A step-by-step derivation · two strands
We start from the most ordinary statement in all of physics — kinetic energy plus potential energy equals total energy — and watch it turn, by a single mechanical substitution, into both the time-dependent and time-independent Schrödinger equations. Every step comes in two strands: an intuitive one you can hold in your head, and a rigorous one where no equation is skipped and every symbol earns its place.
The whole idea, in one sentence
The Schrödinger equation is just the classical bookkeeping \(E = \tfrac{p^2}{2m} + V\) — except we let it act on a wave, and replace "multiply by energy" with "differentiate in time" and "multiply by momentum" with "differentiate in space."
One honest caveat before we start. You cannot logically deduce quantum mechanics from classical mechanics — if you could, it would not have taken the twentieth century to find it. What follows is the path of plausibility: we take the classical skeleton, add exactly one new physical input (that matter has a wavelength), and discover that there is essentially only one equation those two ideas can fit into. The "derivation" is really a guided guess that turns out to be right.
How to read the page: each step opens with the intuitive strand — the picture, in plain words. Directly below it sits the rigorous strand — the same claim with every equation written out, nothing waved through. Read the intuition first, play with the widget, then check the rigor; or read both in parallel. They always assert the same thing.
Step 1 · the classical starting point
Strip classical mechanics down and you are left with one quantity that never changes: the total energy. For a single particle it is split into two accounts — motion (kinetic) and position (potential). As the particle rolls through a valley the two accounts trade back and forth, but the sum holds fixed. That fixed total, and how it splits, is the only thing we will carry across into the quantum world.
Drag the particle through the well below. The total \(E\) is a fixed line; the kinetic and potential bars trade off to always add up to it. Where the well is deepest, the particle moves fastest; at the turning points the kinetic account hits zero and the particle (classically) cannot go further.
A particle of mass \(m\) at position \(x(t)\) with velocity \(v=\dot x\) carries momentum \(p = mv\), kinetic energy \(T\), and potential energy \(V(x)\). The kinetic energy in terms of momentum:
\[ T \;=\; \tfrac12 m v^2 \;=\; \tfrac12 m \left(\frac{p}{m}\right)^{\!2} \;=\; \frac{p^2}{2m}. \]
The total energy is the Hamiltonian, a function of position and momentum:
\[ H(x,p) \;=\; \underbrace{\frac{p^2}{2m}}_{\text{kinetic}} \;+\; \underbrace{V(x)}_{\text{potential}} \;=\; E. \]
Why is \(E\) constant? Differentiate along the trajectory and use Newton's second law \(\dot p = F = -\partial_x V\):
\[ \frac{dE}{dt} \;=\; \frac{p\,\dot p}{m} + \frac{\partial V}{\partial x}\,\dot x \;=\; \dot x\,\dot p + \frac{\partial V}{\partial x}\,\dot x \;=\; \dot x\!\left(-\frac{\partial V}{\partial x}\right) + \frac{\partial V}{\partial x}\,\dot x \;=\; 0, \]
where we used \(p/m = \dot x\). Equivalently, Newton's law is repackaged as Hamilton's equations — the Hamiltonian generates the motion:
\[ \dot x = \frac{\partial H}{\partial p} = \frac{p}{m}, \qquad \dot p = -\frac{\partial H}{\partial x} = -\frac{\partial V}{\partial x}. \]
For the widget's well \(V(x)=\tfrac12 x^2\), the classically allowed region is where \(T = E - V(x) \ge 0\), i.e. \(|x| \le x_{\text{turn}} = \sqrt{2E}\). At \(E=0.8\) that gives \(x_{\text{turn}} = \sqrt{1.6} \approx 1.26\) — check it against the widget.
Step 2 · the symmetry idea (still classical)
We are still entirely in classical physics — no waves yet. Before we can promote energy and momentum into a wave theory, we need to know what they fundamentally are, and classical mechanics already has the deep answer through Noether's theorem: every continuous symmetry owns a conserved quantity. Momentum is defined as the quantity belonging to the symmetry of sliding space sideways — conserved precisely when the physics does not care where you are. Energy is the quantity belonging to sliding in time — conserved when it does not care when.
The concrete machine behind "slide sideways" turns out to be the humble derivative: an infinitesimal slide of a function is a derivative, and a finite slide is derivatives stacked up — an exponential of the derivative. The familiar \(p = mv\) is merely the value momentum takes on a moving particle; "generator of slides" is what momentum is. Hold onto one more fact: the only shapes a slide cannot reshape — only re-phase — are the pure waves \(e^{ikx}\). Those are exactly the plane waves we meet next. Walk the slide machine below.
