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A Bell test compares experimental probabilities or expectation values with a Bell inequality. In a two-setting photon polarization test, each station chooses a polarizer orientation and records an outcome. A coincidence circuit identifies candidate pairs using an agreed timing rule. The choice of which trials enter the statistic is part of the experiment: correlations calculated only from detected pairs can differ from correlations of all emitted pairs. The Polarization CHSH count model separates ideal expectation values from this experimental selection problem.

Choose the qubit computational basis $|0⟩,|1⟩$, also called the $Z$ basis, and the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, also called the $X$ basis. Compare a coherent preparation of $|+⟩$ with choosing $|0⟩$ or $|1⟩$ at random with equal probabilities and discarding that label:These are original ideal preparation models. Their $Z$-basis probabilities are both $(1/2,1/2)$, because their quantum state populations agree. Their $X$-basis probabilities are respectively $(1,0)$ and $(1/2,1/2)$. Indeed $⟨+|{\rho }_{+}|+⟩=1$, whereas $⟨+|I/2|+⟩=1/2$. The missing off-diagonal quantum coherence therefore changes a prediction that the common populations cannot determine.

For a relative quantum phase $\varphi$, the coherent family instead hasThe last formula follows by adding or subtracting the two probability amplitudes before taking the squared modulus. At $\varphi =\pi$, the state is $|-⟩$ and the $X$ outcomes switch to $(0,1)$, while the $Z$ outcomes remain $(1/2,1/2)$. In contrast, $-|+⟩$ changes both components by the same global phase and retains the $X$ outcomes $(1,0)$. In photon polarization, the identification $|0⟩=|H⟩$, $|1⟩=|V⟩$ makes $|+⟩$ and $|-⟩$ the two orthogonal diagonal linear polarizations; these labels describe measurement bases, not spatial axes for photon spin.

Even perfect statistics in one basis cannot certify a preparation. The above $Z$ data fit both density matrices. Conversely, distinct preparation histories can yield the same quantum state:The off-diagonal terms cancel in the right-hand ensemble. Every measurement on the qubit alone then has identical probabilities for those two histories. A retained classical label would allow conditioning on different members of the quantum state ensemble; it is extra information beyond the unlabelled density operator.

Comb self-referencing compares two parts of one optical frequency comb. Coherent broadening supplies an octave, a factor of two in frequency. Second-harmonic generation doubles a red tooth; a photodiode detects its beat with an original blue tooth.^[ref] Applying the tooth formula givesThe doubled branch carries $2f_0$, while the original high-frequency branch carries $f_0$: their difference retains exactly one offset.

The displayed difference uses the order doubled tooth minus original tooth. An ordinary electronic frequency counter returns a nonnegative beat magnitude; spectral selection and known electronic shifts must identify whether it represents $f_0$, $f_r-f_0$, or a shifted version. Coherent broadening, overlapping photon polarization, spatial alignment and temporal overlap are required. Merely producing a broad patch of colored light does not establish a usable offset signal.

In the stationary-state terminology used here, an excited state is a normalized energy eigenstate with energy eigenvalue $E>E_0$, where $E_0$ is an attained ground-state energy. Its excitation cost above the ground state is $\Delta E=E-E_0>0$. Several energy eigenstates can share one excited energy eigenvalue through degenerate energy levels. An arbitrary quantum superposition with energy expectation value above $E_0$ need not be a single stationary excited state; its energy content is described by its Born rule probabilities or, for a continuum, its scalar spectral measure.

An excited state can be a stationary state: the second infinite square well mode is an immediate example. Spontaneous photon emission is a transition caused by coupling matter to the electromagnetic field; it is absent from an isolated matter-only Hamiltonian operator. The relevant energy eigenstate of that isolated model is therefore not automatically an energy eigenstate of the full interacting system. This distinction explains how an ideal stationary excited state can approximate a state with a finite experimental lifetime. See electric-dipole interaction for the coupling used in the simplest optical model.

For a one-photon transition between stationary internal states, neglecting recoil, conservation of energy requiresHere $h$ is the Planck constant and $\nu$ is the photon frequency. Energy matching alone does not guarantee a transition: in the electric-dipole interaction approximation its strength also involves the matrix element $⟨f|\mathbf{d}\cdot \mathbf{ϵ}|i⟩$, where $\mathbf{d}$ is the electric dipole moment operator and $\mathbf{ϵ}$ specifies photon polarization. A zero matrix element gives an electric-dipole selection rule. Thus an absent optical line does not prove the corresponding energy eigenvalue is absent; a different interaction can couple different states. ^[ref]. The Franck-Hertz experiment gives a particularly instructive comparison between collision excitation and photon emission.

The first excited state terminology assumes a least energy eigenvalue above $E_0$. Such a next level is not guaranteed by quantum mechanics: a continuous spectrum can start immediately above the spectral bottom, or energy eigenvalues can accumulate there. For example, a Hamiltonian operator with energy eigenvalues $0,\epsilon ,\epsilon /2,\epsilon /3,\dots$, where $\epsilon >0$, has a ground-state energy zero but no least positive energy eigenvalue: every candidate $\epsilon /n$ is exceeded downward by $\epsilon /(n+1)$. A free particle on the whole line has neither a ground state nor a first stationary excited state. Each use of a discrete excitation ladder therefore needs a specified Hamiltonian operator and its boundary conditions.

A polarizing beam splitter converts a photon polarization qubit to two paths. Rotating one arm's photon polarization to match the other and recombining at a balanced beam splitter gives $P_1(\theta )=1/2+\mathrm{Re}({\alpha }^{\ast }\beta e^{i\theta })$. Measurements at phases zero and $\pi /2$ recover the transverse Bloch vector components; direct counts in the two separated arms recover the population difference. All three components, rather than a fringe alone, determine an arbitrary photon polarization density matrix.

// Target: quantum-theory.bigb

Interferometric visibility measures the contrast of a scanned interference fringe. For the mutually coherent, equally polarized two-beam model in wave interference, with fixed individual wave intensities and $I_1+I_2>0$, the constructive interference intensity and destructive interference intensity giveEqual nonzero wave intensities give $\mathcal{V}=1$; $I_1=1,I_2=0.25$ give $\mathcal{V}=0.8$. Unequal wave intensities can therefore reduce interferometric visibility even with perfect mutual optical coherence. Random phase differences or distinguishable wave polarizations can reduce it further; zero contrast alone does not identify the cause. Both intensities zero leave the ratio undefined.

For normalized paths with complex amplitudes $\alpha ,\beta$ and identical internal states, the output probability is $1/2+\mathrm{Re}({\alpha }^{\ast }\beta e^{i\theta })$, so $\mathcal{V}=2|\alpha \beta |$. A photon polarization fringe alone determines transverse quantum coherence but not the sign of the population difference.

// Target: optics.bigb

Let $\Sigma =m_e^2+m_t^2$, $Q_e=-1$, $Q_t=2/3$ and $N_c=3$. The single tree-level Feynman diagram is annihilation through a virtual photon. Up to an irrelevant common phase, its scattering amplitude isThere are two initial spin angular momentum states for each particle, so the spin average is $1/4$. The final color charge sum gives ${\sum }_{c,d}|{\delta }_{cd}{|}^2=N_c$. There is no additional photon polarization average for a virtual photon and no separate polarization label beyond the Dirac spinor spin sums.

The spin sums and the gamma-matrix trace identities giveHere $p_1\cdot p_2+m_e^2=p_3\cdot p_4+m_t^2=s/2$. Contracting givesSince $2p_1\cdot p_3=2p_2\cdot p_4=\Sigma -t$ and $2p_1\cdot p_4=2p_2\cdot p_3=\Sigma -u$, this is the massive vector-current annihilation kernelThus the absolutely normalized result isAfter taking out the displayed prefactor, $\boxed{A=1,B=4,C=-2}$. Because the source uses a proportionality sign, only their ratio is fixed until that prefactor is specified. Equivalently, after taking out just $e^4/s^2$, their values are $8/3,32/3,-16/3$. No mass has been neglected. At $m_e=0$ and $s=4m_t^2$, the bracket yields the finite isotropic threshold value of the scattering amplitude; the vanishing threshold scattering cross-section comes from relativistic two-body phase space.

In a large periodic quantization volume $\mathcal{V}$, one wavevector state occupies $(2\pi {)}^3/\mathcal{V}$ in wavevector space. There are two transverse photon polarizations, and each oscillator contributes $\hslash ck/2$. A lower wavelength cutoff ${\lambda }_{\min }$ corresponds to $k_{\max }=2\pi /{\lambda }_{\min }$, so the energy density isFor ${\lambda }_{\min }=10^{-10}\mathrm{m}$,The factor two is polarization, not an extra independent count of positive and negative frequencies. The continuum wavevector integral already counts independent travelling-wave modes in the usual field quantization convention. The quartic cutoff dependence also shows why an assumed shortest wavelength is crucial.

