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A parameter-dependent system of differential relations maps a solution of one differential equation to a solution of another, or to a new solution of the same equation. Compatibility of the relations supplies the equations of motion. Integration constants select a particular transformed field. The Sine-Gordon Bäcklund transformation is an example of an auto-transformation and also generates conserved currents.

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.

An interaction $-\lambda j^{\mu }{\partial }_{\mu }\varphi$ with an exactly conserved current is redundant: integration by parts gives $\lambda \varphi {\partial }_{\mu }{j}^{\mu }$ up to a boundary term. For the Dirac current, the local phase redefinition $\psi =e^{-i\lambda \varphi }\chi$ removes the coupling, so its on-shell scattering amplitudes vanish.

In a local quantum field theory obeying special relativity, with a positive physical state space, a conserved continuous internal global symmetry in field theory and a selected Lorentz-invariant ground state, a broken generator requires a massless Goldstone boson. For $G\to H$, there are $\dim G-\dim H$ symmetry-required Goldstone bosons in this setting; additional accidental massless excitations are possible. Jeffrey Goldstone's institutional biography identifies his work with Abdus Salam and Steven Weinberg on the zero-mass theorem.^[ref]

The Goldstone directions in the scalar mass matrix give a short classical proof at a smooth minimum. The quantum argument uses a conserved current $j^{\mu }$ and a scalar local operator $O$ whose expectation changes under the broken generator. The Ward identity for the current–local operator correlation function then has a nonzero contact term. After a Fourier transform, its divergence remains nonzero as four-momentum tends to zero. Lorentz invariance forces the vector–scalar correlator to have the form $p^{\mu }F(p^2)$, so $F$ must have a pole at $p^2=0$. The spectral representation associates that pole with a physical zero-mass excitation. A properly regulated spatial integral of the conserved current implements the broken generator; a well-defined global charge acting on the broken ground state cannot simply be assumed.

The hypotheses explain the boundaries of the result. A discrete symmetry has no infinitesimal generator and need not give a Goldstone boson. A gauge redundancy does not supply the physical global generator assumed here. In nonrelativistic matter the type-B Goldstone boson shows why one cannot universally count one independent mode per broken generator. Long-wavelength fluctuations can also prevent an assumed broken state from existing, as in the Mermin-Wagner theorem.

Consider one massive gauge field with $m>0$ after Higgs mechanism. For this comparison perform a Wick rotation and take a real nonzero momentum in Euclidean space $k_E$. The transverse and longitudinal orthogonal projectors areHere “longitudinal” means parallel to the off-shell four-momentum, not the physical longitudinal polarization of an on-shell massive gauge boson. The R-xi gauge quadratic vector operator and its inverse areMultiplication verifies $K_{\xi }{D}_{\xi }=P^T+P^L=I$, since $P^T{P}^L=0$ and $(P^{T,L}{)}^2=P^{T,L}$. The unitary gauge vector operator lacks the gauge fixing derivative term in the longitudinal sector, givingThus every eigenvalue of $D_{\xi }$ falls as $1/k_E^2$ at fixed finite positive $\xi$, whereas $D_U$ has the constant longitudinal eigenvalue $1/m^2$. Unitary gauge consequently obscures ordinary power counting; this is not a failure of the physical theory.

For a worked comparison choose $|k_E|=10m$. The dimensionless longitudinal eigenvalue $m^2{D}_L$ is $1$ in unitary gauge, $1/101$ at $\xi =1$, and $1/11$ at $\xi =10$. Its crossover is $|k_E|\sim \sqrt{\xi }m$: taking $\xi \to \infty$ first gives a constant, whereas taking $|k_E|\to \infty$ first gives zero. The two limits do not commute. Contracting the propagator between conserved external conserved currents projects away $P^L$, but one cannot impose that simplification separately on every internal line of an interacting Non-Abelian gauge theory.

For $\delta \varphi =\epsilon X$ and a variational symmetry of a Lagrangian density, write ${\Pi }^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )$ and $\mathcal{E}={\mathcal{L}}_{\varphi }-{\partial }_{\mu }{\Pi }^{\mu }$. The product rule gives $\delta \mathcal{L}/\epsilon =\mathcal{E}X+{\partial }_{\mu }({\Pi }^{\mu }X)$. Therefore $j^{\mu }={\Pi }^{\mu }X-K^{\mu }$ obeys ${\partial }_{\mu }{j}^{\mu }=-\mathcal{E}X$, so it is a conserved current on shell. Spatial integration gives a conserved Noether charge only when the boundary flux vanishes. The formula extends to several components by summation.

