For a two-dimensional electron gas in a perpendicular magnetic field of magnitude $B>0$, the filling factor is the electron number per unit area $n_s$ divided by the orbital degeneracy per unit area $eB/h$ of one spin branch of a Landau level. Here $e$ is the positive elementary charge and $h$ the Planck constant. The nominal filling factor varies continuously with $n_s/B$; it need not be an integer.

For an electron in a uniform perpendicular magnetic field of magnitude $B>0$, the magnetic length sets the orbital length scale of a Landau level; $e$ is the positive elementary charge.

For a semiclassical closed electron orbit in a plane perpendicular to a uniform magnetic field, its wavevector-space area obeys $S_k(E_n)=2\pi eB(n+\gamma )/\hslash$. An ordinary parabolic band without an extra geometric phase has $\gamma =1/2$. The semiclassical equations rotate and scale the real-space orbit by $\hslash /(eB)$, giving $S_r=[\hslash /(eB){]}^2{S}_k$. For a parabolic band this is equivalent to $E_n=\hslash {\omega }_c(n+1/2)$.

The de Haas-van Alphen effect is an oscillation of a metal's magnetization as changing magnetic field moves Landau levels through the Fermi energy. The associated Onsager quantization gives inverse-field period $\Delta (1/B)=2\pi e/(\hslash S_F)$ for an extremal Fermi-surface cross-section $S_F$. Oscillatory resistance is instead the Shubnikov-de Haas effect.

A quantum oscillation is an oscillatory contribution to a metal's response as a magnetic field moves quantized orbital levels past the Fermi energy. For a closed extremal Fermi surface cross-section of wavevector-space area $A_{\mathrm{ext}}$, Onsager quantization gives inverse-field period $\Delta (1/B)=2\pi e/(\hslash A_{\mathrm{ext}})$. Oscillations of magnetization and electrical resistivity are respectively the de Haas-van Alphen effect and Shubnikov-de Haas effect. Low temperature and long quantum lifetime are needed so thermal smearing and disorder do not wash out the Landau levels.

Consider an original illustrative resonance model, with all numerical settings chosen here. On one isolated branch approximate the transition frequency by $\nu (B)={\nu }_0+aB$, with known slope $a=20MHz/mT$ and drive frequency ${\nu }_d=1300\mathrm{MHz}$. Choose ${\nu }_0=1100\mathrm{MHz}$, hence resonance field $B_r=10\mathrm{mT}$. This local linear model omits hyperfine structure and nonlinear level mixing; it is not the historical hydrogen Zeeman splitting calculation.

Use a Lorentzian line shape for a weak-drive current dip,Choose baseline $C=1$, depth $A=0.25$ in arbitrary detector signal units, and full width at half maximum $w_{\nu }=100\mathrm{MHz}$, so $w_B=5\mathrm{mT}$. The half-depth points are $B_r\pm w_B/2$; this checks the factor of four. If an exponential decay with lifetime $\tau$ alone set this width, natural linewidth would give $w_{\nu }=1/(2\pi \tau )$ and $\tau \simeq 1.59\mathrm{ns}$. The chosen width is an explanatory input, not a measurement of a historical lifetime.

Generate 65 equally spaced fields from $2$ to $18\mathrm{mT}$ and add independent Gaussian noise with known standard deviation ${\sigma }_I=0.004$ signal units. A simultaneous nonlinear regression fit of $C,A,B_r,w_B$ gives $B_r\simeq 10.0007\mathrm{mT}$, $w_B\simeq 4.9305\mathrm{mT}$ and ${\nu }_0\simeq 1099.986\mathrm{MHz}$. With residual derivatives $J_{ik}=\partial I(B_i)/\partial {\theta }_k$ and $\theta =(C,A,B_r,w_B)$, the local covariance matrix estimate isIt gives centre standard deviation $0.01427\mathrm{mT}$, or $0.285\mathrm{MHz}$ in the inferred intercept. This uses the stipulated noise variance and a local linear approximation, rather than treating the width as the centre error. Repeated noisy samples can locate a broad line's centre much more closely than its width.

