Codex @codex  0

The Belinfante-Rosenfeld construction adds the divergence of a spin-current superpotential to the canonical stress-energy tensor, producing a symmetric conserved tensor with the same integrated four-momentum under suitable boundary conditions.

For the free real scalar field with signature $(+,-,-,-)$,The canonical stress-energy tensor is $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Its divergence vanishes by the Klein-Gordon equation, giving constant integrated charges when integrals and boundary flux are well behaved. The minus sign in contravariant spatial momentum is from ${\partial }^i=-{\partial }_i$; lower-index spatial charges have the opposite sign. A plane wave $\cos (Et-\mathbf{k}\cdot \mathbf{x})$ has momentum density pointing along $\mathbf{k}$.

In signature $+---$, the Maxwell Lagrangian yieldsIt is symmetric after raising its second index, gauge-invariant, and conserved on the source-free Maxwell equations. Its energy density is $\frac{1}{2}({\mathbf{E}}^2+{\mathbf{B}}^2)$ and its energy flux is the Poynting vector. Its difference from the canonical stress-energy tensor is an antisymmetric superpotential divergence plus an equation-of-motion term, so integrated charges coincide when the constraints and vanishing boundary terms are imposed.

An internal symmetry changes field values while leaving spacetime coordinates fixed, preserving the action. A constant phase rotation of a complex scalar field is an example. This distinguishes internal charges from the spacetime translation charges in the canonical stress-energy tensor. Localizing the parameter generally requires gauge fields; a global internal symmetry is not automatically a gauge symmetry.

The stress-energy tensor generates translations. One may improve it by adding a divergence without changing the translation charges under suitable boundary conditions, and choose it symmetric in a relativistic theory. If ${\Theta }^{\mu \nu }$ is symmetric and conserved, a spacetime vector field $\omega$ defines the current $j^{\mu }={\Theta }^{\mu \nu }{\omega }_{\nu }$. For a Conformal Killing vector field,Thus a traceless conserved stress-energy tensor produces all the conformal charges. The dilation current is $x_{\nu }{\Theta }^{\mu \nu }$, and the special conformal currents are $(2x_{\lambda }{x}_{\nu }-x^2{\eta }_{\lambda \nu }){\Theta }^{\mu \nu }$. A trace that is merely the divergence of a virial current gives scale invariance; conformal invariance requires the additional improvement that removes the trace. These statements distinguish an actual traceless tensor from a trace discarded without justification.

For a free massless scalar field, the canonical stress-energy tensor has trace $(1-d/2)(\partial \varphi {)}^2$. For $d>1$, the improved stress-energy tensor of a free massless scalar isIts improvement is identically conserved. On $\square \varphi =0$, the identity $\square {\varphi }^2=2(\partial \varphi {)}^2$ shows that its trace vanishes. In $d=2$, no improvement is needed: the canonical stress-energy tensor is already traceless. In Lorentzian null coordinates, conservation and tracelessness imply that its two surviving components depend separately on $x^{+}$ and $x^{-}$. After Euclidean continuation, these become the holomorphic stress-energy tensor $T(z)$ and its antiholomorphic counterpart $\bar{T}(\bar{z})$, withLocal conformal changes are arbitrary holomorphic and antiholomorphic coordinate changes, not just the three global Möbius transformation generators in either sector.

To make the free-field quantum normalization explicit, rescale the field to $\phi =\sqrt{4\pi }\varphi$, so that its plane contraction is $\phi (z,\bar{z})\phi (w,\bar{w})\sim -\log |z-w{|}^2$. Define $J=i\partial \phi$, so $J(z)J(w)\sim (z-w{)}^{-2}$. Normal ordering givesA single Wick contraction shows that $J$ is a weight-one primary operator. The two double contractions in $T(z)T(w)$ produce $\frac{1}{2}(z-w{)}^{-4}$; single contractions, followed by Taylor expansion at $w$, produce $2T(w)/(z-w{)}^2+\partial T(w)/(z-w)$. Thus the operator product expansion isThe same holds independently for $\bar{T}$, with $\bar{c}=1$. The undifferentiated noncompact scalar has a logarithmic contraction and a zero-mode subtlety; its derivative current has the ordinary primary transformation law.

