For height $x$ and horizontal position $y$, Fermat principle in a positive index profile $n(x)=1+\beta x$, $\beta >0$, conserves $n(x)y^{\prime }/\sqrt{1+(y^{\prime }{)}^2}$ on a graph $y(x)$. For positive magnitude $c$ of the horizontal optical momentum, integration gives the displayed catenary. Its minimum height is $x_0=(c-1)/\beta$, at $y=y_0$. Thus $c>1$ produces a smooth turning ray wholly above a road at $x=0$. The two graph branches have opposite signs of $dy/dx$; $x(y)$ continues smoothly through their common horizontal tangent. For a signed negative graph constant, use $|c|$ in the prefactor; $c=0$ gives a vertical ray separately.

Let $G=(V,E)$ be a finite undirected simple graph, with $n=|V|$ vertices and $m=|E|$ edges. A cut of a graph assigns each vertex a label zero or one; an edge crosses precisely when its endpoints have different labels. Its size is the number of crossing edges. A maximum cut maximizes that size.

For the analysis, assign independent fair bits $X_v$ to the vertices. For each edge $uv$, the indicator random variable $I_{uv}=1_{\{X_u\ne X_v\}}$ has expected value $1/2$: exactly two of the four equally likely endpoint assignments cross. Thus linearity of expectation gives the random cut lower bound:Linearity of expectation does not require the indicator random variables to be independent. What matters here is the stipulated distribution of the endpoint bits.

Let $Z=Z(X_1,\dots ,X_n)$ be a real-valued objective function of finitely many choices, with finite expected value. Suppose a prefix $P=(X_1=x_1,\dots ,X_i=x_i)$ has positive probability. Write $q(P)=\mathbb{E}[Z\mid P]$. If the next choice is a fair bit, the law of total expectation givesAt least one child is at least this average. Choose a child with the larger value, using a fixed rule for ties. The resulting loop invariant is $q(P)\ge \mathbb{E}Z$. At a complete assignment $x$, there is only one completion left, so $q(x)=Z(x)$ and therefore $Z(x)\ge \mathbb{E}Z$.

For a biased choice or more than two possibilities, use the corresponding weighted average:Some positive-probability child is at least the weighted average. For minimization, choose a child at most the current conditional expectation and reverse the invariant. The averaging identity is the same in both directions.^[ref]

The word "probabilities" in the method's name also covers an indicator random variable for an undesirable outcome: keeping its conditional expectation, which is its conditional probability, strictly below one eventually forces that indicator to be zero. Optimizing an ordinary numerical objective function is the method of conditional expectations formulation used below.

Alongside the 1927 Davisson–Germer experiment, George Paget Thomson developed an independent transmission test. Alexander Reid first used celluloid; Thomson used thin metal films. A narrow electron beam crossed a film about $10^{-6}\mathrm{cm}$ thick, reaching a photographic plate perpendicular to the beam. Rings appeared. Accelerating voltages of $20$–$60\mathrm{kV}$ set the de Broglie wavelengths; metal structures known from X-ray diffraction allowed quantitative comparison. A downstream magnetic field deflected ring-forming electrons like those passing through holes, identifying the carriers as electrons rather than X-rays. Celluloid's uncertain structure limited its quantitative interpretation.^[ref]

The Davisson–Germer experiment uses crystal reflection and angular electric current scans; this transmission camera records a spatial pattern behind the specimen. Both probe interference in electron diffraction, with different specimen preparations and readouts. A developed photographic plate records accumulated detections, not a direct image of the wavefunction.

Why rings? In an ideal film containing many small crystals with uniformly distributed orientations, a family of lattice planes with spacing $d$ can satisfy Bragg's law at every azimuth about the incident beam. First-order elastic scattering then forms a cone of opening half-angle $2\theta$, where $2d\sin \theta =\lambda$. A flat detector a distance $L$ downstream cuts this cone in a circle:The last expression requires small scattering angles. It explains a useful control: hold the specimen and camera distance fixed, increase accelerating voltage, and the ring radii should shrink with the electron accelerating-voltage wavelength. It predicts positions, not intensities: preferred crystal orientations, film thickness and repeated quantum scattering can alter a real pattern.

