The circumference of a circle is its boundary arc length. For radius $r$, it equals $2\pi r$. This growing boundary length accounts for the radial factor in the Rayleigh distribution.

A circular sector is the portion of a disc between two radii and the connecting circle arc. If its radius is $R$ and its central angle is $\theta$ radians, its area isIndeed, polar coordinates give $A={\int }_0^{\theta }{\int }_0^{R}rdrd\varphi$. The full disc corresponds to $\theta =2\pi$, a semicircle to $\theta =\pi$, and a quarter-disc to $\theta =\pi /2$. In a solid with horizontal sector cross-sections, radius $R(z)$ and angle $\theta (z)$, its volume is $\frac{1}{2}\int \theta (z)R(z{)}^{2}dz$. Holding the angles and heights fixed while scaling every radius by a factor $s$ scales that volume by $s^2$.

A circular arc is a connected portion of a circle. For radius $R>0$ and subtended central angle $0<\theta <2\pi$ measured in radians, its arc length is $R\theta$. Two distinct endpoints determine two circular arcs: when they are not opposite, the minor arc has $\theta <\pi$ and the major arc has $\theta >\pi$. Opposite endpoints determine two semicircles. The straight line segment joining the endpoints is a chord, rather than the curved circular arc itself.

Let $H,V$ be orthogonal linear polarizations, and prepare the Bell state $|{\Phi }^{+}⟩=(|HH⟩+|VV⟩)/\sqrt{2}$. The $+1$ analyzer vector is $|a⟩=\cos a|H⟩+\sin a|V⟩$; its orthogonal companion has outcome $-1$. Direct inner products with the Bell state give $P_{++}=P_{--}={\cos }^2(a-b)/2$ and $P_{+-}=P_{-+}={\sin }^2(a-b)/2$. Thus the expectation value of the outcome product is $E(a,b)=\cos 2(a-b)$. These are physical polarizer angles; their corresponding Bloch sphere angles are doubled.

For a stipulated isotropic admixture, use the density matrixThe maximally mixed state contributes zero correlation, so $P_{st}(a,b)=[1+stv\cos 2(a-b)]/4$ for $s,t\in \{-1,1\}$. Each local marginal is $1/2$, independently of the other setting, illustrating quantum no-signalling. In this model $v$ is the interferometric visibility of the coincidence fringe; other noise models need not have one common visibility for every basis.

Choose $a_0=0^{\circ }$, $a_1=45^{\circ }$, $b_0=22.5^{\circ }$ and $b_1=-22.5^{\circ }$. The four correlations in the displayed CHSH inequality are $v/\sqrt{2},v/\sqrt{2},v/\sqrt{2},-v/\sqrt{2}$. Consequently $S=2\sqrt{2}v$, exceeding the local bound exactly when $v>1/\sqrt{2}$. Perfect correlation at a single common setting would not yield this comparison between incompatible quantum measurements.

To connect probabilities with expected counts, stipulate $M$ emitted pairs per setting pair and independent detection efficiencies ${\eta }_A,{\eta }_B$, independent also of settings, outcomes and hidden variables. The expected joint counts are $n_{st}=M{\eta }_A{\eta }_B{P}_{st}$. Their normalized correlation isFor original teaching inputs $M=1000$, ${\eta }_A=0.8$, ${\eta }_B=0.75$ and $v=0.8$, the $(a_0,b_0)$ expected counts are $n_{++}=n_{--}\simeq 234.85$ and $n_{+-}=n_{-+}\simeq 65.15$. They total $600$ and give $E_{\det }\simeq 0.56569$, hence $S\simeq 2.26274$. Fractional expected counts are not measured data. Efficiency cancels here because the loss model was specified to be independent; this algebra supplies no justification for discarding failures in an arbitrary Bell test with a detection loophole.

For the displayed metric tensor, its inverse has components $g^{rr}=e^{-2\chi }$ and $g^{\theta \theta }=e^{-2\chi }/r^2$. The nonzero second-kind Christoffel symbols are ${\Gamma }^r{}_{rr}={\chi }^{\prime }$, ${\Gamma }^r{}_{\theta \theta }=-r-r^2{\chi }^{\prime }$ and ${\Gamma }^{\theta }{}_{r\theta }={\Gamma }^{\theta }{}_{\theta r}=1/r+{\chi }^{\prime }$. They follow by differentiating the two diagonal metric entries and raising the first index of the first-kind connection coefficients. Polar coordinates require treating $r=0$ separately.

