= Solution
Treat $\phi,\phi^\dagger$ as independent variables when taking the <Legendre transform>. Their <canonical momenta> are
$$
\boxed{\pi=\frac{\partial\mathcal L}{\partial\dot\phi}=\dot\phi^\dagger,\qquad
\pi^\dagger=\frac{\partial\mathcal L}{\partial\dot\phi^\dagger}=\dot\phi.}
$$
The <canonical quantization of a complex scalar field> imposes, at equal times,
$$
\begin{aligned}
[\phi(t,\mathbf x),\pi(t,\mathbf y)]&=i\delta^3(\mathbf x-\mathbf y),\\
[\phi^\dagger(t,\mathbf x),\pi^\dagger(t,\mathbf y)]&=i\delta^3(\mathbf x-\mathbf y),
\end{aligned}
$$
with $[\phi,\phi^\dagger]=[\phi,\phi]=[\pi,\pi^\dagger]=[\pi,\pi]=0$, together with their adjoints, and $[\phi,\pi^\dagger]=[\phi^\dagger,\pi]=0$. 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 $\pi\dot\phi+\pi^\dagger\dot\phi^\dagger-\mathcal L$, giving
$$
\boxed{H=\int d^3x\left(\pi^\dagger\pi+\nabla\phi^\dagger\cdot\nabla\phi+m^2\phi^\dagger\phi\right).}
$$
The <Heisenberg equation of motion> is $\dot O=i[H,O]$. For example, $i[\int\pi^\dagger\pi,\phi]=\pi^\dagger$, while commuting $\pi$ with the gradient term and integrating its derivative of a <Dirac delta distribution> gives $\nabla^2\phi^\dagger$. Thus
$$
\boxed{\dot\phi=\pi^\dagger,\quad\dot\phi^\dagger=\pi,\quad
\dot\pi^\dagger=(\nabla^2-m^2)\phi,\quad\dot\pi=(\nabla^2-m^2)\phi^\dagger.}
$$
Both 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 $d\Pi_p=d^3p/[(2\pi)^3 2E_p]$, its mode expansion is
$$
\phi(x)=\int d\Pi_p\left[a(p)e^{-ip\cdot x}+b^\dagger(p)e^{ip\cdot x}\right],\qquad p^0=E_p>0.
$$
One can recover the mode operators at any fixed time using <complex scalar mode inversion>:
$$
a(p)=\int d^3x\,e^{ip\cdot x}(E_p\phi+i\dot\phi),\qquad
b^\dagger(p)=\int d^3x\,e^{-ip\cdot x}(E_p\phi-i\dot\phi).
$$
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,
$$
[a(p),a^\dagger(q)]=(E_p+E_q)(2\pi)^3\delta^3(\mathbf p-\mathbf q)
=(2\pi)^3 2E_p\delta^3(\mathbf p-\mathbf q).
$$
Likewise $[b(p),b^\dagger(q)]$ 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 $a^\dagger b^\dagger$ 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 leaves
$$
H_{\rm bare}=\int d\Pi_p\,E_p\left[a^\dagger(p)a(p)+b(p)b^\dagger(p)\right]
=H_{\rm normal}+E_0I.
$$
In 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>:
$$
\boxed{H_{\rm normal}=\int d\Pi_p\,E_p[a^\dagger(p)a(p)+b^\dagger(p)b(p)].}
$$
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>: $[H,a^\dagger(p)]=E_pa^\dagger(p)$ and $[H,b^\dagger(p)]=E_pb^\dagger(p)$. 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 $\phi\mapsto e^{-i\alpha}\phi$, has the indicated <Noether current>. Taking its divergence, the two first-derivative terms cancel, giving
$$
\partial_\mu J^\mu=i\left(\phi^\dagger\Box\phi-(\Box\phi^\dagger)\phi\right)
=i[-m^2\phi^\dagger\phi+m^2\phi^\dagger\phi]=\boxed{0}.
$$
The 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 $i\int(\phi^\dagger\dot\phi-\dot\phi^\dagger\phi)d^3x$ yields
$$
Q_{\rm bare}=\int d\Pi_p\left[a^\dagger(p)a(p)-b(p)b^\dagger(p)\right].
$$
The 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 $Q_{\rm bare}=N_a-N_b-N_{\rm modes}$. Thus the requested finite charge formula uses \b[zero vacuum charge], or equivalently the <vacuum subtraction of a complex scalar charge>. With that conventional subtraction,
$$
\boxed{Q=\int d\Pi_p\left[a^\dagger(p)a(p)-b^\dagger(p)b(p)\right],\qquad Q|0\rangle=0.}
$$
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 give
$$
[Q,a^\dagger(p)]=a^\dagger(p),\qquad [Q,b^\dagger(p)]=-b^\dagger(p),
$$
or $\boxed{Qa^\dagger=a^\dagger(Q+1),\quad Qb^\dagger=b^\dagger(Q-1)}$. 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 \b[$+1$ 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.
Back to article page