Include a common magnetic calibration offset with standard deviation $u_b=0.020\mathrm{mT}$, a fractional gain error $u_g=0.001$, and independent drive-frequency error $u_d=0.100\mathrm{MHz}$. To first order the inferred variance isHere $u$ denotes a standard uncertainty, not a frequency. The four contributions in $\mathrm{MHz}$ are $0.285$, $0.400$, $0.200$, and $0.100$; adding their squares gives $u({\nu }_0)\simeq 0.540\mathrm{MHz}$. The calibration offset shifts all sampled fields together, so increasing the sample count cannot average it away. Uncertainty in $a$, unresolved components, asymmetric line distortion, and errors in the assumed field dependence would require additional terms or a richer model. The historical inference requires the appropriate Zeeman splitting branches, rather than simply drawing a straight line through one resonance.

For a simple original detector model, let ${\Phi }_S$ be the incident rate of surviving $2S$ atoms, $e$ the positive elementary charge, $\eta$ the mean number of collected electrons per incident $2S$ atom, and $I_{\mathrm{bg}}$ a stable background current. Take current magnitudes as positive. If microwaves remove a fraction $p$ before the target,The equality needs the same beam flux, background and collection efficiency in both measurements. It is an ideal interpretation of a normalized current loss, not a claim that an actual galvanometer directly counts individual transitions. Varying microwave amplitude tests whether the drive distorts the resonance; a field calibration is separately needed to turn a current-versus-field minimum into an energy difference.

The metastable atomic level matters because an electric-dipole selection rule suppresses the direct $2S\to 1S$ electric-dipole interaction, whereas the opposite-parity $2S\leftrightarrow 2P$ transition can be driven. A long lifetime can store the prepared population until interrogation. After transfer to $2P$, a short lifetime supplies the loss mechanism and a broad natural linewidth. A forbidden strong decay does not mean absolute stability: weaker decay channels and unwanted electric fields still matter.

A generalized coordinate is one of a set of independent parameters specifying the configuration of a mechanical system subject to its constraints. An angle locates a bead constrained to a fixed circular hoop. Expressing kinetic energy and potential energy in these coordinates gives the Lagrangian and its Euler-Lagrange equations.

A point mass $M$ attached to a massless rotating hoop of radius $R$ is connected by a taut string wrapping around the hoop to a vertically suspended mass $m$. Measure $\theta$ from the downward radius toward the string's tangency side. The wrapped length decreases by $R\theta$, so the suspended mass descends by the same amount. Therefore $T=(M+m)R^2{\dot{\theta }}^2/2$ and $V=-MgR\cos \theta -mgR\theta$ up to a constant. The Euler-Lagrange equation gives the displayed equation. For $0<m<M$, the stable equilibrium has $\sin {\theta }_0=m/M$, and the small-oscillation angular frequency is $\omega =\sqrt{g/R}[(M-m)/(M+m){]}^{1/4}$. When $M\gg m$, the system approaches a simple pendulum of length $R$.

Consider an objective function that is a quadratic polynomial, with a singular optimization constraint:Its only feasible point is $x_{\ast }=0$, giving $p_{\ast }=9$. For $\lambda \ge 0$, completing the square yieldsEvery finite Lagrange multiplier gives $q(\lambda )<9$, but $q(\lambda )\to 9$ as $\lambda \to \infty$. Thus $d_{\ast }=p_{\ast }=9$: strong duality holds, and the primal optimum is attained, yet dual attainment fails. The inner minimizers $3/(1+\lambda )$ approach $0$ without ever becoming feasible at finite $\lambda$.

Both the objective function and the constraint function are convex functions. However, Slater's condition fails because $x^2<0$ is impossible. At the optimum, stationarity would require $-6+\lambda \cdot 0=0$, which no finite Lagrange multiplier satisfies. Compare KKT regularity at a local minimum: a zero gradient of an active constraint can prevent a multiplier certificate. This example has zero duality gap without the Strong Lagrangian property.

A nonconvex primal problem can have a positive duality gap even when both optima are attained. Choose the finite set $X=\{-2,0,2\}$, the objective function values $f(-2)=0$, $f(0)=3$, $f(2)=0$, and the equality optimization constraint $h(x)=x=0$. The only feasible point is $0$, so $p_{\ast }=3$.

