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An integral of the time component of a conserved current is a conserved charge when the spatial boundary flux vanishes. Charges need not come from an immediately apparent elementary symmetry; a field theory can have an infinite higher-derivative current hierarchy. A Noether charge is obtained from the symmetry current supplied by the Noether theorem.

For the local variation $\delta \psi =i\alpha (x)\psi$, $\delta \bar{\psi }=-i\alpha (x)\bar{\psi }$, the terms proportional to $\alpha$ cancel and the differentiated phase givesThus, for a compactly supported $\alpha$, integration by parts givesThis Dirac current is the Noether current in the convention $\delta S=-\int j^{\mu }{\partial }_{\mu }\alpha$. Using the convention with a plus sign reverses the Noether current and the conserved charge together. There is one independent global symmetry in field theory, with four spacetime components of its Noether current. The associated conserved charge is $Q=\int d^{3}x{j}^0$, with vacuum subtraction when required.

An adiabatic change of box size preserves the mode labels, whose momenta scale as $p\propto L^{-1}$. Since $V=L^3$, differentiation gives $dp/dV=-p/(3V)$. Occupation numbers are fixed under this slow mechanical change, so summing the changes in the individual particle energies givesComparison with $dE=-PdV$ at fixed entropy and particle number yieldsThe integration notation means summing over the original momentum-mode labels; it does not assume that the density in momentum space is unchanged after rescaling. In the ultrarelativistic limit $E(p)=cp$, so $p{E}^{\prime }(p)=E(p)$ and $P=E/(3V)=\rho c^2/3$.

Adiabatic radiation obeys $d(E{a}^3)=-Pd(a^3)$, giving energy density proportional to $a^{-4}$. Equilibrium photon energy density is proportional to $T^4$, so $T\propto a^{-1}$. Therefore the photon number in a comoving volume, $n_{\gamma }{a}^3\propto T^3{a}^3$, is constant. Its entropy is also constant: photon entropy density is $({ϵ}_{\gamma }+P)/T\propto T^3$. Photon number is not a microscopic conserved charge, but equilibrium processes maintaining the spectrum do not alter this comoving-number scaling in adiabatic expansion; entropy-producing processes would invalidate the assumptions.

The degeneracy factor $g_A$ counts internal states with the same energy, including the allowed spin states. The chemical potential ${\mu }_A$ measures the free-energy cost of changing the number of particles of species $A$; in this convention the rest energy remains in the exponential of the nonrelativistic Maxwell--Boltzmann number density.

The chemical potential is fixed by the densities of conserved charges and by chemical equilibrium between reactions. For a reaction ${\sum }_A{\nu }_{A}A=0$, chemical equilibrium requires ${\sum }_A{\nu }_A{\mu }_A=0$. Equivalently, when these are the only constraints, ${\mu }_A$ is a linear combination of the chemical potentials of the conserved charges carried by $A$. An unconstrained species has zero chemical potential. In particular, the photon chemical potential vanishes in thermal equilibrium when photon-number-changing reactions are effective.

Use units $\hslash =c=1$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. Treat the complex scalar field and its adjoint as independent variables when varying the action. Their canonical momenta areCanonical quantization in the Heisenberg picture imposes, at the same time $t$,All other independent equal-time commutators vanish: in particular $[\varphi ,{\varphi }^{†}]=[\pi ,{\pi }^{†}]=[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$, together with the same-type field and momentum commutators. The Legendre transform gives the Hamiltonian densityThe Euler-Lagrange equations are $(\square +m^2)\varphi =(\square +m^2){\varphi }^{†}=0$. The positive- and negative-frequency plane waves form a complete basis of solutions of the Klein-Gordon equation. Since the field is complex, their operator coefficients are independent rather than being adjoints of each other. Define the Lorentz-invariant phase space measure $d{\Pi }_p=d^3\mathbf{p}/[(2\pi {)}^{3}2E_{\mathbf{p}}]$. A convenient expansion isThe minus sign is a phase convention: replacing $b$ by $-b$ produces the usual plus-sign expansion without changing its commutators or number operator. The normalization is fixed by the canonical commutation relations. For example,when $[a(\mathbf{p}),a^{†}(\mathbf{q})]=[b(\mathbf{p}),b^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_{\mathbf{p}}{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$. The cross commutators vanish and the other canonical relation follows by taking adjoints with the operator order reversed. Conversely these mode commutators can be obtained by Fourier inversion of the equal-time relations. This is canonical quantization of a complex scalar field.

Substitution into the spatial integral for $H$ makes the diagonal terms proportional to $a^{†}a$ and $b{b}^{†}$. Terms creating or annihilating a pair have $\mathbf{q}=-\mathbf{p}$ and a coefficient $E_{\mathbf{p}}^2-{\mathbf{p}}^2-m^2=0$, so they cancel. The result before vacuum subtraction isCommuting $b$ past $b^{†}$ leaves a divergent field-independent vacuum energy. The displayed zero-vacuum-energy form must therefore be understood as a normal-ordered Hamiltonian:A regulator can be introduced before subtracting the constant. This is the normal-ordered Hamiltonian of a free complex scalar field; the unrenormalized Hamiltonian is not literally equal to this expression without that convention.

The scalar-field vacuum is annihilated by both annihilation operators. Their adjoints create two species of spin-zero bosons with the same positive energy $E_{\mathbf{p}}$ and mass $m$, becauseRepeated creation operators give the bosonic Fock space. The negative-frequency part does not create a negative-energy physical state; it creates the antiparticle of positive energy. The charge distinguishes the two species.

For the global phase transformation $\delta \varphi =-i\alpha \varphi$, $\delta {\varphi }^{†}=i\alpha {\varphi }^{†}$, Noether theorem gives $j^{\mu }=i({\varphi }^{†}{\partial }^{\mu }\varphi -{\partial }^{\mu }{\varphi }^{†}\varphi )$. Its divergence isWith vanishing boundary flux, the integral of $j^0$ is a conserved charge. Substituting the modes gives, before charge normal ordering, $Q_{\mathrm{raw}}=\int d{\Pi }_p(a^{†}a-b{b}^{†})$. The pair terms cancel because their frequency difference is zero when their spatial momenta sum to zero. Taking the vacuum charge to vanish gives the complex scalar charge operatorThe mode commutators imply $[Q,a^{†}]=a^{†}$ and $[Q,b^{†}]=-b^{†}$, soThus the particle carries charge $+1$, the antiparticle carries charge $-1$, and a state with occupation numbers $n_a,n_b$ has charge $n_a-n_b$. Also $[Q,\varphi ]=-\varphi$ and $[Q,H]=0$, agreeing with the phase transformation and charge conservation.

Use $\hslash =c=1$ in the field-theory calculations. Work with signature $(+,-)$ and $a\ne 0$. Since $x=(x_{+}+x_{-})$ and $t=(x_{+}-x_{-})$, ${\partial }_{+}={\partial }_x+{\partial }_t$, ${\partial }_{-}={\partial }_x-{\partial }_t$, and the Sine-Gordon equation is ${\varphi }_{+-}=\sin \varphi$. Introduce $w=(\rho +\kappa )/2$ and $d=(\rho -\kappa )/2$. The specified Sine-Gordon Bäcklund transformation becomes $d_{+}=a\sin w$, $w_{-}=a^{-1}\sin d$. Its compatibility givesAdding and subtracting these equations establishesThus a compatible transformed field satisfies the same Sine-Gordon equation. Smoothness is needed to interchange mixed derivatives. The statement cannot literally include $a=0$, since one defining equation contains $1/a$.

Generating the kink. With zero seed, the two first-order equations give ${\rho }_{+}=2a\sin (\rho /2)$ and ${\rho }_{-}=2a^{-1}\sin (\rho /2)$. On a nonconstant branch put $p=\tan (\rho /4)$; then $p_{+}=ap$, $p_{-}=a^{-1}p$. Integration givesFor $a>0,C>0$, this is the Sine-Gordon kink, after absorbing $C$ in the center position. Its speed and Lorentz factor obeyChanging the sign of $a$ or of the exponential coefficient selects the corresponding antikink orientation; $|v|<1$ whenever $a$ is finite and nonzero. Constant vacuum branches, omitted by division by $\sin (\rho /2)$, can be added separately.

The continuously parametrized Bäcklund transformation does more than produce this one solution. Successive compatible transformations and Bianchi permutability for sine-Gordon Bäcklund transformations construct multisoliton solutions; expansion of the generating relations gives the local conserved-charge hierarchy of sine-Gordon theory. Together with its Lax pair and inverse scattering transform, this is the structure behind classical integrability, elastic scattering and the absence of generic radiative energy loss in soliton collisions. Existence of a transformation alone is not a proof that every generated conserved charge is independent and in involution; those are additional properties of the integrable system.

Transforming the static kink at $a=1$. The seed obeys ${\kappa }_x=2\mathrm{sech}x$, $\sin (\kappa /2)=\mathrm{sech}x$, $\cos (\kappa /2)=-\tanh x$. Adding and subtracting the two Sine-Gordon Bäcklund transformation equations yieldsAgain set $p=\tan (\rho /4)$. These simplify to $p_x=-p\tanh x$, $p_t=\mathrm{sech}x$. The first gives $p=f(t)/\cosh x$, and the second gives $f^{\prime }=1$. ThereforeContinuous inverse-tangent branches may be chosen without creating artificial jumps. At spatial infinity the field approaches the same vacuum, so its net topological charge is zero. For large $|t-t_0|$ there are two well-separated transitions near $x_{\pm }=\pm \mathrm{arcosh}|t-t_0|$, with speeds tending to zero. The two transitions exchange their kink/antikink orientations as they pass through the collision, while the field remains smooth.

This is the Sine-Gordon threshold kink-antikink solution, a separatrix between finite-speed scattering and the bounded Sine-Gordon breather motion, rather than a finite-period breather. In fact it is the $v\to 0$ limit of the scattering field in question 3. At $t=t_0$, $\rho =0$ and ${\rho }_t=4\mathrm{sech}x$, so the reduced energy is $\int {\rho }_t^{2}dx/2=16$, exactly twice the reduced single-kink energy $8$.

