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Treat $\varphi ,{\varphi }^{†}$ as independent variables when taking the Legendre transform. Their canonical momenta areThe canonical quantization of a complex scalar field imposes, at equal times,with $[\varphi ,{\varphi }^{†}]=[\varphi ,\varphi ]=[\pi ,{\pi }^{†}]=[\pi ,\pi ]=0$, together with their adjoints, and $[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$. These are canonical commutation relations for two real scalar degrees of freedom, written in a complex basis. Local fields are operator-valued distributions, so these identities are understood after smearing or with a regulator.

The Hamiltonian density is $\pi \dot{\varphi }+{\pi }^{†}{\dot{\varphi }}^{†}-\mathcal{L}$, givingThe Heisenberg equation of motion is $\dot{O}=i[H,O]$. For example, $i[\int {\pi }^{†}\pi ,\varphi ]={\pi }^{†}$, while commuting $\pi$ with the gradient term and integrating its derivative of a Dirac delta distribution gives ${\nabla }^2{\varphi }^{†}$. ThusBoth fields satisfy the Klein-Gordon equation. Their spatial Fourier modes therefore have frequencies $\pm E_p$, $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. A complex scalar field has independent coefficients for these two frequency sectors, so, defining the invariant measure $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, its mode expansion isOne can recover the mode operators at any fixed time using complex scalar mode inversion:Substitution of the expansion verifies these inverses: the spatial integral selects the appropriate momentum, while the two frequency sectors acquire factors $E_p\pm E_q$. The unwanted sector vanishes when the delta function sets $E_p=E_q$. The canonical commutation relations then give, for example,Likewise $[b(p),b^{†}(q)]$ has the same value. The cross commutators vanish; their coefficients contain $E_p-E_q$ on the corresponding momentum delta function. Conversely, these oscillator brackets reproduce the equal-time field brackets, confirming the normalization of the measure.

Insert the mode expansion into $H$ and integrate over space. The terms $a^{†}{b}^{†}$ and $ba$ enforce opposite spatial momenta; their coefficient is $-E_p^2+{\mathbf{p}}^2+m^2=0$. For the diagonal terms the coefficient is $E_p^2+{\mathbf{p}}^2+m^2=2E_p^2$. Combining this with the two invariant measures leavesIn a box, $E_0={\sum }_{\mathbf{p}}{E}_p$, one half-quantum for each of the two real oscillator species. It diverges as the cutoff is removed. Normal ordering removes this vacuum energy, yielding the normal-ordered Hamiltonian of a free complex scalar field:In this nongravitational free theory an additive constant in $H$ changes only an overall state phase, not Heisenberg equations of motion, energy differences or scattering probabilities. It is therefore consistent to choose the Fock vacuum energy as zero. Both creation operators raise the energy: $[H,a^{†}(p)]=E_p{a}^{†}(p)$ and $[H,b^{†}(p)]=E_p{b}^{†}(p)$. Repeated creation builds a bosonic Fock space of spin-zero particles and antiparticles, with the same positive mass and energy. The negative-frequency part of the field does not describe negative-energy physical particles.

The global phase symmetry of a complex scalar field, with orientation $\varphi \mapsto e^{-i\alpha }\varphi$, has the indicated Noether current. Taking its divergence, the two first-derivative terms cancel, givingThe same calculation holds for its normal-ordered version. Integrating its temporal component gives a conserved charge when spatial boundary flux vanishes.

Here the ordering convention is important. Direct substitution into the literally ordered product $i\int ({\varphi }^{†}\dot{\varphi }-{\dot{\varphi }}^{†}\varphi )d^{3}x$ yieldsThe pair terms again vanish, now because their coefficient is a difference of equal energies. Reordering the $b$ term produces a divergent vacuum constant: with a finite-mode regulator $Q_{\mathrm{bare}}=N_a-N_b-N_{\mathrm{modes}}$. Thus the requested finite charge formula uses zero vacuum charge, or equivalently the vacuum subtraction of a complex scalar charge. With that conventional subtraction,The subtraction changes neither current conservation nor the action of $Q$ on fields; without it, the unrenormalized product does not literally equal the displayed normal-ordered operator.

Finally the oscillator commutators giveor $\boxed{Q{a}^{†}=a^{†}(Q+1),Q{b}^{†}=b^{†}(Q-1)}$. Acting on a charge eigenstate, the first creator raises its charge by one and the second lowers it by one. Starting from the neutral Fock vacuum, their one-particle states therefore have charges $+1$ and $-1$ respectively. This is the complex scalar charge operator $N_a-N_b$, whereas the energy counts the sum of the two occupations. A multiparticle state has additive charge equal to the number of particles minus the number of antiparticles.