Free Particle and Plane Waves

Take away the walls. No well, no barrier, no force — just a particle of mass $m$ alone on an infinite line. Classically this is the dullest problem in mechanics: the particle coasts at constant velocity, forever. Quantum mechanically it is one of the strangest. The energy eigenstates turn out to be waves that fill all of space, each carrying a perfectly definite momentum and absolutely no location. They cannot be normalized, so — read that again — the free particle has no physical energy eigenstates at all. Every allowed energy from zero upward is permitted, with nothing to quantize, and yet no single one of those states could ever describe an electron actually flying down a beamline. You already have the tools to see why. The Time-Independent Schrödinger Equation in One Dimension told you that where $E>V$ the wavefunction oscillates; with $V=0$ everywhere it oscillates everywhere, and Bound States versus Scattering States told you what that means: every free state belongs to the scattering continuum. What is new here is learning to work with a basis of impossible states — plane waves, normalized not to one but to a Dirac delta — because every real state is a superposition of them. The payoff is the vocabulary of all of scattering theory: the $e^{ikx}$ that arrives at every barrier and well in this chapter, the $\lambda = h/p$ that made electrons diffract, and a puzzle — crests that travel at half the particle's speed — whose resolution is one article away.

Key takeaways

With $V=0$ the TISE becomes $\psi''=-k^2\psi$ : solutions are plane waves $e^{\pm ikx}$ with $E=\hbar^2k^2/2m$ , and every $E\ge 0$ is allowed — a continuous spectrum with nothing to quantize.
A plane wave has $|\psi|^2=\text{const}$ , so $\int|\psi|^2\,\mathrm{d}x=\infty$ : it is not a physical state, only a basis element, "normalized" to a Dirac delta, $\langle\psi_k|\psi_{k'}\rangle=\delta(k-k')$ .
Plane waves are simultaneous eigenstates of momentum ( $p=\hbar k$ ) and energy — definite momentum is bought at the price of total delocalization, the extreme of the uncertainty principle.
Physical free particles are superpositions — wave packets — built in Wave Packets and Their Construction; the plane wave supplies the alphabet, not the sentence.
The crests of $e^{i(kx-\omega t)}$ move at $\omega/k=\hbar k/2m$ , exactly half the classical velocity $p/m$ — a paradox stated here and resolved in Group Velocity versus Phase Velocity.
Key idea: freedom kills quantization — with no boundary conditions every $k$ survives, and the price of the continuum is a basis of unnormalizable states.

The core insight

For a free particle the stationary states are plane waves $e^{ikx}$ with $E=\hbar^2k^2/2m$ : a continuous spectrum of momentum eigenstates that fill all space and cannot be normalized — they form the Fourier basis out of which every genuine, localizable free-particle state must be built.

Where this lives today

Every transmission electron microscope is a plane-wave factory: a 300 keV gun prepares electrons with $\lambda\approx1.97$ pm — about fifty times finer than an atom — and cryo-EM images protein structures with those waves. Neutron beamlines run the same physics at 1.8 Å. And the box-normalization count of Example 2, $L/2\pi$ states per unit $k$ , is the opening line of every calculation of electron gases, semiconductor carrier densities, and blackbody spectra — the free-particle continuum, tamed by counting.

The physical picture

Picture an ocean swell of a single, perfect wavelength: crest, trough, crest, trough, identically, from horizon to horizon with no beginning and no end. That is a plane wave. Ask it "what is your wavelength?" and it answers with infinite precision — and through de Broglie's $p=h/\lambda$ that means a momentum known exactly. Ask it "where are you?" and the question fails: it is equally everywhere. Its probability density $|e^{ikx}|^2 = 1$ is a flat line across the universe. Now recall what quantized the Infinite Square Well: the walls. A wave pinned at two ends can only form standing patterns that fit, and the fitting condition selected discrete energies. Out here there are no walls, no pins, no fitting condition — so nothing gets selected. Every wavelength is as good as every other, every $k$ is allowed, and the energy $E=\hbar^2k^2/2m$ sweeps through a continuum from zero upward. Quantization was never a quantum axiom; it was always geometry, and without geometry it evaporates.