For $x^{\prime \mu }=x^{\mu }+\epsilon {\xi }^{\mu }$ and ${\varphi }_a^{\prime }(x^{\prime })={\varphi }_a(x)+\epsilon {\Delta }_a$, the fixed-coordinate field variation is $Q_a={\Delta }_a-{\xi }^{\nu }{\partial }_{\nu }{\varphi }_a$. Suppose action invariance is expressed as ${\delta }_0\mathcal{L}/\epsilon +{\partial }_{\mu }(\mathcal{L}{\xi }^{\mu })={\partial }_{\mu }{K}^{\mu }$. Writing ${\Pi }_a^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }{\varphi }_a)$ and $E_a={\mathcal{L}}_{{\varphi }_a}-{\partial }_{\mu }{\Pi }_a^{\mu }$, the product rule gives ${\partial }_{\mu }{j}^{\mu }=-{\sum }_a{E}_a{Q}_a$. Thus $j^{\mu }$ is a conserved current on shell. A vanishing spatial boundary flux is additionally required for its Noether charge to be conserved. This formula includes internal transformations, translations and Lorentz transformations.

At order $e^2$, scalar particle–antiparticle scattering has a $t$-channel photon exchange and an $s$-channel annihilation and recreation diagram. They are shown below, with incoming momenta $p,q$ and outgoing momenta $p^{\prime },q^{\prime }$. The seagull vertex has only two scalar legs and does not yield a four-scalar tree diagram at this order.

For the exchange graph, let $k=p-p^{\prime }=q^{\prime }-q$. Apart from overall charge signs, the two scalar currents are $J=p+p^{\prime }$ and $K=q+q^{\prime }$. The external mass shell conditions giveFor the annihilation graph, $k=p+q=p^{\prime }+q^{\prime }$, with currents $J=p-q$ and $K=p^{\prime }-q^{\prime }$, again up to vertex signs. Their contractions are $p^2-q^2=0$ and $p^{\prime 2}-q^{\prime 2}=0$. Thus each graph couples the photon to conserved currents, the relevant scalar quantum electrodynamics Ward identity.

Write $k=(\omega ,\mathbf{k})$ and ${\kappa }^2=|\mathbf{k}{|}^2$. conserved current means $\mathbf{k}\cdot \mathbf{J}=\omega J^0$ and $\mathbf{k}\cdot \mathbf{K}=\omega K^0$. For $\kappa \ne 0$, contraction with the Coulomb-gauge photon propagator gives, displaying the rational identity before the pole boundary limit,We used $k^2={\omega }^2-{\kappa }^2$. Restoring the common Feynman i-epsilon prescription yields the Coulomb-gauge propagator between conserved currents resultHence each on-shell tree amplitude can use the Lorentz-invariant propagator. This is equality after contraction, not equality of the gauge-dependent tensors. At $\mathbf{k}=0$, the separate Coulomb-gauge terms are ill-defined; combine them first and take the conserved-current limit. The combined expression has only the physical photon pole.

For fields ${\Phi }_a$ with a first-derivative Lagrangian density, suppose an infinitesimal global transformation at fixed coordinates is $\delta {\Phi }_a=\alpha {\Delta }_a$ and $\delta \mathcal{L}=\alpha {\partial }_{\mu }{K}^{\mu }$, with constant infinitesimal $\alpha$. Noether theorem gives the on shell conserved currentThis follows by varying the Lagrangian, integrating the terms with ${\partial }_{\mu }{\Delta }_a$ by parts, and setting the Euler-Lagrange expressions to zero. If spatial boundary flux vanishes, $\int d^{3}x{J}^0$ is a conserved Noether charge. Spacetime symmetries are covered by including the induced field variation at fixed coordinates and the corresponding total derivative $K^{\mu }$.

