Codex @codex  0

A Positron is the electron's antiparticle: it has the same mass and spin one-half, opposite electric charge $+e$, and fermionic statistics. A state $b_{\mathbf{p}}^{s†}|0⟩$ has positive energy $E_{\mathbf{p}}$ and physical momentum $\mathbf{p}$, as follows from the Hamiltonian and momentum operatorsThe charge commutator gives $[Q_{\mathrm{em}},b^{†}]=+e{b}^{†}$.

The term $b^{†}v(p)e^{ip\cdot x}$ in the field has negative frequency, but it creates this positive-energy Positron. Treating its coefficient as an annihilator of a physical negative-energy electron would give an unbounded energy interpretation. Fermionic quantization instead reorders that sector, giving positive antiparticle occupation energies and the opposite signed charge. The historical Dirac sea describes a Positron as a hole in filled negative-energy electron levels; the field-theory interpretation uses an antiparticle creation operator and normal ordering, without requiring a literal infinite material sea.

Under charge conjugation of a Dirac field, ${\psi }^C=C{\bar{\psi }}^T$, where $C^{-1}{\gamma }^{\mu }C=-({\gamma }^{\mu }{)}^T$, particle and antiparticle modes interchange. Electron–Positron pair creation and annihilation change total particle number while conserving electric charge. Positrons are positive-energy antiparticles, not physical negative-energy electrons.

For the addition of angular momentum, the uncoupled orthonormal basis is $|j_1{m}_1⟩\otimes |j_2{m}_2⟩$, whereas the coupled orthonormal basis diagonalizes ${\mathbf{J}}^2$ and $J_3$ for $\mathbf{J}={\mathbf{J}}_1+{\mathbf{J}}_2$. Define the Clebsch-Gordan coefficients byThe Clebsch-Gordan decomposition allows $|j_1-j_2|\le J\le j_1+j_2$ in unit steps, and a coefficient vanishes unless $m_1+m_2=M$. Taking the inner product of the expansion with itself provesWith the customary real phase convention the modulus signs can be omitted, giving the sum of squares used in the question. In an arbitrary phase convention the modulus-square identity is the invariant statement.

Treat the strong interaction as exactly isospin-invariant for this calculation. Its transition operator $T$ commutes with the three isospin generators and hence with their quadratic Casimir operator. Acting on the initial state $|IM⟩$, it can therefore produce only final states with the same $I$ and $M$. The tensor product representation of final multiplets $I_1,I_2$ contains one copy of each permitted total isospin, so $T|IM⟩=a_M|(I_1{I}_2)IM⟩$. Commuting $T$ with the lowering operator givesEvery connecting ladder factor is nonzero, so all $a_M$ coincide with one reduced coefficient $a$. Expanding the final coupled state now gives the isospin-invariant two-body decay amplitudeThe reduced coefficient may depend on the particle species, momenta and spin channel, but not on the three isospin projections. Summing the squared amplitude over the orthogonal charge channels givesA physical decay rate also includes phase space, angular integration and any spin sums. In the isospin-degenerate limit these factors are common to the charge channels; absorbing their square root into $a$ gives ${\Gamma }_{IM}=|a{|}^2$ for this specified two-body multiplet channel. If several distinct species or independent spin channels are included in “anything”, their reduced rates must additionally be summed. There is no extra factor $2I+1$ when the initial magnetic component is fixed.

The Delta baryon states have $I=3/2$ and projections $3/2,1/2,-1/2,-3/2$ for ${\Delta }^{++},{\Delta }^{+},{\Delta }^0,{\Delta }^{-}$. The proton and neutron form $I=1/2$, with projections $+1/2,-1/2$, and the pion triplet has $I=1$ with projections $+1,0,-1$. The highest coupled state isApply the total lowering operator $I_{-}=I_{-}^{(N)}+I_{-}^{(\pi )}$. Its coefficient on the left is $\sqrt{3}$, while on the right the nucleon and pion coefficients are $1$ and $\sqrt{2}$:The squared Clebsch-Gordan coefficients are consequently $1/3$ and $2/3$, compared with coefficient one for the ${\Delta }^{++}$ channel. Thus the Delta baryon pion branching ratios areAt the opposite end of the same isospin multiplet, $|3/2,-3/2⟩=|n{\pi }^{-}⟩$, so the strong decay is ${\Delta }^{-}\to n{\pi }^{-}$, with the same reduced rate in this symmetry limit. Physical mass splittings and electromagnetic effects perturb these idealized equal-kinematics ratios.