The generator of translations. Shift a function by \(a\): \(f(x) \to f(x-a)\). Taylor-expand around \(x\):
\[ f(x-a) \;=\; \sum_{n=0}^{\infty} \frac{(-a)^n}{n!}\,\frac{\partial^n f}{\partial x^n} \;=\; f(x) - a\,\partial_x f + \frac{a^2}{2}\,\partial_x^2 f - \cdots \;=\; e^{-a\,\partial_x} f(x). \]
Here \(e^{-a\,\partial_x}\) is shorthand for the operator power series \(\sum_n \tfrac{(-a\,\partial_x)^n}{n!}\) — the same construction as the matrix exponential \(e^{At}\). The derivative \(\partial_x\) is therefore the generator of translations: every term of a finite slide is built from it.
Eigenfunctions of the slide. Feed the pure wave \(e^{ikx}\) to the slide operator, term by term:
\[ e^{-a\,\partial_x}\, e^{ikx} \;=\; \sum_{n=0}^{\infty} \frac{(-a)^n}{n!}\,\underbrace{\partial_x^n\, e^{ikx}}_{=\,(ik)^n e^{ikx}} \;=\; \left(\sum_{n=0}^{\infty} \frac{(-ika)^n}{n!}\right) e^{ikx} \;=\; e^{-ika}\, e^{ikx}. \]
The slide returns the same wave multiplied by the constant phase \(e^{-ika}\): plane waves are the eigenfunctions of translation, with eigenvalue \(e^{-ika}\). No other shape survives a slide unchanged.
Noether, in one line. If the Hamiltonian is translation-invariant — \(H(x+a,p) = H(x,p)\) for all \(a\), i.e. \(\partial_x H = -\,\dot p = 0\) — then momentum is conserved. Switch on a potential and the symmetry breaks by exactly the force law:
\[ \text{no } V(x): \;\; \dot p = -\partial_x V = 0 \;\;(\text{conserved}); \qquad \text{with } V(x): \;\; \dot p = -\partial_x V \neq 0. \]
Energy plays the identical role for time: if \(H\) has no explicit \(t\)-dependence, \(\tfrac{dE}{dt} = \partial_t H = 0\) — time-translation symmetry \(\Rightarrow\) energy conserved. The dictionary "space-slide \(\leftrightarrow\) momentum, time-slide \(\leftrightarrow\) energy" is the Ansatz we carry into the quantum theory unchanged.
Notice none of this used a wavefunction, \(\hbar\), or anything quantum — it is just classical symmetry. The quantum theory will change exactly one thing: what the generator acts on. Once the particle becomes a complex amplitude (next step), this same translation generator, acting on it, becomes the momentum operator — and that is the move that finally produces \(p=\hbar k\).
Step 3 · the one new ingredient
Here is the single physical input that classical mechanics does not have. De Broglie's proposal: a particle is not a point — it is a wave. The simplest travelling wave is the plane wave: a corkscrew of phase gliding through space. Two knobs control it — how tightly wound it is in space (the wavenumber \(k\)) and how fast it spins in time (the frequency \(\omega\)). For now these are purely wave-geometric labels; they earn their physical names — momentum and energy — in Step 5.
To see it: freeze one location and the wave is just a point bobbing up and down at rate \(\omega\); its position along \(x\) only sets where in its cycle the bob starts. Stitch all locations together and the staggered starting phases make the pattern appear to travel. Set \(k=0\) and every point shares the same phase — the whole line bobs in unison. Build it step by step below.
The plane wave is the complex exponential
\[ \psi(x,t) \;=\; e^{\,i(kx - \omega t)} \;=\; \cos(kx-\omega t) \;+\; i\,\sin(kx-\omega t), \]
by Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\). The symbols: \(k = 2\pi/\lambda\) is the wavenumber (radians of phase per metre, \(\lambda\) the wavelength), and \(\omega = 2\pi/T_{\text{per}}\) is the angular frequency (radians of phase per second, \(T_{\text{per}}\) the period).
Why it "travels": the phase velocity. Follow one crest — a point of constant phase \(\phi_0\). Constant phase means
\[ kx - \omega t = \phi_0 \;=\; \text{const} \quad\Longrightarrow\quad x(t) = \frac{\phi_0}{k} + \frac{\omega}{k}\,t \quad\Longrightarrow\quad v_p \;=\; \frac{dx}{dt} \;=\; \frac{\omega}{k}. \]
The pattern moves rigidly at the phase velocity \(v_p = \omega/k\). At a fixed location \(x_0\), \(\psi(x_0,t) = e^{ikx_0}\,e^{-i\omega t}\): a unit arrow rotating clockwise in the complex plane at rate \(\omega\), with fixed phase offset \(kx_0\).
