Conserved Charge as Symmetry Generator

Statement

In a canonically quantized field theory, the conserved Noether charge \( Q=\int d^3x\, j^0 \) associated with a continuous internal symmetry acts on the fields as the generator of that symmetry: for an infinitesimal transformation \( \varphi\to\varphi+\delta\varphi \) (with the parameter \( \alpha \) scaled out, so \( \delta\varphi \) is the change per unit \( \alpha \)), the equal-time commutator satisfies \( [\,Q,\varphi(x)\,]=-\,i\,\delta\varphi(x) \). Equivalently, the finite unitary \( U(\alpha)=e^{\,i\alpha Q} \) implements the symmetry by conjugation, \( U\,\varphi\,U^{-1}=\varphi+\alpha\,\delta\varphi+\mathcal{O}(\alpha^2) \).

Why it matters

Noether's first theorem hands us a conserved number \( Q \) whenever the action is invariant, but says nothing about what \( Q \) does to states. Quantization turns \( Q \) into an operator, and this result identifies that operator with the geometric object it came from: \( Q \) is not merely conserved, it is the infinitesimal generator that rotates, shifts, or phases the fields. This is what makes "charge" and "symmetry" two views of one structure.

The relation underwrites almost everything downstream: Ward–Takahashi identities, the classification of states by charge sectors, the spontaneous-breaking criterion \( Q|0\rangle\neq 0 \), and the very statement that gauge redundancies are generated by (would-be) charges. It is the bridge from the classical Noether current to the quantum algebra of observables.

Assumptions

Derivation

Result

Limiting cases

• Free complex scalar, U(1): \( \delta\varphi=-i\varphi \) gives \( [Q,\varphi]=-\varphi \), so \( Q \) counts particle-minus-antiparticle number, \( Q|k\rangle=+|k\rangle \).
• Spacetime translations: \( \delta\varphi=\partial_\mu\varphi \) promotes \( Q\to P_\mu \) and reproduces \( [P_\mu,\varphi]=-i\partial_\mu\varphi \), the Heisenberg equation for \( \mu=0 \).
• Free real scalar, no continuous internal symmetry: \( \delta\varphi=0 \), the charge is trivial and \( [Q,\varphi]=0 \) — the theorem is empty but consistent.
• Classical limit \( \hbar\to0 \): \( \tfrac{1}{i\hbar}[Q,\varphi]\to\{Q,\varphi\}_{\mathrm{PB}}=\delta\varphi \), recovering the Poisson-bracket generator of the classical symmetry.

Breaks when

Failure modes

• Sign flips: writing \( [Q,\varphi]=+i\delta\varphi \) by forgetting the antisymmetry \( [\pi,\varphi]=-[\varphi,\pi] \) in Step 6. The exponentiation in Step 8 is the sign check.
• Dropping the \( i \): conflating the quantum commutator with the classical Poisson bracket \( \{Q,\varphi\}=\delta\varphi \); the factor \( i\hbar \) from \( [\;,\;]=i\hbar\{\;,\;\} \) is mandatory.
• Using \( j^0=\pi\dot\varphi \) instead of \( \pi\,\delta\varphi \): confusing the Hamiltonian density (energy) with the Noether charge density; only \( \delta\varphi \) belongs here.
• Forgetting the \( K^\mu \) term: for symmetries with \( \delta\mathcal{L}=\partial_\mu K^\mu \) (e.g. translations), omitting \( -K^0 \) gives the wrong charge and a wrong commutator.
• Equal-time slip: commuting \( Q(t) \) with \( \varphi(t') \) at unequal times and still using the naive delta function; the canonical relation is equal-time only.
• Ordering \( \pi\,\delta\varphi \) vs. \( \delta\varphi\,\pi \): assuming they are interchangeable when \( \delta\varphi \) is not ultralocal, silently discarding Schwinger terms.

Discussion

The content of this result is that quantization is a homomorphism from the classical symmetry algebra (Poisson brackets of charges) to the operator algebra (commutators). Noether's theorem gives us the map object \( Q \); canonical quantization gives us the bracket; and their marriage says \( \mathrm{ad}_Q\equiv[Q,\cdot] \) is precisely the infinitesimal symmetry action \( -i\,\delta \). Charges of a symmetry group therefore close into the same Lie algebra as the group generators — \( [Q_a,Q_b]=i f_{ab}{}^c Q_c \) — which is why isospin, flavour, and Poincaré charges obey the algebras they do.

Because \( Q \) generates the transformation, its eigenvalues label superselection sectors: states of definite charge, and matrix elements \( \langle\beta|\varphi|\alpha\rangle \) that vanish unless \( q_\beta-q_\alpha \) matches the charge \( \varphi \) carries. This is the operator origin of selection rules. It also makes the invariance of the vacuum, \( Q|0\rangle=0 \), the sharp criterion separating a linearly-realized (Wigner) symmetry from a spontaneously broken (Nambu–Goldstone) one.

At the level of correlation functions the same identity, inserted inside a path integral, becomes the Ward–Takahashi identity: \( \partial_\mu\langle j^\mu(x)\,\varphi(y_1)\cdots\rangle=-i\sum_k\delta(x-y_k)\langle\varphi(y_1)\cdots\delta\varphi(y_k)\cdots\rangle \). The contact terms on the right are nothing but the equal-time \( [Q,\varphi]=-i\delta\varphi \) commutator re-expressed as delta-function singularities when the divergence of the current hits an operator insertion. Anomalies are precisely the statement that this identity picks up an extra, non-classical term from the measure.

Common misconceptions. "Conserved" and "generator" are not the same claim: a quantity can be conserved (commute with \( H \)) yet be a c-number that generates nothing. The theorem's force is that the Noether charge is simultaneously conserved and the generator, and that these two facts have the same origin (invariance of the action). Also, the relation is exact, not a leading-order approximation: it holds to all orders in \( \alpha \) through the finite conjugation \( e^{i\alpha Q}\varphi e^{-i\alpha Q} \).

Worked examples

Problems

1. Show that for the U(1) complex scalar, \( [Q,\varphi^\dagger]=+\varphi^\dagger \), and hence that \( \varphi^\dagger \) creates a quantum of charge \( +1 \) while \( \varphi \) lowers charge by one.
   Solution

   \( [Q,\varphi^\dagger(\mathbf{y})]=-i\int d^3x\,[-\pi^\dagger\varphi^\dagger,\varphi^\dagger(\mathbf{y})] \). Only the \( \pi^\dagger \) term contributes: \( =+i\int d^3x\,[\pi^\dagger(\mathbf{x}),\varphi^\dagger(\mathbf{y})]\varphi^\dagger(\mathbf{x})=+i\int d^3x\,(-i\delta^3)\varphi^\dagger=+\varphi^\dagger(\mathbf{y}) \). This equals \( -i\,\delta\varphi^\dagger \) with \( \delta\varphi^\dagger=+i\varphi^\dagger \). Since \( Q(\varphi^\dagger|q\rangle)=(q+1)\varphi^\dagger|q\rangle \) follows from \( [Q,\varphi^\dagger]=+\varphi^\dagger \), \( \varphi^\dagger \) raises charge by one.
2. Two commuting charges \( Q_1,Q_2 \) generate variations \( \delta_1\varphi,\delta_2\varphi \). Use the Jacobi identity to show that \( [\,[Q_1,Q_2],\varphi\,]=-i\big(\delta_1(\delta_2\varphi)-\delta_2(\delta_1\varphi)\big) \), and interpret when \( [Q_1,Q_2]=0 \).
   Solution

   Jacobi: \( [[Q_1,Q_2],\varphi]=[Q_1,[Q_2,\varphi]]-[Q_2,[Q_1,\varphi]] \). Insert \( [Q_a,\varphi]=-i\delta_a\varphi \): \( =[Q_1,-i\delta_2\varphi]-[Q_2,-i\delta_1\varphi]=-i\big(-i\delta_1\delta_2\varphi\big)+i\big(-i\delta_2\delta_1\varphi\big) \). Wait, track carefully: \( [Q_1,-i\delta_2\varphi]=-i\,[Q_1,\delta_2\varphi]=-i(-i\,\delta_1(\delta_2\varphi))=-\delta_1\delta_2\varphi \), and similarly the second term gives \( +\delta_2\delta_1\varphi \). So \( [[Q_1,Q_2],\varphi]=-\delta_1\delta_2\varphi+\delta_2\delta_1\varphi=-\big(\delta_1\delta_2-\delta_2\delta_1\big)\varphi \). Writing this as \( -i\,\delta_{[1,2]}\varphi \) shows the charge commutator generates the commutator of transformations. If the two transformations commute (abelian), \( [Q_1,Q_2]=0 \) and the charges are simultaneously diagonalizable.
3. A free real scalar of mass \( m \) has Hamiltonian \( H=\int d^3x\,\tfrac12(\pi^2+(\nabla\varphi)^2+m^2\varphi^2) \). Verify by direct commutation that \( [H,\varphi(\mathbf{y})]=-i\pi(\mathbf{y})=-i\dot\varphi(\mathbf{y}) \), consistent with time translation being generated by \( H \).
   Solution

   Only the \( \tfrac12\pi^2 \) term has a nonzero commutator with \( \varphi \): \( [H,\varphi(\mathbf{y})]=\int d^3x\,\tfrac12[\pi(\mathbf{x})^2,\varphi(\mathbf{y})] \). Use \( [\pi^2,\varphi]=\pi[\pi,\varphi]+[\pi,\varphi]\pi=2\pi(-i\delta^3) \). So \( [H,\varphi(\mathbf{y})]=\int d^3x\,\tfrac12\cdot 2\pi(\mathbf{x})(-i\delta^3(\mathbf{x}-\mathbf{y}))=-i\pi(\mathbf{y}) \). Since the Heisenberg equation gives \( \dot\varphi=i[H,\varphi]=\pi \), we confirm \( [H,\varphi]=-i\dot\varphi \), i.e. \( \delta\varphi=\dot\varphi \) for time translation.
4. For SU(2) isospin acting on a doublet \( \varphi_a\ (a=1,2) \) as \( \delta\varphi_a=-i\,(T^A)_{ab}\varphi_b \) with \( T^A=\tfrac12\sigma^A \), write the charges \( Q^A \) and show \( [Q^A,\varphi_a]=-(T^A)_{ab}\varphi_b \). Confirm the charges satisfy \( [Q^A,Q^B]=i\epsilon^{ABC}Q^C \).
   Solution

   \( Q^A=\int d^3x\,\pi_a\,\delta\varphi_a/(\text{per }\alpha^A)=-i\int d^3x\,\pi_a(T^A)_{ab}\varphi_b \). Then \( [Q^A,\varphi_c(\mathbf{y})]=-i\int d^3x\,(T^A)_{ab}[\pi_a(\mathbf{x}),\varphi_c(\mathbf{y})]\varphi_b(\mathbf{x})=-i\int d^3x\,(T^A)_{ab}(-i\delta_{ac}\delta^3)\varphi_b=-(T^A)_{cb}\varphi_b(\mathbf{y}) \), matching \( -i\delta\varphi_c \). For the algebra, apply the previous problem's result: \( \delta^A\delta^B-\delta^B\delta^A \) acting on \( \varphi \) equals \( -[T^A,T^B]\varphi=-i\epsilon^{ABC}T^C\varphi \), i.e. \( \delta_{[A,B]}=\epsilon^{ABC}\delta^C \), which forces \( [Q^A,Q^B]=i\epsilon^{ABC}Q^C \) — the su(2) algebra.
5. Estimate the "size" of a symmetry rotation: for a coherent-like field configuration with \( \varphi\sim v \) (a VEV) under U(1), how does \( e^{i\alpha Q} \) act, and why does the charge fail to exist when \( v\neq0 \)? Give the Fabri–Picasso argument schematically.
   Solution

   Formally \( e^{i\alpha Q}\varphi e^{-i\alpha Q}=e^{-i\alpha}\varphi \), rotating the phase of the VEV. Compute the norm of \( Q|0\rangle \): \( \langle0|Q^2|0\rangle=\int d^3x\,d^3y\,\langle0|j^0(x)j^0(y)|0\rangle \). By translation invariance the integrand depends only on \( \mathbf{x}-\mathbf{y} \), so one integral gives an overall volume factor \( V\to\infty \) times a nonzero constant (nonzero precisely because \( \langle0|j^0|0\rangle \) probes the broken order parameter \( v\neq0 \)). Hence \( \|Q|0\rangle\|^2\propto V\to\infty \): \( Q|0\rangle \) is non-normalizable, so \( Q \) does not exist as an operator on the Hilbert space and cannot be exponentiated to a unitary — the symmetry is spontaneously broken, realized non-linearly by massless Goldstone bosons. The local relation \( [Q,\varphi]=-i\delta\varphi \) survives only as a statement about the current.