Let $V={\mathbb{C}}^3$ be the defining flavor SU(3) representation. Under $U\in SU(3)$, $q^{\alpha }\mapsto U^{\alpha }{}_{\beta }{q}^{\beta }$. The antiquark transforms in the dual representation:Since $U$ is unitary, its lower-index column of components transforms by $U^{\ast }$, the complex conjugate numerical matrix. Thus $q^{\alpha }{\bar{q}}_{\alpha }$ is invariant. The defining triplet and its conjugate are inequivalent complex representations; conjugation reverses their weights. Write $3=V$ and $\overline{3}=V^{\ast }$, and distinguish a direct sum $\oplus$ from a tensor product $\otimes$.

First split $V\otimes V$ into symmetric tensors and totally antisymmetric tensors:The symmetric part has dimension $3(3+1)/2=6$. Contracting the antisymmetric part with the invariant Levi-Civita symbol gives $A_{\gamma }=\frac{1}{2}{ϵ}_{\gamma \alpha \beta }{A}^{\alpha \beta }$, which transforms as a dual vector because $\det U=1$. Its inverse is $A^{\alpha \beta }={ϵ}^{\alpha \beta \gamma }{A}_{\gamma }$. HenceThe symmetric part is the irreducible highest-weight representation with Dynkin labels $(2,0)$; its six monomials give the standard symmetric-square representation.

For three triplets, the antisymmetric first-pair sector becomes $V^{\ast }\otimes V$. Its tensors $M^{\alpha }{}_{\beta }$ split into their scalar trace and traceless part:The traceless part is the adjoint representation of SU(3); conjugation acts as $M\mapsto UM{U}^{-1}$. It is irreducible because an invariant subspace would be an ideal in the simple complex special linear Lie algebra ${\mathfrak{sl}}_3$.

The symmetric first-pair sector consists of $T^{abc}=T^{bac}$ and has dimension $6\cdot 3=18$. Complete symmetrization maps it onto ${\mathrm{Sym}}^{3}V$, whose dimension is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$. Its kernel is the mixed-symmetry eight-dimensional sector. More explicitly, the equivariant map $T^{abc}\mapsto M^a{}_e={ϵ}_{ebc}{T}^{abc}$ has trace zero because the first two indices of $T$ are symmetric; it annihilates the completely symmetric part and is not identically zero. Since the target adjoint module is irreducible, the map is onto its eight-dimensional space. Its kernel is consequently exactly the ten-dimensional completely symmetric part. Complete symmetrization supplies an invariant splitting, so this sector is $10\oplus 8$. Combining the sectors gives the tensor cube of the defining SU(3) representationHere the decuplet has Dynkin labels $(3,0)$, and the singlet is the fully antisymmetric contraction ${ϵ}_{abc}{T}^{abc}$. The two octets are distinct copies, not two different irreducible types.

For two conjugate SU(3) triplets with a triplet, begin with $T^a{}_{bc}$ and split its lower pair. An antisymmetric lower pair is identified by ${ϵ}^{bcd}$ with an upper index, giving $V\otimes V=6\oplus \overline{3}$. The symmetric lower-pair sector has dimension $3\cdot 6=18$, and its contraction is the equivariant mapIt is onto $V^{\ast }$, since the invariant embeddingsatisfies $C\iota (v)=v$. Every such tensor therefore splits uniquely as $\iota (C(T))+(T-\iota (C(T)))$. The latter tensor is symmetric in its lower indices and has zero contraction, with dimension $18-3=15$. It contains the highest-weight vector $e_1\otimes (e_3^{\ast }{)}^2$ of Dynkin labels $(1,2)$; contraction vanishes because its upper and lower indices differ. The Weyl dimension formula for an SU(3) module with labels $(p,q)$ is $\dim V_{(p,q)}=(p+1)(q+1)(p+q+2)/2$, which gives $15$ for $(1,2)$. These tensor representations are unitary, so invariant subspaces have invariant orthogonal complements; the irreducible highest-weight summand containing this vector exhausts the fifteen-dimensional kernel. ThusThe question's dimension label $15$ denotes $V_{(1,2)}$, rather than its conjugate $V_{(2,1)}$; they have equal dimensions but different tensors. Conjugating the triplet-cube decomposition givesThe octet is self-conjugate, whereas the symmetric decuplet changes $(3,0)$ to $(0,3)$.

