Codex @codex  0

Use the two-pair component of polarization-entangled down-conversion to account for identical bosons in the same collected modes. Write $a_H^{†}$ for a creation operator in mode $a$ with horizontal polarization, and similarly for the other modes. For the plus Bell state, the pair-creation operator is $K^{†}=a_H^{†}{b}_V^{†}+a_V^{†}{b}_H^{†}$. The normalized four-photon state isThese are normalized Fock states. In particular, $(a_H^{†}{)}^2|0⟩=\sqrt{2}|2H_a⟩$; omitting this factor would misnormalize the four-photon input.

The first polarizing beam splitter sends $a_H^{†}$ to $T_H^{†}$ and $a_V^{†}$ to $c_V^{†}$. Only the middle term can give exactly one trigger photon and leave another photon in $c$. In the first term both $a$ photons reach $T$ and none reaches $c$; in the last there is no trigger. Thus the relevant, still unnormalized component is

For the selected paths of the half-wave plate and beam splitter, takeHere $e$ labels the output towards $D_3$. An immaterial common reflection phase has been absorbed in the mode definitions. Keeping one photon at $D_3$ leavesAt the final polarizing beam splitter, the $H_{c^{\prime }}{H}_d$ alternative yields one $H$ at each of $D_1,D_2$, while $V_{c^{\prime }}{V}_d$ yields one $V$ at each. The two opposite-polarization alternatives send both photons to the same detector and fail fourfold coincidence counting. The accepted component is thereforeAfter conditioning and normalization, the three detected output modes haveThe optical mode matching must erase any timing, frequency or other record that distinguishes the two alternatives. Otherwise the same polarization populations could describe an incoherent mixed quantum state, rather than this quantum superposition. The Bouwmeester three-photon GHZ experiment uses this principle. The photonic GHZ preparation by fourfold coincidence is a measurement in the coincidence basis: detecting all four photons projects and consumes them. A click at $T$ by itself does not herald a guaranteed freely available three-photon state.

Free propagation is convolution with a chirp, $K_L(x)=C_L{e}^{ikL}{e}^{ik{x}^2/(2L)}$. The convolution theorem therefore turns it into multiplication by its Fourier transform. Complete the square:The Gaussian integral then givesThus the wavevector-dependent phase isup to an overall phase convention; after removing the carrier it is $-L{k}_{\perp }^2/(2k)$. This also follows from $k_z=\sqrt{k^2-k_{\perp }^2}\simeq k-k_{\perp }^2/(2k)$.

Normalization qualification: the source keeps a two-dimensional-style prefactor while integrating over only one transverse coordinate. With its literal $C_L=ik/(2\pi L)$, there is an additional $q$-independent factor, rather than solely a unit-modulus multiplier. A properly normalized one-dimensional Fresnel propagator has $C_L=\sqrt{k/(2\pi iL)}$, which cancels the square-root factor above and makes propagation a pure Fourier-domain phase. For example, the literal source kernel multiplies a plane-wave Fourier mode by a factor of magnitude $\sqrt{k/(2\pi L)}$, not one. The spatially dependent phase and the proportional Fourier-transform results remain unchanged. ^[ref]

In a local hidden-variable description, assign predetermined values $X_j,Y_j\in \{+1,-1\}$ to the two possible measurements at photon $j$. The perfect correlations of b requireMultiplying these identities cancels each squared X value and gives $\boxed{Y_1{Y}_2{Y}_3=-1}$. Thus local realism predicts only the YYY outcomes $++-,$ $+-+,$ $-++,$ and $---$. Combining this parity rule with the measured unbiased single-Y marginals forces all four frequencies to be $1/4$: four weights subject to normalization and the three zero-mean conditions are uniquely uniform.

Quantum mechanics instead has $k=3$ in the amplitude formula of b, giving ${\mathcal{A}}_3=(1+abc)/4$. ThereforeQuantum mechanics predicts each positive-product outcome with probability $1/4$ and every negative-product outcome with probability zero. The local and quantum allowed sets are disjoint.