An example of a theory with a Lorentz four-vector is the source-free Maxwell field with electromagnetic four-potential $A_{\mu }$. Under an Abelian gauge transformation,The change in $F_{\mu \nu }$ is ${\partial }_{\mu }{\partial }_{\nu }\chi -{\partial }_{\nu }{\partial }_{\mu }\chi =0$, so the Maxwell Lagrangian is gauge-invariant. For $A_{\mu }=(\Phi ,-\mathbf{A})$ define $E_i=F_{0i}$ and $F_{ij}=-{ϵ}_{ijk}{B}_k$. With signature $+---$,The minus sign in $-F^2/4$ gives the positive electric kinetic term and positive physical field energy; reversing it would reverse the Hamiltonian's sign for the physical modes. The normalization $1/4$ accounts for antisymmetry and yields the conventional equations. Varying $A_{\nu }$ gives ${\partial }_{\mu }{F}^{\mu \nu }=0$.

For time translation at fixed coordinates take ${\Delta }_{\nu }={\partial }_0{A}_{\nu }$. Since $\alpha$ is constant, $\delta F_{\mu \nu }=\alpha {\partial }_0{F}_{\mu \nu }$, and hence $\delta \mathcal{L}=\alpha {\partial }_0\mathcal{L}$. Thus $K^{\mu }={\delta }^{\mu }{}_0\mathcal{L}$. The momentum derivative isNoether theorem therefore gives the canonical stress-energy tensor's time-translation currentHere ${\dot{A}}_i=E_i+{\partial }_i{A}_0$ was used. The canonical momentum of $A_0$ vanishes, while that of the lower spatial component $A_i$ is $E_i$: $A_0$ acts as a multiplier for Gauss law, not a propagating degree of freedom. The density above is not manifestly gauge-invariant, but its charge agrees with the physical energy on the constraint surface and with appropriate boundary conditions.

The gauge-covariant time translation instead has ${\Delta }_{\nu }={\partial }_0{A}_{\nu }-{\partial }_{\nu }{A}_0=F_{0\nu }$. It is time translation combined with a gauge transformation of parameter $\chi =-\alpha A_0$. Directly,using the definition of $F$, equivalently its Bianchi identity. The total derivative in $\delta \mathcal{L}$ is unchanged. The improved current isIt is the gauge-invariant Maxwell stress-energy tensor. In particular,The spatial current is the Poynting vector $\mathbf{E}\times \mathbf{B}$, so conservation gives ${\partial }_{t}u+\nabla \cdot (\mathbf{E}\times \mathbf{B})=0$. This is the familiar positive local electromagnetic field energy density, expressed entirely in measurable fields.

To compare the currents,The last term vanishes on the source-free Maxwell equations, leaving a stress-energy tensor improvement generated by a stress-energy superpotential. At the charge level,when the surface term vanishes and Gauss law holds. Thus the symmetries yield the same conserved energy under these conditions while the second supplies a gauge-invariant local density. Boundary terms would have to be retained if those boundary conditions were not imposed.

Use the Minkowski metric $\mathrm{diag}(1,-1)$ and the angular field $u=\beta \varphi$. The Euler-Lagrange equation becomes $u_{tt}-u_{xx}+m^2\sin u=0$. For the dimensionless light-cone coordinatesthis is $u_{\xi \eta }=\sin u$. This normalization keeps the coupling in the relation between the physical field and its angle, rather than silently setting $\beta =1$.

A Bäcklund transformation is a system of first-order differential relations that maps a solution to another solution. One convention for the Sine-Gordon Bäcklund transformation uses a nonzero parameter $a$ and defines $v={\mathcal{B}}_a[u]$ byA compatible initial value or integration constant selects a particular transformed solution. Put $w=(v+u)/2$ and $d=(v-u)/2$. Differentiating gives $w_{\xi \eta }=\cos d\sin w$ and $d_{\xi \eta }=\cos w\sin d$. Their sum and difference yieldThus compatibility of the first-order relations contains the field equations for both fields, and the transformed physical field $v/\beta$ also solves Sine-Gordon theory.

The transformation provides a generating conservation law. On the branch close to the original field write $v=u+2d$. Its first equation isIt recursively determines a formal small-$a$ expansion of $d$ in local derivatives of $u$:At every subsequent order the new coefficient occurs linearly, so this recursion continues indefinitely. The second Bäcklund equation and the first imply the exact identityIndeed ${\partial }_{\eta }\cos d=-a\sin d\sin (u+d)$ and ${\partial }_{\xi }\cos (u+d)=-a^{-1}\sin d\sin (u+d)$. Comparing powers of $a$ therefore gives local conserved currents. If ${\partial }_{\eta }{P}_j={\partial }_{\xi }{Q}_j$, our coordinate convention givesFor localized fields approaching vacua at spatial infinity, the boundary flux vanishes and ${\mathcal{I}}_j$ is conserved. The formal expansion need not converge: each coefficient is a separately exact local conservation law.

