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Write $K={\sigma }_A(x)$ and equip $\mathcal{O}(U)$ with the compact-open topology. All algebra homomorphisms here are complex linear. Choose a finite positively oriented boundary cycle $\Gamma$ in $U\setminus K$ with winding number one on a neighborhood of $K$ and zero outside $U$. Such a cycle can be obtained as the boundary of a finite union of sufficiently small squares whose interior contains $K$ and whose closure is contained in $U$; holes are included with negative boundary orientation. Define the holomorphic functional calculus by the norm-convergent contour integralThe resolvent identity proves that $R$ is a Banach-space-valued holomorphic function off $K$. The Cauchy integral theorem, applied after each bounded linear functional, makes the integral independent of the chosen boundary cycle. The Hahn-Banach theorem then gives equality of the algebra-valued integrals. Moreover,which proves complex linearity and continuity for the compact-open topology.

For the constant function $1$ and the coordinate function $Z$, the contour can be deformed to a large circle because their integrands are holomorphic everywhere outside $K$. There the Neumann series gives $R(z)={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$. Integrating term by term showsTo prove multiplicativity, take nested cycles ${\Gamma }_o,{\Gamma }_i$ around $K$, with the outer cycle winding once on the inner cycle and the inner winding zero on the outer cycle. Use the outer cycle for $f$ and the inner for $g$. The resolvent identity readsIn the resulting double integral, the term containing $R(w)$ gives $f(w)g(w)R(w)$ after the $z$ integral, by the Cauchy integral formula. The term containing $R(z)$ vanishes after the $w$ integral, because $w\mapsto g(w)/(z-w)$ is holomorphic on and inside the inner cycle's region. ThereforeThe integrals are over finite compact curves and have continuous integrands, so exchanging their order is legitimate. This proves the existence of the required continuous unital algebra homomorphism.

For uniqueness, the Runge approximation theorem gives density in $\mathcal{O}(U)$, for the compact-open topology, of rational functions with poles outside $U$. More explicitly, exhaust $U$ by compact sets that are holomorphically convex relative to $U$, and approximate on those sets by rational functions whose poles lie in the complement of $U$; this gives convergence on every compact subset. Any unital algebra homomorphism taking $Z$ to $x$ must take $(Z-\alpha {)}^{-1}$, for $\alpha \notin U$, to $(x-\alpha 1{)}^{-1}$, since it preserves inverse equations. Partial fractions and polynomial algebra therefore fix its value on every such rational function. Continuity fixes its value on their limits. Thus the continuous unital homomorphism with the prescribed coordinate value is unique. Restricting a function to a smaller neighborhood of $K$ gives the same value, either by the contour definition or this uniqueness argument.

To prove the holomorphic spectral mapping theorem, first let $\mu \notin f(K)$. There is an open neighborhood $V\subseteq U$ of $K$ on which $f-\mu$ has no zeros. The holomorphic function $h=1/(f-\mu )$ on $V$ supplies a two-sided inverse throughThus ${\sigma }_A({\Theta }_x(f))\subseteq f(K)$. Conversely, if $\lambda \in K$ and $\mu =f(\lambda )$, the divided differenceis holomorphic on $U$, and multiplicativity givesThe two factors commute. A commuting product $uv$ can be invertible only if both factors are invertible: its inverse commutes with each factor, and $v(uv{)}^{-1}$ is a two-sided inverse for $u$. Invertibility of the displayed product would therefore contradict $\lambda \in K$. This proves

For the root, let $D=\mathbb{C}\setminus (-\infty ,0]$ and $q(z)=\exp (\mathrm{Log}z/3)$, using the principal complex logarithm. Define $y={\Theta }_x(q)$. Since $q(z{)}^3=z$, multiplicativity and the holomorphic spectral mapping theorem giveThis is the principal cube root of a Banach-algebra element.

For uniqueness, suppose $v^3=x$ and ${\sigma }_A(v)\subset S$. Cubing maps the whole open sector $S$ into $D$, sois a continuous unital algebra homomorphism: pullback by cubing is continuous for the compact-open topology, since images of compact sets are compact. It takes the coordinate function to $v^3=x$, and uniqueness already proved gives $F={\Theta }_x$. For $z\in S$, the scalar principal cube root satisfies $q(z^3)=z$, because $3\arg z\in (-\pi ,\pi )$. HenceThis proves uniqueness with the stated spectral sector, without assuming the ambient Banach algebra is commutative or that algebra characters separate its elements.