Figure

A plane wave fills all space with one wavelength — a packet localizes by mixing many

The two panels are the whole story of this article. The top wave is mathematically perfect and physically impossible; the bottom lump is what an actual electron in a beam looks like. The rules of the game: solve for the impossible states because they are simple, then superpose them into possible ones. The recipe for the mixing — which $k$ 's, with which amplitudes — is the business of Wave Packets and Their Construction. Here we build the alphabet.

Figure

Every point of the parabola is allowed — the free continuum against the well's discrete rungs

Solving the free Schrödinger equation

Set $V=0$ in the time-independent Schrödinger equation. There is nothing left but kinetic energy:

-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}=E\psi \qquad (1)

where: $\hbar=1.055\times10^{-34}\ \mathrm{J\,s}$ — the reduced Planck constant; $m$ — the particle's mass; $E$ — the energy eigenvalue. Divide through and the equation says something you can read at sight: $\psi''=-(2mE/\hbar^2)\psi$ , curvature proportional to minus the function itself. For $E>0$ that is the harmonic-oscillation equation in space — the wavefunction wiggles forever, as the curvature dial of the Time-Independent Schrödinger Equation in One Dimension promised for a classically allowed region that never ends. Define the wavenumber $k$ to absorb the constants, and the general solution is two travelling waves:

\psi(x)=A\,e^{ikx}+B\,e^{-ikx},\qquad k\equiv\frac{\sqrt{2mE}}{\hbar} \qquad (2)

where: $A$ and $B$ — complex amplitudes of the right-moving ( $e^{ikx}$ ) and left-moving ( $e^{-ikx}$ ) components. Attach the stationary-state clock factor $e^{-iEt/\hbar}$ and $e^{ikx}$ becomes $e^{i(kx-\omega t)}$ , a waveform whose crests genuinely slide to the right; $e^{-ikx}$ slides left. Nothing restricts $k$ : no wall to reach, no boundary condition to satisfy, no transcendental equation to solve. The spectrum is the entire half-line $E\ge0$ — continuous, doubly degenerate (both directions for each $E$ ), exactly as Bound States versus Scattering States classifies for a particle with $E$ above the potential at infinity. The travelling wave carries its dispersion relation with it. Writing the full time-dependent solution:

\Psi_k(x,t)=A\,e^{i(kx-\omega t)},\qquad \omega=\frac{E}{\hbar}=\frac{\hbar k^2}{2m} \qquad (3)

where: $\omega$ — the angular frequency of the quantum phase. The relation $\omega\propto k^2$ is called quadratic dispersion, and it is the single most consequential fact about free matter waves: components of different $k$ move at different speeds. Light in vacuum has $\omega = ck$ — linear, dispersionless — and that difference is why laser pulses hold together while electron packets smear (the full account lives in Wave Packet Spreading and Dispersion). And the momentum? Apply $\hat p = -i\hbar\,\partial/\partial x$ to a plane wave and it comes straight back:

\hat p\,e^{ikx}=\hbar k\,e^{ikx}\qquad\Longrightarrow\qquad p=\hbar k,\quad E=\frac{p^2}{2m} \qquad (4)

where: $p=\hbar k$ — de Broglie's relation in operator clothing, since $p = \hbar k = h/\lambda$ . The plane wave is an exact momentum eigenstate: measure the momentum of a particle in state $e^{ikx}$ and you get $\hbar k$ , certainly. It is simultaneously an energy eigenstate because the free Hamiltonian is just $\hat p^2/2m$ — a function of $\hat p$ alone, so the two operators share every eigenstate. Definite momentum, definite energy, and — as we now make precise — a total, irreparable loss of position.