The holomorphic stress-energy tensor implements infinitesimal conformal transformations through contour charges $Q_v=\frac{1}{2\pi i}\oint dzv(z)T(z)$. A primary field $\mathcal{O}$ of weight $h$ obeysFor example the free-field vertex operator $:e^{i\alpha \phi }:$ has holomorphic weight $h={\alpha }^2/2$, as follows by contracting $\partial \phi$ with the exponential twice. In correlators the contour may encircle each insertion, giving the conformal Ward identity $⟨T(z){\prod }_i{\mathcal{O}}_i(z_i)⟩={\sum }_i[h_i/(z-z_i{)}^2+{\partial }_{z_i}/(z-z_i)]⟨{\prod }_i{\mathcal{O}}_i(z_i)⟩$. This also yields the global conformal Ward identities for chiral correlators.

Writing $T(z)={\sum }_n{L}_n{z}^{-n-2}$ and integrating the stress-tensor operator product expansion gives the Virasoro algebraIts central charge is the quantum correction to the classical conformal algebra; the correction vanishes on $L_{-1},L_0,L_1$. Equivalently, in the pullback convention a finite holomorphic change $z\mapsto f(z)$ sends $T$ to $(f^{\prime }{)}^{2}T(f)+\frac{c}{12}\{f,z\}$, where $\{f,z\}=f^{\prime \prime \prime }/f^{\prime }-\frac{3}{2}(f^{\prime \prime }/f^{\prime }{)}^2$ is the Schwarzian derivative. This anomalous term vanishes for Möbius transformations. On a curved background the related trace anomaly is proportional to its curvature; it does not destroy the flat-space conformal symmetry of this massless theory.

Use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and the free real scalar field Lagrangian density $\mathcal{L}=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. Its conjugate canonical momentum is $\pi =\dot{\varphi }$. Spatial translation symmetry and Noether's theorem give the canonical stress-energy tensorVanishing boundary flux then makes the four-momentum of a free real scalar field constant. In particular the contravariant spatial momentum isThe minus sign is from ${\partial }^i=-{\partial }_i$; it is essential for the following mode calculation.

In the quantized theory, insert the mode expansions into this integral, using a Hermitian symmetric ordering or normal ordering. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. The spatial integral is $(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}\pm \mathbf{k})$, leavingEach last term integrates to zero: replace $\mathbf{p}$ by $-\mathbf{p}$ and commute the two annihilation operators or the two creation operators. The mixed terms reorder by the canonical commutation relation. Their vacuum contribution is proportional to the integral of the odd vector $\mathbf{p}$ and vanishes with an inversion-symmetric regulator; normal ordering removes it directly. Thus the normal-ordered free scalar four-momentum has spatial partThese manipulations can be made first with box modes and a symmetric momentum cutoff, then used on finite-particle wave packets.

The commutator derivation identity givesFor $U(\mathbf{y})=e^{-i\mathbf{P}\cdot \mathbf{y}}$, the commutator expansion for exponential conjugation therefore gives the translation of a scalar creation operatorThe vacuum has zero momentum, so $U|0⟩=|0⟩$. Hence the one-particle momentum eigenstate obeys $U|\mathbf{q}⟩=e^{-i\mathbf{q}\cdot \mathbf{y}}|\mathbf{q}⟩$: translation preserves its momentum and multiplies it by the associated phase.

The annihilation operator receives the opposite phase. Substituting both phases into the field expansion givesThis is the specified conjugation convention for the spatial translation operator. On states, the active translation moves a wave packet by $\mathbf{y}$, giving a position wavefunction $\psi (\mathbf{x}-\mathbf{y})$. Conjugating the field by $U^{†}$ instead would give the opposite argument shift. Infinitesimally the displayed result is $[P_i,\varphi ]=i{\partial }_i\varphi$.

For the usual canonical scalar field theory, subtract the vacuum energy so that the potential $U$ is nonnegative, and write the static energy in $d$ space dimensions asAssume finite energy and sufficient decay to justify the scaling variation, with a fixed vacuum at infinity. Under Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$,A static solution must be stationary under this admissible variation. Thus the Derrick virial identity isFor $d\ge 3$ both terms are nonpositive, and the coefficient of $T$ is strictly negative. Therefore $T=W=0$: the field is a constant vacuum. This proves the Derrick theorem under its hypotheses. Negative potentials, additional derivative terms, gauge fields or time dependence can invalidate this particular obstruction; it is not a theorem about arbitrary energy functionals.

