E):                         # TISE as first-order system, y = (psi, psi')
    return [y[1], 2.0 * m * (V(x) - E) / hbar**2 * y[0]]

def shoot(E, x_max=6.0):
    """Integrate from the left forbidden region, starting on the decaying branch."""
    kappa = np.sqrt(2.0 * m * abs(E)) / hbar
    sol = solve_ivp(rhs, [-x_max, x_max], [1e-8, kappa * 1e-8], args=(E,),
                    max_step=0.02, rtol=1e-10)
    return sol.t, sol.y[0]

for E in [-3.4, -3.0, -2.2, -1.8, -1.4]:
    x, psi = shoot(E)
    print(f"E = {E:5.2f}   psi(x_max) = {psi[-1]:+.2e}")
# E = -3.40   psi(x_max) = +9.79e+02   <- diverges upward
# E = -3.00   psi(x_max) = -3.92e+01   <- sign flip: eigenvalue E0 ~ -3.09 crossed
# E = -2.20   psi(x_max) = -3.09e+00
# E = -1.80   psi(x_max) = -1.91e-01
# E = -1.40   psi(x_max) = +5.98e-03   <- second flip: E1 ~ -1.50 crossed

Every trial $\psi$ diverges — except arbitrarily close to an eigenvalue, where the growing exponential is starved. Bracketing the flips and bisecting is the full shooting method of Lesson 3. Now the curvature rules, on one figure:

E0 = -3.094                               # ~ ground-state energy, from the flip above
x, psi = shoot(E0)
psi = psi / np.max(np.abs(psi))
xt = np.sqrt(-2.0 * np.log(-E0 / V0))     # classical turning points, V(x) = E
plt.plot(x, psi, label=r"$\psi$ at $E \approx E_0$")
plt.plot(x, V(x) / V0, "--", label=r"$V/V_0$")
plt.axhline(E0 / V0, color="gray", lw=0.8)
for s in (-xt, xt):
    plt.axvline(s, color="k", ls=":", lw=0.8)
plt.xlabel("x"); plt.legend(); plt.show()
# Between the dotted turning points psi curves TOWARD the axis (one nodeless hump:
# the ground state); beyond them it curves AWAY, decaying exponentially.

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Exercises

E1 (easy). An electron has kinetic energy $4\ \mathrm{eV}$. Compute $k$, $\lambda$, and $p$, and verify $\lambda = h/p$.

Solution

$k = \sqrt{2\times511000\times4}/197.3 \approx 10.25\ \mathrm{nm^{-1}}$, so $\lambda = 2\pi/k \approx 0.613\ \mathrm{nm}$, $p = \hbar k \approx 2022\ \mathrm{eV}/c$, and $h/p = 2\pi\times197.3/2022 \approx 0.613\ \mathrm{nm}$. ✓ (Handy: $\lambda \approx 1.226\ \mathrm{nm}/\sqrt{E[\mathrm{eV}]}$.)

E2 (easy). Show that $\cos kx$ is an energy eigenfunction of the free particle but not a momentum eigenfunction, and write it in terms of momentum eigenfunctions.

Solution

$-\frac{\hbar^2}{2m}(\cos kx)'' = \frac{\hbar^2k^2}{2m}\cos kx$: energy eigenfunction. But $\hat p\cos kx = i\hbar k\sin kx \neq \text{const}\times\cos kx$. Indeed $\cos kx = \tfrac12(e^{ikx} + e^{-ikx})$: an equal superposition of momenta $\pm\hbar k$. Energy $\propto p^2$ cannot tell right-movers from left-movers; momentum can.

E3 (medium). A proton faces the same $1\ \mathrm{eV}$ deficit as the electron in the text. Find its penetration depth and explain why tunneling is a light particle's game.

