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In a fixed fermion-parity sector use lattice spacing one and the unitary discrete Fourier transformThe allowed momenta satisfy $e^{ikN}=-P$: they are $k=2\pi (\ell +1/2)/N$ for $P=+1$ and $k=2\pi \ell /N$ for $P=-1$, with $N$ distinct values modulo $2\pi$. The hopping and number sums become ${\sum }_{k}2\cos k{c}_k^{†}{c}_k$ and ${\sum }_k{c}_k^{†}{c}_k$. ThusHere $E$ is the energy of the all-up spin configuration, not generally the ground-state energy. The dispersion has minimum $h-J$ at $k=0$, maximum $h+J$ at $k=\pi$, and positive curvature at its minimum. Since the fermions do not interact, each mode can be occupied zero or one time. A negative-energy occupation lowers the many-body energy, so in the thermodynamic limitThis is a filled Fermi sea containing many fermions; it is not obtained by putting a single particle into the band minimum. For $h>J$ every mode has positive energy and the vacuum is the ground state. Its spin interpretation is complete longitudinal polarization. At $h=J$ the zero-momentum one-particle excitation first becomes gapless.

For an exact finite ring use parity-consistent filling of a free-fermion band. In each momentum grid fill negative-energy levels, check that their particle number has the chosen parity, and if necessary toggle a mode with the smallest $|{ϵ}_k|$. Compare the two sector energies. A zero-energy mode allows degenerate occupations. This restores the physical spin Hilbert space without changing the stated band.

For one coordinate with Hamiltonian $H={\hat{p}}^2/(2m)+V(\hat{x})$, the propagator is $K(x_f,t_f;x_i,t_i)=⟨x_f|e^{-iHT/\hslash }|x_i⟩$, where $T=t_f-t_i>0$. Divide $T$ into $M$ intervals $\delta t$, use the Trotter product formula to separate kinetic and potential evolution, and insert complete position and momentum bases at each slice. The momentum integral is Gaussian:Multiplying the short-time kernels gives the time-sliced configuration-space path integral,with $x_0=x_i$, $x_M=x_f$. Continuum notation abbreviates this limit as ${\int }_{x_i}^{x_f}\mathcal{D}x{e}^{iS[x]/\hslash }$, where $S=\int dt[m{\dot{x}}^2/2-V(x)]$. The measure includes the displayed normalization; it is not an unweighted count of paths. A small damping or analytic-continuation prescription defines the oscillatory integrals. The paths contribute complex amplitudes, so alternative histories interfere before probabilities are calculated. This is the central difference from a classical statistical mixture of trajectories.^[ref]

When actions vary rapidly relative to $\hslash$, stationary phase selects classical paths. Put $x=x_{\mathrm{cl}}+\eta$, with $\eta$ zero at the endpoints. Vanishing first variation gives $m{\ddot{x}}_{\mathrm{cl}}+V^{\prime }(x_{\mathrm{cl}})=0$. The quadratic term isThe resulting Gaussian fluctuation determinant is the semiclassical propagator prefactor. In one dimension, away from caustics,The branch convention makes $\nu =0$ for a direct free-particle path before any caustic. The Maslov index tracks the phase changes as fluctuations pass through a focal singularity. In several dimensions replace the endpoint derivative by the Van Vleck determinant. The approximation sums all relevant classical paths; real-time action need only be stationary, not a minimum. It fails in this simple form at a caustic or coalescing saddles, where a uniform approximation is needed. Quadratic actions have no higher fluctuation terms, making their Gaussian evaluation exact. For example $S_{\mathrm{cl}}=m(x_f-x_i{)}^2/(2T)$ gives the exact free-particle kernel $\sqrt{m/(2\pi i\hslash T)}{e}^{i{S}_{\mathrm{cl}}/\hslash }$. Tunnelling involves classically forbidden regions and can be described by imaginary-time saddles rather than real classical trajectories.

Quantum statistical mechanics follows by Wick rotation $t=-i\tau$. The imaginary-time evolution operator is $e^{-\tau H/\hslash }$, with Euclidean actionTaking a thermal trace identifies and integrates the endpoint positions:This thermal trace as a closed Euclidean path integral explains the periodic imaginary-time direction. A density-matrix element instead retains open endpoints. For many identical particles the trace also sums over endpoint permutations, with a plus sign for bosons and permutation parity for fermions. In coherent-state language this becomes periodic bosonic fields and antiperiodic Grassmann fields, with corresponding Matsubara frequencies.