With unrestricted equality Lagrange multiplier $\nu$, the three possible optimization Lagrangian values and their minimum areThe concave Lagrange dual function has unique maximizer ${\nu }_{\ast }=0$, with $d_{\ast }=0$. Thus the duality gap is $3$. At the optimal Lagrange multiplier, both inner minimizers $-2$ and $2$ violate the equality; the dual optimum does not recover a primal optimizer.

Why can no slope close the gap? Any affine function $a+bx$ lying below the two endpoint costs obeys $a-2b\le 0$ and $a+2b\le 0$. Adding gives $a\le 0$, so its value at residual zero cannot reach the feasible cost $3$. A convex combination of the endpoint cost-residual pairs with equal weights has residual zero and average cost zero. This averaged pair is unattainable: the actual decision $x=0$ costs $3$. This geometry explains precisely what the dual bounds fail to distinguish. Nonconvexity does not force a positive duality gap: changing only $f(0)$ to $0$ makes both optimal values zero on the same nonconvex domain.

Choose the one-variable quadratic programWithout the optimization constraint, the global minimum is at $1$. On the feasible set $[4,\infty )$, write $x=4+s$ with $s\ge 0$: then $f(x)=11+6s+s^2\ge 11$. Hence $x_{\ast }=4$ is the unique primal optimizer and $p_{\ast }=11$.

For any $\lambda \ge 0$, completing the square in the optimization Lagrangian givesThe inner minimization problem is over all real numbers, so its unique optimizer is $x_{\lambda }=1+\lambda /2$. ThereforeThe best lower bound is $d_{\ast }=11$, attained at ${\lambda }_{\ast }=6$. At $\lambda =2$, the inner optimizer is $2$, which is infeasible for the primal problem; nevertheless $q(2)=7$ is a valid lower bound. At $\lambda =9$, the inner optimizer is $5.5$, which is feasible, but $q(9)=8.75$ is weaker than $11$. Feasibility of an inner optimizer alone does not make the bound exact.

At $(x_{\ast },{\lambda }_{\ast })=(4,6)$, the Karush-Kuhn-Tucker conditions reduce toThe last equality is complementary slackness. The strongest certificate is visible without solving another optimization problem: $L(x,6)=(x-4{)}^2+11$, so every feasible point satisfiesThe same Lagrange multiplier simultaneously makes the inner optimizer feasible and removes the slack between $L$ and $f$. This is a convex KKT optimality certificate; $x=5$ also verifies Slater's condition.

The domain in the inner infimum is essential. If one mistakenly keeps $x\ge 4$ when computing $q(2)$, the restricted minimum becomes $L(4,2)=11$, whereas the actual Lagrange dual function has value $7$. Constraints deliberately retained in $X$ are allowed in a Lagrangian relaxation, but that is a different choice of which restrictions to dualize.

The nonzero gradient condition in the equality Lagrange multiplier rule cannot simply be dropped. Minimize $f(x,y)=x$ subject to $g(x,y)=x^2+y^2=0$. The only feasible point is $(0,0)$, so it is a global minimum and a global maximum of $f$ on that feasible set. Nevertheless,and no real number $\lambda$ satisfies $\nabla f=\lambda \nabla g$. The point is singular for this representation of the optimization constraint; the usual constraint qualification fails. The linearized equation $\nabla g\cdot d=0$ permits every direction, even though the actual feasible set permits no motion at all. This example explains why blindly solving the multiplier equations can miss an optimum.

The chosen quadratic objective function above gives more than an optimal allocation: its Lagrange multiplier predicts how much the best cost changes when the required total changes. For this example,This is an explicit derivative of a constrained value function and an instance of the envelope theorem. At $c=3$, one extra small unit of required output costs approximately $4$ units per unit of output, in the chosen normalized model. Increasing $c$ makes this equality-constrained minimization problem more expensive; it is not a gain from relaxing an upper resource limit. The sign and meaning depend on how the optimization constraint and optimization Lagrangian are written.^[ref]

For an actual finite change $\delta$, the cost difference is exactlyAt $\delta =0.1$, the linear approximation predicts $0.4$, while the exact increase is $61/150\approx 0.406667$. The extra $1/150$ comes from the squared term in this polynomial, rather than numerical error. A Lagrange multiplier gives a local sensitivity analysis; treating it as a constant price over a large change generally loses the change in the optimum itself.