The argument principle states that if $h$ is meromorphic on a neighbourhood of a positively oriented simple closed contour and its interior, with neither zeros nor poles on the contour, thenwhere $N$ and $P$ count its interior zeros and poles with multiplicity. This integer is also the winding number of $h(\gamma )$ about zero.

Fix $z_0\in U$. Injectivity prevents $f$ from being constant on a neighbourhood of $z_0$, so its Taylor expansion has the formChoose a small closed disk $D\subset U$, centred at $z_0$, such that $h$ and $mh+(z-z_0)h^{\prime }$ are nonzero throughout $D$. Then $f-f(z_0)$ has its sole zero at $z_0$ in $D$, of multiplicity $m$, while $f^{\prime }$ has no zero in $D\setminus \{z_0\}$. On its boundary let $\delta =\min |f-f(z_0)|>0$.

Choose $0<|w|<\delta$. For every $0\le t\le 1$, the holomorphic function $f-f(z_0)-tw$ is nonzero on $\partial D$. By the argument principle, its number of zeros is an integer given by a contour integral varying continuously with $t$, hence remains $m$. At $t=1$ none of these zeros is $z_0$, so their derivatives are nonzero and they are simple zeros. Thus there are $m$ distinct points taking the value $f(z_0)+w$. Injectivity forces $m=1$, proving $f^{\prime }(z_0)\ne 0$ everywhere in $U$.

Use the half-line transform $\hat{u}(k,t)={\int }_0^{\infty }{e}^{-ikx}u(x,t)dx$ and let $g_j(t)={\partial }_x^{j}u(0,t)$. Integrating the heat equation twice by parts givesHere ${\hat{u}}_0(k)=(1+ik{)}^{-2}$ andwith removable singularities at the apparent denominator zeros. Let $D^{+}=\{\mathrm{Im}k>0,\mathrm{Re}{k}^2<0\}$, and orient its boundary from infinity on the $3\pi /4$ ray to zero and then out on the $\pi /4$ ray. Transform inversion and contour deformation giveTo eliminate the unknown boundary derivative, evaluate the transform identity at $-k$: ${\tilde{g}}_1={\hat{u}}_0(-k)+ik{\tilde{g}}_0-e^{k^{2}t}\hat{u}(-k,t)$. The resulting extra contour integral of $e^{ikx}\hat{u}(-k,t)$ vanishes by analyticity and decay in the upper half-plane. This proves that the boxed expression involves only the prescribed data. It is intended for $x,t>0$, with boundary and initial values obtained as limits.

Use ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =c=1$. In the interaction picture, fields evolve with the free Hamiltonian, while interactions enter the state evolution through the Dyson series. Write the Klein-Gordon field as $\varphi ={\varphi }^{(+)}+{\varphi }^{(-)}$, with the positive-frequency part of a quantum field containing annihilation operators and ${\varphi }^{(-)}=({\varphi }^{(+)}{)}^{†}$ containing creation operators. The mode normalization in the PDF corresponds to $[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$ and $E_{\mathbf{p}}=\sqrt{|\mathbf{p}{|}^2+m^2}$.

The time-ordered product places fields with the latest time on the left. For two bosonic insertions,The normal-ordered product $:\varphi (x_1)\cdots \varphi (x_n):$ instead moves every creation operator to the left of every annihilation operator, without adding the commutators generated by that rearrangement. Its vacuum expectation value is zero whenever a nonempty set of free-field factors remains.

The mode commutator gives the positive-frequency Wightman functionDefine the Feynman propagator here to be the two-point function itself, rather than that function divided by $i$. ThenEquivalently,The positive-energy pole is below and the negative-energy pole above the real axis. Closing the energy contour integral below for positive time separation and above for negative time separation gives the two terms above. Thus this is the scalar Feynman propagator pole prescription, and $(\square +m^2){\Delta }_F=-i{\delta }^4$. In particular, $T\{\varphi (x)\varphi (y)\}=:\varphi (x)\varphi (y):+{\Delta }_F(x-y)$: the difference between time ordering and normal ordering is the Wick contraction.

For $n$ free bosonic fields, the Wick theorem stateswhere $P$ runs over all sets of disjoint pairs of indices, including the empty set, and $V(P)$ is the set of paired indices. The empty normal-ordered product equals one. This is an operator identity for free fields, understood with smearing or a regulator when products of operator-valued distributions require it.