For a nondegenerate spin-one-half ideal gas with energies $\mp {\mu }_{B}B(\mathbf{r})$, the two component chemical potentials are $k_{B}T\ln (n_{\pm }/n_Q)\mp {\mu }_{B}B$. Diffusive and spin-conversion equilibrium require one spatially constant common chemical potential, provided spin conversion is allowed. Hence the displayed densities give total number density $n=2n_Q{e}^{\beta \mu }\cosh (\beta {\mu }_{B}B)$ and magnetization $M=2{\mu }_B{n}_Q{e}^{\beta \mu }\sinh (\beta {\mu }_{B}B)=n{\mu }_B\tanh (\beta {\mu }_{B}B)$. The high-temperature fixed-density limit is $M\simeq n{\mu }_B^{2}B/(k_{B}T)$, a Curie law. At fixed chemical potential the density itself varies with temperature, so the unnormalized magnetization need not have a pure $1/T$ dependence.

For noninteger real $\gamma$, expand a smooth forcing as $F={\sum }_{n\ge 0}{F}_n{T}_n$ in Chebyshev polynomials. Because $(1-x^2)T_n^{\prime \prime }-x{T}_n^{\prime }=-n^2{T}_n$, division of each coefficient by ${\gamma }^2-n^2$ gives the regular solution. The coefficients use the weighted inner product with weight $(1-x^2{)}^{-1/2}$. Choosing the same endpoint domain as the regular eigenfunctions of the Sturm-Liouville problem is essential: finite values alone also allow homogeneous solutions $\cos (\gamma \arccos x)$ and $\sin (\gamma \arccos x)$, which generally have divergent endpoint derivatives. Setting $x=\cos \theta$ converts the regular endpoint condition into the Neumann boundary conditions $Z_{\theta }(0)=Z_{\theta }(\pi )=0$. The homogeneous equation then has no nonzero solution when $\gamma$ is noninteger, proving uniqueness in this domain.

For the Chebyshev differential equation $(1-x^2)y^{\prime \prime }-x{y}^{\prime }+k^{2}y=0$, a power series about the ordinary point zero splits into even and odd solutions with the displayed recurrence relation. Every nonterminating parity series has radius of convergence one. If $k^2=m^2$ for a nonnegative integer $m$, the parity series containing $x^m$ terminates, giving a polynomial with infinite radius of convergence; the other independent series still has radius one. Normalized solutions are $\cos (k\arcsin x)$ and $\sin (k\arcsin x)/k$, with the latter replaced by $\arcsin x$ when $k=0$.

The Chebyshev differential equation $-(1-x^2)y^{\prime \prime }+x{y}^{\prime }=\lambda y$ on $-1<x<1$, with zero endpoint values, has the Sturm-Liouville formThus $p=\sqrt{1-x^2}$ and the weight defining the weighted inner product is $w=(1-x^2{)}^{-1/2}$. The substitution $x=\cos \theta$ maps the problem to $-Y^{\prime \prime }=\lambda Y$ on $0<\theta <\pi$ with homogeneous Dirichlet boundary conditions. It also transforms $w(x)dx$ into $-d\theta$. Consequently the normalized eigenfunctions areTheir weighted inner product is the ordinary sine inner product on $[0,\pi ]$. Equivalently, $y_n=\sqrt{2/\pi }\sqrt{1-x^2}{U}_{n-1}(x)$, where $U_{n-1}$ is a Chebyshev polynomial of the second kind. These are not the cosine solutions $T_n$, which have different endpoint values. The original endpoints are singular because $p$ vanishes and $w$ diverges there, but the transformed Dirichlet problem specifies the self-adjoint realization. The eigenfunctions vanish at both endpoints and their quasi-derivatives $p{y}_n^{\prime }$ have finite limits, so the boundary form vanishes.

George Elwood Smith reports fabricating MOS capacitors, thermally generating the first packet, and clocking it to an output that pushed it into the substrate for electric current detection. A later integrated eight-bit device demonstrated imaging. Dark or reversed-clock control traces are not specified; feasibility alone does not calibrate charge-transfer efficiency.^[ref]

A discriminating repeat would compare identical cycles with and without initial carrier injection, then suppress or reverse the transfer sequence while retaining output switching. An electric current transient synchronized with gate switching can include capacitor feedthrough; only the difference tied to initial electric charge and the expected transfer delay tests packet transport. These are proposed controls, not controls documented in the lecture. For an optical version, a blocked exposure estimates dark current and read noise; a spatially isolated illuminated optical detector pixel exposes any deferred charge in CCD readout trail. Photon shot noise persists even with ideal transfer.

Deferred charge in CCD readout reaches later output samples than its originating packet. George Elwood Smith describes surface states at the silicon–silicon dioxide interface capturing electrons and releasing them later, smearing the image. A buried channel keeps the stored packet away from that interface.^[ref]

An original comparison separates two interpretations of the same main-packet deficit. If a single transfer defers fraction $ϵ$ by exactly one sample and loses nothing, an isolated input packet $N$ produces samples ${\eta }_{\mathrm{tr}}N$ followed by $ϵN$: summing both recovers $N$. A truly lossy transfer yields only ${\eta }_{\mathrm{tr}}N$, with no later recovery. For $N=10000$ and $ϵ=0.01$, both give $9900$ in the intended sample, but only the deferred case gives a $100$-electron trail. Such a trail redistributes position information even when total electric charge is conserved. Repeated delays require a time-response model; $N_{\mathrm{in}}{\eta }_{\mathrm{tr}}^n$ alone cannot predict trail shape. Electron-hole recombination, escape to another contact, or release beyond the recorded window can instead remove signal from the measured sum.

