= Solution
For the <complex scalar field>, the <canonical momenta> are $\pi=\dot\phi^\dagger$ and $\pi^\dagger=\dot\phi$. Split $H=H_0+H_I$. The <interaction picture> is obtained from the <Schrödinger picture> by $|\Psi_I(t)\rangle=e^{iH_0t}|\Psi_S(t)\rangle$ and $O_I(t)=e^{iH_0t}O_Se^{-iH_0t}$, choosing a common time origin. The fields therefore obey the free <Klein-Gordon equation>, while the states evolve through $H_I(t)$. The <canonical quantization of a complex scalar field> gives
$$
\phi_I(x)=\int d\Pi_p\,[a(p)e^{-ip\cdot x}+b^\dagger(p)e^{ip\cdot x}],\qquad d\Pi_p=\frac{d^3p}{(2\pi)^3 2E_p}.
$$
The adjoint expansion contains $a^\dagger$ and $b$. Particle and <antiparticle> modes are independent; their nonzero <commutators> are
$$
\boxed{[a(p),a^\dagger(q)]=[b(p),b^\dagger(q)]=(2\pi)^3 2E_p\delta^{(3)}(\mathbf p-\mathbf q).}
$$
All cross-species <commutators>, and all pairs of <creation operators> or of <annihilation operators>, vanish. These formulas reproduce $[\phi(\mathbf x),\pi(\mathbf y)]=[\phi^\dagger(\mathbf x),\pi^\dagger(\mathbf y)]=i\delta^{(3)}(\mathbf x-\mathbf y)$ and the remaining vanishing equal-time <commutators>.
There are no derivative interactions, so the <interaction Hamiltonian> is $H_I(t)=-\int d^3x\,\mathcal L_I(x)$. Differentiating the <interaction picture> state gives $i\partial_t|\Psi_I\rangle=H_I(t)|\Psi_I\rangle$. Its <time-evolution operator> satisfies
$$
U(t,t_0)=I-i\int_{t_0}^t dt_1\,H_I(t_1)U(t_1,t_0).
$$
Iteration gives ordered integration regions $t_1>\cdots>t_n$. Each is $1/n!$ of the full region with a <time-ordered product>, yielding the <Dyson series>
$$
U(t,t_0)=\sum_{n=0}^\infty\frac{(-i)^n}{n!}\int_{t_0}^t dt_1\cdots dt_n\,T[H_I(t_1)\cdots H_I(t_n)].
$$
With the usual asymptotic scattering prescription, taking $t_0\to-\infty$ and $t\to+\infty$ proves
$$
\boxed{S=T\exp\!\left(i\int d^4x\,\mathcal L_I(x)\right).}
$$
<Time ordering> is essential because different interaction Hamiltonians need not commute.
At first order, use the connected tree <scattering amplitude>, equivalently the <normal-ordered> quartic vertex with the vacuum and one-particle mass contributions removed. This qualification matters if the local product is interpreted literally without <normal ordering>: single-vertex <tadpole diagrams> and <vacuum diagrams> are not the requested connected two-particle amplitude. The <complex scalar quartic contact vertex> is obtained from
$$
S-I=-\frac{i\lambda}{4}\int d^4x\,:\!\phi^{\dagger 2}(x)\phi^2(x)\!:+O(\lambda^2).
$$
In a fully connected matrix element, an incoming particle and outgoing antiparticle attach to the two $\phi$ factors; an outgoing particle and incoming antiparticle attach to the two $\phi^\dagger$ factors. Each pair has two assignments, giving $2!2!=4$. The external <relativistic normalization of a one-particle state> cancels each mode measure: for example $[\phi(x),a^\dagger(p_1)]=e^{-ip_1x}$ and $[b(q_2),\phi(x)]=e^{iq_2x}$. Consequently the connected integrand is $4e^{-i(p_1+q_1-p_2-q_2)\cdot x}$, and
$$
\langle p_2q_2|(S-I)|p_1q_1\rangle_{\rm conn}
=-i\lambda(2\pi)^4\delta^{(4)}(p_1+q_1-p_2-q_2)+O(\lambda^2).
$$
Thus \b[the invariant connected amplitude is]
$$
\boxed{\mathcal T=-\lambda+O(\lambda^2).}
$$
The <Dirac delta function> expresses <four-momentum conservation>.
For the identity contribution, commute $a(p_2)$ through $a^\dagger(p_1)$ and independently $b(q_2)$ through $b^\dagger(q_1)$; all terms leaving a <creation operator> against the left <Fock vacuum> or an <annihilation operator> against the right <Fock vacuum> vanish. Hence
$$
\boxed{\langle p_2q_2|I|p_1q_1\rangle=(2\pi)^6(2E_{p_1})(2E_{q_1})\delta^{(3)}(\mathbf p_2-\mathbf p_1)\delta^{(3)}(\mathbf q_2-\mathbf q_1).}
$$
There is no exchange term between the two distinct particle and <antiparticle> species.
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