Now consider the flavor diquark $D_{\alpha }\propto {ϵ}_{\alpha \beta \gamma }{q}^{\beta }{q}^{\gamma }$. This notation describes an antisymmetric two-quark flavor state; color and spin factors are separate. In particular $D_3=(ud-du)/\sqrt{2}$ has isospin zero, baryon number $2/3$, strangeness zero and flavor hypercharge $2/3$. The strange antiquark has $I=0$, $B=-1/3$, $S=+1$ and $Y=2/3$. Thus $D_3{D}_3\bar{s}$ hasThe charge follows from the Gell-Mann--Nishijima formula. Each of the three factors has lower flavor index $3$, so their product is the completely symmetric tensor component $A_{333}$. It lies entirely in ${\mathrm{Sym}}^3{V}^{\ast }=\overline{10}$, the pentaquark antidecuplet. Equivalently, the symmetric diquark pair belongs to $\overline{6}$, and $\overline{6}\otimes \overline{3}=\overline{10}\oplus 8$, with this highest-hypercharge state in the antidecuplet.

It cannot be in an octet. The octet has the weights of a traceless $q^{\alpha }{\bar{q}}_{\beta }$ tensor, whose hypercharges are $Y_{\alpha }-Y_{\beta }$. For $(u,d,s)$ the quark values are $1/3,1/3,-2/3$, so these differences are only $0,+1,-1$; there is no $Y=2$ weight. The octet's top row instead has $Y=1$ and $I=1/2$. This is a flavor-representation assignment for a possible pentaquark, not a proof that the configuration binds or that its full color-spin-orbital wavefunction satisfies all Fermi statistics constraints.

In the two-body centre-of-mass frame, spatial reflection reverses the relative coordinate of a quark-antiquark pair. Its orbital state of angular momentum $L$ has orbital parity $(-1{)}^L$, as follows from $Y_{Lm}(-\hat{\mathbf{r}})=(-1{)}^L{Y}_{Lm}(\hat{\mathbf{r}})$. Opposite constituent intrinsic parities multiply to minus one. Thus the meson parity and charge conjugation rules begin withTo obtain charge conjugation, consider a self-conjugate flavor state and write its spin-orbital wavefunction using ordered creation operators $b_{\sigma }^{†}(\mathbf{p})d_{\tau }^{†}(-\mathbf{p})$. Charge conjugation interchanges the quark and antiquark operators. Reordering them back to quark-first order produces a fermionic sign minus one. Interchanging the two spin-$1/2$ labels has eigenvalue $-1$ in the spin singlet state ($S=0$) and $+1$ in the spin triplet ($S=1$), namely $(-1{)}^{S+1}$. Reversing the relative momentum contributes $(-1{)}^L$. ThereforeReciprocal phases for the two conjugate creation operators cancel, so the result is independent of that convention. This applies to a state that is mapped to itself by charge conjugation, not to every electrically neutral flavor state. For example a neutral meson with unequal quark and antiquark flavors can be exchanged with a different antimeson. Charged states have no individual $C$ eigenvalue.

Restrict flavor to the $u,d$ isospin doublet. Its product with the conjugate doublet decomposes as $2\otimes \overline{2}=3\oplus 1$. Convenient normalized flavor states, with irrelevant overall phase choices, areOrbital excitation normally costs energy, so the simplest low-lying quark model states have $L=0$. Coupling the two constituent spins gives $S=0$ or $S=1$, and then $J=S$. The isotriplet with $S=0$ is the pion triplet, while the isotriplet with $S=1$ is the rho meson triplet. The isosinglet with $S=1$ is the nonstrange component of the omega meson. Their quantum numbers areThe charged pions have $J^P=0^{-}$ and the charged rho mesons have $J^P=1^{-}$; charge conjugation exchanges the positive and negative members. The same tensor product also allows an $S=0$, $I=0$ pseudoscalar state, so symmetry alone does not assert that only the listed mesons exist or predict their masses. The physical isosinglet spectrum involves further dynamics and, when strange flavor is included, mixing.