Derivation AContinuous spectrum and the failure of normalization

We are establishing the three defining facts of the free particle in one sweep: every energy $E \ge 0$ is allowed, no energy $E<0$ is allowed, and none of the eigenstates can be normalized. The strategy is to solve the $V=0$ eigenvalue problem for both signs of $E$ and then confront each solution with the two admissibility demands — boundedness everywhere, and square-integrability. The point to watch is that the two demands act differently: boundedness kills the negative energies outright, while square-integrability fails for every survivor, which is what forces the basis interpretation.

Step 1. Rewrite the eigenvalue equation with all constants in one place:

\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}=-\frac{2mE}{\hbar^2}\,\psi

Step 2. Suppose first $E>0$ and set $k^2 = 2mE/\hbar^2$ . The characteristic roots are $\pm ik$ , giving oscillatory exponentials:

\psi(x)=A\,e^{ikx}+B\,e^{-ikx}

Why oscillation was inevitable: with $E>V$ everywhere, the curvature $\psi''$ always opposes $\psi$ — the function perpetually bends back toward the axis and can never run away.

Step 3. Suppose instead $E<0$ and set $\kappa^2=-2mE/\hbar^2>0$ . The roots are now real, $\pm\kappa$ :

\psi(x)=C\,e^{\kappa x}+D\,e^{-\kappa x}

On the full line both pieces are fatal: $e^{\kappa x}$ blows up as $x\to+\infty$ and $e^{-\kappa x}$ blows up as $x\to-\infty$ . Keeping either violates boundedness; keeping neither leaves nothing. There is no admissible solution — negative energies are simply absent, so the spectrum starts exactly at $E=0$ . Why this differs from a well: inside the Finite Square Well the growing exponential is discarded on each side separately and the pieces are stitched at the well — that stitching is what selects discrete energies. With no well there is nowhere to stitch.

Step 4. Now test the $E>0$ survivors for square-integrability. For a single plane wave,

\int_{-\infty}^{\infty}|A\,e^{ikx}|^2\,\mathrm{d}x=|A|^2\int_{-\infty}^{\infty}\mathrm{d}x=\infty

The density is a constant, and a constant integrated over the line diverges. No choice of $A$ fixes this: the plane wave is not a normalizable state, hence not a physical one.

Step 5. Conclude. Every $E>0$ admits bounded solutions (all real $k$ , both signs), so the spectrum is the continuum $E\ge0$ ; none of the eigenfunctions is square-integrable, so the free particle possesses no physical stationary states at all. What survives is their role as a basis: superpositions $\int\phi(k)e^{ikx}\mathrm{d}k$ can be — and are — normalizable.

A basis of impossible states: delta normalization

If plane waves cannot be normalized, what replaces $\int|\psi|^2\mathrm{d}x=1$ ? The answer is one of the quiet masterstrokes of the formalism. Attach a $1/\sqrt{2\pi}$ and compute the overlap of two different plane waves — the calculation is one line, and its output is the central identity of Fourier analysis:

\psi_k(x)=\frac{1}{\sqrt{2\pi}}e^{ikx},\qquad \langle\psi_k|\psi_{k'}\rangle=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(k'-k)x}\,\mathrm{d}x=\delta(k-k') \qquad (5)

where: $\delta(k-k')$ — the Dirac delta function: zero whenever $k\ne k'$ , infinite at $k=k'$ , with unit integral. Read it as the continuum's answer to orthonormality. For the Infinite Square Well the bound states obeyed $\langle\psi_m|\psi_n\rangle=\delta_{mn}$ , a Kronecker delta; here the discrete index has melted into a continuous label and the Kronecker delta has melted into a Dirac delta. Different momenta are still orthogonal — the overlap vanishes for $k \ne k'$ — and the infinity at $k=k'$ is precisely the divergence of Step 4 above, wearing its formal uniform. This is what "the plane waves form a basis" means in practice: any state can be written as $\psi(x) = \frac{1}{\sqrt{2\pi}}\int\phi(k)e^{ikx}\mathrm{d}k$ , and the delta identity extracts the coefficients, $\phi(k) = \langle\psi_k|\psi\rangle$ . The expansion coefficient $\phi(k)$ turns out to be a full-fledged wavefunction in its own right — the Momentum-Space Wavefunction introduced back in the foundations chapter — and building states this way is the entire subject of Wave Packets and Their Construction. There is a second, humbler bookkeeping scheme that every solid-state and statistical-mechanics course leans on. Put the particle in a fictitious box of length $L$ with periodic boundary conditions, $\psi(x+L)=\psi(x)$ :

