For the amputated one-particle-irreducible vertex, every internal bridge is forbidden. Define the Feynman integralsFor $E=3,L=1$, the counting identity in part b gives $V_3+2V_4=3$. The only one-particle-irreducible Feynman diagrams are a triangle of three cubic vertices and a two-propagator bubble joining one cubic and one quartic vertex. Write ${\mathcal{V}}_3$ for the amputated Feynman vertex, including its overall factor of $i$. With all external momenta incoming,The triangle has $S=1$ for labelled external legs. The mixed bubble has $S=2$ because its two internal lines may be interchanged. There are three distinct terms, according to which external leg is attached to the cubic vertex; the other two attach to the quartic vertex. No tadpole attached by a bridge is one-particle irreducible.

The requested position-space three-point correlation function is obtained by restoring the external propagators and a Fourier transform:The tree-level contribution would replace the last factor by $v_3$. This explicitly distinguishes the amputated vertex from the correlation function while giving all loop integrations in momentum space.

Let $D_{xy}=⟨0|T\varphi (x)\varphi (y)|0{⟩}_0$ be the free Feynman propagator, and abbreviate $v_3=-i\kappa$, $v_4=-i\lambda$. Expand the normalized Dyson series, so vacuum bubbles cancel between numerator and denominator, and retain the connected correlation function. Wick's theorem yieldsThese are respectively the free line, the quartic tadpole diagram, the two-cubic-vertex bubble diagram, and an tadpole contribution to a cubic-theory two-point function, attached by a bridge to the vertex carrying both external legs. The last graph is connected but is a one-particle-reducible Feynman diagram; removing its bridge disconnects it. It must be included when all connected graphs are requested. It disappears if an additional vanishing-one-point-function renormalization condition and its corresponding tadpole subtraction are imposed; that convention was not specified here.

The factors can be counted directly. The quartic term has $4\cdot 3=12$ attachments for the two labelled external legs, divided by $4!=24$. At second order in the cubic interaction the expansion denominator is $2!(3!{)}^2=72$. There are 36 Wick contractions for the bubble, and 36 for the attached-tadpole graph, giving $1/2$ for each. Thus each loop topology shown has symmetry factor $S=2$. Disconnected products of one-point functions and vacuum bubbles are not part of $G_c$.^[ref]

The one-loop cubic-quartic scalar vertex classification follows without truncating arbitrarily in powers of the couplings. A connected graph with $E$ external legs, $V_3$ cubic vertices, $V_4$ quartic vertices and loop number $L$ satisfiesFor $E=2,L=1$, the only possibilities are $(V_3,V_4)=(2,0)$ or $(0,1)$, exactly those just listed.

In $d$ spacetime dimensions, $[d^{d}x]=-d$ and a dimensionless action requires $[\mathcal{L}]=d$. Since $[{\partial }_{\mu }]=1$, the canonical kinetic term fixes $2+2[\varphi ]=d$. The cubic and quartic terms then giveThese brackets denote mass dimensions. The cubic coupling is dimensionless at $d=6$, whereas the quartic coupling is dimensionless at $d=4$. Hence there is no single dimension in which both nonzero couplings are dimensionless. For example, at $d=4$ the cubic coupling still has mass dimension one.^[ref]

The two oscillator families commute with each other. Their canonical commutation relations giveBoth terms in the mode expansion of a free field therefore acquire the same factor, givingThe Noether charge is the generator of the internal symmetry. In the active-operator convention $U(\alpha )=e^{i\alpha Q}$, repeated commutators giveThe sign agrees with part b. A negative sign for $[Q,\varphi ]$ would correspond to the opposite current convention, not to the current used here. Subtracting a multiple of the identity from $Q$ to fix the vacuum Noether charge has no effect on this generator relation.