For example $P_0=u_{\xi }^2/2$, $Q_0=1-\cos u$, a light-cone combination of energy and momentum. The next coefficient is a derivative improvement, so it contributes no independent charge under the same decay conditions. At order $a^2$, remove the improvement generated by ${\partial }_{\xi }(u_{\xi }{u}_{\xi \xi })$ and multiply by $-2$. One obtains the genuinely higher conservation lawIt can also be checked directly using $u_{\xi \eta }=\sin u$. Continuing the recursion, and using the opposite light-cone construction, produces the local conserved-charge hierarchy of sine-Gordon theory, with infinitely many nontrivial higher-spin charges after derivative improvements are removed. This is the Bäcklund generating current for sine-Gordon conserved charges. The hierarchy is the characteristic field-theory form of classical integrability; an ordinary energy conservation law alone would not supply these constraints.

The same transformation constructs solutions rather than only currents. Starting from the vacuum $u_0=0$, its two first-order equations integrate toFor $a=e^{\rho }>0$ the exponent is $m\cosh \rho x-m\sinh \rho t+{\delta }_a$. This is a Sine-Gordon kink with velocity $\tanh \rho$, center set by ${\delta }_a$ and classical rest mass $8m/{\beta }^2$. Negative parameters can supply the corresponding opposite-orientation seeds.

The allowed Bianchi permutability for sine-Gordon Bäcklund transformations gives the two-step field algebraically. For vacuum seed and $a\ne b$,Branches of the inverse tangent must be continued smoothly; its principal value alone does not specify the vacuum labels of a multi-kink field. All angles here are $u=\beta \varphi$. In the physical-field version each field difference in the superposition formula carries $\beta$; the formula printed without it implicitly uses the $\beta =1$ angular-field convention.

To display two real scattering solutions, take $0<v<1$, set $\Gamma =(1-v^2{)}^{-1/2}$, and put $X=m\Gamma x$, $Y=m\Gamma vt$. Choosing $a=e^{\rho }$, $b=e^{-\rho }$ with $v=\tanh \rho$ and zero phase constants yields the Sine-Gordon kink-antikink scattering solutionChoosing instead $b=-e^{-\rho }$ yields the Sine-Gordon two-kink solutionThe latter has net angular winding $4\pi$, while the former has zero net winding. At large positive or negative time they separate into localized kinks with velocities $\pm v$. Repeated commuting transformations give general multi-soliton fields. Continuing the kink-antikink velocity to an imaginary value also gives a real Sine-Gordon breather:up to the irrelevant overall field sign and translations.

The scattering interpretation ties the construction to the conserved hierarchy. In an exact multi-soliton sector, the incoming species and rapidities reappear after the collision, up to permutation: the solitons change positions, not their asymptotic shapes or velocities, and the collision emits no radiation. In the two-kink example the large-time centers obey $|X|=|Y|-\log v+o(1)$, so a right-moving trajectory acquires a shift $-2\log v/(m\Gamma )$. These shifts encode the interaction even though the collision is elastic. The commuting construction makes the net displacement in a many-soliton collision the sum of its pairwise displacements, independent of how the collisions are ordered; this is pairwise additivity of soliton shifts and the classical counterpart of factorized scattering.

Conservation of the entire hierarchy is much stronger than conservation of energy and momentum: its independent rapidity-weighted sums constrain the whole asymptotic soliton data and rule out particle production in this sector. Generic initial fields may also contain radiation scattering data, so this statement is about the exact soliton collisions, not a claim that every initial field is a pure soliton. The Bäcklund map unifies the conserved hierarchy, explicit soliton construction and elastic, pairwise scattering picture of classical integrability.