\psi_n(x)=\frac{1}{\sqrt{L}}\,e^{ik_nx},\qquad k_n=\frac{2\pi n}{L},\quad n=0,\pm1,\pm2,\dots \qquad (6)

where: $L$ — the box length, destined to be sent to infinity at the end of every honest calculation. Now the states are genuinely normalized ( $\int_0^L|\psi_n|^2\mathrm{d}x=1$ ), the spectrum is discrete, and states can be counted: exactly $L/2\pi$ of them per unit interval of $k$ . That count — the density of states — is the quantity that survives the $L\to\infty$ limit and powers every electron-gas and blackbody calculation you will ever do. Box normalization and delta normalization are two dialects for the same continuum; fluency in both is the goal.

The two speeds of a matter wave

How fast does a plane wave move? Watch one crest of $e^{i(kx-\omega t)}$ : it sits where the phase $kx-\omega t$ is constant, so it travels at $x/t=\omega/k$ . Insert the dispersion relation (3):

v_{\text{phase}}=\frac{\omega}{k}=\frac{\hbar k}{2m}=\frac{p}{2m}=\frac{v_{\text{classical}}}{2} \qquad (7)

where: $v_{\text{classical}}=p/m$ — the speed a classical particle of this momentum actually has. Sit with the discomfort: the crests of the matter wave move at half the particle's speed. If you naively identified "speed of the wave" with "speed of the particle," quantum mechanics would be off by a factor of two in its very first prediction. It is not off. The resolution is that a single plane wave carries no thing whose arrival you could time — its density is the same everywhere, forever — so the crest speed is not the transport speed of anything physical. What does move at $p/m$ is the envelope of a superposition, and demonstrating that — the group velocity theorem — is the whole business of Group Velocity versus Phase Velocity. Here we only flag the factor of two as real, and as a warning against reading kinematics off a single mode.

Go deeper · Derivation B · 1 minMomentum eigenstates and the uncertainty extreme

Derivation BMomentum eigenstates and the uncertainty extreme

We are pinning down the operator credentials of the plane wave: that it is a simultaneous eigenstate of momentum and energy, why such simultaneous eigenstates exist at all, and what the momentum sharpness costs in position. The strategy is direct operator action followed by a commutator argument, ending with the uncertainty-principle reading. Watch how the argument leans on the Hamiltonian being a function of $\hat p$ alone — the moment a potential $V(x)$ appears, this entire structure collapses, which is precisely why bound states are never momentum eigenstates.

Step 1. Act with the momentum operator:

\hat p\,e^{ikx}=-i\hbar\frac{\partial}{\partial x}e^{ikx}=-i\hbar\,(ik)\,e^{ikx}=\hbar k\,e^{ikx}

An eigenstate, eigenvalue $p=\hbar k$ . Right-movers ( $k>0$ ) carry positive momentum, left-movers negative — the sign of $k$ is the direction of travel.

Step 2. Act with the free Hamiltonian, which is built from $\hat p$ :

\hat H=\frac{\hat p^2}{2m}\qquad\Longrightarrow\qquad \hat H\,e^{ikx}=\frac{(\hbar k)^2}{2m}\,e^{ikx}

Why both at once: since $\hat H$ is a function of $\hat p$ , the two commute, $[\hat H,\hat p]=0$ , and commuting observables admit a common eigenbasis. For the free particle the plane waves are that basis.