The Euclidean weight resembles a classical Boltzmann weight. A time-sliced local quantum field in $d$ spatial dimensions becomes a classical-looking field model in $d+1$ dimensions, with anisotropic couplings and extra extent $\beta \hslash$. Thus zero-temperature quantum criticality can be treated using an infinitely extended imaginary-time direction, while nonzero temperature imposes a finite thickness. The dynamical scaling of time need not equal the spatial scaling: the dynamical critical exponent specifies their relationship. This quantum-to-classical imaginary-time mapping requires qualification when weights have signs or imaginary terms; such a functional integral is not automatically an ordinary positive classical probability distribution. Even with a positive weight, the mapping concerns equilibrium statistics, not literal dissipative real-time trajectories. For a free particle the Euclidean kernel is a diffusion kernel with diffusion coefficient $\hslash /(2m)$, illustrating the mathematical correspondence.

The classical high-temperature limit is a further approximation, rather than simply another name for imaginary time. If quantum level spacings and exchange effects are negligible, a one-particle partition function in $d$ dimensions becomesFor indistinguishable particles there is a factor $1/N!$ in the dilute classical gas, and exchange is negligible when the number density times the thermal de Broglie wavelength to power $d$ is small. For a harmonic oscillator, the exact quantum result $Z=[2\sinh (\beta \hslash \omega /2){]}^{-1}$ tends to $1/(\beta \hslash \omega )$ as $\beta \hslash \omega \to 0$, recovering the classical phase-space integral. stationary phase explains classical mechanics; the short imaginary-time, weak-exchange limit explains classical statistical mechanics. These are related but distinct limits of the same quantum formulation.

Take a uniform nearest-neighbour harmonic chain with displacement $u_j$, conjugate momentum $p_j$, mass $m_0$, spring stiffness $\kappa$, lattice spacing $a$ and periodic length $\ell =Na$. These specify a standard model for the requested notes, not additional numerical exam data. Its Lagrangian and Hamiltonian areVariation gives $m_0{\ddot{u}}_j=\kappa (u_{j+1}-2u_j+u_{j-1})$. For a mode $e^{i(kja-\omega t)}$, periodicity requires $k=2\pi n/\ell$ andThe lattice makes $k$ periodic modulo $2\pi /a$, so choose one Brillouin zone. At long wavelength ${\omega }_k\simeq v|k|$, where $v=a\sqrt{\kappa /m_0}$ is the sound speed. The continuum limit of a harmonic chain expands $u_{j+1}-u_j\simeq a{\partial }_{x}u$ and replaces ${\sum }_j$ by $a^{-1}\int dx$. Defining mass density $\rho =m_0/a$ and one-dimensional elastic coefficient $K=\kappa a$ givesThe momentum density is $\Pi =\rho {\partial }_{t}u$, and the Hamiltonian density is ${\Pi }^2/(2\rho )+K({\partial }_{x}u{)}^2/2$. Translational invariance of $u$ makes total momentum conserved. The continuum approximation describes acoustic wavelengths large compared with $a$; it is not a justification for extending the lattice dispersion and zero-point sums to arbitrarily large wavevector.^[ref]

canonical quantization promotes the coordinates and momenta to operators with $[u_j,p_l]=i\hslash {\delta }_{jl}$ and the other equal-time commutators zero. In the continuum, $[u(x),\Pi (x^{\prime })]=i\hslash {\delta }_{\mathrm{per}}(x-x^{\prime })$. The nonzero normal modes can be represented by independent phonon operators obeying $[b_k,b_q^{†}]={\delta }_{kq}$. A useful lattice expansion iswhere $X=N^{-1}{\sum }_j{u}_j$ is the centre-of-mass displacement. The corresponding momentum isIncluding the centre-of-mass term restores the full site commutator: the nonzero-mode sum alone gives $i\hslash ({\delta }_{jl}-1/N)$. Substitution diagonalizes the Hamiltonian,A phonon is one quantum of a collective normal mode, carrying energy $\hslash {\omega }_k$ and lattice momentum $\hslash k$ modulo a reciprocal-lattice vector. Reality of $u_j$ relates its Fourier coordinates by $u_k^{†}=u_{-k}$; it does not remove the independent oppositely travelling phonon modes. Using one complete Brillouin zone avoids double-counting real standing-wave coordinates.^[ref]