For a chosen example, take $L=0.200\mathrm{m}$ and $d=0.200\mathrm{nm}$. The exact relativistic electron accelerating-voltage wavelength gives $\lambda =8.589\mathrm{pm}$ at $20\mathrm{kV}$ and $4.176\mathrm{pm}$ at $80\mathrm{kV}$. The corresponding exact ring radii are $8.594\mathrm{mm}$ and $4.176\mathrm{mm}$. These are calculated examples with chosen spacings and camera distance, not Thomson's measurements. The higher voltage lies outside the range quoted from his lecture, deliberately illustrating a prediction rather than reconstructing his data.

For a one-dimensional free particle with $m>0$ and $p>0$, take the positive total energy from the relativistic energy-momentum relation:A fixed plane wave quantum phase obeys $kx-\omega t=\text{constant}$. Its phase velocity is thereforeFor a narrow wave packet centered at $p_0$, expand its dispersion relation as $E(p)\simeq E(p_0)+E^{\prime }(p_0)(p-p_0)$. In the Fourier inversion integral building the wave packet, the envelope then depends on $x-E^{\prime }(p_0)t$. Thus its group velocity isThe superluminal phase velocity follows a repeating phase marker; it does not describe a localized signal or superluminal transport of energy. Higher derivatives of the dispersion relation produce wave packet spreading, so the translated-envelope argument is a narrow-band approximation. For $v=0.6c$, the Lorentz factor is $1.25$, $p=0.75mc$ and $E=1.25m{c}^2$: the group velocity is $0.6c$ while the phase velocity is $5c/3$.

The total-energy convention matters. Define $\psi =e^{im{c}^{2}t/\hslash }\Psi$ to remove the common rest energy from a wavefunction $\Psi$ with total-energy time dependence. ThenThis is exactly the free-particle momentum convention used by the Schrödinger equation here. Removing a common time-dependent global phase changes the reported phase velocity, while leaving the de Broglie wavelength, relative spatial quantum phases and group velocity unchanged. Adding $m{c}^2$ again to a total energy would count the rest energy twice.

In the 1923 note Waves and Quanta, Louis de Broglie proposed a periodic process associated with rest energy, remaining in phase with a wave along the moving particle's path. He distinguished the internal frequency slowed by time dilation from the wave frequency based on total energy.^[ref] The proposal developed into his 1924 thesis.^[ref] In the 1929 lecture The wave nature of the electron, he reconstructed the argument using a Lorentz transformation.^[ref]

Here is that quantum phase argument in modern notation. Set ${\omega }_0=m{c}^2/\hslash$ in the rest frame. A Lorentz transformation to a frame where the particle has velocity $v$ givesAlong $x=vt$, proper time is $\tau =t/\gamma$, soThis is the harmony: evaluating the spatial wave at the moving particle gives the slower clock phase. At fixed $x$, its angular frequency is instead $E/\hslash =\gamma {\omega }_0$. Since $k=p/\hslash$, the de Broglie wavelength follows as $\lambda =2\pi /k=h/p$. The proposed clock was a theoretical construction, not an observed mechanical vibration.

Retain the first $N\ge 2$ number states of a quantum harmonic oscillator. Let ${\Pi }_N$ be the orthogonal projection onto their span and compress $X_N={\Pi }_{N}X{\Pi }_N$, $P_N={\Pi }_{N}P{\Pi }_N$, restricted to this span. The compressed annihilation operator $a_N$ satisfiesFor $n<N-1$, the two products give $n+1$ and $n$; on the top state they give $0$ and $N-1$. Thus, with $\ell =\sqrt{\hslash /(m\omega )}$,On every quantum superposition of $|0⟩,\dots ,|N-2⟩$, this commutator acts exactly as $i\hslash I_N$. For a normalized retained state with coefficients $c_n$,Small probability at the cutoff must therefore be judged relative to $1/N$ for this expectation. For $N=8$, the diagonal of $[X_N,P_N]/(i\hslash )$ is $(1,1,1,1,1,1,1,-7)$, with zero trace.