Let ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. The conjugate momentum fields are ${\pi }_{\varphi }={\dot{\varphi }}^{†}$ and ${\pi }_{{\varphi }^{†}}=\dot{\varphi }$. Inserting the given mode expansion of a free field into $i\int d^{3}x({\dot{\varphi }}^{†}\varphi -{\varphi }^{†}\dot{\varphi })$ produces spatial Dirac delta factors. The $b^{†}{c}^{†}$ and $cb$ terms vanish after integration: they contain ${\omega }_{\mathbf{p}}-{\omega }_{\mathbf{q}}$ on ${\delta }^3(\mathbf{p}+\mathbf{q})$, and ${\omega }_{-\mathbf{p}}={\omega }_{\mathbf{p}}$. The number-conserving terms instead add their two frequency factors, cancelling the mode normalization. The un-subtracted ordered expression isThe last term is an infinite vacuum constant, not a finite physical vacuum charge. Choosing the Fock vacuum to have zero Noether charge, equivalently using normal ordering, givesA state containing $n_c$ quanta created by $c^{†}$ and $n_b$ by $b^{†}$ has Noether charge $n_c-n_b$. Thus the two families describe opposite unit Noether charges; with this source's current the $c^{†}$ quantum has positive Noether charge and the $b^{†}$ quantum negative Noether charge. This is an internal Noether charge; identifying it with electric charge requires a specified electromagnetic coupling. The vacuum subtraction of a complex scalar charge leaves all commutators unchanged.^[ref]

Scope of the expansion: a fixed-frequency expansion with time-independent oscillator coefficients is the free-field or asymptotic-field form. For arbitrary interacting $f$, it can be used as canonical oscillator coordinates at a chosen time, rather than as the exact interacting time dependence. For example, $f(s)=m^{2}s+\lambda s^2$ gives $(\square +m^2)\varphi +2\lambda {\varphi }^{\ast }\varphi \varphi =0$, which a generic free plane-wave superposition does not satisfy. The Noether charge and its generator action remain valid in the interacting theory; the symmetry does not depend on using a free Hamiltonian.

For a constant real parameter, ${\partial }_{\mu }(e^{i\alpha }\varphi )=e^{i\alpha }{\partial }_{\mu }\varphi$. The phases therefore cancel separately in the kinetic term and in ${\varphi }^{\ast }\varphi$, leaving any $f({\varphi }^{\ast }\varphi )$ unchanged. This is the global phase symmetry of a complex scalar field, with circle group $U(1)$.

Use infinitesimal variations per unit parameter, $\delta \varphi =i\varphi$, $\delta {\varphi }^{\ast }=-i{\varphi }^{\ast }$. The Lagrangian density is exactly invariant, so no total-derivative correction is needed in Noether's theorem. The Noether current isIt depends only on derivatives of the Lagrangian density with respect to field derivatives, so the derivative-free scalar potential cannot alter its form. Direct differentiation cancels the mixed first-derivative terms:Thus it is conserved on solutions of the Euler-Lagrange equations. Its spatial integral is a conserved Noether charge when the flux through the boundary vanishes. Its overall sign is fixed here by $\delta \varphi =+i\varphi$; reversing the orientation of the phase transformation reverses the current.

Treat $\varphi$ and ${\varphi }^{\ast }$ as independent variables in the Euler-Lagrange equations. Write $s={\varphi }^{\ast }\varphi$, and assume the scalar potential $f(s)$ is real and differentiable. Variation with respect to ${\varphi }^{\ast }$ givesConsequently the equations for the complex scalar field and its complex conjugate areNo polynomial assumption is needed: only the first derivative of the scalar potential enters. For $f(s)=m^{2}s$, both reduce to the free Klein-Gordon equation.

Because the quartic coupling constant has mass dimension $ϵ=4-D$, introduce a renormalization scale $\mu$ and define the dimensionless renormalized coupling $g$ by minimal subtraction scheme:Here ${\lambda }_0$ is the fixed bare coupling constant. Equivalently the tree-level quartic term is $-{\mu }^{ϵ}g{\varphi }^4/4!$, and its one-loop counterterm is $-{\mu }^{ϵ}3g^2{\varphi }^4/(16{\pi }^2ϵ4!)$. This states explicitly the scale convention implicit in solution.

Put $a=3/(16{\pi }^2)$. Differentiating the bare relation at fixed ${\lambda }_0$ givesWrite $\hat{\beta }=-ϵg+b{g}^2+O(g^3)$. The coefficient of $g^2$ in the last identity is $a+b-2a$, so $b=a$. ThereforeThe first term is the engineering term in a dimensionally continued beta function. In four dimensions it disappears, leaving the positive one-loop quartic scalar beta function $3g^2/(16{\pi }^2)$. Thus the renormalized coupling increases with increasing renormalization scale near weak positive coupling. This quantum running is the obstruction to extending the classical scale invariance in solution to the interacting quantum theory.