The free centre of mass has no restoring force and must not be treated as an oscillator of zero frequency. Fixing it, weakly pinning it, or specifying a finite centre-of-mass volume defines the corresponding zero-mode factor. With a fixed centre of mass, the internal thermal partition function and occupations areThe internal energy is ${\sum }_{k\ne 0}\hslash {\omega }_k(⟨n_k⟩+1/2)$; its high-temperature heat capacity approaches $(N-1)k_B$. For a large one-dimensional chain at temperatures below the lattice cutoff but above its finite-size mode spacing,This one-dimensional acoustic phonon heat capacity shows that quantization turns classical normal waves into bosonic excitations with a linear low-temperature heat capacity in one dimension. The Euclidean field action $S_E=\frac{1}{2}{\int }_0^{\beta \hslash }d\tau dx[\rho ({\partial }_{\tau }u{)}^2+K({\partial }_{x}u{)}^2]$ is Gaussian, with periodic imaginary-time boundary conditions; its Fourier modes reproduce these oscillators. Absolute displacement correlations require the zero-mode convention, and the lattice regulates short-distance fluctuations.

Use the contact-interaction Bose-gas field integral for a homogeneous three-dimensional weakly interacting Bose gas with repulsive short-range interaction as the standard model. Let particle mass be $m_b$, scattering length $a_s>0$, physical coupling $g=4\pi {\hslash }^2{a}_s/m_b$ and gas parameter $n{a}_s^3\ll 1$. These are symbolic model definitions, not supplied numerical parameters. The grand-canonical coherent-state integral has periodic fields and actionThe contact term is the normal symbol of $g{\hat{\psi }}^{†2}{\hat{\psi }}^2/2$. In three dimensions a contact interaction needs a regulator and coupling matched to the measured scattering length; the displayed $g$ is the low-energy physical coupling. This field integral is useful because an ordered saddle and its quantum/thermal fluctuations follow from the same action.

Stationarity with respect to $\bar{\psi }$ gives the imaginary-time Gross–Pitaevskii equation. A uniform saddle exists at $\mu >0$ with ${\psi }_0=\sqrt{n_0}{e}^{i{\theta }_0}$ and condensate density $n_0=\mu /g$. Its arbitrary phase reflects global $U(1)$ symmetry; a selected condensate fixes a phase in the thermodynamic symmetry-breaking description. Write $\psi =\sqrt{n_0}+\eta$ after choosing that phase. Linear terms vanish, and the quadratic part isWith ${\epsilon }_k={\hslash }^2{k}^2/(2m_b)$ and the Fourier convention $e^{i\mathbf{k}\cdot \mathbf{x}-i{\omega }_n\tau }$, the paired quadratic kernel isThe one-half in the Nambu spinor action removes pair double counting. Analytic continuation of the inverse kernel gives the Bogoliubov dispersionAt high momentum it approaches the free-particle kinetic energy plus its mean-field shift. The crossover scale is the healing length $\xi =\hslash /\sqrt{2m_{b}g{n}_0}$. Gaussian integration adds a fluctuation determinant to the saddle free energy. Expressed using the physical three-dimensional coupling, its zero-temperature integrand can be regularized as $\frac{1}{2}[E_k-{\epsilon }_k-\mu +{\mu }^2/(2{\epsilon }_k)]$; the subtraction follows from renormalization of a three-dimensional dilute-gas contact coupling, not discarding an infinite answer arbitrarily. The thermal contribution is $k_{B}T\int d^{3}k/(2\pi {)}^3\log (1-e^{-\beta E_k})$. A calculation beyond mean field must keep that matching and its thermodynamic derivatives consistent.

The density-phase action of a Bose gas makes the physical content clear, $\psi =\sqrt{n_0+\delta n}{e}^{i\theta }$. Up to a periodic density total derivative, the long-wavelength quadratic action containsnumber-phase conjugacy pairs the density fluctuation and phase in the quadratic density-phase action. