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For a Schrödinger-picture state satisfyingdefine the interaction picture byDifferentiation givesThe formal solution from $t_0$ is the Dyson seriesThrough quadratic order,Differentiating giveswhich checks the equation to the requested order.

At the endpoints, $|i{⟩}_I=e^{-i{H}_{0}T/2}|i{⟩}_S$ and $|f{⟩}_I=e^{i{H}_{0}T/2}|f{⟩}_S$. ThereforewhereFor ${\varphi }_1(p_1)\to {\varphi }_2(p_1^{\prime }){\varphi }_2(p_2^{\prime }){\varphi }_2(p_3^{\prime })$, the leading term isThere are $3!$ contractions of the identical outgoing fields. With covariantly normalized states,Equivalently, the interaction vertex from $-\lambda {\varphi }_1{\varphi }_2^3$ is $-i3!\lambda$.

With a nonconstant potential energy, momentum no longer diagonalizes the Hamiltonian operator. The operator derivation must use the energy eigenfunctions of the full Hamiltonian, a Dyson series, or a time-sliced Trotter product formula. In the classical derivation, the straight paths are replaced by every solution of the nonlinear Euler-Lagrange equation with the specified endpoints. The semiclassical propagator becomes a sumwhere the prefactor is the Van Vleck determinant and ${\mu }_{\mathrm{cl}}$ is a Maslov index. Unlike a quadratic theory, the classical-path sum is generally only an asymptotic approximation: the exact path integral includes fluctuations of every order. On the circle, the sum must still include all winding number sectors.

The interaction $-g\varphi {\Psi }_i^{\ast }{\Psi }_i$ gives one cubic vertex $-ig$ joining a real-scalar line to a particle-antiparticle pair of either complex scalar species. The second-order term of the Dyson series isWick theorem contracts the incoming ${\Psi }_1{\overline{\Psi }}_1$ pair at one vertex, the outgoing ${\Psi }_2{\overline{\Psi }}_2$ pair at the other, and the two $\varphi$ fields with each other. The two assignments of $x,y$ cancel the factor $2!$. Thus there is one connected tree-level Feynman diagram, the $s$-channel exchangeWith standard relativistic external-state normalization,up to the physically irrelevant common sign convention for $\mathcal{M}$.

In the center-of-momentum frame, $q=(E_1,-\mathbf{p})$ and $p=(E_1,\mathbf{p})$, soPair creation is kinematically possible exactly when $s\ge 4M_2^2$. The threshold incident momentum is therefore

With interaction density $\mu {\varphi }^3/3!-\lambda {\varphi }^4/4!$, the Dyson series contains $i{\mathcal{L}}_{\mathrm{int}}$ and hence cubic vertex $i\mu$ and quartic vertex $-i\lambda$. Wick contractions into the labelled slots cancel each factorial. Position integration gives the four-momentum delta function, and equivalent unlabelled contractions leave the Feynman-diagram symmetry factor. The free line carries $i/(p^2-m^2+i0)$ in the mostly-minus convention.

The time-ordered exponential solves $i\hslash {\partial }_{t}U=H(t)U$ with initial condition $U(t_0,t_0)=I$. Its expansion is the Dyson series. For bounded continuous Hamiltonian operators the series converges, as in differentiation of the Dyson time-ordered exponential. If the generators commute at all times, time ordering is unnecessary and an ordinary operator exponential suffices.

If a Hermitian interaction-picture Hamiltonian satisfies $\hat{T}V(t){\hat{T}}^{-1}=V(-t)$ for an antiunitary operator, its Dyson series conjugates $(-i{)}^n$ to $i^n$. The substitution $t_j\mapsto -t_j$ converts ordered integration to anti-ordered integration, giving $\hat{T}S{\hat{T}}^{-1}=S^{†}$. Antiunitarity then gives the displayed reciprocity: the time-reversed process has the reversed outgoing state as its incoming state and vice versa. It does not merely reverse the momenta while leaving their incoming/outgoing roles fixed.

Wick theorem expresses a time-ordered product of free fields as its normal-ordered product plus the sum over every possible contraction. It turns the Dyson series into Feynman diagrams.