Past exam of the physics A course of the University of Cambridge / 2015

Original source: Released examination PDF, physical pages 100–105. All alternatives are treated below.

The Born rule uses the squared modulus, so the imaginary term contributes positively: $|a+ix{|}^2=a^2+x^2$. For $a>0$, wavefunction normalization would multiply the supplied expression by $N=\sqrt{3/(4a^3)}$. Equivalently, take the ratio of the integrals directly:Using ${\psi }^2$ instead of $|\psi {|}^2$ would fail to produce a real, nonnegative probability density.

For the quantum harmonic oscillator, put ${\ell }_0=\sqrt{\hslash /(m\omega )}$ and $\xi =x/{\ell }_0$. The normalized energy eigenstates arewhere $H_n$ is a Hermite polynomial. Successive energies are equally spaced; the ground state retains zero-point energy $\hslash \omega /2$. The $n$th wavefunction has $n$ nodes of a wavefunction and parity $(-1{)}^n$. In particular, the first three shapes are proportional to $e^{-{\xi }^2/2}$, $\xi e^{-{\xi }^2/2}$, and $(2{\xi }^2-1)e^{-{\xi }^2/2}$. The classical turning points satisfy $x=\pm {\ell }_0\sqrt{2n+1}$; the wavefunctions extend beyond them and decay, rather than ending abruptly.

The wavefunctions are plotted with arbitrary vertical scale about their energy levels. Their signed excursions are amplitudes, not additional energies or probability densities.

In the position representation in quantum mechanics, $\hat{\mathbf{p}}=-i\hslash \nabla$. Therefore the orbital angular momentum operator givesAt fixed $r,\theta$, the spherical coordinates obey ${\partial }_{\varphi }x=-y$, ${\partial }_{\varphi }y=x$, and ${\partial }_{\varphi }z=0$. The chain rule consequently gives ${\partial }_{\varphi }=-y{\partial }_x+x{\partial }_y$, henceThe derivative holds the other two spherical coordinates fixed; it is not a derivative along an arbitrary spatial curve.

For a fixed magnitude zero point, photometric magnitude uncertainty follows from propagation of uncertaintyThis is the absolute standard uncertainty of the apparent magnitude. The requested fractional uncertainty is insteadThis is the requested fractional standard uncertainty as a function of $m$; there is no single numerical fractional error independent of the magnitude. At $m=0$ it is undefined. Moreover, its value changes when the magnitude zero point changes; the absolute magnitude uncertainty is the useful invariant of such a zero-point shift. Treating the stated percentage as a finite perturbation rather than a standard uncertainty gives changes $-0.1035$ mag for $f\mapsto 1.1f$ and $+0.1144$ mag for $f\mapsto 0.9f$, consistent with the linear estimate.

negative feedback and the ideal operational amplifier's virtual short hold the inverting node at zero potential, with no current entering either input. The parallel feedback resistor and capacitor have electrical impedanceKirchhoff's current law at the inverting node gives $V_{\mathrm{in}}/R_i=-V_{\mathrm{out}}/Z_f$. For the component values read from the circuit,This inverting low-pass amplifier has a DC gain magnitude of 1000 and cutoff frequency $f_c=(2\pi R_{f}C{)}^{-1}=0.15915$ Hz. At the corner, $|G|=707.1$. Above it the gain approaches $1/(2\pi f{R}_{i}C)=159.15/f$, with $f$ in Hz: a slope of $-1$ on logarithmic axes, or $-20$ dB per decade on a Bode plot. At $0.01,1,10,100$ Hz the magnitudes are approximately $998,157.2,15.91,1.592$. These are voltage ratios, not decibels.

Single travelling wave. Differentiation of the plane wave gives ${\partial }_x\psi =ik\psi$, so its probability current isThis is probability density times the free-particle velocity $\hslash k/m$.

Counterpropagating waves. Remove the common time phase and set $z=B^{\ast }A{e}^{2ikx}$. Direct multiplication givesConsequentlyThe interference term $z-z^{\ast }$ is purely imaginary in the last real-part operation. The probability density can have interference fringes even though this net probability current is spatially constant.