Why this wave and not another. The plane wave is exactly the translation eigenfunction of Step 2: \(e^{-a\partial_x} e^{ikx} = e^{-ika}\,e^{ikx}\). A slide cannot reshape it, only re-phase it. If momentum is to be "the handle translations grab," the state of definite momentum must be the state translations cannot deform — the plane wave is forced on us.
Step 4 · reading the wave
We can now draw the plane wave — but what does it mean? Two features demand interpretation. First, \(\psi\) is complex — a rotating arrow, real and imaginary parts a quarter-cycle apart. That hidden phase is exactly what will encode momentum and energy once we apply the translation generator in Step 5. Second, it is spread out over all of space — so where is the particle? Born's rule answers: the squared length of the arrow at each point is the probability of finding the particle there.
A plane wave's arrow has the same length everywhere, so its probability is flat — the particle is equally likely to be anywhere. That is not a bug: it is the price of a perfectly definite wavelength. Watch the point particle dissolve into a complex amplitude below.
The state of the particle is a complex-valued function \(\psi(x,t)\), the wavefunction. Born's rule: the probability density for finding the particle at \(x\) is the squared modulus, and total probability is one:
\[ \rho(x,t) \;=\; |\psi(x,t)|^2 \;=\; \psi^*(x,t)\,\psi(x,t), \qquad \int_{-\infty}^{\infty} |\psi(x,t)|^2\,dx \;=\; 1, \]
where \(\psi^* \) is the complex conjugate (\(i \to -i\)). For the plane wave the density is computed in one line:
\[ |\psi|^2 \;=\; \left(e^{i(kx-\omega t)}\right)^{*} e^{i(kx-\omega t)} \;=\; e^{-i(kx-\omega t)}\, e^{i(kx-\omega t)} \;=\; e^{0} \;=\; 1 \quad \text{everywhere}. \]
Flat, and therefore not normalizable — \(\int 1\,dx = \infty\). A lone plane wave is an idealized limit, not a physical state; the fix (superposition) is Step 6.
Why amplitudes and not probabilities. Amplitudes add before they are squared. For two overlapping waves:
\[ |\psi_1 + \psi_2|^2 \;=\; (\psi_1+\psi_2)^*(\psi_1+\psi_2) \;=\; |\psi_1|^2 + |\psi_2|^2 + \underbrace{2\,\mathrm{Re}\!\left(\psi_1^*\psi_2\right)}_{\text{interference}}. \]
The cross term \(2\,\mathrm{Re}(\psi_1^*\psi_2)\) can be negative — two possibilities can cancel. Classical probabilities can never do that; this single term is the entire difference between quantum and classical randomness.
Step 5 · apply the operators
Now the two threads meet. Step 2 found the generator of translations — the bare derivative \(\partial_x\). Steps 3–4 supplied the thing for it to act on: the complex wavefunction. Dress the generator with two constants and it becomes the momentum operator; its time-twin becomes the energy operator. Feed the plane wave to either one and something remarkable happens: the operator hands back the same wave, just multiplied by a number. That number is the momentum (\(\hbar k\)) or the energy (\(\hbar\omega\)) of the wave.
So the de Broglie relation \(p = \hbar k\) — "more wiggles per metre means more momentum" — is not an extra postulate. It is what the symmetry definition of momentum outputs when you point it at a wave. The widget makes this visible: the operator only rescales the wave, and the scale factor is exactly \(p\). Same wave, louder.
Define the momentum operator and the energy operator as the dressed generators of space- and time-translation:
\[ \hat p \;=\; -i\hbar\,\frac{\partial}{\partial x}, \qquad \hat E \;=\; +\,i\hbar\,\frac{\partial}{\partial t}. \]
Each factor earns its keep. The \(\hbar\) (Planck's constant over \(2\pi\), units \(\mathrm{J\,s}\)) converts "radians of phase per metre" into \(\mathrm{kg\,m\,s^{-1}}\) — it is the single empirically measured constant in the whole construction. The \(-i\) makes \(\hat p\) Hermitian, so its eigenvalues are real, measurable numbers: integrating by parts (boundary terms vanish for normalizable \(\psi\)),
\[ \int \varphi^*\,(-i\hbar\,\partial_x\psi)\,dx \;=\; \Big[-i\hbar\,\varphi^*\psi\Big]_{-\infty}^{\infty} + \int (i\hbar\,\partial_x\varphi^*)\,\psi\,dx \;=\; \int \big(-i\hbar\,\partial_x\varphi\big)^{*}\,\psi\,dx . \]
The \(i\) flips sign under conjugation exactly when the minus sign from integration by parts appears — the two cancel, which is the Hermiticity condition \(\langle\varphi,\hat p\,\psi\rangle = \langle\hat p\,\varphi,\psi\rangle\). Without the \(i\), the bare \(\hbar\,\partial_x\) would be anti-Hermitian, with imaginary eigenvalues.