For thermal photons, the available thermal length is ${\ell }_T=\hslash c/(k_{B}T)$. The number of occupied modes per volume scales as ${\ell }_T^{-3}$, while their characteristic energy is of order $k_{B}T=\hslash c/{\ell }_T$. Thus $n_{\gamma }\propto {\ell }_T^{-3}$ and $u_{\gamma }={\rho }_{\gamma }{c}^2\propto \hslash c{\ell }_T^{-4}$. More explicitly, the two photon polarizations and the Bose-Einstein distribution giveconsistent with the rounded coefficients supplied in the source.

Use its definition $s=n_{\gamma }/n_B$, with $n_B={\rho }_B/m_p={\Omega }_B{\rho }_c/m_p$ and ${\rho }_c$ the mass-equivalent cosmological critical density. Writing $Q=k_{B}T/(\hslash c)$, the photon energy formula gives $Q^4={\Omega }_{\gamma }{\rho }_{c}c/(0.66\hslash )$. ThereforeThis photon-to-baryon ratio from density parameters has leading coefficient $0.328\simeq 0.33$, as required. For an illustrative estimate, take the stated ${\rho }_c=10^{-26}\mathrm{kg}{\mathrm{m}}^{-3}$, ${\Omega }_{\gamma }=5\times 10^{-5}$ and ${\Omega }_B=0.05$. Then $\boxed{s\sim 1.4\times 10^9}$, i.e. of order a billion photons per baryon. The source alone supplies no numerical density parameters, so an estimate must state such additional inputs.

Strictly, this $s$ is the photon-to-baryon ratio, rather than the thermodynamic entropy per baryon itself. For a thermal photon gas, $S_{\gamma }/(k_B{N}_B)=[4u_{\gamma }/(3k_{B}T{n}_{\gamma })]s\simeq 3.6s$ with the exact blackbody coefficients. The ratio used in the source is proportional to the radiation entropy per baryon under adiabatic evolution.

In quantum dense coding, a shared Bell pair lets Alice send two classical bits by transmitting one qubit. Use photon polarization as the computational basis, $|0⟩=|H⟩$, $|1⟩=|V⟩$, and define the Bell statesAlice and Bob initially share $|{\varphi }^{+}{⟩}_{AB}$. Alice encodes a two-bit message $(m,n)$ by applying the Pauli operator $X^n{Z}^m$ to her qubit:The last minus sign is a global phase. Alice transmits her photon to Bob, who now holds both qubits and uses a Bell-basis measurement to identify the message. The four Bell states are orthogonal, so an ideal Bell state analyzer decodes them without error. The resource cost is one previously shared Bell pair and one transmitted qubit per two classical bits; the shared quantum entanglement is consumed. Before transmission Bob's reduced density matrix is $I/2$ for every message, consistent with the no-communication theorem.

The BBO crystal supplies the polarization-entangled photon pair. An ultraviolet laser pumps this nonlinear optical crystal; spontaneous parametric down-conversion produces signal and idler photons with ${\omega }_p={\omega }_s+{\omega }_i$ and phase matching ${\mathbf{k}}_p={\mathbf{k}}_s+{\mathbf{k}}_i$. In a noncollinear type-II phase matching arrangement, collect the intersections of the two emission cones. The alternatives $|H{⟩}_A|V{⟩}_B$ and $|V{⟩}_A|H{⟩}_B$ must be indistinguishable in arrival time, spectrum and spatial mode. Compensation of birefringence and adjustment of the relative quantum phase then give $(|HV⟩+|VH⟩)/\sqrt{2}$; a local Pauli X gate converts it to $|{\varphi }^{+}⟩$. Half-wave plates and other waveplates implement Alice's Pauli gates. The BBO is the pair source; the message is encoded subsequently by the local optical elements.^[ref]

A basic passive linear optics Bell state analyzer with no extra resources identifies only two of the four polarization Bell states unambiguously. Thus the ideal two-bit capacity presupposes complete Bell-basis measurement; that basic optical implementation alone does not attain it deterministically.