Scattering amplitudes. Locate the step at $x=0$ and first take $E>V$. With $k_1=\sqrt{2mE}/\hslash$ and $k_2=\sqrt{2m(E-V)}/\hslash$, useThe common factor $e^{-iEt/\hslash }$ has been suppressed. A finite step with the same mass on both sides requires continuity of $\psi$ and ${\psi }^{\prime }$, giving $1+r_a=t_a$ and $k_1(1-r_a)=k_2{t}_a$. HenceIf $0<E<V$, choose $k_2=i\kappa$, $\kappa =\sqrt{2m(V-E)}/\hslash$, so the transmitted wave decays. The same amplitude formulae continue to hold, but the semi-infinite forbidden region carries no outgoing probability current: $R=1$ and $T=0$.

Flux balance. In the propagating case,In particular,The velocity ratio in $T$ is essential: the squared transmission amplitude alone is not the transmission probability when the two velocities differ.

Numerical transmission. The mass cancels from the ratio, and the energies can remain in the same electronvolt units:Thus about $92.8\%$ is transmitted and $7.18\%$ reflected, despite the incident energy exceeding the step.

Stationary states. Inside the infinite square well, the time-independent Schrödinger equation has sinusoidal solutions. The boundary conditions $\varphi (0)=\varphi (a)=0$ select $k_{n}a=n\pi$; normalization then givesEach wavefunction vanishes outside the well. There is no nonzero $n=0$ state.

Immediate energy measurement after expansion. For the one-sided sudden expansion of an infinite square well, take the original wall at $x=0$ as fixed, so the expanded well occupies $0<x<2a$. In the sudden approximation, the initial wavefunction remains $\sqrt{2/a}\sin (\pi x/a)$ on $0<x<a$, extended by zero elsewhere. The new ground state is ${\chi }_1=a^{-1/2}\sin (\pi x/(2a))$. Its probability amplitude isThe Born rule therefore givesThe motion of the walls matters: a symmetric expansion about the old midpoint would instead give $64/(9{\pi }^2)$. The stated width alone does not explicitly identify the final endpoints. The first result uses the natural continuation of the specified left boundary; if both boundaries are moved, the final location must be supplied. For a new well $(b,b+2a)$ containing $(0,a)$, the overlap is $c_1(b)=(4\sqrt{2}/(3\pi ))[\cos (\pi b/(2a))-\sin (\pi b/(2a))]$.

Evolution of one energy eigenstate. For a time-independent Hamiltonian operator, write $|\psi (t)⟩=f(t)|{\varphi }_n⟩$. Substitution into the Time-dependent Schrödinger equation gives $i\hslash \dot{f}=E_{n}f$, soOnly a global phase changes, leaving all measurement probabilities unchanged.

Delayed energy measurement. Expand the unchanged initial state in the complete new energy eigenbasis,For the fixed-left-wall geometry, direct overlap integration givesorthogonality isolates the new ground-state amplitude as $c_1{e}^{-i{E}_1^{\prime }t/\hslash }$. Its phase cancels on squaring, so the ground-state probability remains $32/(9{\pi }^2)$ for every delay, provided the expanded well remains fixed and no intervening measurement or environment changes the state. The position probability density generally evolves because different eigenstates acquire different relative phases.

Factorization. Put $\alpha =\sqrt{m\omega /(2\hslash )}$ and use the canonical commutation relation $[x,p]=i\hslash$:It follows thatThe $-1/2$ arises from quantum operator ordering; treating $x$ and $p$ as commuting would lose the zero-point energy.

commutators. Direct use of $[x,p]=i\hslash$ gives $[a,a^{†}]=1$. The product rule for commutators then yields $[a^{†}a,a]=-a$ and $[a^{†}a,a^{†}]=a^{†}$, hence

Raising the energy. If $H|\varphi ⟩=E|\varphi ⟩$, thenThus the creation operator produces an energy eigenstate with energy $E+\hslash \omega$. It is nonzero: for a normalized physical oscillator eigenstate, $\Vert a^{†}\varphi {\Vert }^2=⟨a{a}^{†}⟩=E/(\hslash \omega )+1/2>0$. Normalize by this norm if a normalized state is wanted.