For mass comparisons below, the original PDF really prints $m^2/\beta$. Restoring physical coordinates $X=x/m$, $T=t/m$ gives the angular-field action ${\beta }^{-1}\int dtdx[\frac{1}{2}(\partial \varphi {)}^2+\cos \varphi -1]$ and literal kink mass $8m/\beta$, for $\beta >0$. The usual Sine-Gordon theory instead has prefactor $m^2/{\beta }^2$ and mass $8m/{\beta }^2$. The equation of motion and every field above are unchanged by this overall normalization; the quantum scattering comparison in question 3 is not.

For nonzero real $g$, the continuous internal symmetry of a classical field theory is the global U(1) symmetryThe kinetic terms and mass terms are invariant separately, while the phases in ${\varphi }^2{\Phi }^{†}$ cancel. Independent phase rotations would instead require $2{\alpha }_{\varphi }-{\alpha }_{\Phi }=0$ modulo $2\pi$, leaving one connected circle. Simultaneous complex conjugation is also a discrete symmetry, reversing the continuous charge; including it gives $U(1)⋊{\mathbb{Z}}_2\cong O(2)$ for generic parameters. At $g=0$ the continuous symmetry enlarges to $U(1)\times U(1)$.

The Noether current with particles assigned positive charges isFor example, put $A={\varphi }^2{\Phi }^{†}$ and $B={\varphi }^{†2}\Phi$. The Euler-Lagrange equations give ${\partial }_{\mu }{j}_{\varphi }^{\mu }=ig(A-B)$ and ${\partial }_{\mu }{j}_{\Phi }^{\mu }=ig(B-A)$, with the factor $2$ already included in $j_{\Phi }^{\mu }$. Their sum vanishes. The Noether charge $Q=\int d^{3}x{j}^0$ is conserved and, after normal ordering, countsThus a $\Phi$ particle carries two units of the same conserved charge carried by a $\varphi$ particle; their antiparticles carry the negatives. The interaction permits conversion of one $\Phi$ particle into two $\varphi$ particles when allowed by conservation of energy, while preserving this weighted number.

The two complex scalar fields have scalar propagatorscontractions of two unconjugated fields or two conjugated fields vanish. Draw an arrow on each line in the direction of positive conserved charge flow: a solid line carries one unit and a dashed line two units. A particle arrow follows its physical momentum; an antiparticle arrow opposes it.

There are two conjugate three-leg interaction vertices, both with Feynman rule $-ig$: one has a charge-two arrow entering and two charge-one arrows leaving, and the other reverses every arrow. These describe the two conjugate terms ${\varphi }^2{\Phi }^{†}$ and ${\varphi }^{†2}\Phi$. The two identical $\varphi$ attachments cancel the $1/2$ in the Lagrangian density. All crossed assignments are obtained by moving external lines between initial and final states while maintaining the arrows. There is no interaction vertex with three equal-sign charges.

Each interaction vertex conserves four-momentum, supplying $(2\pi {)}^4{\delta }^4(\sum p_j)$ with all momenta entering. Integrate each independent loop momentum with $d^{4}k/(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. No minus sign is assigned to a closed scalar loop. These rules are for $i\mathcal{M}$ after amputating external lines and removing the overall four-momentum conservation factor.

Label the incoming four-momenta $p_1,p_2$ and the outgoing ones $p_3,p_4$. Define the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$. Two incoming $\varphi$ particles have conserved charge $+2$, so the only tree-level Feynman diagram is an $s$-channel $\Phi$ exchange, shown in panel (b) of b's figure. Each end converts two charge-one lines into one charge-two line. There are no $t$- or $u$-channel exchanges: an incoming and an outgoing particle at such an end would transfer zero charge, for which this theory has no cubic interaction vertex.The single Feynman diagram is already symmetric under exchanging the two outgoing particles. This does not add another copy of the scattering amplitude; the identical final-state symmetry factor enters the integrated scattering cross-section instead.

For fields ${\phi }_A$ and a continuous infinitesimal internal symmetry $\delta {\phi }_A=ϵ\Delta {\phi }_A$, suppose the density changes by a divergence, $\delta \mathcal{L}=ϵ{\partial }_{\mu }{K}^{\mu }$. Varying the action and using the Euler-Lagrange equations gives the Noether currentThe conserved charge is $Q=\int j^0{d}^{3}x$ when the flux through the boundary vanishes; more generally $dQ/dt=-\oint \mathbf{j}\cdot d\mathbf{S}$. If the Lagrangian is exactly invariant, $K^{\mu }=0$. Treat a complex field and its conjugate as independent variables during variation, imposing conjugacy on physical solutions. Time and space translations give energy and momentum through the corresponding spacetime version of the theorem. Continuous global symmetry gives an ordinary charge; local gauge redundancy requires additional care and is not an independent conserved charge for every arbitrary function.