The charge-transfer efficiency is the fraction of a packet delivered into its intended next site or time slot in one transfer, under stated conditions. It is distinct from quantum efficiency, which concerns conversion of incident photons into collected charge carriers. Define the complementary inefficiency as $ϵ=1-{\eta }_{\mathrm{tr}}$.

In an original model with constant $0\le {\eta }_{\mathrm{tr}}\le 1$, let $N_j$ denote the mean number of original packet electrons still delivered on time after $j$ transfers. Assuming no reinjection into that packet,Repeated substitution proves the result; for different efficiencies the fraction is ${\prod }_{j=1}^n{\eta }_j$. For $ϵ\ll 1$, a Taylor expansion of $\ln (1-ϵ)$ gives ${\eta }_{\mathrm{tr}}^n\simeq e^{-nϵ}$, while the simpler $1-nϵ$ requires $nϵ\ll 1$.

Choose $N_{\mathrm{in}}=10000$ and $n=2000$. For ${\eta }_{\mathrm{tr}}=0.9999$, the expected on-time output is $8187.23$ electrons; for ${\eta }_{\mathrm{tr}}=0.99999$, it is $9801.99$. These are model means, so fractional electron counts do not describe an individual packet. Even a small per-transfer imperfection can imprint distance-dependent image errors.

If every original electron independently stays on time with probability $p={\eta }_{\mathrm{tr}}^n$, a fixed initial count gives a binomial distribution with variance $N_{\mathrm{in}}p(1-p)$. An initial Poisson distribution with mean $\mu$ instead gives an on-time Poisson distribution with mean and variance $\mu p$, by Poisson thinning. A deterministic attenuation formula alone does not specify noise. Real trapping can correlate packets and depends on timing, illumination and temperature; the independent constant-efficiency model then fails.

For an electron packet, a sufficiently positive gate electric potential attracts electrons beneath the insulating layer of a MOS capacitor. The corresponding potential energy is $U=-e\varphi$, where $e>0$ is the magnitude of elementary charge: a larger electric potential gives a deeper potential well for electrons. A drawing of potential energy must therefore reverse the vertical sense of a drawing of electric potential.

A simplified three-phase transfer raises the adjacent gate's electric potential before lowering the old holding gate. This overlaps the accessible regions, then leaves the packet under the new gate. Repeat with the next gate to move it forward. Isolated wells with an intervening barrier would store the packet without transferring it.^[ref]

Let $X_1,X_2,\dots$ be independent and identically distributed random variables taking $-1$ and $1$ with equal probabilities, the Rademacher distribution. The two opposite complex exponentials give ${\phi }_X(t)=(e^{it}+e^{-it})/2=\cos t$. Their expected value is zero and their variance is one. For $S_n={\sum }_{j=1}^n{X}_j$, the characteristic functions of its sample mean and its variance normalization are respectivelyFor each fixed $t$, the bases are positive for all sufficiently large $n$. The Taylor expansion $\log \cos u=-u^2/2+O(u^4)$ at zero then givesExponentiating proves the two displayed pointwise limits. The first limit is the characteristic function of the constant zero; the second is that of the standard normal distribution. The Lévy continuity theorem gives the corresponding convergence in distribution statements. Averaging has variance $1/n$, whereas dividing the sum by $\sqrt{n}$ preserves variance one. This sign calculation illustrates why the law of large numbers and the central limit theorem use different scales; it is not a proof for arbitrary probability distributions.

Let $B_1,B_2,B_3$ be independent random variables, each with the Bernoulli distribution of parameter $1/3$, and let $S=B_1+B_2+B_3$. Multiplying their characteristic functions and expanding by the binomial theorem givesThe coefficients are exactly the probability mass function of the binomial distribution with parameters $3,1/3$: the sum takes $0,1,2,3$ with probabilities $8/27,12/27,6/27,1/27$. Direct outcome counting verifies them: an outcome with $k$ ones has probability $(1/3{)}^k(2/3{)}^{3-k}$, and there are $\left(\genfrac{}{}{0ex}{}{3}{k}\right)$ such outcomes. The coefficients sum to one, recovering ${\phi }_S(0)=1$.

At $t=\pi$, ${\phi }_S(\pi )=(1/3{)}^3=1/27$. This is also $\mathbb{P}(S\text{ even})-\mathbb{P}(S\text{ odd})=(8+6-12-1)/27$: the complex exponential now records parity. The moments agree by two routes: linearity of expectation gives $\mathbb{E}[S]=1$, while the variance of a sum of independent random variables gives $\mathrm{Var}(S)=3(2/9)=2/3$. Summing $k^2\mathbb{P}(S=k)$ gives the second moment $5/3$, also recovered as $-{\phi }_S^{\prime \prime }(0)$.

Let $A$ be a fair bit, let $N$ be an independent Bernoulli random variable with $\mathbb{P}(N=1)=p$, and set $B=A\oplus N$, where $\oplus$ is exclusive or. Thus $N=1$ flips the first bit. This constructs a binary symmetric channel with a fair input, for $0\le p\le 1$. The joint probability mass function isThe column sums show that $B$ is also a fair bit, whatever $p$ is. Both marginal distributions therefore have information entropy one bit. Given $A$, only the flip decision remains unknown, giving conditional entropy $H(B\mid A)=h_2(p)$, whereis the binary entropy, with $h_2(0)=h_2(1)=0$. The chain rule for information entropy and the definition of mutual information now give

Take $p=1/10$. The four joint probabilities, in row order, are $(0.45,0.05,0.05,0.45)$. Directly from the definition of joint entropy,Conditioning on either value of $A$ gives probabilities $0.9$ and $0.1$ for the two values of $B$, henceAdding $H(A)+H(B)=2$ would overcount by exactly $I(A;B)$. These are average uncertainties under the stipulated probability distribution; the four possible pairs are still represented by four distinct outcomes.

There is another way to see the decomposition: $(A,B)\mapsto (A,A\oplus B)=(A,N)$ is a bijection. It merely relabels the four outcomes, so it preserves joint entropy. The independence of $A$ and $N$ then gives $H(A,B)=H(A,N)=H(A)+H(N)=1+h_2(p)$. Describing the initial bit and the flip decision separates shared uncertainty from new uncertainty.

At $p=0$, the bits are identical: $B=A$, so $H(B\mid A)=0$ and $I(A;B)=1$. At $p=1$, they are complementary: $B=1-A$, with exactly the same conditional entropy and mutual information. In either case, observing one determines the other. At $p=1/2$, all four joint probabilities are $1/4$; the bits are independent, $H(A,B)=2$, and $I(A;B)=0$. Perfect disagreement is therefore just as informative as perfect agreement.

For $p>1/2$, guessing $A=1-B$ succeeds with probability $p$, better than guessing $A=B$. Changing $B$ to $1-B$ replaces $p$ by $1-p$ and preserves mutual information, since the relabeling is invertible. What matters is how well the observation distinguishes the input, rather than how often its symbol matches the input. The same symmetry makes the binary entropy curve peak at independence, while the mutual information curve reaches its minimum there.

If $U$ is a unitary matrix with no eigenvalue equal to one, the displayed inverse Cayley transform of a Hermitian matrix is Hermitian. Since $U^{†}=U^{-1}$, taking the Hermitian conjugate gives $B^{†}=-i(I-U^{-1}{)}^{-1}(I+U^{-1})=i(I-U{)}^{-1}(I+U)=B$, using commutativity of rational functions of $U$. An eigenvalue $e^{i\theta }\ne 1$ is mapped to the real number $-\cot (\theta /2)$. Conversely $U=(B-iI)(B+iI{)}^{-1}$, so this correspondence identifies Hermitian matrices with unitary matrices whose spectrum avoids one.

Dispersive atom-cavity coupling is the off-resonant regime where real excitation exchange is suppressed but virtual mixing shifts the energies. For a two-level system describing an chemical atom coupled to one mode, the scale $\chi \simeq g^2/\Delta$ appears when $|\Delta |$ is large compared with the relevant occupation-enhanced coupling $g\sqrt{n+1}$.

An effective term proportional to $\hslash \chi a^{†}a{\sigma }_z$ then makes the atomic phase depend on photon number and the cavity frequency depend on atomic state. Measuring that phase can infer field occupation without resonantly absorbing a photon in the ideal approximation. Finite detuning, losses and probe backaction still matter; dispersive does not mean completely uncoupled or universally disturbance-free. A model must state detuning sign and the retained perturbative order.

If $s>0$ and $\Theta$ has the continuous uniform distribution on $(-\pi /2,\pi /2)$, then $Y=s\tan \Theta$ has the Cauchy distribution with location zero and scale $s$. The tangent function is strictly increasing on this interval, so its cumulative distribution function isDifferentiating gives the probability density function $s/[\pi (s^2+y^2)]$. The probability of either endpoint is zero, so including endpoints in the uniform interval does not alter this probability distribution. The slow quadratic decay makes the expected value undefined: its positive and negative parts both have infinite integrals. Its second moment is also infinite, so it has no finite standard deviation.