Use gamma matrices satisfying the Clifford algebra relation $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, with ${\gamma }^{0†}={\gamma }^0$ and ${\gamma }^{i†}=-{\gamma }^i$. The Dirac adjoint is $\bar{\Psi }={\Psi }^{†}{\gamma }^0$. Take the Dirac actionIt differs from the manifestly Hermitian form with $(i/2)\bar{\Psi }{\gamma }^{\mu }{\overleftrightarrow{\partial }}_{\mu }\Psi$ only by a boundary term. Treat $\Psi$ and $\bar{\Psi }$ as independent variables when applying the principle of stationary action. Varying $\bar{\Psi }$ givesVarying $\Psi$ and integrating by parts gives the adjoint Dirac equation, $i({\partial }_{\mu }\bar{\Psi }){\gamma }^{\mu }+m\bar{\Psi }=0$. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ also gives $(\square +m^2)\Psi =0$, so its dispersion relation is Lorentz invariant.

For a Lorentz transformation $x^{\prime }=\Lambda x$, the field transforms in the Spinor representation of the Lorentz group:Since ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$, this identity gives the Lorentz covariance of the Dirac operator:Thus every solution is carried to another solution. The field is a spinor rather than a four-vector; the transformation of the gamma matrices supplies the necessary covariance. The Dirac action is Lorentz invariant because its integrand is a scalar and $d^{4}x$ is invariant.

For electric charge $q$, use the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$. The minimal electromagnetic coupling of a Dirac field isThe local gauge transformations in this convention areDirect substitution gives $D_{\mu }^{\prime }{\Psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\Psi$, which proves gauge covariance of the charged Dirac equation. The mass and kinetic terms are invariant, and $F_{\mu \nu }^{\prime }=F_{\mu \nu }$, so the full action has gauge invariance. The conserved current is $j^{\mu }=q\bar{\Psi }{\gamma }^{\mu }\Psi$. It transforms as a four-vector under Lorentz transformations, and varying $A_{\mu }$ gives ${\partial }_{\nu }{F}^{\nu \mu }=j^{\mu }$.

To make the infinitesimal spinor transformation explicit, write ${\Lambda }^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }+{\omega }^{\mu }{}_{\nu }$ with ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. DefineThe Lorentz generators from gamma-matrix commutators have precisely the needed algebra: $[{\sigma }^{\mu \nu },{\gamma }^{\rho }]=2i({\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu })$, which verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }$ to first order. The infinitesimal transformation of a Dirac field is thereforeAt the same coordinate argument, the orbital change must also be included:Both formulas describe the same transformation; their arguments differ.

Finally, the gamma adjoint identities imply ${\sigma }^{\mu \nu †}={\gamma }^0{\sigma }^{\mu \nu }{\gamma }^0$ and hence the pseudo-unitarity of the spinor Lorentz representation, $S^{†}{\gamma }^{0}S={\gamma }^0$. It follows thatThus the Dirac scalar bilinear $\bar{\Psi }\Psi$ is a Lorentz scalar. Infinitesimally the spin terms in $\delta (\bar{\Psi }\Psi )$ cancel; at fixed coordinates only the ordinary scalar orbital transformation remains.

Let ${\mathcal{G}}_R=⟨{\mathcal{O}}_{1,R}(x_1)\cdots {\mathcal{O}}_{n,R}(x_n)⟩$ at separated points, with composite operators chosen in a basis without operator mixing. Write ${\mathcal{O}}_{i,B}=Z_i{\mathcal{O}}_{i,R}$, ${\beta }^A=\Lambda d{g}^A/d\Lambda$, and ${\gamma }_i=\Lambda d\log Z_i/d\Lambda$, all derivatives taken at fixed bare data. Bare correlations are independent of the arbitrary intermediate renormalization scale. Hence the Callan–Symanzik equation isThe explicit scale derivative holds positions and renormalized parameters fixed; the renormalization-group beta functions describe their running; and the anomalous dimensions account for renormalization of each insertion. Dimensionful parameters add their own running-parameter derivatives. With composite operator mixing, each ${\gamma }_i$ is a matrix acting on its operator slot. The displayed sign follows the stated convention for $Z_i$. Coincident products can additionally need contact terms in a correlation function, which are important below.

In four dimensions, a conserved current arising as a Noether current has classical mass dimension three. Its two-point product has dimension six, and integration over $d^{4}x$ reduces that to two. The transverse tensor $k_{\mu }{k}_{\nu }-{\delta }_{\mu \nu }{k}^2$ already has dimension two. Thus the scalar coefficient has dimension zero:The assumed vanishing anomalous dimension protects the individual current, but does not remove the logarithmic short-distance divergence of its product with another current.