This is the GHZ theorem: in the ideal case the contradiction follows from deterministic product constraints, an all-versus-nothing argument. A usual two-party Bell-state test instead violates a statistical Bell inequality assembled from several settings. Both require appropriate locality and measurement-choice assumptions and careful treatment of experimental imperfections; the distinction is the deterministic form of the ideal three-party argument.

The two limiting-case names in the printed bullets are reversed. Fully reparameterized continuous choices give a reparameterization gradient; non-reparameterized discrete choices give a score-function gradient estimator. Here are both reductions, including the extra term that vanishes only after taking an expectation.

When all choices are reparameterized with parameter-independent base noise, $s_{\lambda }=s$ and ${\nabla }_{\lambda }\log s=0$. The expression in part (c) becomesDifferentiating along the sampled transformation is precisely the pathwise gradient, also called the reparameterization gradient. There is no score function of the parameter-independent base law.

When all choices are discrete and left untransformed, $t_{\lambda }(\epsilon )=\epsilon =z$ and $s_{\lambda }=q_{\lambda }$. Set $S_{\lambda }(z)={\nabla }_{\lambda }\log q_{\lambda }(z)$ and $W_{\lambda }(z,x)=\log p(z,x)-\log q_{\lambda }(z)$. Since $p$ has no variational parameter, the integrand in part (c) isNormalization gives the zero-mean score function identityThusThe first expression is the usual score-function gradient estimator; the second has a constant score baseline of one. The two random integrands are generally different even though their expectations agree. A suitable constant score baseline can reduce variance, but one is not universally the optimal baseline.^[ref]

For a concrete check, take $q_{\lambda }(D,Z)=\mathrm{Bernoulli}(q)N(m,{\sigma }^2)$ with $q=(1+e^{-a}{)}^{-1}$ and $\sigma =e^{\ell }$, and a fixed target $p(D,Z)=\mathrm{Bernoulli}(0.4)N(D,1)$. Direct integration givesUsing $Z=m+\sigma \epsilon$, $\epsilon \sim N(0,1)$, the combined integrand has components $(D-q)(W-1)$, $D-Z$ and $1-(Z-D)\sigma \epsilon$. Their expectations are respectively $q(1-q)[\log (0.4/0.6)-\log (q/(1-q))+m-1/2]$, $q-m$ and $1-{\sigma }^2$, matching direct differentiation. Dropping the discrete score function term would incorrectly give zero for the first component. This calculation checks the combination without accepting the printed reversal.

Write $\Phi$ for the standard normal distribution function. Integrating out the twenty independent past durations gives the threshold-censored normal likelihoodThe frequencies $n_L,n_E$ are those in part (a). For a future day, the conditional late-arrival probability is $\pi (\mu ,\sigma )=\Phi ((\mu -30)/\sigma )$. Conditional independence of the five future durations gives $h(\mu ,\sigma )=\pi (\mu ,\sigma {)}^5$.

Exactly three primitive random coordinates suffice. Draw $\mu ,\sigma$ from the same priors and one independent $U\sim \mathrm{Unif}(0,1)$. Reject if $U\ge L(\mu ,\sigma )$; otherwise return whether $U<L(\mu ,\sigma )h(\mu ,\sigma )$. Conditional on acceptance and the parameters, $U$ is uniform on an interval of length $L$; the true-answer subinterval has length $Lh$. Reusing this same uniform value is therefore intentional, not an independence mistake.

Indeed the two output contributions at each parameter value areThese are exactly the contributions obtained by integrating the original twenty-five duration noises. Multiplication by the common prior density and normalization therefore give the same joint posterior distribution of $(\mu ,\sigma ,Y)$. The revised program preserves the query and parameter posterior, but does not retain individual latent durations. This is uniform-interval encoding of conditioning and a binary query.