Here is an inductive proof of the scalar Wick theorem. Sort the arguments chronologically so that $x_1^0$ is the latest. For any remaining set $I$, commuting the annihilation part of ${\varphi }_1$ through the creation parts in a normal-ordered product givesThe creation part of ${\varphi }_1$ is already in the correct position. Every commutator on the right is a scalar, so no further operator rearrangement is needed for it. Because $x_1$ is latest, that commutator is ${\Delta }_F(x_1-x_j)$. Assume the Wick theorem for the other $n-1$ fields and multiply its expansion by ${\varphi }_1$. The first term of the displayed identity gives every pairing in which index 1 is unpaired; the sum gives every pairing in which 1 is paired with exactly one $j$. Each final partial pairing has exactly one such construction. The one-field identity supplies the base case, proving the theorem by mathematical induction. Applying time ordering restores arbitrary argument order.

Taking the vacuum expectation value retains only complete pairings. Odd free correlators vanish; for example the free scalar four-point function is ${\Delta }_{12}{\Delta }_{34}+{\Delta }_{13}{\Delta }_{24}+{\Delta }_{14}{\Delta }_{23}$. These facts explain why propagators are the edges of Feynman diagrams.

For the interacting quartic scalar field theory, the interaction Hamiltonian density is $\lambda {\varphi }^4/4!$. With a single time ordering acting on external insertions and interaction vertices, the unnormalized numerator expands asApply the proved Wick theorem term by term. Each complete pairing connects external insertions and four-legged vertices by Feynman propagators. There are $4!$ assignments of four distinct incident legs at a vertex, canceling its factorial. Permuting identical integrated vertices cancels the corresponding part of $v!$. Permutations still leaving the labeled graph unchanged give its Feynman-diagram symmetry factor. For example,whereas the two-point tadpole diagram has coefficient $-i\lambda /2$, because choosing the two vertex legs contracted with the labeled external fields gives $4\cdot 3$ possibilities and the other two contract with each other.

Thus the position-space rules attach ${\Delta }_F$ to every edge, $-i\lambda \int d^{4}z$ to every four-legged interaction vertex, and divide by the graph's Feynman-diagram symmetry factor. Fourier transformation gives the momentum-space Feynman rules:Enforce momentum conservation at each vertex, retaining the overall four-momentum delta function. External lines of these correlation functions remain unamputated. The unnormalized expression includes Vacuum Feynman diagrams; dividing by $N_0=⟨0|S|0⟩$ cancels those disconnected vacuum factors. They should not be discarded without that normalization.

There is a time-ordering convention in the printed correlation-function notation worth making explicit. The usual interacting correlator uses the single global time ordering in $N_n$ above. If the displayed $T\{{\varphi }_1\cdots {\varphi }_n\}S$ is read literally with $S$ outside that ordering, the external block and the vertex block are separately time ordered: their cross-contractions are $D^{+}(x_i-z)$ because each external field stands to the left of every vertex field. The same pairing construction then uses ${\Delta }_F$ for lines within either block and $D^{+}$ for lines between blocks; in momentum space the latter is $2\pi \theta (p^0)\delta (p^2-m^2)$. This literal product is different from the conventional globally time-ordered interacting correlator, and does not have the ordinary external Feynman propagators derived above.

Take the holomorphic function $f(z)=1/z$ on a neighborhood of the annulus. Every polynomial has integral zero around the unit circle by the Cauchy integral theorem, while ${\int }_{|z|=1}{z}^{-1}dz=2\pi i$. Uniform convergence on the annulus would imply convergence of these contour integrals, since their difference is bounded by $2\pi$ times the uniform error. Thus uniform polynomial approximation on the full annulus need not exist.

For $z=R+iy$, $0\le y\le \pi$, the numerator modulus is $e^{-2Ry/\pi }$ and the denominator modulus is at least $1-e^{-2R}$. Hence the magnitude of the right-side contour integral is at mostOn $z=-R+iy$ the corresponding bounds are $e^{2Ry/\pi }$ and $e^{2R}-1$, giving exactly the same bound $\pi /(2R)$. Thus the total vertical contribution is $O(R^{-1})$ and vanishes. This integral estimate handles the endpoints where a pointwise uniform-decay argument would fail.

The dispersion symmetry elimination of a boundary trace uses the symmetry of the dispersion relationFor $k\in \overline{D^{+}}$, $\mathrm{Im}\nu (k)\le 0$, so the global relation is valid at $\nu (k)$. Since $i\nu +\alpha =-ik$, it givesSubstitution into the contour integral representation produces an unwanted integral ${\int }_{\partial D^{+}}{e}^{ikx}\hat{u}(\nu (k),t)dk$. Its integrand is analytic in $D^{+}$ and decays on closing the contour upwards; Jordan lemma makes this integral zero for $x>0$. Hence the Fokas method eliminates the unknown normal derivative:All quantities here are determined by the prescribed initial and Dirichlet boundary data up to time $t$.

For verification and for numerical evaluation it is useful to evaluate the spectral contour integrals, giving a half-line drift reflection kernel. Putand define the half-line drift boundary kernelFubini's theorem, the Gaussian Fourier transform and contour deformation give the equivalent causal formulaFor the reflected initial term, ${\hat{u}}_0(i\alpha -k)={\int }_0^{\infty }{e}^{\alpha y+iky}{u}_0(y)dy$ on the contour; the evaluated Gaussian supplies the necessary large-$y$ decay. If $e^{\alpha y}{u}_0(y)$ is not integrable, truncate the initial conditions first, evaluate, and pass to the limit using Gaussian bounds. No extra exponential-decay assumption on the original data is needed for this kernel formula.

The boundary kernel follows particularly simply fromThis calculation independently checks both the sign and the coefficient of the boundary forcing.

Work first with $q$ in the Schwartz space, so that all Fourier manipulations and spectral contour integrals are justified; weaker classes follow by the usual density or distribution arguments. DefineThe phase is purely imaginary. Since ${\partial }_{\bar{z}}E=kE$, setting $F=E\Phi$ reduces the spectral equation to ${\partial }_{\bar{z}}\Phi =E^{-1}q$. The whole-plane Cauchy-Pompeiu formula therefore constructs the solution decaying spatially at infinity:The freedom to add $E$ times an entire function is removed by this decay condition. For every fixed $k$, the integral is a spatial Cauchy-Green operator applied to a modulated source.

Now differentiate in the conjugate spectral parameter. The two exponential derivatives produce $\zeta -z$, canceling the Cauchy denominator, soThis is the spectral dbar equation: its right-hand side is the forward transform of $q$ multiplied by a known plane wave. Apply the whole-plane Cauchy-Pompeiu formula again, now in $k$:The spatial spectral equation also gives $F(z,k)=-q(z)/k+o(k^{-1})$ as $|k|\to \infty$. One way to justify this is to integrate by parts in the first Cauchy integral: $F=-q/k+T_k(q_{\bar{z}})/k$, where the modulated Cauchy integral $T_k(q_{\bar{z}})$ tends to zero by the Riemann-Lebesgue lemma. Comparing the $1/k$ coefficient of the spectral contour integral therefore givesThis derives the transform pair from two uses of the Cauchy-Pompeiu formula, not from an assumed inversion formula.

To identify the usual normalization, write $z=x+iy$ and $k=k_1+i{k}_2$. Then $\bar{k}z-k\bar{z}=2i(k_{1}y-k_{2}x)$. Set ${\xi }_1=2k_2$, ${\xi }_2=-2k_1$, so $d{\xi }_{1}d{\xi }_2=4dA(k)$. The result is exactly the two-dimensional Fourier transform pairThe factor four in the real-frequency change of variables is essential.

Let $T$ be a closed triangle contained, together with a neighbourhood, in the domain of a holomorphic function $f$. We prove Cauchy theorem for a triangle without assuming continuity of $f^{\prime }$.

Join the edge midpoints to subdivide $T$ into four triangles, each with half its diameter and perimeter. Their positively oriented contour integrals sum to the integral around $T$, because the internal edges cancel. At least one subtriangle has integral of modulus at least one quarter of the original. Repeat that choice to obtain nested closed triangles $T_n$ satisfyingTheir intersection is a single point $z_{\ast }$, by compactness and shrinking diameter. complex differentiability at a point at $z_{\ast }$ givesThe first two terms have polynomial primitives, hence integrate to zero around each triangle. For sufficiently large $n$, $|r|<\epsilon$ on $T_n$, so the complex line integral estimate givesCombining the bounds gives $|{\int }_{\partial T}f(z)dz|\le \epsilon d_0{P}_0$. Since $\epsilon$ is arbitrary,The orientation can be reversed with no change to the conclusion.

Use Fourier inversion with the stated Fourier transform convention:For $x<0$, close the contour integral in the lower half-plane, where $|e^{ikx}|=e^{-x\mathrm{Im}k}$ decays. Jordan lemma eliminates the large semicircle. The contour is clockwise, so the residue theorem supplies $-2\pi i$ times the residue at $k=-ip$. That simple pole hasConsequentlyThe minus sign uses both the clockwise orientation and the residue $-i{e}^{px}$; it is consistent with the full odd function $\mathrm{sgn}(x)e^{-p|x|}$ away from zero.

Use ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =1$. The free real scalar field has actionThe canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Hamiltonian operator is obtained by the Legendre transform of the density:Canonical quantization promotes $\varphi$ and $\pi$ to Hermitian operator-valued fields and imposes the equal-time canonical commutation relationsTheir Heisenberg equation of motion gives $\dot{\varphi }=\pi$, $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, hence the Klein-Gordon equation $(\square +m^2)\varphi =0$.

Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. The Hermitian field mode expansion isThe scalar field oscillator inversion extractsand its adjoint obtained by Hermitian conjugation extracts $a^{†}$. Substitute these expressions into the equal-time canonical commutation relation. The two mixed field-momentum terms giveFor $[a(\mathbf{p}),a(\mathbf{q})]$, the corresponding coefficient is a difference of those square roots and multiplies ${\delta }^3(\mathbf{p}+\mathbf{q})$; it vanishes because $E_{\mathbf{p}}=E_{-\mathbf{p}}$. Taking the adjoint gives $[a^{†},a^{†}]=0$. Thus these are bosonic annihilation operators and creation operators. The vacuum satisfies $a(\mathbf{p})|0⟩=0$, and normal ordering gives $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}(\mathbf{p})a(\mathbf{p})/(2\pi {)}^3$, after removing the constant zero-point energy. A one-particle excitation has energy $E_{\mathbf{p}}$ and spin zero.

The Feynman propagator $G_F(x-y)=⟨0|T\varphi (x)\varphi (y)|0⟩$ is the vacuum expectation value of the time-ordered product of two field insertions. For $x^0>y^0$, the field at $y$ creates a one-particle excitation from the vacuum and the field at $x$ annihilates it; the opposite time ordering reverses the roles. It is a propagation amplitude and correlation function of vacuum fluctuations, rather than a transition probability. From the oscillator expansion, only the $a$-$a^{†}$ contraction survives, so with $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$,Here $\theta$ is the Heaviside step function, and changing $\mathbf{p}$ to $-\mathbf{p}$ in the second term gives the last line.

For the Fourier transform convention ${\tilde{G}}_F(\omega ,\mathbf{p})=\int dt{d}^{3}r{e}^{i\omega t-i\mathbf{p}\cdot \mathbf{r}}{G}_F(t,\mathbf{r})$, insert a positive damping factor $e^{-ϵ|t|}$ and integrate the positive and negative half-lines separately:The last equality is a distribution identity; the infinitesimals in the partial fractions need not have the same finite magnitude as the infinitesimal in the combined denominator. The Feynman i-epsilon prescription means that the positive-frequency pole is just below the real energy axis and the negative-frequency pole is just above it. Closing the contour integral below for $t>0$, and above for $t<0$, reproduces the oscillator result by the residue theorem. This prescription fixes which homogeneous solutions are added to the Green function. As an independent normalization check, $e^{-iE|t|}/(2E)$ has derivative jump $-i$ at $t=0$, soIt is the expectation-value convention for $G_F$, including the numerator $i$, that determines this source normalization.

The vacuum is a centered free Gaussian state. The Wick theorem expresses the four-field time-ordered product as the sum of all pair Wick contractions plus terms containing a normal-ordered product. The latter terms have zero vacuum expectation value. There are exactly three complete pairings, with no fermionic signs for this bosonic field. Thus the free scalar four-point function isThese formulas describe canonical quantization of a real scalar field; in particular, a real scalar field uses a single oscillator family rather than independent charged-particle and antiparticle families.

Choose a star centre $a$ of the open star-shaped set $D$ and define the candidate antiderivative by a straight-segment contour integral:For each fixed $z\in D$ and all sufficiently small $h$, the filled triangle with vertices $a,z,z+h$ lies in $D$. Indeed, the compact segment $[a,z]$ has a neighbourhood of some positive radius contained in the open set $D$, and every point of this triangle lies within $|h|$ of that segment. The zero triangle integral consequently givesBy continuity, the last integral tends to $f(z)$ as $h\to 0$. HenceIn particular, this proves the existence of the antiderivative without incorrectly assuming the star-shaped set is convex.

On a general domain, the conclusion can fail. Take $D=\mathbb{C}\setminus \{0\}$ and $f(z)=1/z$. Every filled triangle contained in $D$ has zero boundary integral by Cauchy integral theorem, but the unit circle has integral $2\pi i$. An antiderivative would make every closed contour integral zero. This is the period obstruction to a holomorphic antiderivative: the local conclusion of Morera's theorem does not remove the global obstruction.

With the mode normalization given, the oscillator canonical commutation relations are $[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$, with the other two commutators zero. Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p^0=E_{\mathbf{p}}$. The convention compatible with the requested numerator iswithout an additional factor of $i$ outside the vacuum expectation value. Put $\tau =x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$. The Fock vacuum is annihilated by $a_{\mathbf{p}}$, so only $a{a}^{†}$ contributes to the two Wightman functions. Consequently,In the second term we changed $\mathbf{p}\mapsto -\mathbf{p}$ to give the same spatial exponential.

Now perform the energy contour integralHere the residue step uses $E>0$; the massless zero-momentum point is interpreted through the smeared distribution limit, not as an isolated normalized oscillator. The Feynman i-epsilon prescription puts the positive-energy pole below the real axis and the negative-energy pole above it: $p^0=E-i0$ and $p^0=-E+i0$. For $\tau >0$, close in the lower half-plane, clockwise; the residue theorem gives $-i$ times the residue $i{e}^{-iE\tau }/(2E)$, hence $e^{-iE\tau }/(2E)$. For $\tau <0$, close in the upper half-plane, counterclockwise; the negative-energy residue is $-i{e}^{iE\tau }/(2E)$, again giving a positive $e^{iE\tau }/(2E)$. These are exactly the two time-ordered terms. Restoring the spatial integral proves the scalar Feynman propagator pole prescription:The prescription in this formula supplies the pole convention left unspecified in the printed display; an unprescribed ordinary real-axis integral would not be well-defined. It is a distribution limit after smearing, not an absolutely convergent four-dimensional integral. With this normalization the derivative jump of a free scalar time-ordered two-point function gives $(\square +m^2){\Delta }_F=-i{\delta }^4(x-y)$, an independent check of both the numerator and the sign.

The pole displacement is exaggerated in this original schematic. The contour orientation and selected pole reproduce the time ordering of the Feynman propagator.

First make explicit the usual method of steepest descent hypotheses: the saddle point is simple, $G^{\prime }(z_0)=0$ and $g_2=G^{\prime \prime }(z_0)\ne 0$; a contour deformation leads to the two descent arms with one clockwise half-indentation around the pole; and endpoints or other contour portions contribute smaller terms. The sign of the exponential matters: $|e^{-i\lambda G}|=e^{\lambda \mathrm{Im}G}$, so decay is toward valleys of $\mathrm{Im}G$.

Use a local analytic function $w(z)$ as coordinate. Nondegeneracy allows an analytic square root after factoring the double zero of $G-G(z_0)$, so the coordinate can be chosen withThe sign of $c$ is selected by the orientation of the contour integral, with $w$ running from negative to positive real values. Matching cubic coefficients gives $d/c=-c{G}^{\prime \prime \prime }(z_0)/(6g_2)$. The transformed amplitude therefore has the Laurent seriesThe Cauchy principal value contribution of the pole on the two real arms is zero by oddness. Its clockwise semicircular indentation contributes $-i\pi F(z_0)$, since $\int dw/w=-i\pi$ there. The regular term contributes the Gaussian integral $B\sqrt{2\pi /\lambda }$. Consequently the simple saddle coinciding with a simple pole has the additive expansionThe remainder written here assumes the other contour contributions are negligible to that order. Otherwise the local result has to be supplemented by them. The next local correction is generically $O({\lambda }^{-1/2})$ times the displayed exponential. If $B=0$, it can vanish; odd Taylor terms integrate to zero between the symmetric arms. When $F(z_0)\ne 0$, the leading term gives the printed asymptotic equivalence. When $F(z_0)=0$, asymptotic equivalence to zero is not meaningful: the additive formula is still valid and a regular saddle-point approximation may lead instead.