In entanglement swapping, start with independent Bell pairs $|{\varphi }^{+}{⟩}_{12}|{\varphi }^{+}{⟩}_{34}$. Send qubits 2 and 3 to a central Bell state analyzer, retaining 1 and 4 at separate endpoints. Reordering the tensor product factors to group 23 and 14 gives the identityConsequently each ideal Bell-basis measurement outcome has probability $1/4$ and projects 14 onto the corresponding Bell state, even though those two photons have never interacted. Communicating the two-bit result permits an endpoint Pauli gate to convert every outcome into the same chosen Bell pair. Without that record, the endpoints have ${\rho }_{14}=I_4/4$, a product state, rather than a known entangled state.

The optical Bell state analyzer overlaps photons 2 and 3 at a balanced beam splitter, with matched spatial modes, spectra and arrival times, and resolves the output photon polarizations using polarizing beam splitters and single-photon detectors. Two-photon interference maps Bell states to different output patterns. The antisymmetric $|{\psi }^{-}⟩$ gives one photon in each output with opposite polarizations. The symmetric $|{\psi }^{+}⟩$ gives two opposite-polarization photons in the same output; the two polarization detectors there can resolve this pattern. The $|{\varphi }^{\pm }⟩$ states give same-polarization pairs and remain mutually ambiguous in this basic setup. The detected pattern therefore heralds a particular endpoint Bell state. With ideal detection this arrangement succeeds unambiguously on two of four equally likely inputs, with probability $1/2$; the successful events can be selected. A complete analyzer would accept all four outcomes. The endpoint quantum entanglement is conditional on the measurement record, so it enables no instantaneous signalling.^[ref]

For quantum key distribution with Bell pairs, a source distributes one photon to Alice and its partner to Bob. One implementation is the Ekert protocol, using the spin singlet state $|{\psi }^{-}⟩=(|01⟩-|10⟩)/\sqrt{2}$. Each party independently chooses a polarization measurement setting after distribution. Equal settings give opposite outcomes; Bob reverses his outcome-to-bit assignment so these rounds give matching random bits. Other setting pairs test a Bell inequality, with ideal spin singlet state correlation $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$ for the corresponding Bloch sphere axes. The optical polarizer angles correspond to twice those angles on the appropriate Bloch sphere great circle.

Over an authenticated public classical channel, Alice and Bob announce settings and detection times, select coincident detection rounds, and perform key sifting. They reveal a test sample to estimate the quantum bit error rate and the chosen Bell inequality signal, discarding all revealed bits. Excess disturbance causes an abort. Otherwise information reconciliation makes the retained strings agree and privacy amplification shortens them to remove an adversary's residual information, including information disclosed during reconciliation. The tested quantum entanglement supplies correlations unavailable to an independently prepared classical mixture; a security claim also requires stated detector and sampling assumptions.^[ref]

Each photon polarization analyzer uses a setting-dependent waveplate followed by a polarizing beam splitter. A single-photon detector on each output converts the routed photon into an electrical click, recording the outcome and its time. For example, a single-photon avalanche diode is a reverse-biased photodiode operated above breakdown: absorption creates a charge carrier, impact ionization triggers an avalanche, and the resulting pulse is registered. A quenching circuit lowers the bias to stop the avalanche and then resets the device. The detector dead time during recovery limits the count rate; dark counts produce false clicks, and imperfect detection loses rounds. The polarization information comes from the preceding analyzer, while the detector registers which output fired. Coincidence timing associates Alice's and Bob's records with the same emitted pair. ^[ref]

Vary the Maxwell Lagrangian with respect to $A_{\nu }$. Since $\delta F_{\mu \nu }={\partial }_{\mu }\delta A_{\nu }-{\partial }_{\nu }\delta A_{\mu }$ and $F$ is antisymmetric,After integration by parts, the Euler-Lagrange field equation isThe gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. This gauge invariance means that potentials related in this way represent the same electromagnetic field; all four potential components are not independent physical photon degrees of freedom. In particular, the ungauged kinetic operator is singular and its temporal canonical momentum vanishes.

Choose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. The field equation then becomes $\boxed{\square A_{\mu }=0}$. Residual gauge transformations with $\square \chi =0$ preserve this condition. The correctly named Lorenz gauge condition concerns the divergence of the potential; Lorentz covariance is the separate transformation property of the equations.