The eigenvalue computation, in full. Differentiate the plane wave in space:
\[ \frac{\partial}{\partial x}\, e^{i(kx-\omega t)} \;=\; ik\, e^{i(kx-\omega t)} \;=\; ik\,\psi \quad\Longrightarrow\quad \hat p\,\psi \;=\; -i\hbar\,(ik)\,\psi \;=\; -\,i^2\,\hbar k\,\psi \;=\; \hbar k\,\psi, \]
using \(i^2 = -1\). And in time:
\[ \frac{\partial}{\partial t}\, e^{i(kx-\omega t)} \;=\; -i\omega\,\psi \quad\Longrightarrow\quad \hat E\,\psi \;=\; i\hbar\,(-i\omega)\,\psi \;=\; \hbar\omega\,\psi. \]
The plane wave is a simultaneous eigenfunction of both operators, with eigenvalues
\[ p \;=\; \hbar k, \qquad E \;=\; \hbar\omega \]
— the de Broglie and Planck–Einstein relations, arriving as outputs of the symmetry definition rather than postulates. (Honesty note: symmetry fixes the structure — \(p \propto k\), \(E \propto \omega\). That the proportionality constant is the same \(\hbar\) for all matter is the empirical content, confirmed by electron diffraction.)
The substitution is less arbitrary than it looks. Write a wavefunction in polar form \(\psi = A\,e^{iS/\hbar}\) and feed it into the equation we are about to build. Expanding in powers of \(\hbar\), the leading term is \(\;\partial_t S + \tfrac{1}{2m}(\partial_x S)^2 + V = 0\) — the Hamilton–Jacobi equation of classical mechanics, with \(S\) the classical action and \(p=\partial_x S\). So the phase of the quantum wave is the classical action divided by \(\hbar\), and classical mechanics is the \(\hbar\to 0\) limit (geometric optics) of a wave theory. Schrödinger found his equation precisely by running Hamilton's optical–mechanical analogy in reverse: if mechanics is the ray-limit of some wave equation, what wave equation is it the limit of?
Each continuous symmetry has a generator, and that generator is the conserved observable: space translations \(\to \hat p=-i\hbar\partial_x\) (momentum); time translations \(\to \hat H\) via \(U(t)=e^{-i\hat Ht/\hbar}\) (energy); rotations \(\to \hat L\) (angular momentum). In each case "symmetry" means the generator commutes with \(\hat H\), and Heisenberg evolution \(\tfrac{d}{dt}\langle\hat A\rangle = \tfrac{i}{\hbar}\langle[\hat H,\hat A]\rangle\) turns commuting into conserved. The canonical pair structure \([\hat x,\hat p]=i\hbar\) is the same statement read sideways: \(\hat p\) generates shifts of \(x\), and \(\hat x\) generates shifts of \(p\). Ehrenfest's theorem then closes the loop with classical mechanics: \(\tfrac{d}{dt}\langle x\rangle = \langle\hat p\rangle/m\) and \(\tfrac{d}{dt}\langle\hat p\rangle = -\langle\partial_x V\rangle\) — mass times velocity, and Newton's second law, recovered as statements about the packet's averages. Together with the group velocity \(v_g=\hbar k_0/m\) of Step 6, every classical role of momentum is accounted for.
One worry before we assemble anything: an infinitely-extended plane wave cannot be a real, localized electron. That is exactly the puzzle the next step resolves.
Step 6 · resolving the puzzle
A fair objection to Steps 3–5: a plane wave stretches from \(-\infty\) to \(+\infty\) with the same brightness everywhere — not localized at all. Real electrons are somewhere. The key realization: the plane wave was never the electron — it is a building block. It is the state of one perfectly definite momentum, and "equally likely to be anywhere" is exactly the right answer for that. A real electron carries a spread of momenta, so we stack many plane waves on top of each other.
Then interference does the work: near one point the components line up in phase and reinforce; far away their phases fan out and cancel. The result is a localized bump — a wave packet — built entirely from non-localized pieces, exactly as a sharp drum hit is a superposition of pure tones. Build it below: start with one wave (flat — the puzzle), add momenta band by band, and watch the packet sharpen, drift, and slowly spread.
Superposition. Weight each plane wave \(e^{ikx}\) by a complex amplitude \(A(k)\) and add them all:
\[ \psi(x,0) \;=\; \int_{-\infty}^{\infty} A(k)\, e^{ikx}\, dk. \]
This is an inverse Fourier transform; \(A(k)\) is the wavefunction in momentum space and \(|A(k)|^2\) the probability density over momentum (Step 7 makes this exact).
