Noether current ($j^{\mu }$)

A continuous variational symmetry gives a current satisfying ${\partial }_{\mu }{j}^{\mu }=0$ on shell. Integrating $j^0$ over space gives a conserved charge when boundary flux vanishes.

Rotating the two transverse displacement components around the string axis gives density ${\ell }_x=\rho ({\psi }_y{\dot{\psi }}_z-{\psi }_z{\dot{\psi }}_y)$ and flux ${\mathcal{J}}_x=-F({\psi }_y{\psi }_z^{\prime }-{\psi }_z{\psi }_y^{\prime })$. The pair of wave equations yields ${\partial }_t{\ell }_x+{\partial }_x{\mathcal{J}}_x=0$. The total charge is physical angular momentum about the axis; it is conserved only when the net endpoint torque flux vanishes.

For $\mathcal{L}=\rho ({\dot{\psi }}_y^2+{\dot{\psi }}_z^2)/2-F(({\psi }_y^{\prime }{)}^2+({\psi }_z^{\prime }{)}^2)/2$, constant displacement symmetry in $y$ gives $j^0=\rho {\dot{\psi }}_y$ and $j^x=-F{\psi }_y^{\prime }$. The wave equation implies ${\partial }_t{j}^0+{\partial }_x{j}^x=0$. Its integrated charge is physical transverse momentum; endpoint forces can change that charge on a finite segment.

For a complex scalar field multiplet with canonical kinetic term and scalar potential invariant under $\Phi \mapsto e^{i\epsilon T}\Phi$, a constant Hermitian matrix $T$ gives the displayed Noether current. Vary the field and its conjugate independently in the Noether current for a first-derivative scalar field. A common alternative convention reverses both the parameter and current sign. Its divergence vanishes on shell.

For a longitudinal displacement field $\psi (z,t)$ and Lagrangian density $\mathcal{L}=A[\rho {\dot{\psi }}^2-K{\psi }^{\prime 2}]/2$, uniform field shifts give Noether current $j^0=A\rho \dot{\psi }$, $j^z=-AK{\psi }^{\prime }$. Its spatial integral is the rod's longitudinal mechanical momentum. A prescribed position-dependent mass density changes $j^0$ to $A\rho (z)\dot{\psi }$ but does not break this internal shift symmetry. Boundary traction can carry momentum into or out of a finite rod, so action symmetry alone does not guarantee conservation of the integrated charge.

If a Lagrangian density depending on first and second derivatives has no undifferentiated field, the scalar-field shift symmetry $\delta \varphi =ϵ$ gives the displayed Noether current. Its divergence is minus the Euler-Lagrange field equation with second derivatives, so it is conserved on shell. The spatial integral of $j^0$ is constant only if the boundary current vanishes. For $\mathcal{L}=\rho {\varphi }_t^2/2+\alpha {\varphi }_x^2/2-\beta {\varphi }_{xx}^2/2$, it gives $j^0=\rho {\varphi }_t$, $j^x=\alpha {\varphi }_x+\beta {\varphi }_{xxx}$. Fixed transverse endpoint positions break the global shift symmetry of the boundary problem; support forces can then change the total transverse momentum even though the bulk continuity equation still holds.

For a symmetric conserved stress-energy tensor $\Theta$ and a Conformal Killing vector field $\omega$ in dimension $d$, the displayed conformal current has divergence $(\partial \cdot \omega ){\Theta }^{\mu }{}_{\mu }/d$. This follows by contracting the Conformal Killing equation with $\Theta$. A traceless tensor therefore gives a conserved current for every conformal generator.

A virial current $V^{\mu }$ represents the trace of a conserved stress-energy tensor as a divergence. Then the dilation Noether current $D^{\mu }=x_{\nu }{T}^{\mu \nu }-V^{\mu }$ is conserved, since its divergence is $T^{\mu }{}_{\mu }-{\partial }_{\mu }{V}^{\mu }$. This establishes scale invariance; a traceless stress-energy tensor improvement is the extra condition needed for conformal currents.

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.

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.

A conserved current is a vector field $j^{\mu }$ satisfying the continuity equation ${\partial }_{\mu }{j}^{\mu }=0$. Its spatial integral is constant in time when no current crosses the boundary.

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.

In a massless four-dimensional theory, current conservation gives a transverse tensor structure, $\int d^{4}x{e}^{-ikx}⟨J_{\mu }(x)J_{\nu }(0)⟩=(k_{\mu }{k}_{\nu }-{\delta }_{\mu \nu }{k}^2)G(k^2)$. With current mass dimension three, the scalar coefficient has dimension zero.

A Noether charge is the spatial integral $Q=\int j^0{d}^{3}x$ of a Noether current. It is time-independent when the current has vanishing flux through the spatial boundary, and in a Hamiltonian formulation it generates the associated continuous symmetry.

The dilatation current associated with scale transformations is $D^{\mu }=x_{\nu }{T}^{\mu \nu }$ up to improvement terms. Its divergence is the trace of the stress-energy tensor, so classical scale invariance requires a traceless improved tensor on shell.

For $\mathcal{L}=(\partial \varphi {)}^2/2-\lambda {\varphi }^4$ and the displayed-weight dilation variation $\delta \varphi =\varphi +x\cdot \partial \varphi$, the full divergence identity is $(4-n)\mathcal{L}+\delta \varphi (\square \varphi +4\lambda {\varphi }^3)$. The field equation removes the second term. In four spacetime dimensions the density changes by a total divergence, so the Noether current includes the subtraction $-x^{\mu }\mathcal{L}$. This is a classical statement about generic solutions and this particular current.

For a first-derivative real scalar field in one space and one time dimension, put $j_D^{\mu }={\mathcal{L}}_{{\partial }_{\mu }\psi }(z{\psi }^{\prime }+t\dot{\psi })-x^{\mu }\mathcal{L}$. Its on-shell divergence isA quadratic derivative density with arbitrary field-dependent coefficients has zero derivative-homogeneity term. An extra term $-\alpha {\psi }^{\prime 4}$ instead gives $-2\alpha {\psi }^{\prime 4}$. A position-dependent kinetic coefficient $A\rho (z)/2$ gives $-Az{\rho }^{\prime }(z){\dot{\psi }}^2/2$. These source terms distinguish scale invariance from internal field-shift symmetry.

For constant $A,\rho ,K$ on all space and time, the active waveform transformation ${\psi }_b(z,t)=\psi (bz,bt)$ leaves the action of $\mathcal{L}=A(\rho {\dot{\psi }}^2-K{\psi }^{\prime 2})/2$ invariant: the density gains $b^2$ and the integration measure gains $b^{-2}$. With canonical momentum $\Pi =A\rho \dot{\psi }$, its Noether current can be chosen asThe integrated charge $D=\int j_D^{0}dz$ is conserved if its boundary flux vanishes and the weighted integral exists. For a rightgoing wave packet $\psi =f(z-ct)$, $c=\sqrt{K/\rho }$, it becomes $D=-A\rho c\int \xi [f^{\prime }(\xi ){]}^{2}d\xi$. A fixed finite interval generally breaks this scale symmetry unless its boundaries and boundary data are transformed too.

Spacetime translations give the canonical stress-energy tensor

For a real scalar field in one spatial dimension with metric signature $(+,-)$ and potential $V$, the stress-energy conservation law is ${\partial }_t{T}^{01}+{\partial }_x{T}^{11}=0$. The force on the field to the left of a fixed cut is $-T^{11}$ at the cut when the far-left boundary flux vanishes. Finite-energy field configurations also permit this conclusion by spatial cutoffs, without assuming pointwise decay of all derivatives.

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 $\mathbf{k}$.

For a canonically quantized real scalar field, the quadratic Hamiltonian contains $\frac{1}{2}{\sum }_{\mathbf{p}}{E}_{\mathbf{p}}$ in a regulated box. Normal ordering subtracts this zero-point energy, leavingThe vacuum momentum is zero with an inversion-symmetric regulator. The canonical commutation relation then gives $[P^{\mu },a_{\mathbf{p}}^{†}]=p^{\mu }{a}_{\mathbf{p}}^{†}$, identifying the momentum and energy added by one particle. The un-subtracted energy differs by a constant and is not literally the same operator.

The canonical commutation relation gives $[P^{\mu },a_p^{†}]=p^{\mu }{a}_p^{†}$ for a free real scalar field. Differentiate the exponential conjugation: its derivative is a scalar multiple of the same operator, yielding the displayed phase. The annihilation operator receives the opposite phase, so at equal time $e^{-i\mathbf{P}\cdot \mathbf{y}}\varphi (\mathbf{x})e^{i\mathbf{P}\cdot \mathbf{y}}=\varphi (\mathbf{x}+\mathbf{y})$. The spatial translation operator translates a state's position wavefunction in the opposite argument direction.

For signature $(+,-,-,-)$, translation invariance of the Maxwell Lagrangian gives this tensor. It is generally neither symmetric nor gauge invariant. The antisymmetric superpotential improvement $-{\partial }_{\rho }(F^{\rho \mu }{A}^{\nu })$ produces, on the source-free equations of motion, the symmetric electromagnetic stress-energy tensor ${\Theta }^{\mu \nu }=-F^{\mu \rho }{F}^{\nu }{}_{\rho }+{\eta }^{\mu \nu }{F}^2/4$, which is traceless in four spacetime dimensions.

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.

If $B^{\rho \mu \nu }=-B^{\mu \rho \nu }$, commuting partial derivatives gives ${\partial }_{\mu }{\partial }_{\rho }{B}^{\rho \mu \nu }=0$. Such an improvement preserves conservation of a stress-energy tensor and changes integrated charges only by a spatial boundary term. The Belinfante-Rosenfeld stress-energy tensor is obtained through this type of improvement.

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 a symmetric conserved stress-energy tensor, the Lorentz current is $J^{\mu \rho \sigma }=x^{\rho }{T}^{\mu \sigma }-x^{\sigma }{T}^{\mu \rho }$. Its spatial integrals generate rotations and boosts.

For a free massless real scalar field with wave speed set to one and metric $(+,-)$, define $\mathcal{E}=({\varphi }_t^2+{\varphi }_x^2)/2$ and $p=-{\varphi }_t{\varphi }_x$. Its symmetric stress-energy tensor is $T^{\mu \nu }=\left(\begin{array}{cc}\mathcal{E}& p\\ p& \mathcal{E}\end{array}\right)$. The Lorentz current has density $tp-x\mathcal{E}$ and flux $t\mathcal{E}-xp$. Its conserved charge is the displayed expression when boundary flux vanishes. In one spatial dimension it generates boosts, not spatial rotations; there is no independent purely spatial angular-momentum component. Its conservation says that the center of energy moves uniformly. This is a symmetry of the wave-field model and need not survive boundaries or additional material effects.

The global phase symmetry of a Dirac field gives the conserved Dirac current $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$.

For a commuting spinor in mostly-plus signature, orient its Dirac current to the future and write $V^i={ϵ}^{†}{\alpha }^iϵ$, with Hermitian ${\alpha }^i=-{\gamma }^0{\gamma }^i$. For every spatial unit vector $u$, $(u\cdot \alpha {)}^2=1$ implies $|u\cdot V|\le {ϵ}^{†}ϵ=V^0$. Choosing $u$ along the spatial current proves $V^2\le 0$. Thus spinor bilinears supply causal vectors suitable for energy-condition estimates. A Grassmann-valued spinor does not have this ordinary positivity interpretation.

In four dimensions a single commuting Majorana spinor is specified by a two-component complex spinor $\xi$. Its future Dirac current is proportional to $({\xi }^{†}\xi ,{\xi }^{†}\mathbf{\sigma }\xi )$, and the Pauli matrix multiplication law implies $|{\xi }^{†}\mathbf{\sigma }\xi {|}^2=({\xi }^{†}\xi {)}^2$. Thus it is null. A Dirac spinor or a sum of independent Majorana currents can be timelike. This distinction matters when applying a timelike-current classification to minimal real supersymmetry.

The Gordon identity rewrites a vector-current matrix element between equal-mass on-shell Dirac spinors using a momentum term and an antisymmetric spin term. In the mostly-minus convention $\{{\Gamma }^a,{\Gamma }^b\}=2{\eta }^{ab}$, $(/p-m)u=0$, and ${\Gamma }^{ab}=[{\Gamma }^a,{\Gamma }^b]/2$, it reads $2m{\bar{u}}^{\prime }{\Gamma }^{a}u={\bar{u}}^{\prime }[(p^{\prime }+p{)}^a-{\Gamma }^{ab}(p_b^{\prime }-p_b)]u$. Derive it by adding the two external Dirac equation identities. The equivalent formula in the mostly-plus Dirac convention is ${\bar{u}}^{\prime }[{\gamma }^{ab}(p_b^{\prime }-p_b)+2m{\gamma }^a+p^a+p^{\prime a}]u=0$. The identity before division by $m$ remains valid for massless equal-mass states.

The Dirac electromagnetic current is the electric charge times the Dirac current. Its coupling $-e{j}^{\mu }{A}_{\mu }$ to the electromagnetic four-potential is the matter interaction of quantum electrodynamics.

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.