Lowest wavefunctions and orthogonality. In the position representation in quantum mechanics,The annihilation condition therefore gives ${\psi }_0^{\prime }=-(m\omega /\hslash )x{\psi }_0$. Applying $a^{†}$ to its solution givesTheir inner product is proportional to ${\int }_{-\infty }^{\infty }x{e}^{-m\omega x^2/\hslash }dx=0$, since the integrand is odd. These are the even and odd lowest wavefunctions shown in A2, with energies $\hslash \omega /2$ and $3\hslash \omega /2$.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

A density operator describes both a quantum state prepared in one pure state and a statistical preparation whose members occupy different pure states. If normalized vectors $|{\psi }_i⟩$ occur with probabilities $p_i$, defineThe resulting operator is Hermitian, positive semidefinite and has trace one. Positivity follows because $⟨v|\rho |v⟩={\sum }_i{p}_i|⟨{\psi }_i|v⟩{|}^2\ge 0$. Conversely, the spectral theorem gives such a representation of any finite-dimensional operator with these properties, using its eigenvectors and eigenvalues. The matrix representing $\rho$ depends on the chosen basis; the operator and its predictions do not.

One operator gives every measurement probability. For a projective measurement with orthogonal projections ${\Pi }_a$ and outcomes $a$,For example, the first formula follows by averaging $⟨{\psi }_i|{\Pi }_a|{\psi }_i⟩$ over the preparation. The trace expression is basis independent. A general measurement replaces ${\Pi }_a$ by positive measurement effects that sum to the identity, without changing the probability formula. Thus $\rho$ specifies the statistics accessible to measurements on that system.

The distinction between pure and mixed states is especially transparent in the eigenvalues. A pure state has $\rho =|\psi ⟩⟨\psi |$, ${\rho }^2=\rho$, and $\mathrm{Tr}{\rho }^2=1$. A mixed state has at least two positive eigenvalues and $\mathrm{Tr}{\rho }^2<1$. For a qubit,pure states lie on the Bloch sphere; mixed states lie inside the Bloch ball. The maximally mixed qubit $\rho =I/2$ can be prepared by choosing equally between the two $z$ spin eigenstates, or equally between the two $x$ spin eigenstates. These ensembles are different preparation procedures but give the same density operator. No measurement on the resulting qubit alone can identify which ensemble was used.

quantum coherence must be distinguished from merely having two populated levels. In the $z$ basis, the state $|+x⟩$ hasBoth give equal probabilities for the two $z$ outcomes. A measurement along $x$ gives the positive result with probability one for the first state and one half for the second. The off-diagonal entries encode the relative phase information responsible for this difference. Whether a matrix entry is off diagonal is basis dependent; purity is not.

For an isolated system, each component evolves with the same unitary time-evolution operator. Therefore $\rho (t)=U(t)\rho (0)U^{†}(t)$ andUnitary evolution preserves the eigenvalues and purity of $\rho$. A density operator commuting with a time-independent Hamiltonian operator is stationary. A thermal equilibrium example is the canonical ensemble ${\rho }_{\beta }=e^{-\beta H}/Z$, where $Z=\mathrm{Tr}(e^{-\beta H})$ and $\beta =1/(k_{\mathrm{B}}T)$. expectation values then reproduce quantum statistical mechanics without assigning the system a definite energy eigenstate on every run.

A mixed state can also arise without classical randomness in the joint preparation. For a bipartite quantum system, the reduced density operator ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ gives every measurement probability accessible to subsystem $A$. For the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, the joint state is pure, whereas ${\rho }_A=I/2$. Its local mixedness reflects entanglement with the inaccessible system; the partial trace discards correlations that matter for joint measurements. More generally, interaction with an environment can suppress off-diagonal entries in a selected basis, producing decoherence while the combined system still evolves unitarily.