The Gaussian packet, computed in full. Choose a Gaussian band of momenta centred at \(k_0\) with width \(\sigma_k\): \(A(k) = e^{-(k-k_0)^2/2\sigma_k^2}\). Substitute \(u = k-k_0\):
\[ \psi(x,0) \;=\; \int e^{-u^2/2\sigma_k^2}\, e^{i(k_0+u)x}\, du \;=\; e^{ik_0x} \int e^{-u^2/2\sigma_k^2 + iux}\, du. \]
Complete the square in the exponent: \(-\tfrac{u^2}{2\sigma_k^2} + iux = -\tfrac{1}{2\sigma_k^2}\big(u - i\sigma_k^2 x\big)^2 - \tfrac{\sigma_k^2 x^2}{2}\). The shifted Gaussian integrates to \(\sqrt{2\pi}\,\sigma_k\) regardless of the shift, so
\[ \psi(x,0) \;=\; \sqrt{2\pi}\,\sigma_k\; e^{ik_0x}\; e^{-\sigma_k^2 x^2/2}. \]
A Gaussian band of momenta gives a Gaussian bump in space: carrier wave \(e^{ik_0x}\) times envelope \(e^{-\sigma_k^2x^2/2}\). The uncertainty product falls out. The position density \(|\psi|^2 \propto e^{-\sigma_k^2 x^2}\) has standard deviation \(\Delta x = 1/(\sqrt2\,\sigma_k)\); the momentum density \(|A(k)|^2 \propto e^{-(k-k_0)^2/\sigma_k^2}\) has \(\Delta k = \sigma_k/\sqrt2\). Multiply:
\[ \Delta x\,\Delta k \;=\; \frac{1}{\sqrt2\,\sigma_k}\cdot\frac{\sigma_k}{\sqrt2} \;=\; \frac12 \qquad\Longrightarrow\qquad \Delta x\,\Delta p \;=\; \hbar\,\Delta x\,\Delta k \;=\; \frac{\hbar}{2}, \]
the minimum the uncertainty principle allows — Gaussians saturate \(\Delta x\,\Delta p \ge \hbar/2\). Narrow band \(\Rightarrow\) wide packet and vice versa; the lone plane wave is the limit \(\Delta k \to 0\), \(\Delta x \to \infty\).
Group velocity. Each component evolves as \(e^{i(kx - \omega(k)t)}\). Add two neighbouring components \(k_0 \pm \tfrac{\Delta k}{2}\) (real parts, using the sum-to-product identity \(\cos\alpha + \cos\beta = 2\cos\tfrac{\alpha-\beta}{2}\cos\tfrac{\alpha+\beta}{2}\)):
\[ \cos(k_1x-\omega_1 t) + \cos(k_2x-\omega_2 t) \;=\; 2\,\underbrace{\cos\!\Big(\tfrac{\Delta k}{2}x - \tfrac{\Delta\omega}{2}t\Big)}_{\text{slow envelope}}\; \underbrace{\cos\!\big(k_0 x - \omega_0 t\big)}_{\text{fast carrier}}. \]
The envelope travels at \(\tfrac{\Delta\omega/2}{\Delta k/2} = \tfrac{\Delta\omega}{\Delta k} \to \tfrac{d\omega}{dk}\). With the free dispersion \(\omega(k) = \hbar k^2/2m\) (derived in Step 8),
\[ v_g \;=\; \left.\frac{d\omega}{dk}\right|_{k_0} \;=\; \frac{\hbar k_0}{m} \;=\; \frac{p_0}{m}, \]
exactly the classical velocity \(v = p/m\): the packet's envelope moves like the classical particle. And because \(\omega(k)\) is curved (\(d^2\omega/dk^2 = \hbar/m \neq 0\)), components drift out of step and the packet slowly spreads.
For a free particle \(E=\tfrac{p^2}{2m}=\tfrac{\hbar^2k^2}{2m}\), so \(\omega(k)=\hbar k^2/2m\). Each component runs at phase velocity \(v_p=\omega/k=\hbar k/2m\), but the envelope travels at \(v_g=d\omega/dk=\hbar k/m=2v_p\): the packet outruns its own ripples. Spreading is the curvature \(d^2\omega/dk^2=\hbar/m\neq 0\); a Gaussian of initial width \(\sigma_x(0)\) widens as \(\sigma_x(t)=\sigma_x(0)\sqrt{1+\big(\hbar t/2m\sigma_x(0)^2\big)^2}\), so tightly-localized packets (small \(\sigma_x(0)\)) spread fastest — squeezing position now is paid for by faster blurring later.
Step 7 · the Fourier picture
Everything in Step 6 was one theorem in disguise — one you already know from sound. Position is the signal; momentum is its spectrum. "A localized particle is built from many momenta" is the same statement as "a short pulse contains a broad band of frequencies." A pure tone rings forever — a plane wave fills all of space; a sharp click is broadband — a localized packet needs many \(k\). Squeeze the signal in one domain and it stretches in the other; that trade-off, written with de Broglie's \(\hbar\), is the uncertainty principle.
Watch it assemble below. The top window is a target — a localized electron. The bottom window starts from a single plane wave: one momentum, dead-flat probability. Each press adds one more wavenumber from the spectrum; the partial sum sharpens until it reproduces the target. Localization is paid for one momentum at a time.
The transform pair. With the symmetric convention,
\[ \tilde\psi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \psi(x)\, e^{-ikx}\, dx, \qquad \psi(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \tilde\psi(k)\, e^{ikx}\, dk. \]
Parseval's theorem guarantees the two descriptions carry the same total probability, so \(|\tilde\psi(k)|^2\) is a legitimate density over momentum:
\[ \int |\psi(x)|^2\, dx \;=\; \int |\tilde\psi(k)|^2\, dk \;=\; 1. \]
The scaling law — the trade-off made exact. Stretch the wavefunction by a factor \(s\): \(\psi_s(x) = \tfrac{1}{\sqrt s}\,\psi(x/s)\) (the prefactor keeps it normalized). Substituting \(y = x/s\) in the transform:
\[ \tilde\psi_s(k) \;=\; \frac{1}{\sqrt{2\pi}}\int \frac{1}{\sqrt s}\,\psi(x/s)\,e^{-ikx}\,dx \;=\; \frac{\sqrt s}{\sqrt{2\pi}}\int \psi(y)\,e^{-i(sk)y}\,dy \;=\; \sqrt{s}\;\tilde\psi(sk). \]
Widen in \(x\) by \(s\) \(\Rightarrow\) squeeze in \(k\) by the same \(s\): the product \(\Delta x\,\Delta k\) is scale-invariant. The general lower bound (proved from the Cauchy–Schwarz inequality applied to \(x\psi\) and \(\partial_x\psi\)) is
\[ \Delta x\,\Delta k \;\ge\; \tfrac12 \qquad\Longrightarrow\qquad \Delta x\,\Delta p \;\ge\; \frac{\hbar}{2}, \]
with equality exactly for Gaussians — the case computed in Step 6. Heisenberg's principle is Fourier analysis wearing \(p = \hbar k\).
A plane wave \(e^{i(kx-\omega t)}\) carries two labels, but for a free particle they are chained by the dispersion relation: combining \(E=\tfrac{\hbar^2k^2}{2m}\) with \(E=\hbar\omega\) gives \(\omega(k)=\hbar k^2/2m\). Once \(k\) is chosen, \(\omega\) is fixed — so we superpose over \(k\) only, each component dragging its own \(\omega(k)\) along. That is why \(k\) carries the whole conversation: \(k\) owns the statics (the shape, the position–momentum Fourier pair), while \(\omega\) owns the dynamics — the phase \(e^{-i\omega(k)t}\) makes the packet move at the group velocity \(v_g=d\omega/dk=\hbar k_0/m\), and the curvature of \(\omega(k)\) is what makes it spread. An electron in vacuum is simply a dispersive medium for matter waves. There is still a genuine \(t\leftrightarrow\omega\) pair — it gives the energy–time relation \(\Delta E\,\Delta t\ge\hbar/2\), the natural linewidth of a state that lives only for a finite time — but that is a statement about how long the state lasts, not where the particle is, which is why it stayed off-stage while we built the packet.
Step 8 · assemble it
We now have everything. Take the classical energy ledger from Step 1, demand that it hold as an operator equation acting on \(\psi\), and substitute the operators from Step 5. Walk the build below one substitution at a time and watch Newton's energy statement turn into Schrödinger's equation. Read the result back in plain words and it is still Step 1: the rate at which the wave's phase turns (its energy) equals its kinetic part plus its potential part. Same ledger; the accounts are now derivatives.