Step 3. Read off the uncertainties. A momentum eigenstate has $\sigma_p=0$ exactly. The Heisenberg relation $\sigma_x\sigma_p\ge\hbar/2$ then forces $\sigma_x\to\infty$ — and indeed $|e^{ikx}|^2$ is constant, position maximally unknown. The plane wave is the uncertainty principle's extreme case, not a violation of it. Why degeneracy appears: $e^{ikx}$ and $e^{-ikx}$ share the energy $\hbar^2k^2/2m$ but carry opposite momenta — a two-fold degeneracy impossible for one-dimensional bound states. The general non-degeneracy theorem for bound spectra (and why scattering states evade it) is taken up in Non-Degeneracy of One-Dimensional Bound States.

Limiting cases

case	result	meaning
$k\to0$ , $E\to0$	$\psi\to\text{const}$	the bottom of the continuum: a uniform, motionless amplitude — still non-normalizable, and no zero-point energy because nothing is confined
$A=B$	$\psi\propto\cos kx$	equal counter-propagating amplitudes make a standing wave: an energy eigenstate that is no longer a momentum eigenstate
$E<0$	no admissible solution	real exponentials blow up at one infinity or the other; the free spectrum has a hard floor at zero
box normalization, finite $L$	$k_n=2\pi n/L$ , spacing $\Delta k=2\pi/L$	the continuum discretized for counting; $L/2\pi$ states per unit $k$ is the density-of-states seed
$L\to\infty$	$\sum_n\to\frac{L}{2\pi}\int\mathrm{d}k$	box results converge to delta-normalized continuum results; the fictitious box drops out of physical answers
$\hbar k$ large (short $\lambda$ )	$\lambda=h/p\to$ atomic scales	matter-wave optics: electrons at 100 eV have $\lambda\approx1.2$ Å and diffract off crystal planes

Common trapA plane wave is a physical state you could prepare in the lab

It is not, and treating it as one produces nonsense — infinite normalization integrals, particles "equally likely to be found anywhere in the universe," and paradoxes about crests moving at half speed. The plane wave is a basis element: an idealized limit ( $\sigma_p\to0$ , $\sigma_x\to\infty$ ) that no preparation can reach, exactly as no classical experiment prepares a perfectly monochromatic infinite wavetrain. Every particle ever detected was a wave packet — a normalizable superposition with finite spreads in both $x$ and $p$ . Compute with plane waves, but when a question asks about a particle — where it is, when it arrives, how it spreads — answer with the packet built from them in Wave Packets and Their Construction. And never write $\int|Ae^{ikx}|^2\mathrm{d}x=1$ : no $A$ makes that true.

Worked examples

Example 1 — the thermal neutron's wavelength (confidence builder). A neutron ( $m_n=1.675\times10^{-27}$ kg) leaves a reactor moderator with the thermal kinetic energy $E=0.025$ eV. Find its de Broglie wavelength, its wavenumber, and the speed its plane-wave components would need to reproduce classically.

Step 1. Momentum from energy: $p=\sqrt{2m_nE}=\sqrt{2\times(1.675\times10^{-27})\times(0.025\times1.602\times10^{-19})}=3.66\times10^{-24}\ \mathrm{kg\,m/s}$ .

Step 2. Wavelength: $\lambda=h/p=(6.626\times10^{-34})/(3.66\times10^{-24})=\boxed{1.81\ \text{Å}}$ — almost exactly an atomic spacing in a crystal, which is why thermal neutrons are a premier diffraction probe of solids.

Step 3. Wavenumber: $k=2\pi/\lambda=3.47\times10^{10}\ \mathrm{m^{-1}}$ , so the eigenvalue bookkeeping reads $p=\hbar k$ : $(1.055\times10^{-34})(3.47\times10^{10})=3.66\times10^{-24}\ \mathrm{kg\,m/s}$ — consistent.

Step 4. Classical speed: $v=p/m_n=(3.66\times10^{-24})/(1.675\times10^{-27})=\boxed{2.19\ \mathrm{km/s}}$ — the famous "2200 m/s" of reactor physics. Note the crest speed of a single component would be $\hbar k/2m=1.09$ km/s: half. The envelope, not the crest, carries the neutron — the accounting is settled in Group Velocity versus Phase Velocity.