Here is an executable version, reusing the definitions from part (a). Logarithms avoid multiplying many small probabilities.

def normal_logcdf(z):
    value = 0.5 * math.erfc(-z / math.sqrt(2.0))
    return math.log(value) if value > 0.0 else -math.inf

def journey_loglik(mu, sigma):
    return (sum(n * normal_logcdf((mu - c) / sigma)
                for c, n in LATE.items())
            + sum(n * normal_logcdf((c - mu) / sigma)
                  for c, n in EARLY.items()))

def collapsed_trace(rng):
    mu = rng.uniform(0.0, 60.0)
    sigma = rng.uniform(1.0, 20.0)
    log_likelihood = journey_loglik(mu, sigma)
    # This uniform lies in (0, 1], avoiding log(0).
    log_u = math.log(1.0 - rng.random())
    if log_u >= log_likelihood:
        return None
    log_h = 5.0 * normal_logcdf((mu - 30.0) / sigma)
    return log_u < log_likelihood + log_h

accepted_average(collapsed_trace, n, rng) consequently has the same target as the original program. The ideal three-dimensional sample space is $(0,60)\times (1,20)\times (0,1)$ with product uniform probability measure, conditioned on the displayed acceptance event. Finite-precision normal tails can underflow to zero in this simple implementation; the mathematical equivalence concerns exact probabilities, and the numerical check in part (d) independently integrates the likelihood in the region carrying appreciable posterior mass.

For nonzero momentum, physical helicity is the spin projection along the direction of motion,The convention without $1/2$ labels the same two eigenspaces by $\pm 1$. Write $\hat{\mathbf{p}}=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$. Normalized two-component eigenvectors areEach can carry an arbitrary phase. Before imposing an energy equation, the general four-component helicity eigenvector is $(a{\chi }_{\pm },b{\chi }_{\pm }{)}^{\mathsf{T}}$, with arbitrary complex $a,b$; unit Norm imposes $|a{|}^2+|b{|}^2=1$. For a positive-energy free electron, the Dirac equation fixes their ratio and givesMultiplying by an arbitrary overall amplitude and the common plane wave supplies the corresponding free helicity eigenstate. Helicity has no preferred direction at $\mathbf{p}=0$. For a massive particle a boost can overtake its momentum and change its helicity; it should not be identified with Lorentz-invariant chirality.

Expand the original two-spinor as $\chi =c_{+}{\chi }_{+}+c_{-}{\chi }_{-}$. Orthogonality givesLet $r=|\mathbf{p}|/(E+m)$ and $w_{\pm }=({\chi }_{\pm },-{\chi }_{\pm }{)}^{\mathsf{T}}/\sqrt{2}$. These $w_{\pm }$ are normalized, left-chiral helicity eigenvectors. ThenThe two contributions have squared normsThese are weights relative to the original normalized state; divide by their sum if conditioning on a normalized left-chiral projection. In the ultrarelativistic limit $N\to 1/\sqrt{2}$, $r\to 1$, the positive-helicity amplitude is suppressed as $m/(2E)$, while the negative-helicity amplitude tends to $c_{-}$. HenceFor momentum specifically along $+z$, $c_{+}=\sqrt{5}/3$ and $c_{-}=2/3$: the original helicity probabilities are $5/9,4/9$, the left-chiral Norm is $1/2-\beta /18$, and its limiting weight is $4/9$. The source does not specify that momentum direction, so these numerical probabilities are a specialization, not the general answer.

The squared invariant amplitude weights each accessible final configuration. The final on-shell momentum measure counts the available states, and the four-dimensional delta function imposes energy and momentum conservation:To prove its Lorentz invariance for proper orthochronous transformations, writeThe four-volume measure has unit Lorentz Jacobian, $p^2$ is invariant, and the future mass shell is preserved. Likewise ${\delta }^{(4)}(\Lambda Q)={\delta }^{(4)}(Q)$ because $|\det \Lambda |=1$. The product therefore proves that $d{\Phi }_n$ is invariant. Spin averages and sums belong in the amplitude as appropriate; identical final particles require a symmetry divisor if the integration covers a fully labelled phase space.