Finally, state evolution conditional on a measurement outcome differs from isolated evolution. An ideal projective measurement yielding $a$ updates the state by the Lüders rule $\rho \mapsto {\Pi }_a\rho {\Pi }_a/P(a)$. Discarding the outcome gives ${\sum }_a{\Pi }_a\rho {\Pi }_a$, which generally removes coherence between different eigenspaces. The Von Neumann entropy $-\mathrm{Tr}(\rho \ln \rho )$ quantifies mixedness: it vanishes for pure states and reaches $\ln d$ for $I/d$. Multiplying by $k_{\mathrm{B}}$ gives thermodynamic entropy units. These uses make the density operator a common language for preparation uncertainty, measurement, thermal states and entangled subsystems, while retaining their distinct physical origins.

With two coherent paths, the probability amplitude at a screen is ${\psi }_1(x)+e^{i\varphi }{\psi }_2(x)$. The Born rule givesThe final term shifts when the relative phase changes, producing interference fringes. For slit separation $d$ in the far field, the phase difference is approximately $2\pi d\sin \theta /\lambda$; constructive interference occurs at $d\sin \theta =n\lambda$. Finite slit widths add a diffraction envelope, so an ideal equally bright, infinitely narrow slit model does not predict the full observed envelope.

single-electron interference pattern buildup starts with isolated localized detector events at sufficiently low source intensity. Repeated events accumulate the interference probability distribution. Thus detection is discrete while propagation is described by coherent probability amplitudes. The effect cannot be explained by interference between different particles if the apparatus is occupied by at most one particle at a time. A motivating realization used a field-emission electron source, an electron biprism to recombine two paths, electron lenses and a position-sensitive electron counter; it recorded individual arrivals building up the fringes. This was a biprism realization of two-path interference, not a plate with two literal slits.^[ref]

The crucial comparison is between both coherent paths open and a preparation that supplies which-path information. If the paths become correlated with detector states $|D_1⟩,|D_2⟩$, the joint state is ${\psi }_1|D_1⟩+e^{i\varphi }{\psi }_2|D_2⟩$. Ignoring that detector multiplies the interference term by $⟨D_1|D_2⟩$. Orthogonal detector states remove the term completely, while identical detector states preserve it. This is a consequence of entanglement and the partial trace; a human need not read the detector.

In a real test, check the single-path distributions, detector background, source coherence and stability of the relative phase. Classical phase drift can wash out fringes without supplying which-path information. Conversely, localized detections alone do not establish coherent two-path propagation; the controlled phase dependence of their accumulated distribution provides the discriminating evidence.

orbital angular momentum generates spatial rotations and is $\mathbf{L}=\mathbf{r}\times \mathbf{p}$. With $\mathbf{p}=-i\hslash \nabla$, its components satisfyThus $L^2$ and one component, conventionally $L_z$, have simultaneous eigenstates, while all three components cannot generally have sharp values. On ordinary single-valued spatial wavefunctions,The spherical harmonics form an orthonormal angular basis. The integer restriction is specific to orbital wavefunctions; half-integer spin is an intrinsic degree of freedom, not the angular dependence of a scalar spatial wavefunction.

The angular momentum ladder operators obeyThe ladder ends at $m=\pm \ell$, producing $2\ell +1$ states. A state with specified $m$ has a sharp projection $\hslash m$, but the angular momentum magnitude is $\hslash \sqrt{\ell (\ell +1)}$, not $\hslash \ell$. No classical picture of a vector with simultaneously fixed Cartesian components reproduces all these measurement properties.