It is convenient to evaluate the one-particle irreducible vertices using the Euclidean action and then translate the counterterms back to the Minkowski spacetime Lagrangian density. For the tadpole integral, analytic continuation of solution givesIndeed $\Gamma (-1+ϵ/2)=\Gamma (ϵ/2)/(ϵ/2-1)=-2/ϵ+O(1)$. The negative pole is a property of analytic continuation, not the value of a convergent positive integral.

The Euclidean space inverse two-point vertex has the one-loop correctionThe factor $1/2$ is the symmetry factor weight of the tadpole diagram. A mass counterterm $\delta m^2={\lambda }_0{m}^2/(16{\pi }^2ϵ)$ cancels this pole. There is no momentum-dependent pole at this order, and hence no one-loop wave-function renormalization.

For the four-point vertex, the three bubble diagrams giveOne can check the relative minus sign independently from the one-loop effective action: its $\frac{1}{2}\mathrm{Tr}\log (1+G{\lambda }_0{\varphi }^2/2)$ term has a positive term linear in ${\lambda }_0{\varphi }^2$ and a negative term quadratic in it. Thus $\delta \lambda =3{\lambda }_0^2/(16{\pi }^2ϵ)$ cancels the quartic pole.

In the Euclidean action, these two counterterms are $+\delta m^2{\varphi }^2/2+\delta \lambda {\varphi }^4/4!$. Returning to the Minkowski spacetime Lagrangian density givesFor a direct sign check with Feynman rules, the two amputated loop pole insertions are $+i{\lambda }_0{m}^2/(16{\pi }^2ϵ)$ and $+i3{\lambda }_0^2/(16{\pi }^2ϵ)$, while the counterterm insertions are $-i\delta m^2$ and $-i\delta \lambda$.

As usual, the displayed pole residues suppress renormalization scale powers: $[{\lambda }_0]=ϵ$ away from four dimensions. The next part restores these powers explicitly with a dimensionless renormalized coupling. Replacing the tree coupling by its renormalized counterpart inside a one-loop residue changes only higher-loop terms.

The positive denominators in this part are Euclidean. Starting with the Minkowski spacetime Feynman rules, first perform a Wick rotation; a bare Minkowski metric integral without a pole prescription would not equal the displayed positive expression. Work initially with $m^2>0$ and $\Re n>D/2$, where the Euclidean massive loop integral converges.

Use Schwinger parameterization followed by the Gaussian integral:The remaining Gamma function integral givesThe factorial form is for positive integral $n$. Dimensional regularization defines the result elsewhere by analytic continuation.

For the scalar bubble integral, combine the two denominators with a Feynman parameter $x$ and shift the integration variable to $\ell =k-(1-x)p_E$. ThenFor $D=4-ϵ$ this isThe Gamma function recurrence gives $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the parameter integral is $1+O(ϵ)$. ConsequentlyThis ultraviolet divergence is independent of the external momentum. The assumption $m^2>0$ avoids an infrared divergence. In particular the completely massless zero-momentum integral is a scaleless integral and vanishes in dimensional regularization; its ultraviolet and infrared poles must be separated before assigning it the displayed ultraviolet residue.

For incoming four-momenta, the momentum-space Feynman rules in the chosen Minkowski metric convention are:

There is one connected one-loop one-particle irreducible two-point graph: a single quartic vertex with two external lines and one tadpole diagram. Its symmetry factor is $S=2$ from interchange of the two loop half-edges.

There are three connected one-loop one-particle irreducible four-point graphs: two quartic vertices joined by two parallel internal lines, with external partitions $(12|34)$, $(13|24)$ and $(14|23)$. These are the $s$, $t$ and $u$ scattering channels; each bubble diagram has symmetry factor $S=2$. External-branch insertions are excluded by the one-particle irreducible requirement.

For a connected cubic Feynman diagram with $V$ internal vertices, $I$ internal lines and three external lines, $3V=2I+3$. Hence $V$ is odd and the loop order is $L=I-V+1=(V-1)/2$. Through order ${\lambda }^3$, there is one tree-level Feynman diagram at $V=1$ and seven labelled one-loop Feynman diagrams at $V=3$:

For an explicit check, use the propagator notation of solution and define, with $\{i,j,k\}=\{1,2,3\}$,There is no ordering of $j,k$ in this notation. The complete connected correlation function through this order isFor completeness of the classification, a one-loop connected internal graph either has a three-edge cycle, a pair of parallel edges and a bridge, or a self-loop and two bridges. The cubic degree condition forces exactly the external attachments just listed. Direct Wick contraction counting gives 1296 pairings for the triangle and 648 for each of the six other labelled graphs, divided by $3!(3!{)}^3=1296$.

Connected does not mean one-particle irreducible. After amputation of external propagators, only the triangle is one-particle irreducible at this order; the six other loop graphs belong to the requested connected answer. The tadpole diagrams cannot be discarded without an additional subtraction prescription.

Use the notation ${\Delta }_{12}=\Delta (y_1-y_2)$, ${\Delta }_{xz}=\Delta (x-z)$ and ${\int }_{xz}=\int d^{D}x{d}^{D}z$. The order-$\lambda$ term vanishes because it has five fields. At order ${\lambda }^2$, enumerate the eight-field Wick contractions, whose expansion denominator is $2!(3!{)}^2=72$.

If the two external fields contract with each other, the remaining vacuum diagrams are a pair of cubic vertices joined by three lines, with six Wick contractions, and a pair joined by one line with a tadpole diagram at each end, with nine Wick contractions. Define their combined integrated weight byThus $Z/Z_0=1+(-i\lambda {)}^2{V}_2+O({\lambda }^4)$. The numerator isThe three latter terms have respectively 36, 36 and 18 Wick contractions. Together with the six and nine Wick contractions, they exhaust all $7!!=105$ pairings.

Dividing by $Z/Z_0$ cancels the two vacuum diagrams, givingThe first correction is the two-line bubble diagram, with symmetry factor $S=2$. The second is the connected tadpole contribution to a cubic-theory two-point function: both external lines meet one vertex, joined by a single line to a second vertex bearing a tadpole diagram; its symmetry factor is also $S=2$. Interchanging the two integrated vertices is already included in the 36 contractions and gives no extra factor.

The final term is the disconnected Feynman diagram consisting of two separate one-point tadpole diagrams, with coefficient $1/4$. It survives cancellation of vacuum bubbles, since each component touches an external point. Indeed it is precisely the leading $⟨\varphi (y_1)⟩⟨\varphi (y_2)⟩$. It would disappear from the connected two-point correlation function, but the question defines the full correlation function. No normal ordering or tadpole subtraction of the interaction was specified. All terms are depicted in solution, and all spacetime integrals remain unevaluated.

Write $S=S_0+S_I$, where $S_0$ contains the quadratic kinetic term and mass term, and $S_I=-\lambda \int d^{D}x\varphi (x{)}^3/3!$. ThenLet $Z_0=\int D\varphi e^{i{S}_0}$ and let $⟨\cdot {⟩}_0$ denote the unit-normalized free Gaussian vacuum expectation value. The unnormalized numerator ${\mathcal{N}}_N$ is consequentlyThe Minkowski path integral is understood with the usual scalar Feynman propagator pole prescription, as a formal perturbation series. A real cubic potential energy is unbounded below; no nonperturbative stable vacuum is being assumed here.

Set $\Delta (x-z)=⟨T\varphi (x)\varphi (z){⟩}_0$, $K_i(x)=\Delta (y_i-x)$, and $t=\Delta (0)$, with a common regularization understood for coincident points. The zero-order one-point term has one field and vanishes. At first order, Wick's theorem gives three choices for the contraction with the external field, leaving a tadpole diagram:The second-order term contains seven fields, so it vanishes by the vanishing odd correlators of a free Gaussian field. HenceThe coefficient is $3/3!=1/2$, or symmetry factor $S=2$ when each diagram is weighted by $1/S$. No spacetime integral has been evaluated. The one-point tadpole diagram is shown in the first panel of the following figure; its other panels organize solution.

Let brackets denote mass dimension. In natural units, $[x]=-1$ and $[{\partial }_{\mu }]=1$, so a dimensionless action requires $[\mathcal{L}]=D$. The kinetic term fixesThe cubic scalar field theory interaction then requires $[\lambda ]+3[\varphi ]=D$, givingThe coupling constant is dimensionless at this critical dimension of a monomial scalar interaction. For $D<6$ it has positive mass dimension, while for $D>6$ it has negative mass dimension, the usual power counting distinction between super-renormalizable and non-renormalizable cubic interactions.

The Heisenberg picture evolution of the annihilation operator follows from $[H,a_{\mathbf{p}}]=-E_{\mathbf{p}}{a}_{\mathbf{p}}$. With $p^0=E_{\mathbf{p}}$, this gives the mode expansion of a free fieldDefine ${\Delta }^{+}(z)={\int }_{\mathbf{p}}{e}^{-ip\cdot z}/(2E_{\mathbf{p}})$. The time-ordered product then gives the Feynman propagatorThe last expression follows by closing the $p^0$ contour integral below for $z^0>0$ and above for $z^0<0$; the scalar Feynman propagator pole prescription supplies the required positive- and negative-frequency terms.

For $z^0\ne 0$ every term obeys the homogeneous Klein-Gordon equation. At $z^0=0$, the equal-time commutator of the fields vanishes, so the propagator is continuous and there is no derivative of a Dirac delta function. Its first time derivative has the jumpThe second time derivative therefore supplies precisely the source:Thus the defined Feynman propagator is a Green's function with source normalization $-i$; $i{\Delta }_F$ has source normalization $+1$. Keeping this factor avoids confusing the vacuum expectation value convention with conventions which define the propagator after dividing by $i$.^[ref]

The Fock vacuum condition and the creation operator commutator giveEquality of the square-root product and $2E_{\mathbf{p}}$ holds on the support of the Dirac delta function.

The positive-energy mass shell has the Lorentz invariant measureHere both $p^2$ and the four-dimensional volume element are invariant, and positive energy is preserved by a proper orthochronous Lorentz transformation. Consequently the three-dimensional Jacobian determinant is $J=E_{\mathbf{\Lambda }p}/E_{\mathbf{p}}$. Transforming the Dirac delta function yieldsso its factor exactly cancels that of $2E_{\mathbf{\Lambda }p}$. Thus the displayed inner product is invariant under simultaneous transformation of both momenta. This is also why the one-particle completeness measure is $d^{3}p/((2\pi {)}^{3}2E_{\mathbf{p}})$.^[ref]

The normal ordering of both linear maps puts every annihilation operator to the right, soIn particular the stated Hamiltonian has already subtracted the free vacuum energy. For a one-particle state, evaluate the commutators using $[AB,C]=A[B,C]+[A,C]B$:Acting on the Fock vacuum therefore givesThe extra factor $\sqrt{2E_{\mathbf{k}}}$ in the chosen relativistic normalization of a one-particle state is a scalar multiplier and does not change either eigenvalue. Applying the same argument to a multiparticle quantum state gives additive total energy and momentum.

The unit-normalized Fock vacuum is characterized byChoose the relativistic normalization of a one-particle state from the outset:Since any two creation operators commute, any permutation of their order gives exactly the same state vector. Thus these multiparticle quantum states are symmetric under exchange and describe bosons, without a minus sign or any additional exchange phase.

The Hilbert space is the bosonic Fock space, the direct sum of the symmetric $n$-particle Hilbert spaces, including the zero-particle sector. Sharp-momentum kets are distributionally normalized; square-integrable wave packets provide ordinary Hilbert space vectors. For repeated occupation of a discrete mode, a unit-normalized number state uses $(a^{†}{)}^n|0⟩/\sqrt{n!}$. That factorial does not belong in the above labelled continuum-ket convention.

Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$ and $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. Insert the mode expansion of a free field for each quantity into the commutator. Only the mixed creation operator and annihilation operator terms survive. The Dirac delta function in their canonical commutation relation sets the two momenta equal, so the square-root factors multiply to $1/2$:In the second equality, each Fourier transform is the same Dirac delta function, since the integration domain is invariant under $\mathbf{p}\mapsto -\mathbf{p}$. This is the equal-time canonical commutation relation. The analogous calculation gives $[\varphi (\mathbf{x}),\varphi (\mathbf{y})]=[\pi (\mathbf{x}),\pi (\mathbf{y})]=0$, because their two mixed terms have opposite signs.