The substitution, every line. Start from the ledger and let it act on \(\psi\):
\[ E = \frac{p^2}{2m} + V(x) \quad\longrightarrow\quad E\,\psi = \frac{p^2}{2m}\,\psi + V(x)\,\psi. \]
Replace \(E \to \hat E = i\hbar\,\partial_t\) and \(p \to \hat p = -i\hbar\,\partial_x\). The only real algebra is squaring the momentum operator — apply it twice and track the constants:
\[ \hat p^{\,2}\psi \;=\; \left(-i\hbar\,\partial_x\right)\!\left(-i\hbar\,\partial_x \psi\right) \;=\; (-i\hbar)^2\, \partial_x^2 \psi \;=\; i^2 \hbar^2\, \partial_x^2\psi \;=\; -\hbar^2\, \frac{\partial^2 \psi}{\partial x^2}. \]
Substituting both operators gives the time-dependent Schrödinger equation (TDSE):
\[ i\hbar\,\frac{\partial \psi}{\partial t} \;=\; -\frac{\hbar^2}{2m}\,\frac{\partial^2 \psi}{\partial x^2} + V(x)\,\psi \;=\; \hat H\,\psi, \qquad \hat H \;=\; \frac{\hat p^{\,2}}{2m} + V(x). \]
Consistency check — the free plane wave. Set \(V=0\) and insert \(\psi = e^{i(kx-\omega t)}\). Left side: \(i\hbar(-i\omega)\psi = \hbar\omega\,\psi\). Right side: \(-\tfrac{\hbar^2}{2m}(ik)^2\psi = \tfrac{\hbar^2k^2}{2m}\psi\). Equating:
\[ \hbar\omega \;=\; \frac{\hbar^2 k^2}{2m} \qquad\Longleftrightarrow\qquad E \;=\; \frac{p^2}{2m}. \]
The equation admits a plane wave precisely when the wave's labels satisfy the classical ledger — the dispersion relation \(\omega(k) = \hbar k^2/2m\) is Newton's energy bookkeeping, wearing wave clothes. This closes the loop: the TDSE is the unique linear wave equation whose dispersion relation reproduces \(E = p^2/2m + V\).
Probability is conserved. Differentiate the density and substitute the TDSE (\(\partial_t\psi = \tfrac{1}{i\hbar}\hat H\psi\)) and its conjugate:
\[ \frac{\partial}{\partial t}|\psi|^2 = \psi^*\,\partial_t\psi + \psi\,\partial_t\psi^* = \frac{1}{i\hbar}\Big(\psi^*\hat H\psi - \psi\,(\hat H\psi)^*\Big) = -\,\frac{\partial}{\partial x}\underbrace{\left[\frac{\hbar}{2mi}\Big(\psi^*\partial_x\psi - \psi\,\partial_x\psi^*\Big)\right]}_{\text{probability current } j(x,t)} \]
(the \(V\psi\) terms cancel because \(V\) is real; the kinetic terms combine into a total derivative). This is a continuity equation \(\partial_t\rho + \partial_x j = 0\): probability is never created or destroyed, only moved — the wave analogue of "the total in the ledger never changes."
The asymmetry is forced. The energy operator \(\hat E = i\hbar\partial_t\) is first order because energy appears linearly in \(H\). The kinetic term carries \(p^2\), and \(\hat p^2 = -\hbar^2\partial_x^2\) is second order. This single-time-derivative structure is why the wavefunction must be complex: a real first-order-in-time wave equation would just decay or grow, but \(i\) makes \(\partial_t\psi\) a rotation, so \(|\psi|^2\) is conserved and probability is preserved. Solving \(i\hbar\,\partial_t\psi=\hat H\psi\) gives \(\psi(t)=e^{-i\hat H t/\hbar}\psi(0)\): a unitary (norm-preserving) flow generated by \(\hat H\).
Step 9 · freeze the time
The TDSE describes everything, but it is a partial differential equation in two variables. When the potential does not change in time, there is a clean shortcut: look for solutions whose shape is frozen and whose only motion is a spinning phase. For such states the probability cloud \(|\psi|^2\) never moves at all — hence stationary state — and the TDSE collapses into a pure eigenvalue problem: find the shapes the Hamiltonian merely rescales, and the rescaling factors are the allowed energies. That is where quantized levels come from. Pick a level \(n\) below and play: the real part oscillates, but the cloud sits perfectly still.