This is unavoidable for finite matrices: trace of a matrix commutator gives $\mathrm{tr}(XP-PX)=0$, whereas $\mathrm{tr}(i\hslash I_N)=i\hslash N\ne 0$. Compression does not preserve products: ${\Pi }_{N}XP{\Pi }_N$ can use intermediate states outside the retained span, while $X_N{P}_N$ cannot. Accordingly, compressing the full oscillator Hamiltonian operator gives ${\Pi }_{N}H{\Pi }_N=\hslash \omega (a_N^{†}{a}_N+I_N/2)$, but building it from the compressed squares givesThe two prescriptions agree on lower number states and disagree at the boundary. In the infinite oscillator, the canonical commutation relation holds on the common invariant dense subspace of finite number state combinations. Its unbounded position and momentum operators do not admit the finite trace argument; writing the formal identity does not assert that both products act on every vector of the Hilbert space.

In 1926, Erwin Schrödinger constructed arrays from differential operators and complete orthogonal function systems, showing how their products reproduce the array calculus. His paper connected wave mechanics with matrix mechanics; the translation's sections 2–3 develop the product and commutation rules.^[ref]

The modern proof is a change of orthonormal basis. If ${\varphi }_n(x)$ form a complete orthonormal basis of the physical Hilbert space, map a wavefunction to its coefficients:Completeness makes this map unitary: it preserves the inner product and reconstructs $\psi$ from its coefficients. Inserting the complete orthonormal basis proves the matrix multiplication rule. The Schrödinger equation becomesFor a time-independent Hamiltonian operator, the Heisenberg picture transfers the same evolution to the observables, preserving measurement predictions. Unitary equivalence requires matching operator domains as well as formal entries. A finite selection of eigenfunctions gives a compression, not an exact replacement of the full Hilbert space; a continuous spectrum needs its spectral integrals. This equivalence concerns representations of specified operators, and does not remove the ordering question when quantizing an arbitrary classical expression, as Weyl ordering illustrates.

In 1925, Werner Heisenberg proposed replacing classical orbital quantities with arrays of transition amplitudes and frequencies. His paper's opening demanded relations between quantities observable in principle; section 1 built products by summing over an intermediate state and noted their dependence on order. The formulas above express that construction in modern linear operator notation.^[ref]

Pumping and a heated bath prepared Mercury vapor around a hot platinum wire, mesh and collector in the original Franck-Hertz experiment. A galvanometer read collector electric current while accelerating voltage varied at fixed retarding voltage. Retarding scans distinguished outgoing electron energies.^[ref] The losses constrain excitation of chemical atoms; they do not determine every optical coupling. The Franck-Hertz experiment explains the original mistaken ionization interpretation.

Such excitation and spectral line evidence motivated a mechanics of discrete energy levels and transitions. Replacing one representation by another was a theoretical construction; the Franck-Hertz experiment does not measure a matrix product or establish a unique representation of an observable.

Later in 1925, Max Born and Pascual Jordan developed the array calculus systematically. Their section 3 states $pq-qp=hI/(2\pi i)$, equivalently the modern canonical commutation relationThis relates position and momentum through an intrinsically noncommuting product.^[ref]

For the quantum harmonic oscillator, the canonical commutation relation and creation and annihilation operators give $H=\hslash \omega (a^{†}a+I/2)$, recovering both the spacing and zero-point energy algebraically. One must retain infinitely many number states for the exact relation: the finite cutoff below reveals the obstruction explicitly.

For two dimensionless observables on a two-level system, takeBoth are Hermitian matrices: $A$ has eigenvalues $1,0$ and $B$ has eigenvalues $1,-1$. YetFor instance, $AB|0⟩=0$ whereas $BA|0⟩=|1⟩$. In $AB$, $B$ acts first. Neither product is self-adjoint, so the product of two observables is not automatically another observable. The commutator records this dependence on order. These operator products by themselves do not specify the outcomes of consecutive measurements; that additionally requires measurement in quantum mechanics.