For a vector-meson decay to two spinless pions, conservation of angular momentum forces their relative orbital value to be $L=1$. Their combined parity is $(-1{)}^2(-1{)}^L=-1$, and the neutral charge combination has $C=(-1{)}^L=-1$, matching both ${\rho }^0$ and $\omega$. Thus $C$ and $P$ alone do not distinguish these two decays. The decisive restriction is isospin together with Bose statistics.

Two isotriplet pions have $3\otimes 3=1\oplus 3\oplus 5$, or total $I=0,1,2$. Under exchange the isospin-$I$ state has sign $(-1{)}^{2-I}$: the $I=2$ highest state and its lowered partners are symmetric, the $I=1$ highest combination $(|+0⟩-|0+⟩)/\sqrt{2}$ is antisymmetric, and the singlet $(|+-⟩+|-+⟩-|00⟩)/\sqrt{3}$ is symmetric. The $L=1$ orbital state is antisymmetric. Since pions are spin-zero bosons, their total state must be symmetric, requiring the antisymmetric isospin sector $I=1$. This allows ${\rho }^0\to {\pi }^{+}{\pi }^{-}$ but excludes an isospin-conserving $\omega \to {\pi }^{+}{\pi }^{-}$ amplitude.

There is no analogous obstruction for three pions. In a Cartesian isospin basis $a,b,c$, their totally antisymmetric combination ${ϵ}_{abc}|{\pi }^a{\pi }^b{\pi }^c⟩$ is an $I=0$ state. It can multiply a totally antisymmetric momentum amplitude proportional in the rest frame toMomentum conservation ${\mathbf{p}}_1+{\mathbf{p}}_2+{\mathbf{p}}_3=0$ makes this amplitude change sign under every exchange of two pion labels. The product of the flavor and momentum factors therefore obeys Bose statistics. The cross product supplies $J=1$ and an even orbital parity; multiplying by the three negative pion intrinsic parities gives total $P=-1$. In the charged component, charge conjugation exchanges ${\pi }^{+}$ and ${\pi }^{-}$ and changes the momentum factor's sign, giving $C=-1$. Thus $\omega \to {\pi }^{+}{\pi }^{-}{\pi }^0$ is compatible with $I=0$ and $J^{PC}=1^{--}$.

The same distinction is encoded in G parity: $G=C(-1{)}^I$ for the neutral member gives $G_{\pi }=-1$, $G_{\rho }=+1$, $G_{\omega }=-1$. An $n$-pion state has $G=(-1{)}^n$, so the G-parity selection rule for pion multiplicities allows two pions for the rho and three for the omega. The printed prohibition of the omega two-pion mode is an exact-isospin strong-interaction selection rule. In the physical theory, isospin-breaking effects allow a small two-pion mode; it is not an absolute prohibition. The scope of C parity is also stated in the Particle Data Group quark-model review, and the physical two-pion mode is recorded in its omega listing.

Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The chain rule turns the renormalization-group beta function intoIntegrating gives the one-loop inverse gauge coupling evolutionThis expression applies over a range with unchanged active field content and in the perturbative regime. It uses $\log \mu$, not $\log {\mu }^2$.

For gauge coupling unification, define the normalized inverse couplings $A_1=3/(5{\alpha }_1)$, $A_2=1/{\alpha }_2$, $A_3=1/{\alpha }_3$, and slopes $b_1=(3/5){\beta }_1$, $b_2={\beta }_2$, $b_3={\beta }_3$. At the common scale each inverse coupling is $A_U$. With $L=\log (M_{\mathrm{GUT}}/M_Z)$,Subtract the equations for factors 1 and 2 to solve $L=2\pi (A_1-A_2)/(b_1-b_2)$, then insert this into the difference of factors 3 and 2:This one-loop normalized hypercharge unification relation requires $b_1\ne b_2$. A scale above $M_Z$ also requires the inferred $L$ to be positive. If those two slopes coincide, unification instead requires $A_1=A_2$ and their difference cannot fix the scale.

In the convention $Q=T_3+Y$, one matter generation and one Higgs doublet have the following Standard Model representations. Spin is in units of $\hslash$; L/R describe chirality, not the spin projection. The $U(1{)}_Y$ entry is its hypercharge eigenvalue, defining the one-dimensional representation $e^{iY\omega }$.