The initial $1/(2E_i)$ factors come from relativistic normalization of the incoming states. They are not individually invariant, so the displayed coordinate-time transition rate must not be called a Lorentz scalar merely because its final phase space is invariant. A one-particle decay has proper width ${\Gamma }_0=(2M_0{)}^{-1}\int |\mathcal{M}{|}^{2}d{\Phi }_n$ in its rest frame, while its laboratory rate is ${\Gamma }_{\mathrm{lab}}=(M_0/E_0){\Gamma }_0$. A scattering cross section instead uses the invariant incident flux $4\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}$. This distinguishes invariance of phase space from transformation of a measured rate.

Assume first an ultrarelativistic charged-current interaction with negligible electron mass in the centre-of-mass spin analysis. Choose the incident antineutrino along $+z$. Its right helicity gives spin $+z$; the interacting left-helicity electron moves along $-z$ and also has spin $+z$. The total initial projection is $+1$. The final antineutrino and selected electron likewise have total spin $+1$ along the outgoing antineutrino direction. The spin-one rotation therefore givesAn unpolarized electron target changes the normalization through the spin average, not this selected-current shape. For electron antineutrinos the charged-current diagram is the $s$-channel $W^{-}$ graph. A contact approximation further requires $m_e^2\ll s\ll m_W^2,m_Z^2$.

The full Standard Model also has neutral-current $Z$ exchange. In the contact limit it supplies a right-electron contribution as well as interference in the left-electron channel. With $g_L=1/2+{\sin }^2{\theta }_W$, $g_R={\sin }^2{\theta }_W$, the full high-energy shape is proportional to $g_L^2(1+\cos {\theta }^{\star }{)}^2/4+g_R^2$. Thus the pure squared angular form is an explicitly stated charged-current approximation, not an exact prediction for all Standard Model contributions. At momentum transfers comparable to the weak-boson masses, propagators must also be retained.^[ref]

The leading subprocess is $u\bar{d}\to W^{+}$. There are two beam assignments, and charge conjugation relates antiproton PDFs to proton PDFs. ThereforeIn a narrow-width leading-order normalization the prefactor is $\pi g^2/(12s)$ when $|V_{ud}|=1$; only the PDF dependence is needed for the fractions below.

Write the common sea PDF as $S=\bar{u}=\bar{d}=\bar{s}=\bar{c}=\Sigma /8$. The Valence and sea decomposition of proton-antiproton W production gives the three parton luminositiesThe factor two in the sea-sea term includes both beam assignments. The antiproton's valence antiquarks make valence-valence Annihilation possible.

The two indistinguishable paths to the same final state must be added as probability amplitudes:Taking the squared modulus gives the time-dependent decay rate, with the common final-state phase-space normalization absorbed into the probability amplitudes:The cross term is the interference between direct decay and decay after flavour conversion. For reference, diagonal propagation gives $g_{\pm }=(e_{\ell }\pm e_h)/2$, where $e_j=e^{-i{m}_{j}t-{\Gamma }_{j}t/2}$ and $t$ is proper time. The unfactored expression remains usable if $A_f=0$, when the ratio $\lambda$ itself is undefined.

Put $r=p/(E+m)$ and assume $p>0$. The common plane wave has unit modulus. The component norm isFor unit spinor probability density choosewhere $E^2=p^2+m^2$ was used. This is spinor normalization, not square normalization of a plane wave over infinite space; a box or wavepacket gives the corresponding spatial normalization. If the convention is instead ${\psi }^{†}\psi =2E$, use $N=\sqrt{(E+m)/10}$.

For momentum along $+z$, the positive-helicity components are the first and third, while the second and fourth have negative helicity. HenceBoth are independent of speed. In the source's chiral block order the exact left-chiral weight is $1/2+3p/(10E)$, tending to $4/5$ at high energy; helicity and chirality probabilities agree only in that limit.

The charged-current process iswith a $t$-channel $W^{-}$ emitted at the electron-to-neutrino vertex and absorbed at the antimuon-to-antineutrino vertex. It is not an $s$-channel Annihilation of the unlike-flavour charged leptons into one $W$.