Use $\beta (g)=\Lambda dg/d\Lambda$ and let $\mathcal{D}=\Lambda {\partial }_{\Lambda }+\beta (g){\partial }_g$ at fixed momentum. Bare-scale independence gives $\mathcal{D}{G}_0=0$ at fixed ${\Lambda }_0$. Differentiating the additive subtraction instead gives $\mathcal{D}[-a\log ({\Lambda }_0/\Lambda )]=+a$. Hence the renormalized conserved-current two-point function obeys an inhomogeneous equation:This is the Callan-Symanzik equation with additive renormalization. The source comes from a local contact term in a correlation function counterterm, despite the current's zero anomalous dimension. A homogeneous equation would miss this contribution.

For $k=\sqrt{k^2}>0$, run the renormalization scale from $\Lambda$ to $k$, keeping the physical momentum fixed. Along the renormalization-group flow, $d{G}_{\mathrm{ren}}/d\log \mu =a$, soDimensional analysis makes $C$ independent of $k$ in this massless problem. Asymptotic freedom means $g(k)\to 0$ as $k/\Lambda \to \infty$. For the usual current normalization with a regular free-theory matching value, $C(g(k))\to C(0)$. Consequently the leading ultraviolet behaviour at this one-loop subtraction accuracy isIn particular the leading coefficient is $-a/2$, with its sign fixed by the subtraction convention in the question. If $\beta (g)=-b{g}^3+\cdots$, $b>0$, then $g(k{)}^2\sim [2b\log (k/\Lambda ){]}^{-1}$, explaining the approach to free-theory matching. Beyond the stated approximation, a coupling-dependent additive source can give subleading logarithms; neither its coefficients nor the finite matching constant are determined by the supplied constant $a$.

Use a local first-derivative Lagrangian density $\mathcal{L}(\varphi ,{\partial }_{\mu }\varphi ,x)$ with sufficient differentiability, and let variations have compact support or vanish at the spacetime boundary. If the phrase “a function of the field” were read literally as forbidding derivatives, the equation below would reduce to $\partial \mathcal{L}/\partial \varphi =0$; a propagating field requires derivative dependence. Varying the action givesIntegration by parts and arbitrary interior variations yield the Euler-Lagrange field equation,The boundary condition is part of this derivation; if boundary variations are allowed, their separate boundary equations must also be imposed.

A variational symmetry of a Lagrangian density is an infinitesimal field change $\delta \varphi =\epsilon X$ for which $\delta \mathcal{L}=\epsilon {\partial }_{\mu }{K}^{\mu }$ off shell. Equality to zero is sufficient but is not necessary: a divergence changes only the boundary contribution to the action. For fixed-coordinate field variations,Equating the two expressions proves the scalar-field form of the Noether theorem:Thus every differentiable one-parameter variational symmetry gives a conserved current, and its Noether charge $Q=\int d^{3}x{j}^0$ is constant when the spatial boundary flux vanishes. This is the Noether current for a first-derivative scalar field; the same identity applies to several real components by summing over them. Identically conserved improvement terms can change its local expression without changing the charge under the same boundary assumptions.

For an active spacetime translation, ${\varphi }^{\prime }(x)=\varphi (x-a)$ and $\delta \varphi =-a^{\nu }{\partial }_{\nu }\varphi$. If the Lagrangian density has no explicit coordinate dependence, $\delta \mathcal{L}=-a^{\nu }{\partial }_{\nu }\mathcal{L}$, so $K^{\mu }=-a^{\mu }\mathcal{L}$. The Noether current is $j^{\mu }=-a^{\nu }{T}^{\mu }{}_{\nu }$, whereThese four translation currents are the canonical stress-energy tensor, also called the energy-momentum tensor. The sign in $j=-aT$ follows from the chosen active translation; the translation charges themselves can be labelled by $P_{\nu }=\int T^0{}_{\nu }{d}^{3}x$.

For the free real scalar field, take the standard kinetic term and mass term. Its Lagrangian density and Klein-Gordon equation areHere ${\Pi }^{\mu }={\partial }^{\mu }\varphi$, so $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Raising the charge index gives the four-momentum of a free real scalar field,The first is total energy, and the three components of the second are physical spatial momentum. With signature $(+,-,-,-)$, $P_i=-P^i$; this explains the opposite sign if the conserved quantities are instead written with lower spatial indices. For example, a plane wave proportional to $\cos (Et-\mathbf{k}\cdot \mathbf{x})$ has momentum density along $\mathbf{k}$, confirming the sign. All charges require convergence of the integrals and vanishing boundary flux, or periodic boundary conditions in a finite box.

For the complex scalar field, treat $\psi$ and ${\psi }^{\ast }$ as independent variables when varying, equivalently use their two real components. The global phase symmetry of a complex scalar field is $\psi \mapsto e^{-i\alpha }\psi$, ${\psi }^{\ast }\mapsto e^{i\alpha }{\psi }^{\ast }$ for constant $\alpha$. Both the kinetic term and $V(|\psi {|}^2)$ are invariant. This is a global internal symmetry of a classical field theory with circle group $U(1)$; a spacetime-dependent phase would require a gauge connection.

Choosing this orientation for the phase parameter gives the Noether charge of a complex scalar field:To check conservation directly, the Euler-Lagrange field equations are $\square \psi +V^{\prime }(|\psi {|}^2)\psi =0$ and their complex conjugates, assuming a real differentiable potential. Hence ${\partial }_{\mu }{j}^{\mu }=i({\psi }^{\ast }\square \psi -\psi \square {\psi }^{\ast })=0$. Reversing the phase-parameter orientation reverses the current and charge, which is merely a convention. For a scalar carrying electric charge $q$, the physical electric charge is $qQ$ after electromagnetic coupling. In a neutral theory the same global charge can instead label an internal conserved quantum number; the density alone does not identify it automatically with electricity.

Antisymmetry gives $-F^{\rho \mu }{\partial }_{\rho }{A}^{\nu }=F^{\mu \rho }{\partial }_{\rho }{A}^{\nu }$. Combining this with the canonical stress-energy tensor removes the potential in favour of the electromagnetic field tensor:This is the electromagnetic stress-energy tensor in the $(+,-,-,-)$ convention. The product $F^{\mu \rho }{F}^{\nu }{}_{\rho }$ is symmetric in $\mu ,\nu$, since the contracted metric is symmetric. Its expression only in $F$ makes it gauge-invariant. These two properties hold without using the equations of motion.

For conservation, the source-free Maxwell equations are ${\partial }_{\rho }{F}^{\rho \mu }=0$. Hence the given correction equalson shell. This is an antisymmetric superpotential improvement: $B^{\rho \mu \nu }=-F^{\rho \mu }{A}^{\nu }$ is antisymmetric in $\rho ,\mu$, so ${\partial }_{\mu }{\partial }_{\rho }{B}^{\rho \mu \nu }=0$. Conservation of the canonical stress-energy tensor therefore implies ${\partial }_{\mu }{\Theta }^{\mu \nu }=0$. Equivalently, direct differentiation gives ${\partial }_{\mu }{\Theta }^{\mu \nu }=-F^{\nu }{}_{\rho }{\partial }_{\mu }{F}^{\mu \rho }$ after the homogeneous Maxwell equations cancel the field-strength derivative terms.

There are four conserved currents, one for each fixed $\nu$, with charges $P^{\nu }=\int d^{3}x{\Theta }^{0\nu }$. The Belinfante-Rosenfeld stress-energy tensor differs from the canonical tensor only by boundary contributions to these charges when the fields decay suitably.

Finally, in four spacetime dimensions,Thus all requested properties are

A gauge boson couples to the conserved current of a charged fermion throughwith the appropriate chiral projector when the gauge representation is chiral. At momentum transfer $|q^2|\ll M_X^2$, its propagator can be expanded aswhere the longitudinal term drops for a conserved current. The operation of integrating out a field, applied to $X_{\mu }$, gives the local four-fermion interactionup to the normalization used for $J^{\mu }$. Hence the effective Fermi coupling scales as $G\sim g_X^2/M_X^2$. For Standard Model charged currents, the conventional normalization is

The parton distribution function is a number density. Integrating the quark distribution minus the antiquark distribution counts the net flavour charge of the hadron, since every sea quark-antiquark pair contributes zero. The corresponding flavour-vector conserved current counts particles minus antiparticles. For a proton, the net up/down numbers are two and one; for a neutron they are one and two. The rule fixes the difference of distributions, not their sum, and does not imply absence of a sea.