# 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. ### 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. ### 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: $\lambda \approx 1.226\ \mathrm{nm}/\sqrt{E[\mathrm{eV}]}$.)

**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. ∎

--- ## 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).

--- ## 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. --- ← Prev: [Wave Packets, Uncertainty & the Momentum Operator](../c04-schrodinger-equation/03-matter-waves-uncertainty-momentum.md) · Up: [Pre-Term](../README.md) · Next: [Step, Well & Barrier](02-step-well-barrier.md) →