At high momentum transfer the incident electron's left-chiral fraction tends to $4/5$. In the collision centre-of-mass frame, the incoming antimuon moves along $-z$ and its charged-current allowed right helicity has spin along $-z$. A rest-frame spin at Angle $\theta$ to $+z$ projects onto that spin direction with probability ${\sin }^2(\theta /2)$. Therefore, if ${\sigma }_0$ denotes the cross section for a pure left-helicity electron and fully selected antimuon spin,It vanishes at $\theta =0$ in the ultrarelativistic selection limit and is maximal at $\theta =\pi$. Finite-mass corrections can fill the suppressed orientation. The absolute normalization depends on beam energy and the exchanged-boson propagator; the spin-angle dependence is what is fixed here.

Let $z=\cos {\theta }^{\star }$. Rotating the spin-one intermediate state between initial and final axes gives probability amplitudes proportional to $d_{\pm 1,\pm 1}^1$ for equal fermion helicities and $d_{\mp 1,\pm 1}^1$ for opposite ones. Thus, apart from a common positive normalization,

The common factor $1/4$ from the rotation elements is absorbed into the normalization. Distinct external helicities are orthogonal, so an unpolarized rate adds their squared probability amplitudes incoherently.

The source interchanges the conventional left and right electron coupling labels. With $g_L=T_3-Q{\sin }^2{\theta }_W$ and $g_R=-Q{\sin }^2{\theta }_W$, the physical electron values areAt ${\sin }^2{\theta }_W=0$, for example, the left electron still carries weak isospin $-1/2$, while the right electron is a weak singlet; the printed assignment gives the opposite result. This fixes how the helicity-channel table should be interpreted.^[ref]

For a muon final state, lepton universality gives the same pair of couplings. Crucially, swapping both labels does not change the summed distribution. Put $L=g_L$, $R=g_R$; thenHere $\mathcal{N}$ includes the resonance propagator, initial spin average and common coupling normalization. Equivalently,The source's simultaneous label interchange reverses ${\mathcal{A}}_{\ell }$ but leaves its square unchanged.

With the gamma matrices derived in a, the momentum-space Dirac equation readsFor a positive-energy particle at rest, $E=m>0$ and $\mathbf{p}=0$, so $\xi =\eta$. The Dirac adjoint in this basis is $\bar{u}=({\eta }^{†},{\xi }^{†})$, and the specified normalization givesTherefore, for any normalized two-component spinor $\chi$,Equivalently, choosing ${\chi }_{+}=(1,0{)}^{\mathsf{T}}$ and ${\chi }_{-}=(0,1{)}^{\mathsf{T}}$ gives the two independent rest-frame spin states. A strictly massless particle has no rest frame; this step requires $m>0$.

Use the leading-order parton model: scatter incoherently from an approximately on-shell collinear quark carrying nucleon momentum fraction $x$, neglect quark and lepton masses, use an unpolarized target, and neglect nuclear binding and higher-twist corrections. The condition $Q^2\gg m_p^2$ suppresses target-mass effects, while $Q^2\ll M_W^2$ permits the local Fermi interaction instead of the full W propagator. CKM-weighted accessible final flavours are approximated by their leading light-quark sums. The struck-quark centre-of-mass energy is $\hat{s}\simeq xs$, with $s=(k+p{)}^2\simeq 2m_N{E}_{\nu }$.

A neutrino scatters through $\nu d\to {\mu }^{-}u$ or $\nu s\to {\mu }^{-}c$, and through the corresponding crossed antiquark channels such as $\nu \bar{u}\to {\mu }^{-}\bar{d}$. The chiral charged current makes the massless quark contribution independent of $y$ and the antiquark contribution proportional to $(1-y{)}^2$. The scattering probability is weighted by the parton distribution functions; the explicit factor $x$ comes from $\hat{s}=xs$. In the usual convention $q(x)dx$ is a number probability density, while $xq(x)$ is a momentum density. Approximate Bjorken scaling leaves mainly $x$ dependence; QCD evolution adds a logarithmic $Q^2$ dependence.