In a central potential, rotational symmetry gives $[H,L_i]=0$. separation of variables yields $\psi (\mathbf{r})=R_{\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The radial Schrödinger equation includes a centrifugal potential ${\hslash }^2\ell (\ell +1)/(2m{r}^2)$, which suppresses small-radius amplitude for nonzero $\ell$. The energy is independent of $m$ when rotational symmetry is intact, but need not be independent of $\ell$. A magnetic field can break that degeneracy. Spatial inversion gives $Y_{\ell m}(-\hat{\mathbf{r}})=(-1{)}^{\ell }{Y}_{\ell m}(\hat{\mathbf{r}})$, connecting orbital angular momentum to parity and to the even/odd rotational restriction for the hydrogen spin isomers.

For a nonrelativistic spinless particle in a scalar potential, the Time-dependent Schrödinger equation isAn initial wavefunction together with boundary conditions specifies the evolution. The Hamiltonian operator must be self-adjoint on the appropriate domain for closed-system evolution to be unitary. The equation is linear, so a superposition of solutions is another solution. Its complex wavefunction predicts probability amplitudes; the Born rule separately supplies the connection to measurement probabilities.

For a time-independent potential, separation $\psi (\mathbf{r},t)=\varphi (\mathbf{r})e^{-iEt/\hslash }$ gives the time-independent Schrödinger equation $H\varphi =E\varphi$. Normalizability and boundary conditions select the allowed energies of bound states; scattering states instead use incoming and outgoing boundary conditions and generalized normalization. A general state is a superposition of energy eigenstates, with a spectral integral as well if there is a continuous spectrum. Different energies accumulate different phases, allowing time-dependent position densities and oscillatory expectation values even though a single energy eigenstate has a stationary density.

For a real scalar potential, combining the equation with its complex conjugate yields probability conservation:With no current escaping the boundary, $\int |\psi {|}^2{d}^{3}r$ is constant. Matching conditions at a finite potential step follow from the differential equation and enforce the flux balance used in B6.

The Schrödinger equation is a dynamical postulate of nonrelativistic quantum mechanics, not a consequence of classical mechanics alone. The substitutions $E\mapsto i\hslash {\partial }_t$ and $\mathbf{p}\mapsto -i\hslash \nabla$ motivate its free-particle form from the dispersion relation, but do not derive the measurement postulates. The scalar equation omits intrinsic spin and relativistic particle creation. Spinful or relativistic problems require the corresponding extended theory. Nor does its deterministic unitary evolution, by itself, prescribe conditioning on a particular observed measurement outcome.

Graph. The original plot below places distance on the horizontal axis and recession speed on the vertical axis. Both axes include the origin, so the proportional model and departures from it can be judged directly. The displayed line is the fitted through-origin line, not a sequence joining adjacent observations.

Minimization. Write $S_{dd}={\sum }_i{d}_i^2$ and $S_{dV}={\sum }_i{d}_i{V}_i$. Differentiating the specified chi-squared objective givesThus the unique least-squares minimum isConsequentlyAn unweighted average of $V_i/d_i$ would not minimize this chi-squared objective: equal speed errors give greater slope information to the more distant points. No intercept is fitted because the supplied model constrains it to zero.

Estimated error. For the residual estimate of slope uncertainty through the origin, independent equal-variance speed errors implyEquivalently, completing the square gives ${\chi }^2(H)-{\chi }^2(\hat{H})=S_{dd}(H-\hat{H}{)}^2/{\sigma }^2$. The common $\sigma$ is not supplied. Estimate it from the residuals, using eight residual degrees of freedom after fitting one parameter:The estimated standard error is $s/\sqrt{S_{dd}}=0.0281283$ km s${}^{-1}$ kpc${}^{-1}$, givingwith the uncertainty interpreted as one estimated standard error. If the common errors are Gaussian, $(\hat{H}-H)/(s/\sqrt{S_{dd}})$ has a Student's t-distribution with eight degrees of freedom; an exact confidence interval uses the corresponding quantile rather than treating the estimated $s$ as known. The quoted error is conditional on exact distances, independent constant-variance speed scatter and the proportional model. A common distance-calibration bias or correlated velocities would not be captured by this residual calculation.

A systematic error is a reproducible displacement caused by the apparatus, procedure or measurement model. Repeating the same experiment can determine its mean very precisely while retaining that displacement. Reducing random measurement error therefore does not establish accuracy.

Start with a measurement model listing the response, zero offset, calibration scale and relevant environmental corrections. Compare the instrument with traceable standards at several points across its working range; a single zero check cannot reveal a gain error or nonlinearity. Measure blanks and backgrounds using the same acquisition chain as the signal. Repeat calibration before and after the run to detect drift, and retain the uncertainty of every applied correction.

systematic-error cancellation by reversal separates a wanted effect from an offset with a different symmetry. For example, a resistance measurement with thermoelectric offset $V_0$ gives $V_{+}=RI+V_0$ and $V_{-}=-RI+V_0$. The difference givescancelling a stable offset, while $(V_{+}+V_{-})/2$ diagnoses it. Alternate the current direction quickly enough that the offset is approximately unchanged. This does not cancel a current-scale calibration error or resistance changes from Joule heating; vary the current magnitude and monitor temperature to test those effects. Control measurements must target the actual proposed bias.

Randomizing the order of settings prevents slow drift from consistently imitating a physical trend. Interleaving a reference measurement helps track that drift. Reversing geometry, exchanging matched detectors and changing alignment can expose effects tied to one part of the apparatus. Repeating the measurement with an independent method is particularly valuable when its dominant biases differ; agreement between two instruments sharing the same calibration does not test a common calibration error.

Apply justified corrections, propagate their uncertainties and include correlations when the same calibration enters several observations. A systematic component shared by all observations does not shrink as $1/\sqrt{N}$. A measurement uncertainty budget should distinguish the physical origin of each effect from the method used to evaluate its uncertainty: statistical Type A and other Type B evaluations do not simply mean random and systematic, respectively.^[ref]

Bayes' theorem updates a hypothesis using observed data. For mutually exclusive exhaustive hypotheses $H_j$ and data $D$ of nonzero probability,The prior $P(H_j)$ describes information before these data; the likelihood $P(D\mid H_j)$ describes the sampling model; the denominator is the evidence; and the posterior is the updated probability. likelihood and posterior are different conditional quantities. For a continuous parameter, sums become integrals and the posterior is a probability density.

Consider a counter registering independent events at unknown constant rate $\lambda$, observed for known exposure $T$ with negligible background and dead time. The observed count has a Poisson distribution:Take a proper gamma prior in shape-rate convention,Multiplying by the likelihood adds the count to the shape and the exposure to the rate. Normalization therefore givesHere $\beta$ has exposure units. This is Poisson-gamma conjugacy. Successive independent counting runs update the same posterior by adding their counts and exposures, so the result agrees with analysing all the data together.

A credible interval contains a specified posterior probability for $\lambda$, conditional on the chosen prior and model. For low counts, the posterior can be asymmetric and sensitive to the prior, and a symmetric Gaussian error bar can be misleading. For large $n$ and an exposure dominating prior information, the mean approaches $n/T$ and its standard deviation approaches $\sqrt{n}/T$. Background, missed events or time-dependent rates require a richer likelihood; Bayes' theorem cannot compensate for an incorrect physical counting model. The prior must be specified independently of the same data being used in the likelihood, unless its estimation is explicitly included in a larger model.

shot noise arises from fluctuations in discrete transfer events. For equal charges $q$ arriving independently with a constant Poisson rate $\lambda$, the charge collected during $\tau$ is $Q=qN$, withThus the relative counting uncertainty is $1/\sqrt{\lambda \tau }$. Longer integration improves the precision of the mean current even when the microscopic arrivals remain irregular.

In the frequency range where independent arrivals can be treated as instantaneous pulses, the one-sided current power spectral density isFor an ideal unit-gain rectangular bandwidth $B$, this gives $i_{\mathrm{rms}}=\sqrt{2qIB}$. boxcar averaging of Poisson shot noise over duration $\tau$ has effective noise bandwidth $B_{\mathrm{n}}=1/(2\tau )$, reproducing the counting result $i_{\mathrm{rms}}=\sqrt{qI/\tau }$. The two-sided spectral convention instead assigns $qI$ to each sign of frequency. Stating that convention prevents a factor-of-two error.