For the alternative quadratic Lagrangian density, direct differentiation givesso its Euler-Lagrange field equation is also $\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction. Indeed, with $D={\partial }_{\mu }{A}^{\mu }$,The divergence identity follows by expanding ${\partial }_{\mu }{K}^{\mu }$ and commuting its second derivatives. The quadratic density alone describes four wave equations; the gauge constraint must still be incorporated in the physical states. It is not an unconstrained four-positive-polarization replacement for Maxwell equations.

Using this specified boundary-term convention, the canonical momentum conjugate to $A_{\nu }$ isThe canonical commutation relations are thereforewith field-field and momentum-momentum brackets zero. Equivalently $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The temporal component has the opposite sign to the spatial components; replacing the metric by a positive Euclidean one would destroy these brackets.

The wave equation gives massless frequencies ${\omega }_{\mathbf{k}}=|\mathbf{k}|$. Hermiticity pairs its positive- and negative-frequency solutions, so the Heisenberg picture field has the expansionThis is a real field expansion for each component, but its oscillator metric is indefinite. The assumed brackets, with the dagger's vector index restored, areThey verify the photon oscillator completeness and canonical brackets. Inserting ${\Pi }^{\nu }=-{\dot{A}}^{\nu }$ gives at equal timesThe other brackets cancel between the two frequency contributions. Thus both the $2\omega$ normalization and the sign $-g_{\mu \nu }$ are needed.

The Gupta-Bleuler formalism imposes the Lorenz gauge condition weakly on states:Here $(+)$ is the annihilation, positive-frequency part. Setting the full divergence to zero as an operator would conflict with the four-component canonical commutation relations. The subsidiary condition instead selects a physical pre-space inside the covariant photon Fock space, followed by a quotient of its null states.

Apply it to $|k,ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(k)|0⟩$. Commuting $k^{\mu }{a}_{\mu }$ through the creator gives a multiple of $-k_{\nu }{ϵ}^{\nu }|0⟩$. Hence the allowed photon polarization vectors obeyTake $k=(\omega ,0,0,\omega )$. The complete solution iswith complex coefficients. The last two vectors are transverse linear photon polarizations, and $(0,1,\pm i,0)/\sqrt{2}$ gives the two circular photon polarizations. The first vector is $k/\omega$, the allowed temporal-longitudinal combination. A purely spatial longitudinal vector $(0,0,0,1)$ by itself does not satisfy the subsidiary condition.

The oscillator brackets give, apart from the positive continuum normalization factor,Momentum eigenstates are distributionally normalized; a box or a wave packet makes the norm statement literal. The allowed longitudinal combination has zero norm because $k^2=0$, and it is orthogonal to every constrained state since $k\cdot ϵ=0$. It also gives a pure-gauge field wave: the field strength of $A_{\mu }\propto k_{\mu }{e}^{-ik\cdot x}$ vanishes. Thus $ϵ$ and $ϵ+ck$ represent the same physical photon polarization. The Gupta-Bleuler null-state quotient removes this null direction and leaves two positive-norm transverse photon photon polarizations. The negative-norm temporal oscillator of the unrestricted space is not an extra physical state. This is the transverse one-photon physical quotient.

Finally verify the two-photon fermion Ward identity directly for electron-positron annihilation into two photons. Define $D(r)=/r-m$ and the reduced Dirac propagatorThe equality follows from the Clifford algebra, since $(/r-m)(/r+m)=(r^2-m^2)I$. The incoming Dirac spinors obey $D(p)u(p)=0$ and $\overline{v}(q)(/q+m)=0$. On replacing one inserted photon polarization vector by $k$, the spinor factor in the two diagrams isAt the electron end, $/k=D(p)-D(p-k)$, soFor the other diagram, four-momentum conservation gives $p-k^{\prime }=k-q$. HenceThe two contributions therefore cancel exactly:Individual diagrams need not vanish; their sum implements the Ward identity. Linearity then gives invariance under $ϵ\mapsto ϵ+ck$, as required by gauge invariance. In particular, the null longitudinal photon polarization proportional to $k$ has zero physical production amplitude. Physical scattering produces the transverse photon classes, not an independent longitudinal massless photon.

A full fringe scan is unnecessary. Use the same components to make three fixed projective measurements, on separate batches of identically prepared photons. Direct path detection after the polarizing beam splitter measures the $H/V$ populations. Recombine after matching the arm photon polarizations, once with $\theta =0$ and once with $\theta =\pi /2$, to measure the two quadratures of the quantum coherence.

In the $H/V$ basis, write the density matrix in terms of its Bloch vector as $\rho =(I+r_{x}X+r_{y}Y+r_{z}Z)/2$. The previous formula becomesThus three measured relative frequencies giveand determineThe two interferometric settings analyze the diagonal linear polarizations $(|H⟩\pm |V⟩)/\sqrt{2}$ and the circular polarizations $(|H⟩\pm i|V⟩)/\sqrt{2}$, with detector labels fixed by the calibrated quantum phase convention. This is informationally complete three-observable qubit tomography: direct measurements in three bases replace fitting an entire phase-dependent fringe. It also reconstructs a mixed input, for which $|r|\le 1$, although the source here is assumed pure.

Simply measuring transmission through one fixed polarizing beam splitter would determine only $r_z$. Rotating a linear analyzer through different angles can additionally determine $r_x$, but cannot distinguish the two signs of $r_y$ without a relative-phase measurement. For arbitrary, possibly elliptical photon polarization, all three settings above are needed; no extra analyzer element beyond the described path-phase control and photon polarization rotator is required.

Write the unknown pure photon polarization qubit asThe polarizing beam splitter sends the $H$ component down the lower path and the $V$ component up the upper path. After rotating the upper photon polarization to $H$, both paths have the same internal quantum state and can interfere at BS2. Absorb fixed apparatus phases into a calibrated zero for the tunable quantum phase. With the preceding reflection convention, the quantum state before BS2 is $(\alpha |\ell ⟩+i\beta e^{i\theta }|u⟩)|H⟩$, so the two output amplitudes areConsequentlyRepeated counts while scanning $\theta$ determine the complex quantum coherence $\gamma$. The interference visibility is $2|\gamma |$, and the fringe position determines the relative quantum phase $\arg \beta -\arg \alpha$ whenever both components are nonzero.

A fringe alone is not enough to determine an arbitrary photon polarization quantum state: it fixes $|\alpha \beta |$ but not which of $|\alpha {|}^2$ and $|\beta {|}^2$ is larger. Indeed, $|H⟩$ and $|V⟩$ both give the flat fringe $P_1=1/2$. For a general pure quantum state, the two possible population differences areResolve this ambiguity by temporarily detecting the two paths before BS2. Their relative counts give $h=|\alpha {|}^2$ and $1-h=|\beta {|}^2$ using the same polarizing beam splitter and detectors. Taking $\alpha =\sqrt{h}$ as a global-phase choice, the reconstructed quantum state for $h>0$ isIf $h=0$, it is $|V⟩$ up to a global phase. This completes interferometric polarization tomography. The counts are obtained from many identically prepared photons; they do not identify an arbitrary quantum state from one photon.

Without the upper-arm rotation, the photon polarization still labels the path. Before BS2 the quantum state isAfter the balanced beam splitter, the polarization-valued output amplitudes areThe detectors do not resolve photon polarization. Since $⟨H|V⟩=0$, the cross terms vanish in the Born rule:These probabilities are independent of the tunable quantum phase and the input photon polarization, so there is no usable wave interference fringe. Equivalently, taking the partial trace over photon polarization removes the quantum coherence between the two path quantum states. The which-path information is retained in orthogonal photon polarizations even without anybody measuring it. Rotating the two occupied photon polarization quantum states to the same quantum state removes that distinction and allows the cross term in part (ii).

A cubic reflecting cavity supports standing waves with positive integer mode indices and wavevector components $k_i=\pi n_i/L$. A cell in mode space has volume $(\pi /L{)}^3$. In the continuum approximation, a spherical shell of radius $k$ and thickness $dk$ has one-eighth of its volume in the positive octant. There are two independent transverse photon polarizations for a generic wavevector. HenceThe photon energy is $ϵ=\hslash \omega =\hslash ck$, so $dk=dϵ/(\hslash c)$. ThereforeThis is the leading bulk density of states, valid when the cavity is large compared with the relevant wavelengths and many modes lie in the counting interval. A finite cavity has a discrete mode spectrum, with boundary corrections beyond this approximation. Modes with vanishing indices require separate boundary counting but do not change the leading $L^3$ term. Here $\hslash$ is the reduced Planck constant, as printed in the original PDF; replacing it by $h$ would introduce an erroneous factor of $(2\pi {)}^3$.