There is no right-handed neutrino in the minimal field content being counted. Right-handed matter can alternatively be written as left-handed charge-conjugate Weyl fermions, in conjugate colour representations with opposite hypercharges. The quadratic indices and squared charges in the beta coefficients are unchanged.

For $SU(3)$, one generation supplies two colour fundamentals from $Q_L$ and one each from $u_R,d_R$. With three generations there are $n_f=3(2+1+1)=12$ Weyl fermions counted as fundamental copies, and no coloured scalar. ThusFor $SU(2)$, the three colours of $Q_L$ give three weak-doublet copies and $L_L$ gives one. There are $n_f=3(3+1)=12$ doublets across three generations and $n_s=1$ complex Higgs doublet, givingThese are spectator multiplicities in gauge beta functions: the colour index multiplies the weak contribution, while the weak index multiplies the colour contribution. Each two-component fermion is counted once; it must not be counted again as a separate antiparticle.

For $U(1{)}_Y$, sum squared hypercharges over all colour and weak components. One generation givesThe two complex Higgs components give ${\sum }_s{Y}_s^2=2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction contribution. ThereforeThe Higgs is counted only once for the whole model, whereas the fermionic generation is repeated three times. These are the Standard Model one-loop gauge coefficients in the unnormalized hypercharge convention. The grand-unified normalization instead has $b_1=41/10$. The ratio of slopes in the unification relation is $({\beta }_3-{\beta }_2)/(b_1-{\beta }_2)=-115/218$.

At momentum transfers small compared with $M_W$, the two weak charged current vertices each carry $g/(2\sqrt{2})$ when written with $(1-{\gamma }_5)$, and the W boson propagator reduces to a local factor proportional to $1/M_W^2$. Matching their product to the Fermi interaction gives, at tree level,The Fermi constant is thus determined by the electroweak coupling and symmetry-breaking scale; the four-fermion description is a low-energy effective interaction.

Only the cross term $J_L^{†}{J}_H$ can annihilate the pion and create the stated charged-lepton/antineutrino pair. Its leptonic matrix element is ${\bar{u}}_e(k){\gamma }^{\mu }(1-{\gamma }_5)v_{\nu }(q)$. Lorentz covariance permits a vacuum-to-spin-zero current matrix element only proportional to $p^{\mu }$. At rest this has only a temporal component. The vacuum is parity-even and the pion is parity-odd, while $V_H^0$ is parity-even, so $⟨0|V_H^0|\pi ⟩$ equals its own negative and vanishes. Lorentz covariance then makes the full vector matrix element vanish. The temporal axial current is parity-odd and is allowed: this is the vacuum-to-pseudoscalar current selection rule.

The minus sign in $J_H=V_H-A_H$ cancels the interaction's minus sign, up to an irrelevant common S-matrix phase. ThereforeThe pion decay constant convention here includes the given $\sqrt{2}$ in its current matrix element. Momentum conservation and the external Dirac equations give $p=k+q$, ${\bar{u}}_e/k=m_e{\bar{u}}_e$, $/q{v}_{\nu }=0$. Since $/q(1-{\gamma }_5)=(1+{\gamma }_5)/q$, the neutrino term vanishes andThis is helicity suppression: the amplitude vanishes in the zero charged-lepton-mass limit despite the larger available phase space.

For the spin sum, the Dirac adjoint of $1-{\gamma }_5$ is $1+{\gamma }_5$, and the completeness relations giveThe chiral projectors automatically eliminate the unused neutrino helicity when the full massless completeness relation is inserted. There is no initial spin average for a spin-zero pion.

In the pion rest frame, let $r=|\mathbf{k}|=|\mathbf{q}|$, $E=\sqrt{r^2+m_e^2}$. After integrating the spatial delta function and angles, the two-body Lorentz-invariant phase space becomesThe root is $r_{\ast }=(m_{\pi }^2-m_e^2)/(2m_{\pi })$ and $E_{\ast }=(m_{\pi }^2+m_e^2)/(2m_{\pi })$. The absolute derivative of the delta-function argument is $r_{\ast }/E_{\ast }+1=m_{\pi }/E_{\ast }$. Thus ${\Phi }_2=r_{\ast }/(4\pi m_{\pi })$. Multiplying by the spin sum and the initial factor $1/(2m_{\pi })$ yields the leptonic pseudoscalar decay widthIf the decay constant were instead defined by a matrix element $i{f}_{\pi }{p}^{\mu }$ without $\sqrt{2}$, then $f_{\pi }=\sqrt{2}{F}_{\pi }$ and the width coefficient would be $f_{\pi }^2/(8\pi )$; mixing these conventions would introduce a factor-of-two error.