A photodiode illuminated at a stable rate provides a useful measurement: subtract the mean current, amplify its fluctuations through a calibrated linear filter, and compare the variance with the mean photocurrent. Independent detected photoelectrons predict a variance proportional to mean current. Measure the dark and amplifier backgrounds separately, check the filter's effective noise bandwidth and vary illumination to distinguish this term from classical intensity noise, whose variance can scale quadratically with mean current. The shot noise model concerns counting statistics, whereas thermal electrical noise can exist in an unbiased resistor. Walter Schottky's motivating analysis treated amplifier limits and independent electron transfers in thermionic discharge tubes.^[ref]

The Poisson result is a model, not a universal consequence of charge quantization. Correlated arrivals, detector dead time, avalanche multiplication or a non-Poisson light source can change the fluctuations. Filtering reduces integrated noise at the cost of temporal resolution. Increasing the detected event rate reduces relative shot noise but raises its absolute variance; balanced photodetection cancels shared classical fluctuations, not the independent shot noise of two uncorrelated detectors.

Original paper: PHYSICS A (2), 28 May 2015; physical PDF pages 106–111.

The photon relation gives $\lambda =hc/E=2.4797\times 10^{-12}\mathrm{m}$. The diffraction-limited angular scale of an aperture of diameter $D$ is thereforeFor a uniformly illuminated circular aperture, the Rayleigh criterion instead gives $1.22\lambda /D=3.03\times 10^{-11}\mathrm{rad}$. Both express the same requested estimate. This is an ideal diffraction limit, conditional on the effective diameter behaving as a coherent imaging aperture; photon energy and diameter alone do not specify the performance of a practical gamma-ray telescope.

At distance $L$, the small-angle approximation gives $\ell =L\theta$. Thus the transverse resolution is approximately $2.5\times 10^{10}\mathrm{m}$, or $3.0\times 10^{10}\mathrm{m}$ with the circular-aperture Rayleigh criterion. This is a transverse separation at the source, rather than a determination of its distance along the line of sight.

Let $T$ be the common tension and ${\mu }_j$ the linear mass densities. The characteristic impedance of a stretched string is $Z_j=T/c_j=\sqrt{T{\mu }_j}$. Continuity of displacement and transverse force at a massless junction gives $1+r=t$ and $Z_1(1-r)=Z_{2}t$. ConsequentlyThe reflected power is $7.18\%$. The minus sign of $r$ represents a phase reversal, not negative power. Both incident and reflected waves travel in the first string, so their power ratio is simply the squared displacement-amplitude ratio. As a check, $\mathcal{T}=(Z_2/Z_1)|t{|}^2=0.92820$ and $\mathcal{R}+\mathcal{T}=1$.

Compare a directly transmitted ray emerging at $A$ with a ray that makes one extra round trip inside the Fabry–Pérot interferometer and emerges at $C$. Both emerge parallel at angle $\theta$ to the plate normal. The extra internal distance is $2d/\cos \theta$, and their emergence points differ laterally by $AC=2d\tan \theta$. The first ray has already travelled $AC\sin \theta$ towards a common outgoing wavefront when the second reaches $C$. The air-gap optical path difference in a Fabry–Pérot interferometer follows because optical path length equals geometric path length in the air-gap approximation:The subtraction is essential: comparing distances only inside the plates would compare different outgoing wavefronts. The corresponding propagation phase difference is $2\pi \Delta /\lambda$. Coating reflection phases can add a separate phase offset; they do not change this geometric optical path difference.

Count two electron spin states in each reciprocal-space cell of volume $(2\pi {)}^3/V$. Filling a Fermi sphere at zero temperature givesThe free electron model then givesThe energy zero is the bottom of the model's conduction band. The supplied carrier density already incorporates the one-electron-per-atom assumption; it must not be multiplied by another atomic count.