Example 2 — is the continuum really continuous? (exam level). A 1 eV electron flies inside an apparatus of length $L=1$ cm. Using box normalization, find the spacing between adjacent allowed energies and decide whether treating the spectrum as continuous is justified. Then count the states with energy below 1 eV.

Step 1. Wavenumber at 1 eV: $k=\sqrt{2m_eE}/\hbar=\sqrt{2\times(9.109\times10^{-31})\times(1.602\times10^{-19})}/(1.055\times10^{-34})=5.12\times10^{9}\ \mathrm{m^{-1}}$ .

Step 2. Allowed wavenumbers are $k_n=2\pi n/L$ with spacing $\Delta k=2\pi/L=6.28\times10^{2}\ \mathrm{m^{-1}}$ — seven orders of magnitude below $k$ itself.

Step 3. Energy spacing between neighbours near 1 eV: $\Delta E\approx\frac{\mathrm{d}E}{\mathrm{d}k}\Delta k=\frac{\hbar^2k}{m_e}\,\Delta k=\frac{(1.055\times10^{-34})^2(5.12\times10^9)}{9.109\times10^{-31}}\times628=3.9\times10^{-26}\ \mathrm{J}=\boxed{2.5\times10^{-7}\ \mathrm{eV}}$ .

Step 4. Compare to anything physical — thermal smearing at room temperature is $k_BT\approx0.025$ eV, five orders of magnitude coarser. No experiment in this box could resolve the discreteness: the continuum idealization is not an approximation you apologize for, it is the physics.

Step 5. State count below 1 eV: $n_{\max}=k L/2\pi=(5.12\times10^9)(0.01)/(6.28)=8.2\times10^{6}$ , and both signs of $k$ double it: $\boxed{\approx1.6\times10^{7}\ \text{states}}$ in one centimetre, one dimension, one eV. This counting — states per unit $k$ equals $L/2\pi$ — is the density-of-states backbone of statistical and solid-state physics.

The wave made real

Origins · 1 minMax von Laue and how this was discovered

Max von Laue

1879–1960 · University of Munich

Von Laue is the physicist who first caught a plane wave in the act. In 1912, against the advice of senior colleagues who thought thermal vibrations would wash out any effect, he reasoned that if X-rays were waves of ångström wavelength, the regularly spaced atoms of a crystal should diffract them like a three-dimensional grating — and the Friedrich–Knipping photographs of copper sulfate proved it within weeks, earning him the 1914 Nobel Prize. The experiment established the twin pillars this article rests on: that radiation of definite wavelength behaves as an extended plane wave sampling many lattice sites at once, and that sharp wavelength means sharp momentum. When Davisson and Germer saw the same diffraction spots using electrons in 1927, de Broglie's $\lambda=h/p$ — equation (4) of this page — graduated from hypothesis to laboratory fact, with von Laue's crystal method as the referee. He lived to see matter-wave diffraction become a standard tool of the science he had started.

History

1912

— Max von Laue predicts, and Friedrich and Knipping observe, X-ray diffraction by a crystal: plane waves of ångström wavelength made experimentally real

1924

— Louis de Broglie's thesis proposes

\lambda=h/p

: every free particle carries a wave

1926

— Schrödinger's equation gives the matter wave its dynamics; plane waves emerge as the free stationary states

1927

— Davisson and Germer diffract electrons off nickel; G.P. Thomson through metal foil — the electron plane wave confirmed

1930

— Dirac's Principles formalizes the delta-function normalization that lets non-normalizable states serve as a basis

1999

— Arndt and Zeilinger's group diffracts C₆₀ fullerenes: sixty-atom molecules behave as de Broglie plane waves

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §2.4 (The Free Particle) · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 5 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §1.7, §2.4 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Ch. I and Complement · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 6 Key relations used throughout this article: (i) $E=\hbar^2k^2/2m$ from the TISE; (ii) $v_{\text{group}}=2v_{\text{phase}}=\hbar k/m$ , which follows from $\omega=\hbar k^2/2m$ ; (iii) the Gaussian packet spreading formula $\Delta x(t)=\Delta x(0)\sqrt{1+(\hbar t/2m\Delta x(0)^2)^2}$ ; (iv) the delta-function normalization $\int e^{i(k-k')x}\mathrm{d}x=2\pi\delta(k-k')$ ; (v) $\hbar=1.054572\times10^{-34}\,\mathrm{J\,s}$ .