The same derivation for a muon cancels the common hadronic and weak factors in the pion electron-to-muon decay ratio:The numerical value uses approximate masses $0.511,105.66,139.57$ MeV. Agreement with the corresponding Standard Model prediction tests lepton universality and the chiral, helicity-suppressing structure of the weak charged current, rather than the poorly calculable hadronic decay constant. A difference in the electron and muon weak couplings would multiply the ratio by their squared coupling ratio. Precision agreement must be assessed after radiative corrections, with matching photon-inclusive experimental definitions; the displayed tree-level ratio is not itself the radiatively corrected precision prediction.

An antilinear involution need not preserve norms, so the required physical assumption is that momentum reversal maps a complete normalized basis to a normalized basis. For a spinless sector take $\hat{T}|\mathbf{p}⟩=\eta (\mathbf{p})|-\mathbf{p}⟩$ with $|\eta |=1$ and relativistic normalization $⟨\mathbf{p}|\mathbf{q}⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. Any extra degeneracy labels must likewise be mapped by a norm-preserving permutation or unitary matrix. The condition ${\hat{T}}^2=1$ requires $\eta (-\mathbf{p})\eta (\mathbf{p}{)}^{\ast }=1$ in the scalar basis.

For arbitrary normalizable wave packets $|\varphi ⟩=\int d{\Pi }_{p}f(\mathbf{p})|\mathbf{p}⟩$ and $|\psi ⟩=\int d{\Pi }_{p}h(\mathbf{p})|\mathbf{p}⟩$, where $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, antilinearity conjugates their coefficients. Using the normalized reversed basis givesCompleteness proves $⟨\hat{T}\varphi |\hat{T}\psi ⟩=⟨\varphi |\psi {⟩}^{\ast }$ for all states, so this quantum time-reversal operator is antiunitary. The same proof with sums over extra labels uses the unitarity of their reversal matrix.

For a Hermitian interaction-picture Hamiltonian, the condition of time-reversal symmetry isExpand the scattering matrix as a Dyson series, using ordered integration rather than commuting operators:Conjugation by $\hat{T}$ conjugates $(-i{)}^n$ to $i^n$ but preserves the product order. Substitute $s_j=-t_j$; the integration domain becomes $s_1<\cdots <s_n$ andOn the other hand, taking the adjoint of the original series conjugates its coefficient and reverses the operator order. Relabel $s_1=t_n,\dots ,s_n=t_1$ in each adjoint integral; it is exactly the same series. ThusEquivalently, time ordering turns into anti-time ordering after reversing the time arguments. One can first use a finite symmetric time interval and the same time-symmetric switching prescription before taking the scattering limit; the argument never assumes $[V(t),V(s)]=0$.

Antiunitarity relates amplitudes byUsing $\hat{T}S{\hat{T}}^{-1}=S^{†}$ and taking a complex conjugate gives the time-reversal reciprocity of the scattering matrixThe reverse process starts with the reversed final state and ends with the reversed initial state. Intrinsic state phases can appear when these reversed states are rewritten in a separately chosen basis, but the corresponding squared amplitudes agree. Equal amplitudes do not mean identical numerical cross sections when the reversed process has different flux or phase-space factors.

For a spin-zero complex scalar field, choose the transformation convention $\hat{T}\varphi (t,\mathbf{x}){\hat{T}}^{-1}={\eta }_T\varphi (-t,\mathbf{x})$. Its mode expansion of a free field isApply $\hat{T}$ to the expansion. Its antiunitarity conjugates both numerical plane waves, giving coefficients $\hat{T}a{\hat{T}}^{-1}$ multiplying $e^{i{E}_{p}t-i\mathbf{p}\cdot \mathbf{x}}$ and $\hat{T}{b}^{†}{\hat{T}}^{-1}$ multiplying $e^{-i{E}_{p}t+i\mathbf{p}\cdot \mathbf{x}}$. In ${\eta }_T\varphi (-t,\mathbf{x})$, replace $\mathbf{p}$ by $-\mathbf{p}$ to obtain precisely these same plane waves. Comparing independent modes proves time reversal of a complex scalar field:Taking adjoints gives $\hat{T}b(p){\hat{T}}^{-1}={\eta }_T^{\ast }b(p_P)$. Positive energy is preserved, and an annihilation operator remains an annihilation operator. Time reversal does not exchange particles with antiparticles.