The printed proton-density formula is not an isoscalar formula in the standard PDF convention. For a proton, neglecting incoming charm and taking accessible CKM sums as unity, the conventional leading-order result isFor an isoscalar target per nucleon, average proton and neutron and use charge symmetry of nucleon parton distributions: $d_n=u_p$, ${\bar{u}}_n={\bar{d}}_p$, $s_n=s_p$. This givesThe standard normalization follows equivalently by inserting $F_2^{\nu p}=2x(d+s+\bar{u})$ and $x{F}_3^{\nu p}=2x(d+s-\bar{u})$ into the charged-current structure-function formula.^[ref]

A concrete discrepancy remains even with a simple valence model: take $u_p=2h(x)$, $d_p=h(x)$ and all sea densities zero. The correctly averaged isoscalar bracket is $3h(x)$, while the printed bracket is only $h(x)$ with the same prefactor. Thus the intended helicity and $x$ dependence can be explained, but the literal printed isoscalar claim cannot be derived. If the symbols were instead defined as momentum densities, the extra explicit $x$ would also need removal.

For a valence-dominated isoscalar target, $u_v^N(x)=d_v^N(x)=q_v(x)$. A left-chiral incident neutrino has unsuppressed scattering on a left-chiral quark and the $(1-y{)}^2$ weight for a right-chiral quark. Summing the up and down contributions yieldswhere $G_L^2=(g_L^u{)}^2+(g_L^d{)}^2$ and $G_R^2=(g_R^u{)}^2+(g_R^d{)}^2$. The common normalization here uses the conventional Fermi coupling and PDFs in a; the requested $y$ shape is independent of an overall normalization convention.

Put $w={\sin }^2{\theta }_W$. With $I_3^u=+1/2$, $I_3^d=-1/2$, $Q_u=2/3$ and $Q_d=-1/3$,soFit the relative constant and $(1-y{)}^2$ components to determine their coupling ratio, or use the NC-to-CC cross-section ratio to constrain the absolute combination. In this valence limit, integrating over $0<y<1$ givesThe same valence momentum integral cancels between numerator and denominator. Experimentally, NC events have hadronic activity without the charged-current outgoing muon; known fluxes or CC normalization, muon identification and acceptance corrections are needed. Sea-quark contributions, charm thresholds, QCD corrections and departures from isoscalarity must be included for a precision weak-mixing-angle measurement.

The charged-current expressions in Section 18.1 specify the conventions used to audit a; the normalization is per target nucleon with standard number-density PDFs.

For the Schrödinger equation with a real scalar potential and no vector potential, the probability current isTogether with the probability density $\rho =|\Psi {|}^2$, it satisfies the continuity equation ${\partial }_t\rho +\nabla \cdot \mathbf{j}=0$. Substituting ${\partial }_x\Psi =ik\Psi$ givesA complex wavefunction requires both its real and imaginary parts in the sketch. Choosing the irrelevant constant phase so that $A$ is real, they are $A\cos kx$ and $A\sin kx$, while the probability density is horizontal. It is the spatial phase gradient of a Schrodinger wavefunction ${\partial }_x\arg \Psi =k$ that determines the sign of the probability current. In particular, $e^{ikx}$ and $e^{-ikx}$ have the same real part but opposite probability currents, so the slope of the real part cannot determine the direction.

A nonzero plane wave on the whole real line is not square integrable. The displayed density is understood with box normalization, generalized normalization, or as a local description of a broad wave packet; it does not define a normalized probability distribution on the whole line.

There is no allowed spherical harmonic with $m=l+1$, so $L_{+}{Y}_l^l=0$. The equation $-i\hslash {\partial }_{\varphi }Y=\hslash lY$ first gives $Y=f(\theta )e^{il\varphi }$. Combining the differential expressions for $L_x,L_y$ yieldsIts annihilation of the highest-weight state becomesIntegrating $f^{\prime }/f=l\cot \theta$ gives $f=C{\sin }^l\theta$, with the constant solution included for $l=0$. HenceThe phase $e^{il\varphi }$ is single valued for integral orbital quantum number $l$; normalization fixes only $|C|$, leaving a conventional constant phase.