Solution

$\kappa \propto \sqrt{m}$ at fixed $V - E$, so $1/\kappa_p = 0.195\ \mathrm{nm}/\sqrt{1836} \approx 4.6\ \mathrm{pm}$ — forty-three times shorter. Since transmission goes like $e^{-2\kappa a}$ (next lesson), a factor $\sim\!43$ in the exponent means the proton's tunneling probability is roughly the electron's raised to the 43rd power.

E4 (medium). Show that a repulsive spike $V = +\alpha\,\delta(x)$ has no bound state, while $-\alpha\,\delta(x)$ has exactly one.

Solution

A bound state needs $E < 0$ and $\psi = Be^{-\kappa\lvert x\rvert}$ with $\kappa > 0$ (the only normalizable form). The jump condition with $+\alpha$ gives $-2\kappa B = +\frac{2m\alpha}{\hbar^2}B$, i.e. $\kappa = -m\alpha/\hbar^2 < 0$: contradiction. With $-\alpha$: $\kappa = +m\alpha/\hbar^2$, one positive solution — exactly one bound state ($E = -m\alpha^2/2\hbar^2$, Example 2).

E5 (hard). Prove that 1D bound states are non-degenerate, and conclude that bound eigenfunctions of a symmetric potential are automatically even or odd. Hint: Wronskian.

Solution

Let $\psi_1, \psi_2$ solve the TISE with the same $E$, both vanishing at $\pm\infty$. Combining the two TISEs gives $\psi_1''\psi_2 - \psi_2''\psi_1 = 0$, so the Wronskian $W = \psi_1'\psi_2 - \psi_2'\psi_1$ has $W' = 0$: constant — and $W \to 0$ at infinity, so $W \equiv 0$. Then $\psi_1'/\psi_1 = \psi_2'/\psi_2$, i.e. $(\ln\psi_1 - \ln\psi_2)' = 0$, so $\psi_1 = c\,\psi_2$: the same state — non-degenerate. If moreover $V(-x) = V(x)$, then $\psi(-x)$ is an eigenfunction with the same $E$, so $\psi(-x) = c\,\psi(x)$; reflecting twice, $c^2 = 1$, hence $c = \pm1$: even or odd. ∎

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Checkpoint

1. Why is $e^{ikx}$ not a physical state, and what plays its role in honest calculations?
2. State the three matching rules. When exactly may $\psi'$ be discontinuous?
3. What does $\psi''/\psi = 2m(V - E)/\hbar^2$ say about the shape of $\psi$ in allowed and forbidden regions?
4. For an electron $1\ \mathrm{eV}$ below a barrier top, roughly how far does $\psi$ penetrate?
5. In one paragraph: why are bound-state energies discrete?

Answers

1. Infinite norm — only delta-normalizable. Physical states are wave packets, $\int\phi(k)e^{ikx}dk$ with normalizable $\phi(k)$.
2. $\psi$ continuous always; $\psi'$ continuous wherever $V$ is finite; across $\pm\alpha\,\delta(x - x_0)$, $\psi'$ jumps by $\pm(2m\alpha/\hbar^2)\psi(x_0)$. Only an infinite spike or wall breaks $\psi'$-continuity.
3. Allowed: curvature toward the axis — oscillation with $\lambda(x) = h/p(x)$. Forbidden: curvature away — exponentials. Turning points are inflections.
4. Penetration depth $1/\kappa \approx 0.195\ \mathrm{nm}$ — about two atomic radii.
5. A bound state must ride the decaying exponential on both sides. Integrating from the left on the decaying branch, a generic $E$ picks up a growing piece on the right; only at discrete energies does its coefficient vanish (the tail's sign flip is the shooting method's signal).

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Further Reading

• [Gri] Griffiths & Schroeter, §2.4–2.5 — the free particle and the delta-function well; our matching toolkit in action.
• [Sha] Shankar, Ch. 5 — one-dimensional problems, with more formalism per pound.
• [ER] Eisberg & Resnick, Ch. 5–6 — qualitative wavefunction sketching done with old-school patience.

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