Each independent cavity mode is a quantum harmonic oscillator of angular frequency $\omega$. Its occupation numbers $n=0,1,\dots$ represent the numbers of photons in that mode. At thermal equilibrium, photons can be absorbed and emitted by the walls, so their number is not conserved and the photon chemical potential is zero. The Boltzmann distribution of the oscillator excitations then givesThe expression supplied in the paper is the thermal excitation energy. The oscillator's temperature-independent zero-point energy is excluded from the emitted thermal radiation.

Putting $ϵ=\hslash \omega$ and $dϵ=\hslash d\omega$ into the mode count from (a) gives the number of modes per unit cavity volume:Multiplying by the mean thermal energy in each mode yields Planck's law in the angular-frequency convention:The two photon polarizations are already included in the density of states; no further factor of two is needed. $u(T,\omega )d\omega$ has units $\mathrm{J}{\mathrm{m}}^{-3}$, whereas $u(T,\omega )$ has units $\mathrm{J}\mathrm{s}{\mathrm{m}}^{-3}$.

The massless two-body kinematics gives $q_1\cdot q_2=M^2/2$. The scalar scattering amplitude therefore has constant magnitude $|{\mathcal{A}}_{aa}|=|{\lambda }_a|M^2/2$. Use $\int d\Omega =4\pi$ and the identical-particle factor from part d:For the photons, the two-photon vertex of a scalar field-strength interaction following from $\Phi F_{\mu \nu }{F}^{\mu \nu }$ is, up to an irrelevant overall phase,Vertex qualification: the source prints same-index polarization contractions in the second product, whereas the contractions must be crossed as above. The printed expression fails the Ward identity. For a concrete counterexample take $q_1=(E,0,0,E)$, $q_2=(E,0,0,-E)$ and ${ϵ}_2=(1,1,0,-1)$. This is a transverse unit spacelike photon polarization vector, since $q_2\cdot {ϵ}_2=0$ and ${ϵ}_2^2=-1$. Replacing ${ϵ}_1$ by $q_1$ in the printed expression leaves $4{\lambda }_{\gamma }{E}^3$ instead of zero. The corrected crossed expression vanishes under either such replacement.

Choose physical spatial transverse photon polarizations along $x$ and $y$ in the parent rest frame. Then both crossed products vanish; the two equal-polarization combinations have $|{ϵ}_1\cdot {ϵ}_2{|}^2=1$ and the two orthogonal combinations give zero. There is no initial-spin average for a spin-zero parent. HenceThe calculation in this spatial transverse gauge also yields the intended result from the misprinted expression, but only the corrected vertex makes its gauge invariance explicit. For equal-magnitude couplings, the photon decay width is twice the two-scalar decay width because of its two nonzero polarization combinations.

In 1998, Jian-Wei Pan, Dik Bouwmeester, Harald Weinfurter and Anton Zeilinger created two photon pairs by two passages of a pulsed pump through a crystal using spontaneous parametric down-conversion. Interfering the inner photons at a nonpolarizing beam splitter implemented a partial Bell-basis measurement. Narrow optical filters and spatial filtering removed distinguishing information; changing the outer polarizer bases checked the conditional correlations. Fourfold detections selected the entanglement swapping events. The reported interference supported conditional quantum entanglement, but the paper reserved a statistically significant Bell inequality violation for improved visibility; it was not a demonstration closing the detection loophole.^[ref]

The ideal calculation makes the distinction between conditional and unconditional quantum states explicit. Start with spin singlet states $|{\psi }^{-}{⟩}_{12}|{\psi }^{-}{⟩}_{34}$, where $|{\psi }^{-}⟩=(|HV⟩-|VH⟩)/\sqrt{2}$ and $H,V$ denote orthogonal photon polarizations. Taking the inner product with $⟨{\psi }^{-}{|}_{23}$ gives $-|{\psi }^{-}{⟩}_{14}/2$: expanding the four terms leaves precisely the two antisymmetric outer terms. That Bell-basis measurement outcome has probability $1/4$ and leaves the outer pair in a spin singlet state. Ignoring the inner outcome instead leaves the outer density matrix $I_4/4$, a maximally mixed state. A conditional preparation flag is useful for later Bell tests only with the timing and setting conditions in setting-independent heralding preserves Bell locality.