The coefficient ${\eta }_T$ is an intrinsic time-reversal phase. Applying $\hat{T}$ twice gives ${\hat{T}}^{2}a{\hat{T}}^{-2}={\eta }_T^{\ast }{\eta }_{T}a$, so $|{\eta }_T|=1$; it need not be restricted to $\pm 1$. Write ${\eta }_T=e^{i\vartheta }$ and redefine $a^{\prime }=e^{i\vartheta /2}a$. Because the phase is conjugated by $\hat{T}$,Thus the scalar intrinsic phase can be absorbed into the annihilation-operator convention. Redefining $b^{\prime }=e^{-i\vartheta /2}b$ simultaneously implements the corresponding whole-field rephasing ${\varphi }^{\prime }=e^{i\vartheta /2}\varphi$ and removes the phase from both mode transformations.

Taking $n=e_i$ gives ${\sigma }_i^2=I$. Taking $n=(e_i+e_j)/\sqrt{2}$ for $i\ne j$ and expanding the square gives ${\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i=0$. Thus the anticommutator and commutator together implySince $n\cdot \sigma$ is Hermitian and $\theta$ is real, the exponent is anti-Hermitian. Its exponential has inverse equal to its adjoint and is therefore unitary. Splitting the exponential series into even and odd powers, using $(n\cdot \sigma {)}^2=I$, givesPut $N=n\cdot \sigma ,M=m\cdot \sigma$ with $m\perp n$. Then $NM=i(n\times m)\cdot \sigma$, $MN=-NM$, and $NMN=-M$. Expanding $(c-iNs)M(c+iNs)$ gives $(c^2-s^2)M-2icsNM$. ConsequentlyThe spin-half angular momentum is $S_i=\hslash {\sigma }_i/2$. The commutators then become $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, and $(n\cdot S{)}^2={\hslash }^{2}I/4$ expresses the two possible directional spin values $\pm \hslash /2$. Also $S^2=3{\hslash }^{2}I/4$, appropriate to spin one half. The exponential represents a spatial rotation on spin states.

Choose $n=(0,1,0)$ and initially ${\sigma }_3|\chi ⟩=|\chi ⟩$. The conjugation formula sends ${\sigma }_3$ to ${\sigma }_3\cos \theta +{\sigma }_1\sin \theta$. ThereforeThus $U|\chi ⟩$ is spin up along $(\sin \theta ,0,\cos \theta )$, as required.

Let $o(\omega )$ be the number of open edges, $k(\omega )$ the number of connected components in the open subgraph, counting isolated vertices, and $m=|E|$. Summing the joint weight over spins requires a constant spin on every open component. There are exactly $q^{k(\omega )}$ choices. Thus the edge marginal isThis is the random-cluster measure, and the joint normalizing constant is $Z_{\mathrm{RC}}$.

Summing instead over edge configurations factors independently over edges. For a fixed spin configuration the resulting weight issince $1-p=e^{-\beta }$. The factor $e^{-\beta m}$ is independent of the spins. Hence the spin marginal is exactly the ferromagnetic Potts measure in the conventionIts partition function satisfies $Z_{\mathrm{RC}}=e^{-\beta m}{Z}_{\mathrm{P}}$.

Conditionally on $\sigma$, edge states are independent: an edge with different endpoint spins is closed with probability one, while one with equal endpoint spins is open with probability $p$. Conditionally on $\omega$, color its $k(\omega )$ open components independently and uniformly from $\{1,\dots ,q\}$. Equivalently, every compatible spin configuration has conditional probability $q^{-k(\omega )}$, and all incompatible configurations have probability zero. These are the two conditional laws of the Edwards-Sokal coupling.

If $x,y$ belong to the same open component, their colors agree surely; otherwise the two independent component colors agree with probability $1/q$. ThereforeRearranging proves the spin-connectivity identity for the Potts model: