Complex Scalars and Conserved Charge

A free complex scalar is two degenerate real Klein–Gordon fields assembled into one non-Hermitian field. In the standard positive-energy Fock representation it requires two independent oscillator families: $a^\dagger$ creates charge-$+1$ particles, $b^\dagger$ creates charge-$-1$ antiparticles, and both raise the energy by $E_{\mathbf p}>0$. The rigid phase symmetry is generated by the vacuum-normal-ordered charge $Q=N_+-N_-$, whereas the total particle number is $N=N_++N_-$. The difference and the sum answer different physical questions.

This page constructs that model in $d$-dimensional Minkowski spacetime with $m>0$, unit charge normalization, the standard positive-frequency vacuum, and an explicit finite-mode regulator whenever fields are multiplied at the same point. Its scope includes the free action, canonical normalization, particle and antiparticle modes, and charge operator. It does not gauge the phase symmetry, develop general Ward identities or interacting current renormalization, or analyze spontaneous symmetry breaking.

Required background. Quantizing the Real Scalar Field supplies the invariant measure, covariant oscillator normalization, selected vacuum, and finite-box regulator. Classical Symmetries, Currents, and Stress Tensors supplies the localized phase variation, current convention, on-shell conservation statement, and boundary-flux qualification.

Helpful background. Antiparticles and Charge-Conjugate Excitations separates negative frequency from negative state energy and explains the charge-conjugate interpretation. Its locality and charge-conjugation derivations are not repeated here.

A complex field is two real scalar degrees of freedom

The model data are deliberately spare:

• Degrees of freedom. One complex scalar $\Phi=(\phi_1+i\phi_2)/\sqrt2$, equivalently two real scalars $\phi_1,\phi_2$ of the same mass.
• Defining dynamics. A free quadratic action with one parameter $m>0$.
• Internal symmetry. A rigid $U(1)$ phase rotation with charge unit fixed to one. There is no gauge field or covariant derivative.
• State and regulator. The standard Minkowski Fock vacuum, with a periodic box and finite inversion-symmetric mode set used for coincident products.
• Decisive operators. The free Hamiltonian tests energy, while the Noether charge tests the two sectors’ opposite internal charges.

The action is

$$\begin{aligned} S_0[\Phi,\Phi^\dagger] &= \int\mathrm d^d x\,\mathcal L_0, \\ \mathcal L_0 &= \partial_\mu\Phi^\dagger\partial^\mu\Phi -m^2\Phi^\dagger\Phi \\ &= \frac12 \sum_{r=1}^{2} \left( \partial_\mu\phi_r\partial^\mu\phi_r -m^2\phi_r^2 \right). \end{aligned}$$

The complex-field expression for $\mathcal L_0$ has no factor $1/2$: substituting $\Phi=(\phi_1+i\phi_2)/\sqrt2$ produces the two real-field factors of $1/2$ in the last line. This is the quickest normalization check on the model.

Treat $\Phi$ and $\Phi^\dagger$ as independent variables while varying the action, then impose their adjoint relation. The two Euler–Lagrange equations are

$$\begin{aligned} (\Box+m^2)\Phi &= 0, \\ (\Box+m^2)\Phi^\dagger &= 0. \end{aligned}$$

Retain the phase orientation used in the classical prerequisite:

$$\Phi\longmapsto e^{i\alpha}\Phi, \qquad \Phi^\dagger\longmapsto e^{-i\alpha}\Phi^\dagger .$$

Localizing $\alpha$ gives

$$j^\mu = i\left[ \Phi^\dagger\partial^\mu\Phi - (\partial^\mu\Phi^\dagger)\Phi \right],$$

and its divergence is the off-shell identity

$$\begin{aligned} \partial_\mu j^\mu &= i\Phi^\dagger(\Box+m^2)\Phi \\ &\quad- i\bigl((\Box+m^2)\Phi^\dagger\bigr)\Phi. \end{aligned}$$

Both field equations are needed for local conservation. The integrated classical charge

$$Q_{\mathrm{cl}}(t) = \int_\Sigma \mathrm d^{d-1}\mathbf x\, j^0(t,\mathbf x)$$

is time independent only when it exists and the flux through $\partial\Sigma$ vanishes or is included in the balance law. The action, two-real-field decomposition, charge-diagonal modes, canonical current, and free-field normal-ordering qualification are developed together in Coleman 2019, § 6.1, pp. 106–113.

Canonical pairs and independent oscillator families

The momenta are crossed because $\Phi$ and $\Phi^\dagger$ are conjugate:

$$\begin{aligned} \Pi &\equiv \frac{\partial\mathcal L}{\partial\dot\Phi} = \dot\Phi^\dagger, \\ \Pi^\dagger &\equiv \frac{\partial\mathcal L}{\partial\dot\Phi^\dagger} = \dot\Phi. \end{aligned}$$

Canonical quantization therefore requires

$$\begin{aligned} [\widehat\Phi(t,\mathbf x), \widehat\Pi(t,\mathbf y)] &= i\delta^{(d-1)}(\mathbf x-\mathbf y), \\ [\widehat\Phi^\dagger(t,\mathbf x), \widehat\Pi^\dagger(t,\mathbf y)] &= i\delta^{(d-1)}(\mathbf x-\mathbf y), \end{aligned}$$

with every other fundamental equal-time commutator zero. In particular, $\Pi\neq\dot\Phi$: it is the momentum conjugate to $\Phi$ and equals $\dot\Phi^\dagger$.

Use the invariant on-shell measure

$$\begin{aligned} \int_{\mathbf p} &\equiv \int \frac{\mathrm d^{d-1}\mathbf p} {(2\pi)^{d-1}2E_{\mathbf p}}, \\ E_{\mathbf p} &= \sqrt{\mathbf p^2+m^2}. \end{aligned}$$

The quantum field and its adjoint have the mode expansions

$$\begin{aligned} \widehat\Phi(x) &= \int_{\mathbf p} \left[ a(\mathbf p)e^{-ip\cdot x} + b^\dagger(\mathbf p)e^{ip\cdot x} \right], \\ \widehat\Phi^\dagger(x) &= \int_{\mathbf p} \left[ a^\dagger(\mathbf p)e^{ip\cdot x} + b(\mathbf p)e^{-ip\cdot x} \right]. \end{aligned}$$

The two independent oscillator families obey

$$\begin{aligned} [a(\mathbf p),a^\dagger(\mathbf q)] &= (2\pi)^{d-1}2E_{\mathbf p}\, \delta^{(d-1)}(\mathbf p-\mathbf q), \\ [b(\mathbf p),b^\dagger(\mathbf q)] &= (2\pi)^{d-1}2E_{\mathbf p}\, \delta^{(d-1)}(\mathbf p-\mathbf q), \end{aligned}$$

and all mixed commutators vanish. The adjoint operation exchanges the two displayed field expansions; it does not identify $a$ with $b$.

Differentiation gives

$$\begin{aligned} \widehat\Pi(x) &= i\int_{\mathbf p}E_{\mathbf p} a^\dagger(\mathbf p)e^{ip\cdot x} \\ &\quad- i\int_{\mathbf p}E_{\mathbf p} b(\mathbf p)e^{-ip\cdot x} \\ \widehat\Pi^\dagger(x) &= -i\int_{\mathbf p}E_{\mathbf p} a(\mathbf p)e^{-ip\cdot x} \\ &\quad+ i\int_{\mathbf p}E_{\mathbf p} b^\dagger(\mathbf p)e^{ip\cdot x}. \end{aligned}$$

At equal times, put $\mathbf r=\mathbf x-\mathbf y$. The $a$ and $b$ families contribute one half-delta each:

$$\begin{aligned} [\widehat\Phi(t,\mathbf x), \widehat\Pi(t,\mathbf y)] &= \frac{i}{2} \int \frac{\mathrm d^{d-1}\mathbf p} {(2\pi)^{d-1}} e^{i\mathbf p\cdot\mathbf r} \\ &\quad+ \frac{i}{2} \int \frac{\mathrm d^{d-1}\mathbf p} {(2\pi)^{d-1}} e^{-i\mathbf p\cdot\mathbf r} \\ &= i\delta^{(d-1)}(\mathbf r). \end{aligned}$$

The adjoint pair gives the same result. The remaining fundamental commutators vanish by the mixed oscillator algebra or by momentum reversal. This round trip fixes the field measure and the $2E_{\mathbf p}$ in the oscillator commutators as one inseparable normalization package.

Schwartz writes the same two-family expansion using a field coefficient $1/\sqrt{2E_{\mathbf p}}$ and delta-normalized oscillators. The translation to the convention here is

$$\begin{aligned} a_{\mathrm{cov}}(\mathbf p) &= \sqrt{2E_{\mathbf p}}\, a_\delta(\mathbf p), \\ b_{\mathrm{cov}}(\mathbf p) &= \sqrt{2E_{\mathbf p}}\, b_\delta(\mathbf p). \end{aligned}$$

Translating the field coefficient and oscillator algebra together gives the same field and one-particle norms. The independent positive-energy particle and antiparticle modes are displayed in Schwartz 2014, § 9.1, pp. 140–142.

Let $\mathcal H_{1,+}$ and $\mathcal H_{1,-}$ be two copies of the same positive-energy mass-shell Hilbert space. The charged free-field Fock space is

$$\begin{aligned} \mathcal F_{\mathrm{ch}} &= \mathcal F_s( \mathcal H_{1,+}\oplus\mathcal H_{1,-} ) \\ &\cong \mathcal F_s(\mathcal H_{1,+}) \otimes \mathcal F_s(\mathcal H_{1,-}). \end{aligned}$$

Its selected vacuum satisfies

$$a(f)|0\rangle=0, \qquad b(g)|0\rangle=0$$

for all normalizable packets. The free excitation Hamiltonian is

$$H_{\mathrm{exc}} = \mathrm d\Gamma(E\oplus E),$$

so $a^\dagger(f)|0\rangle$ and $b^\dagger(g)|0\rangle$ are both built from future-mass-shell energies. Weinberg derives the covariant causal scalar field and its independent charge-conjugate sector in Weinberg 1995, § 5.2, pp. 201–206.

The normal-ordered U(1) charge

Products in $j^0$ must first be defined. Put the theory in a periodic spatial box of volume $V$ and retain a finite inversion-symmetric set $K_\Lambda$. With $k^0=\omega_{\mathbf k}$,

$$\begin{aligned} \widehat\Phi_\Lambda(x) &= \sum_{\mathbf k\in K_\Lambda} \frac{1}{\sqrt{2\omega_{\mathbf k}V}} \\ &\quad\times \left[ a_{\mathbf k}e^{-ik\cdot x} + b^\dagger_{\mathbf k}e^{ik\cdot x} \right], \end{aligned}$$

where

$$\begin{aligned} [a_{\mathbf k},a^\dagger_{\mathbf l}] &= \delta_{\mathbf k\mathbf l}, \\ [b_{\mathbf k},b^\dagger_{\mathbf l}] &= \delta_{\mathbf k\mathbf l}. \end{aligned}$$

Every mixed commutator is zero. In this regulated system the unsymmetrized classical ordering gives

$$\begin{aligned} Q^{\mathrm{bare}}_\Lambda &= \int_V\mathrm d^{d-1}\mathbf x\, i\left( \widehat\Phi^\dagger_\Lambda \widehat\Pi^\dagger_\Lambda - \widehat\Pi_\Lambda \widehat\Phi_\Lambda \right) \\ &= \sum_{\mathbf k\in K_\Lambda} \left( a^\dagger_{\mathbf k}a_{\mathbf k} - b_{\mathbf k}b^\dagger_{\mathbf k} \right). \end{aligned}$$

The oscillatory cross terms cancel after the spatial integral and momentum reversal. Since $b_{\mathbf k}b^\dagger_{\mathbf k} =b^\dagger_{\mathbf k}b_{\mathbf k}+1$,

$$Q^{\mathrm{bare}}_\Lambda = N_{+,\Lambda} - N_{-,\Lambda} - |K_\Lambda|.$$

Normal ordering relative to the selected vacuum defines the free quantum charge:

$$\begin{aligned} Q_\Lambda &\equiv {:}Q^{\mathrm{bare}}_\Lambda{:} \\ &= \sum_{\mathbf k\in K_\Lambda} \left( a^\dagger_{\mathbf k}a_{\mathbf k} - b^\dagger_{\mathbf k}b_{\mathbf k} \right) \\ &= N_{+,\Lambda}-N_{-,\Lambda}. \end{aligned}$$

The additive shift makes $Q_\Lambda|0\rangle=0$ and does not change any commutator. The divergent continuum analogue of $|K_\Lambda|$ is not an ordinary number to manipulate. Normal Ordering and Vacuum Terms explains the free-vacuum prescription and why it is not a general definition of renormalized composite currents.

The basis-independent continuum statement uses the one-particle charge operator

$$q_1 = I_+\oplus(-I_-), \qquad Q = \mathrm d\Gamma(q_1).$$

In the joint sector decomposition $\Psi=(\psi_{n_+,n_-})_{n_+,n_-\geq0}$,

$$Q\psi_{n_+,n_-} = (n_+-n_-)\psi_{n_+,n_-}.$$

The maximal operator domain is

$$\begin{aligned} \Psi\in\mathcal D(Q) &\Longleftrightarrow \sum_{n_+,n_-=0}^{\infty} (n_+-n_-)^2 \\ &\qquad\times \|\psi_{n_+,n_-}\|^2 < \infty. \end{aligned}$$

On $\mathcal D(N_+)\cap\mathcal D(N_-)$ this is $Q=N_+-N_-$; its self-adjoint closure has the larger maximal domain just displayed. All commutators below are identities on the common finite-particle core.

The charged creators and annihilators satisfy

$$\begin{aligned} [Q,a^\dagger(f)] &= a^\dagger(f), & [Q,a(f)] &= -a(f), \\ [Q,b^\dagger(g)] &= -b^\dagger(g), & [Q,b(g)] &= b(g). \end{aligned}$$

Consequently,

$$\begin{aligned} [Q,\widehat\Phi(x)] &= -\widehat\Phi(x), \\ [Q,\widehat\Phi^\dagger(x)] &= \widehat\Phi^\dagger(x). \end{aligned}$$

For the classical page’s phase orientation, define

$$\begin{aligned} U_{\mathrm{cl}}(\alpha) &= e^{-i\alpha Q}, \\ U_{\mathrm{cl}}(\alpha) \widehat\Phi U_{\mathrm{cl}}(\alpha)^{-1} &= e^{i\alpha}\widehat\Phi. \end{aligned}$$

Equivalently,

$$e^{i\alpha Q} \widehat\Phi e^{-i\alpha Q} = e^{-i\alpha}\widehat\Phi.$$

The second form is the parameter orientation used on the helpful antiparticle page. The two displays are the same $U(1)$ action after $\alpha\mapsto-\alpha$; changing only one sign would be inconsistent.

The regulated Hamiltonian retains both oscillator zero-point terms:

$$\begin{aligned} H_\Lambda &= \sum_{\mathbf k\in K_\Lambda} \omega_{\mathbf k} \left( N_{+,\mathbf k} + N_{-,\mathbf k} + 1 \right) \\ &= E_{0,\Lambda} + H_{\mathrm{exc},\Lambda}, \\ E_{0,\Lambda} &= \sum_{\mathbf k\in K_\Lambda}\omega_{\mathbf k}, \\ H_{\mathrm{exc},\Lambda} &= \sum_{\mathbf k\in K_\Lambda} \omega_{\mathbf k} \left( N_{+,\mathbf k} + N_{-,\mathbf k} \right). \end{aligned}$$

Thus

$$[H_{\mathrm{exc},\Lambda},Q_\Lambda]=0.$$

The plus sign in energy and the minus sign in charge are the central physical contrast. On one-particle packets,

$$\begin{aligned} Q\,a^\dagger(f)|0\rangle &= +a^\dagger(f)|0\rangle, \\ Q\,b^\dagger(g)|0\rangle &= -b^\dagger(g)|0\rangle, \end{aligned}$$

while both wave packets have positive energy support. The $b^\dagger e^{ip\cdot x}$ term is negative-frequency field dependence, not a negative-energy state.

Charge is not total particle number

Within the selected free representation,

$$\begin{aligned} N &= N_++N_-, \\ Q &= N_+-N_-. \end{aligned}$$

The vacuum and a particle–antiparticle pair can both have $Q=0$, although their total particle numbers are respectively zero and two. Likewise, $\widehat\Phi$ lowers charge by one because it either annihilates a charge-$+1$ particle or creates a charge-$-1$ antiparticle; $\widehat\Phi^\dagger$ raises charge by one.

A $U(1)$-invariant Hamiltonian commutes with $Q$, but it need not commute with the free total-number operator. A charge-preserving interaction can create or annihilate particle–antiparticle pairs, mixing $N$ while preserving $Q$. Therefore free conservation of both $N_+$ and $N_-$ is stronger than the single global $U(1)$ conservation law.

Charge eigenspaces are invariant under neutral operators that commute with $Q$. The full charged field algebra also contains $\Phi$ and $\Phi^\dagger$, which connect different eigenspaces. Whether charge labels superselection sectors depends on which observable algebra is being used and is not decided by the free Fock decomposition alone.

Imposing the additional reality condition $\Phi=\Phi^\dagger$ removes one real degree of freedom and collapses the two oscillator families to the single self-conjugate real-scalar family. The continuous phase rotation no longer preserves that reality condition except at its discrete sign subgroup. Merely choosing an electrically neutral coupling does not impose this reality condition, so “neutral” and “self-conjugate” are not synonyms.

What the model establishes and where it stops

The free model supplies several independent checks: the action equals two real-scalar actions, both canonical commutators recover the same delta, $H_{\mathrm{exc}}$ is nonnegative, $Q$ is Hermitian and vacuum-neutral, the charged creators have opposite eigenvalues, and $[H_{\mathrm{exc}},Q]=0$. The current has dimension $d-1$ and the integrated charge is dimensionless.

Those checks do not perform the following extensions:

• Continuous Symmetries, Generators, and Charges develops the general hypersurface generator, boundary, and broken-symmetry qualifications.
• Quantum Currents, Improvements, and Conservation treats renormalized current operators, improvements, and contact terms.
• Background Fields versus Dynamical Gauging distinguishes probing this global current from introducing a dynamical gauge field. Gauging Continuous and Finite Symmetries develops the additional gauging data.
• Symmetry Realization and Breaking develops vacuum order parameters, spontaneous breaking, and the relevant infinite-volume limits.

An interacting Heisenberg field also need not retain the free two-term mode expansion used here. The present construction is the exactly solvable free reference model for those later questions.

Common pitfalls

Inserting a factor $1/2$ in the complex action. The complex kinetic term already equals the sum of two real kinetic terms, each carrying a factor $1/2$. Adding another factor would misnormalize both canonical brackets.

Using the uncrossed canonical momentum. The momentum conjugate to $\Phi$ is $\dot\Phi^\dagger$, not $\dot\Phi$. The wrong pairing makes the intended equal-time commutator vanish and assigns the nonzero bracket to the wrong variables.

Identifying the two oscillator families. A genuinely complex field needs independent $a$ and $b$ operators. Setting $b=a$ imposes an additional reality condition and removes the charged particle–antiparticle structure.

Mixing covariant and delta-normalized oscillators. The invariant measure, field coefficient, oscillator commutator, and one-particle norm must be rescaled together. Changing only one leaves an uncanceled factor of $2E_{\mathbf p}$.

Calling negative frequency negative energy. The sign in $e^{ip\cdot x}$ labels a field component. The Hamiltonian commutator shows that $b^\dagger$ raises the state energy by $E_{\mathbf p}>0$.

Confusing charge with number or electric charge. Here $Q=N_+-N_-$ is the generator of a global internal symmetry with an arbitrarily chosen unit. Calling it electric charge requires extra coupling data, while total number is the sum rather than the difference.

Changing the generator sign without changing the phase convention. The choices $U=e^{-i\alpha Q}$ with $\Phi\mapsto e^{i\alpha}\Phi$ and $U=e^{i\alpha Q}$ with $\Phi\mapsto e^{-i\alpha}\Phi$ are equivalent. Combining the same sign in both places contradicts $[Q,\Phi]=-\Phi$.

Treating free normal ordering as general current renormalization. The finite-regulator subtraction fixes the selected free vacuum’s charge. It does not define interacting composite currents, gauge the symmetry, or settle anomalies.

Check your understanding

1. Retrieval. State the action, the two conjugate momenta, the two-family mode expansion, and the normal-ordered charge.
2. Distinction. Compare the charge, total particle number, and energy of $a^\dagger(f)|0\rangle$, $b^\dagger(g)|0\rangle$, and $a^\dagger(f)b^\dagger(g)|0\rangle$.
3. Normalization and domain. Recover both nonzero equal-time canonical commutators and state $\mathcal D(Q)$ in the $(n_+,n_-)$ decomposition.
4. Failure mode. Diagnose identifying $b$ with $a$, treating a finite box as a UV cutoff, and using the coincident current without a regulator or ordering prescription.
5. Transfer. Starting from two regulated real oscillators $c_{1,\mathbf k}$ and $c_{2,\mathbf k}$, form charge-diagonal oscillator combinations and recover the sum in $H_{\mathrm{exc}}$ and the difference in $Q$.
6. Handoff. Identify the pages that develop general charge generators, renormalized quantum currents, dynamical gauging, and spontaneous symmetry breaking.

Answers and repair routes

1. The action is $S_0=\int\mathrm d^d x\, (\partial_\mu\Phi^\dagger\partial^\mu\Phi-m^2\Phi^\dagger\Phi)$. The momenta are $\Pi=\dot\Phi^\dagger$ and $\Pi^\dagger=\dot\Phi$. The field contains $a e^{-ip\cdot x}+b^\dagger e^{ip\cdot x}$, its adjoint contains $a^\dagger e^{ip\cdot x}+b e^{-ip\cdot x}$, and $Q={:}Q^{\mathrm{bare}}{:}=N_+-N_-$. Revisit Quantizing the Real Scalar Field if the invariant normalization is unclear.

2. The $a^\dagger$ state has $(Q,N)=(+1,1)$, the $b^\dagger$ state has $(Q,N)=(-1,1)$, and their product has $(Q,N)=(0,2)$. All are built from positive-energy packets; for sharp modes the excitation energy is, respectively, $E_{\mathbf p}$, $E_{\mathbf q}$, and $E_{\mathbf p}+E_{\mathbf q}$. Review Antiparticles and Charge-Conjugate Excitations if the frequency sign is being used as an energy test.

3. In $[\Phi,\Pi]$, the $a$ commutator and the $[b^\dagger,b]$ commutator each leave one half of the spatial delta after the time derivative cancels $2E_{\mathbf p}$. The adjoint pair works identically; mixed fundamental brackets vanish. The charge domain is the maximal domain $\mathcal D(Q)$ displayed in the charge section. Return to the canonical-pair section if the crossed momentum is missing.

4. Setting $b=a$ removes the independent conjugate-charge sector and turns the model into a real-field restriction. A periodic box discretizes momenta but leaves infinitely many modes, so a finite $K_\Lambda$ is still required. Finally, $j^0$ is a coincident composite product; the finite regulator and declared free-vacuum ordering make the displayed charge meaningful. Continue to Normal Ordering and Vacuum Terms for the prescription’s limits.

5. Define $a_{\mathbf k}=(c_{1,\mathbf k}+ic_{2,\mathbf k})/\sqrt2$ and $b_{\mathbf k}=(c_{1,\mathbf k}-ic_{2,\mathbf k})/\sqrt2$, with adjoints obtained by conjugation. This unitary change preserves the oscillator algebra and gives $N_++N_-=N_1+N_2$. Mode by mode, the rotation generator is $Q_{\mathbf k}=i(c_{1,\mathbf k}^\dagger c_{2,\mathbf k} -c_{2,\mathbf k}^\dagger c_{1,\mathbf k}) =N_{+,\mathbf k}-N_{-,\mathbf k}$, and $Q_\Lambda=\sum_{\mathbf k\in K_\Lambda}Q_{\mathbf k}$. Hence the degenerate free Hamiltonian is proportional to the sum, while charge is the difference.

6. Use Continuous Symmetries, Generators, and Charges for general generators, Quantum Currents, Improvements, and Conservation for current renormalization, Background Fields versus Dynamical Gauging for gauging, and Symmetry Realization and Breaking for spontaneous breaking.

References

• Coleman, Sidney. Lectures of Sidney Coleman on Quantum Field Theory. Edited by Bryan Gin-ge Chen, David Derbes, David Griffiths, Brian Hill, Richard Sohn, and Yuan-Sen Ting. Singapore: World Scientific, 2019. DOI.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge: Cambridge University Press, 2014. DOI.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge: Cambridge University Press, 1995. DOI.