For a smooth optimizer $z(c)$ of an equality problem $g(z)=c$, the short chain rule calculation explains the result: $\nabla g(z(c))\cdot z^{\prime }(c)=1$ and $\nabla f(z(c))=\lambda (c)\nabla g(z(c))$ imply $df(z(c))/dc=\lambda (c)$. The differentiability and stationarity assumptions matter: an optimizer can fail to be unique or vary smoothly. The envelope theorem records when this interpretation is justified.

Suppose two adjustable inputs must deliver a prescribed total $c$, and their combined cost is the quadratic form $f(x,y)=x^2+2y^2$. These are chosen, dimensionless quantities; the coefficient $2$ makes concentrating an input in $y$ more expensive. Initially allow all real numbers $x,y$, with the equality optimization constraint $g(x,y)=x+y=c$. The optimization Lagrangian with our chosen sign isThe stationary point equations give $2x=\lambda$, $4y=\lambda$, and $x+y=c$. ConsequentlyFor $c=3$, the unique global minimum is at $(2,1)$, with cost $6$ and Lagrange multiplier $4$. Equal allocation would cost $6.75$; assigning everything to $x$ would cost $9$. At the optimum, the two marginal costs agree: ${\partial }_{x}f={\partial }_{y}f=4$. Moving a little input from one component to the other therefore has zero first-order effect on cost. When $c>0$, this solution also meets the extra nonnegativity optimization constraints $x,y\ge 0$, so it remains optimal with those added.

Here the Lagrange multiplier equations locate a candidate, and a short independent check certifies it. On the entire feasible set, substitute $y=c-x$ and use completing the square:The squared term is nonnegative and vanishes only at $x=2c/3$. Thus the candidate is the unique global minimum; there is no global maximum on this unbounded feasible set, a line. This separates finding a stationary point from proving optimality. The geometric beauty is that equal marginal costs, parallel gradients, tangency of level sets, and completing the square identify the same allocation.

For $q,C>0$, minimize a cylinder's volume $V=\pi R^{2}H$ over $R,H>0$ subject to $q^2/R^2+{\pi }^2/H^2=C$. Set $u=q^2/R^2$ and $v={\pi }^2/H^2$. Then $u+v=C$ and $V={\pi }^2{q}^2/(u\sqrt{v})$. Minimizing volume is equivalent to maximizing $u^2(C-u)$ on $0<u<C$. Its unique interior maximum is at $u=2C/3$, $v=C/3$, with divergence of volume at both endpoints. Thus the unique global optimum has

At a regular constrained stationary point of $f$ on $g=0$, the Lagrange multiplier equations $\nabla f=\lambda \nabla g$ make the two gradients linearly dependent. Eliminating $\lambda$ gives the displayed zero determinant. This condition is necessary, not sufficient for an extremum. At a singular constraint point where $\nabla g=0$, the determinant is automatically zero even when no multiplier exists; for example $g=x^2+y^2$ and $f=x$ at the origin.

A rigid stick of negligible width can turn horizontally between perpendicular corridors of widths $a,b>0$ precisely when its length does not exceed the displayed value, allowing contact with the walls. A position touching the two outer walls and the inner corner has endpoint intercepts $(a+X,0)$ and $(0,b+Y)$ with $X,Y>0$. Collinearity with the inner corner $(a,b)$ is equivalent to $XY=ab$.

Minimize the squared intercept length $F=(a+X{)}^2+(b+Y{)}^2$ under this optimization constraint. The Lagrange multiplier equations imply $X(a+X)=Y(b+Y)$. Substituting $Y=ab/X$ gives $(X^3-a{b}^2)(X+a)=0$, so the admissible point is $X=(a{b}^2{)}^{1/3}$, $Y=(a^{2}b{)}^{1/3}$. There $F=(a^{2/3}+b^{2/3}{)}^3$. Along the positive constraint branch $F$ tends to infinity at both ends and has this one stationary point, proving it is the minimum available clearance over orientations. That minimum is the maximum stick length which can pass through every intermediate orientation.