# 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**.
---
## 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.
---
## Theory
### The time-independent Schrödinger equation with constant $V$
From [P.4.2](../c04-schrodinger-equation/02-stationary-states-superposition.md), 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.
Which line applies is set by the sign of $V - E$, and the same sign flip is easy to watch with a
clock attached. A mass in $V(z) = cz^2$ obeys $\ddot z = -(2c/m)\,z$. Rename $t \to x$ and
$-2c/m \to 2m(V - E)/\hbar^2$ and it is the equation above: $c > 0$ is the allowed, oscillating
regime, and $c < 0$ the forbidden, exponential one:
```qm-demo restoring-vs-runaway
```
### 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](../c04-schrodinger-equation/03-matter-waves-uncertainty-momentum.md) 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](../c04-schrodinger-equation/03-matter-waves-uncertainty-momentum.md): 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](02-step-well-barrier.md): 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.
Read backwards, the same identity solves the *inverse* problem. Given one eigenfunction and its energy,
$V(x) = E + \frac{\hbar^2}{2m}\,\psi''(x)/\psi(x)$ wherever $\psi \ne 0$: a single stationary state fixes
the potential that produced it, up to the same additive constant that fixes the zero of energy.
[P.5.3](03-harmonic-oscillator.md) E6 uses this to recover the oscillator from a Gaussian.
### 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](02-step-well-barrier.md) 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](03-harmonic-oscillator.md).
---
## 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](../c03-wave-particle-duality/02-matter-as-waves.md)). 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.
---
## 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*.
```python
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](03-harmonic-oscillator.md). Now the curvature rules, on one figure:
```python
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.
```
---
## 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: $\lambda \approx 1.226\ \mathrm{nm}/\sqrt{E[\mathrm{eV}]}$.)
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.
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.
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).
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. ∎
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).