A Penning trap uses a static electric quadrupole field for axial confinement and a static axial magnetic field for transverse confinement. The Lorentz force bends a moving electric charge's trajectory without doing work. An RF probe may interrogate a trapped particle, but an RF confining voltage is unnecessary in this ideal model. This makes the Penning trap useful for relating cyclotron frequencies and spin precession to a common magnetic field.

The Paul trap alternates focusing and defocusing electric forces. Their time average along a moving trajectory need not vanish, even when the electric field averages to zero at every fixed point. A spatial gradient makes the fast displacement correlate with the driving electric field; the RF-averaged trapping potential calculates the resulting restoring force.

Wolfgang Paul's lecture describes hyperbolic ring/endcap electrodes, loading by electron-impact ionization of low-pressure gas, and resonant ejection through an endcap hole into an electron multiplier. Sweeping RF voltage produces an air mass spectrum.^[ref]

Trapped-ion quantum control prepares an ion's internal quantum state, couples it to quantized motion with laser pulses, and measures the result through state-dependent fluorescence. Its appeal is access to both a controllable two-level system and a quantum harmonic oscillator in one apparatus.

Confinement, preparation, control and readout solve different problems. The Paul trap's RF electric field confines the ion through its RF-averaged trapping potential; it still produces Paul trap micromotion. Doppler cooling and resolved sideband cooling reduce motional excitation. Optical pumping selects an internal quantum state. A coherent pulse then changes amplitudes, while electron shelving converts an internal population into a detectable stream of photons. A bright/dark count by itself does not reveal a quantum superposition's relative phase; an additional controlled rotation before detection can make that phase affect the measured population.

In thermal equilibrium, nondegenerate conduction electrons and valence electron holes share one chemical potential. Multiplying $n=N_c{e}^{-(E_c-\mu )/(k_{B}T)}$ by $p=N_v{e}^{-(\mu -E_v)/(k_{B}T)}$ cancels that chemical potential:This relation remains valid with semiconductor doping under the stated assumptions. Equality of the two individual densities, $n=p=n_i$, additionally requires the intrinsic neutral case. Separate electron and hole quasi-Fermi levels in nonequilibrium need not obey this equilibrium product.

With one parabolic valley and two spin states at each band edge, the prefactor multiplying $(k_{B}T{)}^3$ in the product is $4(m_c{m}_h{)}^{3/2}/(2\pi {\hslash }^2{)}^3$. Both populations must be nondegenerate; the Pauli reduction of the pinned-edge mass-action product quantifies a failure of the classical approximation.

For fixed band effective masses and band gap $E_g$, the intrinsic carrier concentration obeys $n_i=C{T}^{3/2}{e}^{-E_g/(2k_{B}T)}$. ThusA base-ten plot against $x=1000\mathrm{K}/T$ instead has slope $-E_g/(2000k_B\mathrm{K}\ln 10)$ after the prefactor correction. The intrinsic branch must be used, rather than the extrinsic plateau or dopant freeze-out branch. A changing gap or effective mass makes this an activation-energy estimate, and a small gap uncertainty can produce a much larger multiplicative uncertainty in a room-temperature minority density.

In a nondegenerate parabolic-band semiconductor with fixed density-of-states mass, $n(T)=C{T}^{3/2}{e}^{-\Delta (T)/(k_{B}T)}$, where $\Delta (T)=E_c(T)-\mu (T)$. ThusA single measured value of $\Delta$ does not determine its temperature dependence. A fixed-gap intrinsic model instead has $n_i=C_i{T}^{3/2}{e}^{-E_g/(2k_{B}T)}$, with the two effective density of states factors in $C_i$. Identifying a crossover by $n_i(T_c)\simeq n_{\mathrm{ext}}$ gives an order-of-magnitude dopant-density estimate once that activation law is specified. At exact charge neutrality with $n_i=n_{\mathrm{ext}}$, the total mobile-electron density is $(1+\sqrt{5})n_{\mathrm{ext}}/2$, so a crossover is not a sharp equality of every density definition.

Weighted interval scheduling assigns each job a reward $w_i$ and maximizes their sum over compatible jobs. Earliest finish can fail for this different objective function; dynamic programming can compare alternatives.^[ref]

For a separate two-job example, let $X=[0,2)$ have reward one and $Y=[1,3)$ reward ten. Earliest finish accepts $X$ and rejects $Y$, earning one; the optimum accepts $Y$, earning ten. Replacing $Y$ by $X$ preserves job count but loses reward. Thus the interval scheduling exchange argument proves the unit-value theorem only: its preserved quantity must match the intended objective function.

In this separate three-job example, labels $A,B,C$ denote new jobs:Their durations are five, five and two. Consider the rule "accept a shortest remaining job, discard all jobs conflicting with it, and repeat". It selects $C$ first. But $C$ overlaps $A$ on $[4,5)$ and $B$ on $[5,6)$, so both are discarded. The rule returns one job, while the compatible subset $\{A,B\}$ contains two. The common endpoint five is allowed, and no feasible solution contains all three jobs, so two is the maximum.

Duration alone ignores where a job sits. Here the shortest job straddles the useful handover time. Earliest finish instead chooses $A$, then $B$, and attains two. A counterexample can refute a proposed greedy-choice property even when the rule succeeds on many other inputs.

For a well $-a<x<a$, choose normalized energy eigenstates ${\varphi }_n=a^{-1/2}\sin [(n+1)\pi (x+a)/(2a)]$, $n=0,1,\dots$. The diagonal position matrix elements vanish by the parity operator. For $k\ne n$,To derive it, set $y=x+a$ and use $2\sin Ay\sin By=\cos [(A-B)y]-\cos [(A+B)y]$. For unequal positive integer indices $r,s$, the constant part of $x=y-a$ integrates to zero by orthogonality, whileTaking the difference for $j=r-s$ and $j=r+s$ yields the formula after setting $r=k+1,s=n+1$. It vanishes for equal parity and gives $x_{01}=-32a/(9{\pi }^2)$ for this phase convention.

For an infinite square well on $[0,L]$, choose ${\varphi }_n=\sqrt{2/L}\sin (n\pi x/L)$ and $E_n=n^2{E}_1$, $E_1={\pi }^2{\hslash }^2/(2m{L}^2)$. The state $\psi =({\varphi }_1{e}^{-i{E}_{1}t/\hslash }+{\varphi }_2{e}^{-i{E}_{2}t/\hslash })/\sqrt{2}$ has constant expectation value $⟨E⟩=5E_1/2$, butIndeed, the diagonal position matrix elements are $L/2$, whereas the off-diagonal element isA different initial relative phase shifts the oscillation in time. The oscillation comes from quantum interference, while the constant mean energy follows from the fixed populations of the two energy eigenstates.

For a normalized sufficiently regular solution of the Schrödinger equation on $[0,L]$ with Dirichlet walls, the instantaneous outward forces on the walls have magnitudesA wall displacement changes the kinetic-energy quadratic form by the boundary term $-P_{j}d{L}_j$, giving the local force. Equivalently, direct integration by parts in the time derivative of the mean particle momentum givesThe force on the particle is the opposite of the net force $-P_0+P_L$ on the walls. For a stationary energy eigenstate each magnitude equals $2E_n/L$. For the equal superposition of the first two box states, writing $C=E_1/L$ and $\Omega =3E_1/\hslash$ gives $P_0=C(5+4\cos \Omega t)$ and $P_L=C(5-4\cos \Omega t)$. Both time averages equal $2⟨E⟩/L$, the classical mean force on one wall at that energy, even though the quantum forces vary continuously in time.

Consider hard walls at $x=0,3L$, zero potential energy on $0<x<L$ and $2L<x<3L$, and a central rectangular barrier whose height equals the ground state energy $E$. The time-independent Schrödinger equation gives ${\psi }^{\prime \prime }=0$ on the central interval. Reflection symmetry and the nodeless theorem for a one-dimensional ground state require an even, positive state about $3L/2$, hence a constant central segment. Wavefunction matching at a finite potential step makes the derivatives vanish at $L,2L$. The side segments are proportional to $\sin (\pi x/(2L))$ and $\sin (\pi (3L-x)/(2L))$. Their common plateau value is $1/\sqrt{2L}$ after wavefunction normalization, and $E={\pi }^2{\hslash }^2/(8m{L}^2)$. Equality of barrier height and energy is essential: a higher barrier gives positive curvature of a positive wavefunction in the middle, while a lower barrier gives negative curvature.

For two fixed linear coils, mutual inductance $M$ relates magnetic flux linkage in one coil to electric current in the other: ${\Lambda }_{21}=M{I}_1$. The induced electromotive force is ${\mathcal{E}}_2=-Md{I}_1/dt$ with consistent winding orientations. For reciprocal coils, $|M|=k\sqrt{L_1{L}_2}$, where $L_1,L_2$ are self-inductances and $0\le k\le 1$ measures magnetic coupling. Its SI unit is the henry.