Noether's Theorem for Conserved Currents

Derivation

Noether's Theorem for Conserved Currents

D-279 Home PU-304 Threads symmetry · energy · fields Depends on Euler-Lagrange Equations for Fields

Statement

For a classical field theory with action \( S=\int_\Omega \mathcal{L}(\phi_a,\partial_\mu\phi_a,x^\mu)\,d^4x \) whose fields obey the Euler–Lagrange equations, every continuous transformation \( x^\mu\to x^\mu+\epsilon^r X_r^{\mu} \), \( \phi_a\to\phi_a+\epsilon^r\Psi_{r,a} \) that leaves the action invariant produces one locally conserved current per independent parameter, \( \partial_\mu J^{\mu}_r=0 \), with \( J^{\mu}_r=\dfrac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\Psi_{r,a}-T^{\mu}{}_{\nu}\,X_r^{\nu} \), and a corresponding conserved charge \( Q_r=\int J^0_r\,d^3x \) with \( dQ_r/dt=0 \).

Why it matters

Noether's theorem is the exact statement of "symmetry implies conservation law." It turns a geometric or internal invariance of the action — something one can often spot by inspection — into an explicit, computable current whose divergence vanishes on shell. Energy–momentum, angular momentum, and electric charge are not independent postulates: each is the Noether charge of a specific symmetry (time translation, rotation/boost, phase rotation).

The construction is constructive, not merely existential: it hands you the current, the charge, and the density you must integrate. That makes it the engine behind the stress–energy tensor in general relativity, the current operators of quantum field theory, and the Ward identities that survive quantisation.

Assumptions

The action is invariant (or invariant up to a boundary term) under the transformation.If \( \delta S\neq 0 \) even up to \( \partial_\mu K^\mu \), no conserved current follows; the "symmetry" is at most approximate and the charge leaks.

The fields satisfy the Euler–Lagrange equations (the current is conserved only on shell).Off shell \( \partial_\mu J^\mu=\big(\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}-\frac{\partial\mathcal{L}}{\partial\phi_a}\big)\Psi_a\neq0 \); the divergence measures the equation-of-motion residue.

The transformation is continuous, connected to the identity, and depends on constant (global) parameters \( \epsilon^r \).Discrete symmetries (parity, time reversal) give no current; if \( \epsilon^r\to\epsilon^r(x) \) is local you are in gauge territory and the "current" becomes an off-shell identity plus field equations, not a Noether charge.

The Lagrangian is first order in derivatives and the fields plus their variations fall off fast enough at spatial infinity.Higher-derivative theories need extra momentum terms in \( J^\mu \); without the fall-off the surface integral \( \oint J^i\,dS_i \) does not vanish and \( Q \) is not conserved.

Derivation

Work in \( d=4 \) with metric signature \( (+,-,-,-) \). Introduce the total variation \( \delta\phi_a=\phi'_a(x')-\phi_a(x) \) and the form (functional) variation at fixed point \( \bar\delta\phi_a=\phi'_a(x)-\phi_a(x) \); they are related by \( \delta\phi_a=\bar\delta\phi_a+\delta x^\mu\,\partial_\mu\phi_a \). Only \( \bar\delta \) commutes with \( \partial_\mu \).

\[ x'^{\mu}=x^{\mu}+\delta x^{\mu},\qquad \phi'_a(x')=\phi_a(x)+\delta\phi_a,\qquad S'=\int_{\Omega'}\mathcal{L}\big(\phi'_a,\partial'_\mu\phi'_a\big)\,d^4x' \]

Definition of the transformed action; the domain \( \Omega \) is dragged to \( \Omega' \) with the coordinates. A

\[ d^4x'=\left|\det\frac{\partial x'^{\mu}}{\partial x^{\nu}}\right|d^4x=\big(1+\partial_\mu\,\delta x^{\mu}\big)\,d^4x+O(\epsilon^2) \]

Jacobian of the coordinate change to first order; \( \det(\mathbf 1+\epsilon M)=1+\epsilon\,\mathrm{tr}\,M \). B

\[ \delta S=\int_\Omega\!\big[\,\delta\mathcal{L}+\mathcal{L}\,\partial_\mu\delta x^{\mu}\big]d^4x =\int_\Omega\!\big[\,\bar\delta\mathcal{L}+\partial_\mu(\mathcal{L}\,\delta x^{\mu})\big]d^4x \]

Substitute the Jacobian and use \( \delta\mathcal{L}=\bar\delta\mathcal{L}+\delta x^\mu\partial_\mu\mathcal{L} \); the transport term combines with \( \mathcal{L}\,\partial_\mu\delta x^\mu \) into a total divergence. C

\[ \bar\delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi_a}\,\bar\delta\phi_a+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\partial_\mu\bar\delta\phi_a \]

Chain rule for the form variation; because \( \bar\delta \) commutes with \( \partial_\mu \), we may write \( \bar\delta(\partial_\mu\phi_a)=\partial_\mu\bar\delta\phi_a \). A

\[ \frac{\partial\mathcal{L}}{\partial\phi_a}=\partial_\mu\!\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\right)\ \Longrightarrow\ \bar\delta\mathcal{L}=\partial_\mu\!\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\bar\delta\phi_a\right) \]

Insert the Euler–Lagrange field equations (the assumed prior result) and recognise the reverse product rule. B

\[ \delta S=\int_\Omega\partial_\mu\!\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\bar\delta\phi_a+\mathcal{L}\,\delta x^{\mu}\right]d^4x \]

Combine Steps 3 and 5: on shell the whole variation of the action is the integral of a divergence. C

\[ \delta S=\int_\Omega\partial_\mu\!\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\delta\phi_a-\Big(\underbrace{\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\partial_\nu\phi_a-\delta^{\mu}_{\nu}\,\mathcal{L}}_{\displaystyle T^{\mu}{}_{\nu}}\Big)\delta x^{\nu}\right]d^4x \]

Eliminate the form variation with \( \bar\delta\phi_a=\delta\phi_a-\delta x^\nu\partial_\nu\phi_a \); the transport piece assembles into the canonical stress–energy tensor \( T^\mu{}_\nu \). B

\[ \delta S=0\ \text{for every subregion }\Omega\ \Longrightarrow\ \partial_\mu J^{\mu}=0,\qquad J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\delta\phi_a-T^{\mu}{}_{\nu}\,\delta x^{\nu} \]

Invariance forces the integral of the divergence to vanish for arbitrary \( \Omega \); by the fundamental lemma the integrand itself vanishes pointwise. C

\[ \delta x^{\nu}=\epsilon^{r}X_r^{\nu},\quad \delta\phi_a=\epsilon^{r}\Psi_{r,a}\ \Longrightarrow\ J^{\mu}_r=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\Psi_{r,a}-T^{\mu}{}_{\nu}\,X_r^{\nu},\quad \partial_\mu J^{\mu}_r=0 \]

The parameters \( \epsilon^r \) are independent constants, so each coefficient gives a separately conserved current. A

\[ Q_r=\int_{\mathbb{R}^3}J^{0}_r\,d^3x,\qquad \frac{dQ_r}{dt}=\int\partial_0 J^0_r\,d^3x=-\int\partial_i J^i_r\,d^3x=-\oint_{S_\infty}J^i_r\,dS_i=0 \]

Integrate the conservation law over a spatial slice and apply the divergence theorem; the surface term vanishes by the fall-off assumption. B

Result

\[ \boxed{\ \partial_\mu J^{\mu}_r=0,\qquad J^{\mu}_r=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\Psi_{r,a}-T^{\mu}{}_{\nu}\,X_r^{\nu},\qquad Q_r=\int J^0_r\,d^3x=\text{const}\ } \]

Reading. Each continuous symmetry contributes two terms to its current: an internal piece \( \pi^\mu_a\Psi_{r,a} \) built from the canonical momentum \( \pi^\mu_a=\partial\mathcal{L}/\partial(\partial_\mu\phi_a) \) times how the fields change, and a spacetime piece \( -T^\mu{}_\nu X_r^\nu \) built from the stress tensor times how the coordinates move. Pure internal symmetries (\( X=0 \)) give charge/flavour currents; pure translations (\( \Psi=0,\ X_r^\nu=\delta_r^\nu \)) give \( J^\mu{}_\nu=-T^\mu{}_\nu \), i.e. energy–momentum conservation \( \partial_\mu T^\mu{}_\nu=0 \).

Units check. In natural units (\( \hbar=c=1 \), 4D) a scalar has mass dimension \( [\phi]=1 \), \( [\partial_\mu]=1 \), \( [\mathcal{L}]=4 \), so \( [\pi^\mu_a]=3 \). For an internal symmetry \( \Psi \) is dimensionless, giving \( [J^\mu]=3 \) (a number density) and \( [Q]=[J^0]+[d^3x]=3-3=0 \), a pure number as a particle number must be. For translations \( T^\mu{}_\nu \) has dimension 4 (energy density) and \( X^\nu \) dimension \( -1 \), so again \( [J^\mu]=3 \) and the charge \( P_\nu=\int T^0{}_\nu\,d^3x \) has dimension 1 = energy/momentum. Dimensions are consistent.

Limiting cases

Internal symmetry only (\( \delta x^\mu=0 \)): \( J^\mu=\pi^\mu_a\,\Psi_a \). A global U(1) phase gives the familiar \( j^\mu=i(\phi^*\partial^\mu\phi-\phi\,\partial^\mu\phi^*) \).
Spacetime translation \( X_r^\nu=\delta_r^\nu,\ \Psi=0 \): \( J^\mu{}_\nu=-T^\mu{}_\nu \), so \( \partial_\mu T^\mu{}_\nu=0 \) and \( P_\nu=\int T^0{}_\nu d^3x \) (energy for \( \nu=0 \), momentum for \( \nu=i \)).
Lorentz transformation \( \delta x^\mu=\omega^\mu{}_\nu x^\nu \): the angular-momentum current \( M^{\mu\nu\rho}=x^\nu T^{\mu\rho}-x^\rho T^{\mu\nu}\ (+\text{spin}) \) is conserved, \( \partial_\mu M^{\mu\nu\rho}=0 \).
Point mechanics (0+1 dimensions, \( \mathcal{L}(q,\dot q,t) \)): the current collapses to a single number, recovering elementary Noether — time translation \( \to \) energy \( E=p\dot q-L \), spatial translation \( \to \) momentum.

Breaks when

Explicit symmetry breaking. If \( \mathcal{L} \) depends on \( x^\mu \) explicitly (external potential, time-dependent coupling), the corresponding \( X^\mu \) is not a symmetry: \( \partial_\mu T^\mu{}_\nu=-\partial\mathcal{L}/\partial x^\nu\neq0 \) and energy/momentum is not conserved.
Anomalies. A symmetry of the classical action can fail to survive quantisation (e.g. the axial \( U(1)_A \) current): \( \partial_\mu J^\mu_5\propto F\tilde F\neq0 \). Noether's classical theorem still holds for the action; the measure of the path integral breaks it.
Gauge (local) "symmetries." When \( \epsilon^r=\epsilon^r(x) \), Noether's second theorem applies instead: the current is a trivial on-shell identity (a superpotential \( J^\mu=\partial_\nu U^{\mu\nu} \)) and the naive charge is a boundary term, not a bulk conserved quantity.
Non-normalisable configurations. If fields (or their variations) do not decay at spatial infinity — solitons with long-range tails, charged states in unscreened Coulomb fields — the surface term in Step 10 survives and \( dQ/dt\neq0 \).

Failure modes

Confusing form and total variation. Applying Euler–Lagrange to \( \delta\phi \) instead of \( \bar\delta\phi \); only \( \bar\delta \) commutes with \( \partial_\mu \), so the E–L collapse in Step 5 is illegal for the total variation.
Dropping the stress-tensor term for spacetime symmetries. Writing \( J^\mu=\pi^\mu_a\delta\phi_a \) and forgetting \( -T^\mu{}_\nu\delta x^\nu \) — this gives the wrong (and non-conserved) energy current.
Sign/index slip in \( T^\mu{}_\nu \). Using \( \delta^\mu_\nu\mathcal{L} \) with the wrong sign, or contracting the derivative index of \( \pi^\mu_a \) with \( \partial^\mu \) instead of \( \partial_\nu \).
Treating a discrete symmetry as continuous. Trying to build a current for parity or \( \phi\to-\phi \); there is no infinitesimal generator, so no \( J^\mu \).
Assuming conservation off shell. Reporting \( \partial_\mu J^\mu=0 \) for an arbitrary field configuration; it holds only for solutions of the field equations.
Forgetting quasi-symmetry boundary terms. When \( \bar\delta\mathcal{L}=\partial_\mu K^\mu\neq0 \), omitting the \( -K^\mu \) correction gives a current that is not actually conserved.

Discussion

The deepest content of Noether's theorem is that the same object plays two roles. The canonical momentum \( \pi^\mu_a \) that appears in the field equations reappears as the carrier of the conserved current; the stress tensor \( T^\mu{}_\nu \) that sources gravity is exactly the Noether current of translations. Symmetry does not merely permit conservation laws — it identifies which combination of the dynamical variables is conserved and hands you its density.

The relation between the charge and the symmetry becomes an identity at the quantum level: \( Q_r \) is the generator of the transformation. Promoting fields to operators, \( [Q_r,\phi_a]=-i\,\Psi_{r,a} \), so the conserved charge literally implements the symmetry it came from. This is the classical seed of the Ward–Takahashi identities that constrain correlation functions and renormalisation.

Noether's construction also exposes what is ambiguous. The canonical \( T^\mu{}_\nu \) is generally neither symmetric nor gauge-invariant; one is free to add an "improvement" term \( \partial_\rho B^{\rho\mu}{}_\nu \) with \( B^{\rho\mu}=-B^{\mu\rho} \) that changes neither the conservation law nor the charge. The Belinfante–Rosenfeld procedure uses exactly this freedom to build the symmetric, gauge-invariant tensor that couples to the metric — the bridge to general relativity, where \( T^{\mu\nu}=\tfrac{2}{\sqrt{-g}}\,\delta S/\delta g_{\mu\nu} \).

Noether's second theorem, for local (gauge) symmetries, tells a subtler story: infinite-dimensional symmetry groups do not yield independent conserved charges but off-shell identities among the field equations (Bianchi-type constraints), and the associated Noether current is a total derivative \( J^\mu=\partial_\nu U^{\mu\nu} \). Physical charge in gauge theories then lives entirely on the boundary at infinity — the origin of Gauss's law as a surface integral and of asymptotic (BMS) charges in gravity. This is why "conservation of colour charge" is a statement about flux through a surface, not a bulk integral of a gauge-covariant density.

Common misconceptions. (i) "Every conservation law comes from a symmetry of the Lagrangian" — false in general; some conserved quantities (Runge–Lenz, integrability tower) arise from hidden or dynamical symmetries not manifest in \( \mathcal{L} \). (ii) "The Noether current is unique" — it is defined only up to an identically-conserved improvement term. (iii) "Conservation holds always" — it holds on shell; off shell the divergence measures the equation-of-motion residue.

Worked examples

Example 1 — Global U(1) phase symmetry of a complex scalar (internal symmetry, charge current).

\[ \mathcal{L}=\partial_\mu\phi^{*}\partial^{\mu}\phi-m^{2}\phi^{*}\phi,\qquad \phi\to e^{i\alpha}\phi\ \Rightarrow\ \delta\phi=i\alpha\phi,\ \ \delta\phi^{*}=-i\alpha\phi^{*},\ \ \delta x^\mu=0 \]

Identify the transformation: a rigid phase rotation, an internal symmetry so \( X^\mu=0 \). A

\[ \pi^\mu=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}=\partial^{\mu}\phi^{*},\qquad \pi^{*\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^{*})}=\partial^{\mu}\phi \]

Compute the canonical momenta for both independent fields \( \phi,\phi^* \). A

\[ J^{\mu}=\pi^\mu\,\frac{\delta\phi}{\alpha}+\pi^{*\mu}\,\frac{\delta\phi^*}{\alpha}=\partial^{\mu}\phi^{*}(i\phi)+\partial^{\mu}\phi(-i\phi^{*})=i\big(\phi\,\partial^{\mu}\phi^{*}-\phi^{*}\partial^{\mu}\phi\big) \]

Assemble the internal-symmetry current, stripping the constant \( \alpha \). Conventionally one flips the overall sign to define \( j^\mu=i(\phi^*\partial^\mu\phi-\phi\,\partial^\mu\phi^*) \). B

\[ \phi=A\,e^{-ik\cdot x},\ \ k\cdot x=\omega t-\mathbf k\cdot\mathbf x\ \Rightarrow\ \partial^{0}\phi=-i\omega\phi\ \Rightarrow\ j^{0}=i\big(\phi^{*}(-i\omega\phi)-\phi(i\omega\phi^{*})\big)=2\omega|A|^{2} \]

Evaluate the charge density on a plane-wave solution. B

\[ A=0.50\ \text{GeV},\ \ \omega=3.0\ \text{GeV}\ \Rightarrow\ j^{0}=2(3.0)(0.50)^{2}=1.5\ \text{GeV}^{3} \]

Insert numbers; \( j^0 \) is a number density (dimension 3 in natural units). A

\[ \boxed{\,j^{\mu}=i(\phi^{*}\partial^{\mu}\phi-\phi\,\partial^{\mu}\phi^{*}),\qquad j^{0}=2\omega|A|^{2}=1.5\ \text{GeV}^{3}\,} \]

Reading. The U(1) phase symmetry gives a conserved number/charge current; on a plane wave the density is \( 2\omega|A|^2 \), positive and proportional to the mode occupation. A charge \( Q=\int j^0\,d^3x \) is time independent because \( \partial_\mu j^\mu=0 \) on shell.

Example 2 — Time-translation symmetry of a real scalar (energy density from \( T^{00} \)).

\[ \mathcal{L}=\tfrac12\,\partial_\mu\phi\,\partial^{\mu}\phi-\tfrac12 m^{2}\phi^{2},\qquad x^0\to x^0+\epsilon:\ \ X^{0}=1,\ \Psi=0 \]

Time translation is a spacetime symmetry with zero form-change of the field. A

\[ T^{\mu}{}_{\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\nu\phi-\delta^{\mu}_{\nu}\mathcal{L}=\partial^{\mu}\phi\,\partial_\nu\phi-\delta^{\mu}_{\nu}\mathcal{L}\ \Rightarrow\ T^{00}=\dot\phi^{2}-\mathcal{L}=\tfrac12\dot\phi^{2}+\tfrac12(\nabla\phi)^{2}+\tfrac12 m^{2}\phi^{2} \]

Build the canonical stress tensor; the \( 00 \) component is the Hamiltonian (energy) density. B

\[ \phi=\phi_0\cos(\omega t-kx)\ \Rightarrow\ \langle T^{00}\rangle=\tfrac14\phi_0^{2}\big(\omega^{2}+k^{2}+m^{2}\big),\quad \text{with }\omega^{2}=k^{2}+m^{2}\ \Rightarrow\ \langle T^{00}\rangle=\tfrac12\phi_0^{2}\omega^{2} \]

Time-average over a period using \( \langle\sin^2\rangle=\langle\cos^2\rangle=\tfrac12 \) and the on-shell dispersion relation. C

\[ \phi_0=0.10\ \text{GeV},\ m=1.0\ \text{GeV},\ k=2.0\ \text{GeV}\ \Rightarrow\ \omega=\sqrt{5}\approx2.236\ \text{GeV},\ \ \langle T^{00}\rangle=\tfrac12(0.10)^{2}(5.0)=0.025\ \text{GeV}^{4} \]

Substitute numbers; energy density has dimension 4 in natural units. A

\[ \boxed{\,\langle T^{00}\rangle=\tfrac12\phi_0^{2}\omega^{2}=0.025\ \text{GeV}^{4},\qquad E=\int T^{00}\,d^3x=\text{const}\,} \]

Reading. The Noether current of time translation is (minus) the first row of the stress tensor; its density is the energy density, and integrating it gives the conserved total energy. The time-averaged energy density of the mode scales as amplitude-squared times frequency-squared, exactly as for a collection of oscillators.

Problems

(Level A) Derive the Noether current for the global U(1) symmetry of \( \mathcal{L}=\partial_\mu\phi^*\partial^\mu\phi-V(|\phi|^2) \) and verify explicitly that \( \partial_\mu j^\mu=0 \) using the field equations.
Solution
Under \( \phi\to e^{i\alpha}\phi \): \( \delta\phi=i\alpha\phi,\ \delta\phi^*=-i\alpha\phi^* \). With \( \pi^\mu=\partial^\mu\phi^*,\ \pi^{*\mu}=\partial^\mu\phi \), the current is \( j^\mu=i(\phi^*\partial^\mu\phi-\phi\,\partial^\mu\phi^*) \). Then \( \partial_\mu j^\mu=i(\partial_\mu\phi^*\partial^\mu\phi+\phi^*\Box\phi-\partial_\mu\phi\,\partial^\mu\phi^*-\phi\,\Box\phi^*)=i(\phi^*\Box\phi-\phi\,\Box\phi^*) \). The Euler–Lagrange equations give \( \Box\phi=-V'\phi \) and \( \Box\phi^*=-V'\phi^* \) (with \( V'=\partial V/\partial|\phi|^2 \)). Substituting, \( \partial_\mu j^\mu=i(-V'\phi^*\phi+V'\phi\phi^*)=0 \). Conserved on shell.
(Level A/B) For the real scalar \( \mathcal{L}=\tfrac12\partial_\mu\phi\,\partial^\mu\phi-\tfrac12 m^2\phi^2 \), compute the full canonical stress tensor \( T^\mu{}_\nu \) and show that the conserved momentum \( P^i=\int T^{0i}\,d^3x \) for the plane wave \( \phi=\phi_0\cos(\omega t-kx) \) has time-averaged density \( \tfrac12\phi_0^2\omega k \).
Solution
\( T^\mu{}_\nu=\partial^\mu\phi\,\partial_\nu\phi-\delta^\mu_\nu\mathcal{L} \). The momentum density is \( T^{0i}=\partial^0\phi\,\partial^i\phi=\dot\phi\,\partial^i\phi \) (the \( \delta^{0i} \) term vanishes). For \( \phi=\phi_0\cos(\omega t-kx) \): \( \dot\phi=-\phi_0\omega\sin(\cdot),\ \partial_x\phi=\phi_0 k\sin(\cdot) \), so \( \partial^x\phi=-\partial_x\phi=-\phi_0 k\sin(\cdot) \). Then \( T^{0x}=\dot\phi\,\partial^x\phi=(-\phi_0\omega\sin)(-\phi_0 k\sin)=\phi_0^2\omega k\sin^2(\cdot) \). Time-averaging \( \langle\sin^2\rangle=\tfrac12 \) gives \( \langle T^{0x}\rangle=\tfrac12\phi_0^2\omega k \). The ratio to the energy density \( \tfrac12\phi_0^2\omega^2 \) is \( k/\omega \), the phase-space group velocity, as expected for a relativistic wave.
(Level B) A theory of \( N \) real scalars \( \mathcal{L}=\tfrac12\partial_\mu\phi_a\partial^\mu\phi_a-\tfrac12 m^2\phi_a\phi_a \) is invariant under \( \delta\phi_a=\epsilon\,(T)_{ab}\phi_b \) with \( T \) real antisymmetric (an SO(N) rotation). Find the conserved current and its charge, and explain why \( T \) must be antisymmetric.
Solution
Antisymmetry is required for invariance: \( \bar\delta\mathcal{L}=\partial_\mu\phi_a\,\partial^\mu(\epsilon T_{ab}\phi_b)-m^2\phi_a\epsilon T_{ab}\phi_b=\epsilon T_{ab}(\partial_\mu\phi_a\partial^\mu\phi_b-m^2\phi_a\phi_b) \). The bracket is symmetric in \( a\leftrightarrow b \), so contracting with \( T_{ab} \) vanishes iff \( T_{ab}=-T_{ba} \). The current: \( \pi^\mu_a=\partial^\mu\phi_a \), so \( J^\mu=\pi^\mu_a T_{ab}\phi_b=\partial^\mu\phi_a\,T_{ab}\phi_b \). For each generator \( (T^{(ij)})_{ab}=\delta_{ia}\delta_{jb}-\delta_{ja}\delta_{ib} \) this is \( J^{\mu}_{(ij)}=\phi_i\partial^\mu\phi_j-\phi_j\partial^\mu\phi_i \), with \( Q_{(ij)}=\int(\phi_i\dot\phi_j-\phi_j\dot\phi_i)\,d^3x \) — the \( \tfrac12 N(N-1) \) conserved "isospin" charges of SO(N).
(Level B/C) Shift symmetry. The massless free scalar \( \mathcal{L}=\tfrac12\partial_\mu\phi\,\partial^\mu\phi \) is invariant under \( \phi\to\phi+\epsilon \) (constant \( \epsilon \)). Find the Noether current, its charge, and explain what conservation law it encodes. Then consider \( \mathcal{L}=\tfrac12\partial_\mu\phi\,\partial^\mu\phi+c\,\phi \) and show the symmetry becomes a quasi-symmetry, computing the corrected current.
Solution
For the massless case: \( \delta\phi=\epsilon,\ \delta x=0 \), so \( J^\mu=\pi^\mu\,\delta\phi/\epsilon=\partial^\mu\phi \). Conservation \( \partial_\mu J^\mu=\Box\phi=0 \) is exactly the equation of motion, so the "charge" \( Q=\int\dot\phi\,d^3x \) is the total canonical momentum conjugate to the zero mode of \( \phi \). Now add \( c\phi \). Under the shift the Lagrangian changes by a constant, \( \bar\delta\mathcal{L}=c\,\epsilon\neq0 \), so this is a quasi-symmetry: a constant is a total divergence, \( c\,\epsilon=\partial_\mu K^\mu \) with \( K^\mu=\tfrac{c\epsilon}{4}\,x^\mu \) (since \( \partial_\mu x^\mu=4 \) in four dimensions). The Noether recipe for a quasi-symmetry subtracts \( K^\mu \) from the naive current: \( \tilde J^\mu=\partial^\mu\phi-\tfrac{c}{4}x^\mu \). Check with the field equation \( \Box\phi=c \): \( \partial_\mu\tilde J^\mu=\Box\phi-\tfrac{c}{4}\partial_\mu x^\mu=c-\tfrac{c}{4}(4)=0 \). The uncorrected \( \partial^\mu\phi \) alone is not conserved (\( \partial_\mu\partial^\mu\phi=c\neq0 \)); invariance only up to a surface term still yields a conserved current once \( K^\mu \) is removed.
(Level C) Derive the conserved angular-momentum current for a real scalar under an infinitesimal Lorentz transformation \( \delta x^\mu=\omega^\mu{}_\nu x^\nu \) (with \( \omega_{\mu\nu}=-\omega_{\nu\mu} \), and \( \delta\phi=0 \) for a scalar). Show \( M^{\mu\nu\rho}=x^\nu T^{\mu\rho}-x^\rho T^{\mu\nu} \) and that its conservation requires \( T^{\mu\nu} \) to be symmetric.
Solution
With \( \delta\phi=0 \) and \( \delta x^\mu=\omega^\mu{}_\nu x^\nu \), the master current \( J^\mu=-T^\mu{}_\sigma\,\delta x^\sigma=-T^\mu{}_\sigma\,\omega^\sigma{}_\nu x^\nu=-\omega_{\rho\nu}\,x^\nu T^{\mu\rho} \). Because \( \omega_{\rho\nu} \) is an antisymmetric constant, only the antisymmetric part in \( (\rho\nu) \) contributes; stripping the parameter gives the rank-3 current \( M^{\mu\nu\rho}=x^\nu T^{\mu\rho}-x^\rho T^{\mu\nu} \) (antisymmetric in \( \nu\leftrightarrow\rho \)). Its divergence: \( \partial_\mu M^{\mu\nu\rho}=(\partial_\mu x^\nu)T^{\mu\rho}+x^\nu\partial_\mu T^{\mu\rho}-(\partial_\mu x^\rho)T^{\mu\nu}-x^\rho\partial_\mu T^{\mu\nu} \). Using \( \partial_\mu x^\nu=\delta^\nu_\mu \) and translation conservation \( \partial_\mu T^{\mu\rho}=0 \): \( \partial_\mu M^{\mu\nu\rho}=T^{\nu\rho}-T^{\rho\nu} \). Conservation \( \partial_\mu M^{\mu\nu\rho}=0 \) therefore holds iff \( T^{\nu\rho}=T^{\rho\nu} \). For a scalar the canonical \( T^{\mu\nu}=\partial^\mu\phi\,\partial^\nu\phi-\eta^{\mu\nu}\mathcal{L} \) is already symmetric, so orbital angular momentum \( J^{\nu\rho}=\int M^{0\nu\rho}\,d^3x \) is conserved. For fields with spin one must add the Belinfante improvement to symmetrise \( T^{\mu\nu} \).