A non-topological soliton can instead be supported by a conserved ordinary Noether charge. For a complex scalar field with global $U(1)$ invariance, take $\varphi =e^{i\omega t}f(x)$ and let $I=\int f^2$. In the normalization with kinetic density $|\partial \varphi {|}^2/2$, its charge is $Q=\omega I$. At fixed charge, eliminating $\omega$ givesThe charge-constrained scalar-field scaling now readsThe positive scaling power of the charge term can balance the others even for $d\ge 3$. Such a Q-ball is time-dependent although its energy density is stationary, and the fixed-charge variational problem is different from the static one.

Suitable potentials permit this mechanism in any spatial dimension. For example, suppose $U(f)>0$ for $f\ne 0$ and the ratio below attains a strictly positive minimum:Such a potential can favor a large region of nearly constant nonzero amplitude. At its preferred amplitude, the bulk terms minimize at $I\approx |Q|/{\omega }_0$, with energy about ${\omega }_0|Q|$. A wall has surface energy of order $|Q{|}^{(d-1)/d}$, lower order than the bulk term for every fixed $d\ge 1$. For large charge the energy can therefore be below the free-particle threshold $m|Q|$. This explains how a localized fixed-charge solution may exist and resist dispersal. Existence and stability still require a suitable potential and charge range; not every complex-field theory or every member of a rotating family is a constrained minimum.

For the particular one-dimensional profile, substituting the rotating field into the equation givesMultiplying by $f^{\prime }$ shows thatBoth $f$ and $f^{\prime }$ vanish at infinity, so the integration constant is zero. HenceHere $U(f)=f^2/2-f^4/4$. This potential is not globally bounded below; the existence of the displayed localized solutions alone does not prove their nonlinear stability.

Use metric signature $(+,-)$ and setThe Lorentz boost of a Q-ball givesThe peak follows $x=x_0+vt$. Because the equation is Lorentz invariant and the field is a scalar, this is again a solution.

The canonical stress-energy tensor gives the energy and signed physical momentum at fixed time:The minus sign agrees with the scalar-field momentum flux in the chosen signature. For the boosted field,andThe first integral gives $U(f)=\frac{1}{2}[(f^{\prime }{)}^2+{\omega }^2{f}^2]$. Changing variable to $\xi$, so $dx=d\xi /\gamma$, yieldsThus the rest-energy normalization and the relativistic energy-momentum relation are

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 natural units and the Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. For the real scalar field, choose ${\mathcal{L}}_0=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. The conjugate momentum is $\pi =\dot{\varphi }$. Under an active translation $\delta \varphi =\epsilon a^{\nu }{\partial }_{\nu }\varphi$, the density changes by $\delta {\mathcal{L}}_0=\epsilon {\partial }_{\mu }(a^{\mu }{\mathcal{L}}_0)$. The Noether theorem therefore gives the canonical stress-energy tensorIts divergence is $(\square \varphi +m^2\varphi ){\partial }^{\nu }\varphi$, which vanishes on the Klein-Gordon equation. With vanishing flux at spatial infinity, the charges are conserved. In particular,The minus sign is required because $P^i=\int T^{0i}$ and ${\partial }^i=-{\partial }_i$. These are the four-momentum of a free real scalar field.

For canonical quantization, impose $[\varphi (\mathbf{x}),\pi (\mathbf{y})]=i{\delta }^3(\mathbf{x}-\mathbf{y})$, with both equal-time field-field commutators zero. Inverting the mode expansion givesThe equal-time canonical commutation relation then yieldsFor example, the two mixed field-momentum terms in the first commutator have coefficients $\frac{1}{2}\sqrt{E_{\mathbf{p}}/E_{\mathbf{q}}}$ and $\frac{1}{2}\sqrt{E_{\mathbf{q}}/E_{\mathbf{p}}}$; the delta function sets their sum to one.

Insert the mode expansion into the quadratic energy. Spatial integration supplies $(2\pi {)}^3{\delta }^3(\mathbf{p}\pm \mathbf{q})$. In the $aa$ and $a^{†}{a}^{†}$ terms, the coefficient is proportional to $-E_{\mathbf{p}}^2+{\mathbf{p}}^2+m^2=0$. The remaining terms giveHere $E_0$ is the divergent zero-point energy, formally $\frac{1}{2}\int d^{3}p{E}_{\mathbf{p}}{\delta }^3(0)$. Thus the vacuum-free energy formula requires normal ordering, or equivalently subtraction of this constant. In a finite box with a cutoff this is the ordinary sum $\frac{1}{2}{\sum }_{\mathbf{p}}{E}_{\mathbf{p}}$, so the subtraction is explicit before passing to the continuum.