Builds on: Time-Independent Schrödinger Equation in One Dimension · Bound States versus Scattering States

Leads to: Wave Packets and Their Construction · Group Velocity versus Phase Velocity · Wave Packet Spreading and Dispersion · Delta-Function Potential Well and Bound State

If plane waves are unphysical, why does every textbook and every scattering problem use them?

Because they are the simplest elements that carry the right physics, and every honest answer extracted from them survives the upgrade to packets. A scattering calculation with plane waves computes amplitude ratios ( $r$ , $t$ ) that a broad packet reproduces to arbitrary accuracy — a packet 1000 wavelengths wide is monochromatic to one part in a thousand. The plane wave is like the "point mass" of mechanics: an idealization no lab can produce, indispensable because real systems approach it as closely as you please.

What does the Dirac delta normalization actually buy us, in practice?

Completeness with clean bookkeeping. Because $\langle\psi_k|\psi_{k'}\rangle=\delta(k-k')$ , the expansion $\psi=\int\phi(k)\psi_k\,\mathrm{d}k$ can be inverted by a single integral, $\phi(k)=\langle\psi_k|\psi\rangle$ , exactly as Fourier coefficients are extracted for a discrete basis. Every formula you know for discrete bases — expansion, Parseval, probability interpretation of coefficients — carries over with $\sum\to\int$ and $\delta_{mn}\to\delta(k-k')$ . Without it, the continuum would be formally unusable.

Is the k = 0 state a particle at rest?

It is the closest the formalism comes, and it fails instructively. $\psi=\text{const}$ has zero momentum exactly — and therefore no location whatsoever. A real particle at rest is a packet with $\langle p\rangle=0$ and small but nonzero $\sigma_p$ , which necessarily spreads (see Wave Packet Spreading and Dispersion). Perfect rest, like perfect momentum, is a basis-element fiction.

Why is the spectrum doubly degenerate when bound states in 1D are never degenerate?

The non-degeneracy theorem for one-dimensional spectra assumes normalizable solutions vanishing at both infinities — its Wronskian argument then forces any two same-energy solutions to be proportional. Scattering states break the assumption: $e^{ikx}$ and $e^{-ikx}$ vanish nowhere, the Wronskian argument gets no purchase, and both directions of travel coexist at each $E>0$ . The theorem and its precise hypotheses are the subject of Non-Degeneracy of One-Dimensional Bound States.

Does a plane wave move at all? Its probability density is constant everywhere and forever.

The density $|\Psi|^2$ is static, but the state is not inert: the phase sweeps along at $\omega/k$ , and the state carries momentum $\hbar k$ — a probability current flows even though the density never changes, like a river whose level is steady while water streams past. The current-density machinery that makes "flow without accumulation" precise is built in Probability Current in One Dimension, and it is what turns plane-wave amplitudes into measurable reflection and transmission rates in Transmission and Reflection Coefficients.

Why doesn't the free particle have zero-point energy when the infinite square well does?

Zero-point energy is the price of confinement: squeezing $\sigma_x$ down forces $\sigma_p$ (hence kinetic energy) up, and a wavefunction pinned inside a box of width $a$ cannot have less than $E_1=\pi^2\hbar^2/2ma^2$ . The free particle is never squeezed — $\sigma_x$ may be as large as you like — so the spectrum reaches all the way down to $E=0$ . Watch the well's $E_1$ vanish as $a\to\infty$ : confinement energy dies exactly when confinement does. The full story is told in Zero-Point Energy.