Separation of variables, every line. Try a product solution \(\psi(x,t) = \varphi(x)\,T(t)\) in the TDSE:
\[ i\hbar\,\varphi(x)\,\frac{dT}{dt} \;=\; T(t)\left[-\frac{\hbar^2}{2m}\frac{d^2\varphi}{dx^2} + V(x)\,\varphi(x)\right]. \]
Divide both sides by \(\varphi(x)T(t)\):
\[ \underbrace{\,i\hbar\,\frac{1}{T}\frac{dT}{dt}\,}_{\text{depends only on } t} \;=\; \underbrace{\frac{1}{\varphi}\left[-\frac{\hbar^2}{2m}\frac{d^2\varphi}{dx^2} + V\varphi\right]}_{\text{depends only on } x}. \]
A function of \(t\) alone equals a function of \(x\) alone for all \(x,t\) only if both are one constant — call it \(E\) (it has units of energy). The time equation is solved immediately:
\[ i\hbar\,\frac{dT}{dt} = E\,T \quad\Longrightarrow\quad \frac{dT}{T} = -\frac{iE}{\hbar}\,dt \quad\Longrightarrow\quad T(t) = e^{-iEt/\hbar}, \]
a pure phase rotation at rate \(E/\hbar\) — precisely \(E = \hbar\omega\) again. The space equation is the time-independent Schrödinger equation (TISE), an eigenvalue problem for \(\hat H\):
\[ -\frac{\hbar^2}{2m}\frac{d^2\varphi}{dx^2} + V(x)\,\varphi \;=\; E\,\varphi, \qquad\text{i.e.}\qquad \hat H\,\varphi = E\,\varphi. \]
The full solution \(\psi(x,t) = \varphi(x)\,e^{-iEt/\hbar}\) has \(|\psi|^2 = |\varphi|^2\,|e^{-iEt/\hbar}|^2 = |\varphi|^2\): the density is frozen.
Worked example — the particle in a box (the widget below; \(V=0\) inside \([0,L]\), infinite walls, so \(\varphi(0)=\varphi(L)=0\)). Inside the box the TISE reads
\[ \frac{d^2\varphi}{dx^2} = -\frac{2mE}{\hbar^2}\,\varphi \;\equiv\; -k^2\varphi \quad\Longrightarrow\quad \varphi(x) = A\sin(kx) + B\cos(kx). \]
The boundary conditions do the quantizing: \(\varphi(0)=0\) kills the cosine (\(B=0\)); \(\varphi(L)=0\) forces \(\sin(kL)=0\), i.e. \(kL = n\pi\) for integer \(n\ge1\). Only a discrete ladder of wavenumbers fits in the box, and with \(E = \hbar^2k^2/2m\):
\[ k_n = \frac{n\pi}{L}, \qquad E_n = \frac{\hbar^2 \pi^2}{2mL^2}\,n^2, \qquad \varphi_n(x) = \sqrt{\tfrac{2}{L}}\,\sin\!\Big(\frac{n\pi x}{L}\Big), \]
where \(\sqrt{2/L}\) comes from normalization \(\int_0^L \sin^2(n\pi x/L)\,dx = L/2\). With \(\hbar=m=L=1\) (the widget's units): \(E_n = n^2\pi^2/2\), so \(E_1 = \pi^2/2 \approx 4.93\) and \(E_2 = 2\pi^2 \approx 19.74\) — check them against the readout. Quantization is not mysterious: it is the same reason a guitar string has discrete harmonics — only whole half-waves fit between the clamps.
We never invented anything but one idea. Classical mechanics handed us the ledger \(E=\tfrac{p^2}{2m}+V\) (Step 1). Still in the classical world, Noether's theorem told us what energy and momentum are — the conserved quantities of time- and space-translation symmetry — with the derivative \(\partial_x\) as the generator of a spatial slide (Step 2). De Broglie then took the quantum leap, replacing the particle with a plane wave \(e^{i(kx-\omega t)}\) — precisely the state a slide cannot reshape (Step 3) — and Born told us how to read it: \(|\psi|^2\) is the probability of finding the particle, flat for a single plane wave (Step 4). Promoting the generator to act on \(\psi\) as \(\hat p=-i\hbar\partial_x\) and feeding it the wave returned \(\hbar k\) and \(\hbar\omega\): the de Broglie dictionary as an output, not a postulate (Step 5). A real, localized electron turned out to be a superposition of those waves — a wave packet whose width trades against its momentum spread, the uncertainty principle (Step 6) — which is just the Fourier-pair fact that localizing in position forces a spread in momentum, the time–bandwidth limit wearing de Broglie's \(\hbar\) (Step 7). Substituting the operators into the ledger gave the time-dependent equation (Step 8), and separating time from space collapsed it into the eigenvalue problem \(\hat H\varphi=E\varphi\) (Step 9).
So the kernel holds exactly as promised: the Schrödinger equation is the classical energy statement, made to act on a wave, with "multiply by energy/momentum" rewritten as "differentiate in time/space." The formidable-looking PDE is that one plain sentence wearing operator clothes.
Where it shows up. Every spectroscopic line is a difference of TISE eigenvalues \(E_n-E_m\). The "particle in a box" you just played with is the quantum-dot model behind some LED colors. And the unitary flow \(e^{-i\hat H t/\hbar}\) from Step 8 is literally the gate model of quantum computing: choose \(\hat H\), let it run, and you have computed.