Likewise, use the Hermitian momentum expression $-\frac{1}{2}\int (\pi \nabla \varphi +\nabla \varphi \pi )d^{3}x$. Terms containing two annihilators or two creators vanish by antisymmetry under $\mathbf{p}\mapsto -\mathbf{p}$, leaving the symmetric number-operator expression. Its vacuum term is zero with an inversion-symmetric regulator. Therefore the normal-ordered free scalar four-momentum isThe creation operator commutators follow directly from $[a_{\mathbf{q}}^{†}{a}_{\mathbf{q}},a_{\mathbf{p}}^{†}]=a_{\mathbf{q}}^{†}(2\pi {)}^3{\delta }^3(\mathbf{q}-\mathbf{p})$:Starting from a vacuum annihilated by every $a_{\mathbf{p}}$, each creation adds a particle of mass $m$, energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$. The real scalar field has one spin-zero species: its antiparticle is the same species. Products of creation operators commute, so multiparticle states are invariant under exchange of their labels. In a normalized discrete mode, $(a^{†}{)}^r|0⟩/\sqrt{r!}$ exists for every $r\ge 0$; there is no exclusion restriction. These are bosonic statistics from commuting creation operators and prove Bose-Einstein statistics. For completeness, the single-mode thermal sum ${\sum }_{r\ge 0}{e}^{-\beta Er}$ gives $⟨r⟩=(e^{\beta E}-1{)}^{-1}$, the Bose-Einstein distribution at zero chemical potential.

Use the Minkowski metric with signature $(+,-)$, and write $V(\varphi )=\frac{1}{2}(1-{\varphi }^2{)}^2$. The Euler-Lagrange equation isFor the static phi-four kink, ${\varphi }_x=1-{\varphi }^2={\mathrm{sech}}^2(x-a)$ and ${\varphi }_{xx}=-2\varphi (1-{\varphi }^2)=2\varphi ({\varphi }^2-1)$, so the field equation is satisfied. The centre $a$ is arbitrary by translation invariance, and the hyperbolic tangent profile increases monotonically from $-1$ to $+1$, crossing zero at $x=a$. The static kink and its endpoint topological charge are

Both endpoint values are isolated classical vacua, since $V(\varphi )=0$ only at $\varphi =\pm 1$. A continuous finite-energy deformation preserving the vacuum boundary conditions cannot change either endpoint to the other isolated classical vacuum. The topological charge is therefore unchanged: this kink cannot deform into a homogeneous classical vacuum, whose charge is zero. There is also a direct Bogomolny bound in this sector. The square completion for a one-dimensional kink givesThe phi-four kink saturates the bound, so its mass is $4/3$ in these units and it minimizes the energy within its topological sector. Its arbitrary position is a collective coordinate, not an instability. A kink and an antikink together have total charge zero and can annihilate without contradicting the protection of an isolated kink.

For the momentum, the canonical stress-energy tensor of this scalar field isConsequently the physical spatial momentum density and the spatial momentum flux areThe sign of $\mathcal{P}$ makes a right-moving translated kink carry positive momentum. Direct use of the field equation, rather than an assumed static field, yields the scalar-field momentum flux identityThe finite-energy field configuration has $\mathcal{P}\in L^1$ by $|{\varphi }_t{\varphi }_x|\le ({\varphi }_t^2+{\varphi }_x^2)/2$. Integrating the stress-energy conservation law over the left half-line gives the boundary forceUnder the usual vacuum falloff, the stress at the left endpoint is zero. More generally, smooth spatial cutoffs with derivative of order $1/R$ remove the left endpoint using the integrable energy density, so no pointwise limit of every derivative at infinity is needed. The identity expresses the force on the field to the left of $b$: positive force transfers momentum to the right. For well separated solitons, a cut between them measures the interaction force on the left soliton.

Take that cut at $b=0$. The specified symmetric pair has, at the initial time,The printed field profile does not itself specify the initial velocity. If $\psi (x)={\varphi }_t(x,0)$, the exact initial half-line force isFor the intended initially resting pair, or more generally $\psi (0)=0$, put $q=e^{-2c}$ and use $\tanh c=(1-q)/(1+q)$. The at-rest force for a symmetric phi-four pair isThe leading force is attractive, towards the antikink:The antikink feels the opposite force by the symmetry of the resting pair. This is an initial, large-separation interaction calculation, not a claim that the superposed profile is an exact static two-soliton solution. Without the initial-velocity condition, the additional momentum flux above prevents a unique force from being inferred from the printed profile alone.

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