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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 variational symmetry of a Lagrangian density maps admissible field configurations to configurations with the same action, up to an allowed boundary contribution. Thus (i) the action is invariant for suitable boundary conditions, while (ii) the Lagrangian density need only change by a total divergence, rather than be pointwise invariant. For an internal symmetry of a classical field theory, the coordinates are fixed. A spacetime transformation also changes the integration measure and must be treated accordingly.

A conserved current satisfies ${\partial }_{\mu }{j}^{\mu }=0$ on shell. The Noether charge $Q=\int j^0{d}^{3}x$ then obeys $\dot{Q}=-{\int }_{\partial V}\mathbf{j}\cdot d\mathbf{S}$, so conservation of the integrated charge additionally requires vanishing net boundary flux.

For several real field components $q_A$ and a constant infinitesimal parameter, write $\delta q_A=\epsilon X_A$ and $\delta \mathcal{L}=\epsilon {\partial }_{\mu }{K}^{\mu }$. The product rule givesSubtract the specified divergence. The Euler-Lagrange field equations make the first term vanish, proving the Noether theorem withFor a complex scalar field, treat the field and its complex conjugate as independent variables in this calculation; this is equivalent to using its two real components. This is the Noether current for a first-derivative scalar field, summed over components.

Collect the two complex fields into $\Phi =({\varphi }_1,{\varphi }_2{)}^T$. Their Lagrangian density has the radial formFor a constant Hermitian matrix $T$, the unitary transformation $\delta \Phi =i\epsilon T\Phi$ preserves both the kinetic contraction and ${\Phi }^{†}\Phi$. In particular the two specified transformations have generators $T_1={\sigma }_1$ and $T_2={\sigma }_2$, the first two Pauli matrices. Their Noether currents are instances of the unitary scalar multiplet currentExplicitly,An overall change of current sign corresponds to reversing the associated symmetry parameter; the signs here follow the two transformations as written.

The matrix commutator of their infinitesimal matrices is $[i{\sigma }_1,i{\sigma }_2]=-2i{\sigma }_3$. The antisymmetric difference of successive transformations therefore supplies a third symmetry, after a nonzero rescaling of its parameter:Its current isThe three traceless matrices ${\sigma }_a/2$ obey $[{\sigma }_a/2,{\sigma }_b/2]=i{ϵ}_{abc}{\sigma }_c/2$. Their exponentials act as SU(2) on the complex doublet. This identifies both the Lie algebra and the group acting on these fields: the central element $-I$ changes $\Phi$ and is not the identity transformation, so the doublet action is not an action of $SO(3)$ obtained by identifying $I$ with $-I$.

There is also a common global phase symmetry of a complex scalar field, $\Phi \mapsto e^{i\alpha }\Phi$, with current $j_I^{\mu }$. Together with SU(2) this gives $U(2)\simeq (SU(2)\times U(1))/{\mathbb{Z}}_2$ among complex-linear transformations. Independent constant phases of ${\varphi }_1$ and ${\varphi }_2$, and field interchange, are already contained in this group.

In fact this particular scalar potential has a larger real internal symmetry. Write ${\varphi }_1=(q_1+i{q}_2)/\sqrt{2}$ and ${\varphi }_2=(q_3+i{q}_4)/\sqrt{2}$; thenThe O(4) symmetry of a complex scalar doublet includes transformations mixing $\Phi$ with ${\Phi }^{\ast }$, which need not be complex-linear. Its connected part has six rotation generators, so listing $U(2)$ alone would not exhaust the internal symmetries of the displayed isolated theory. Individual field conjugations are examples of its discrete transformations. Finally the spacetime-independent scalar coefficients and Lorentz contractions give Poincaré symmetry, and the real scalar action also permits spatial reflection and time reversal. These statements concern the action; a chosen vacuum or external boundary conditions can preserve fewer symmetries. Special parameter choices can introduce additional symmetries, but are not required here.

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.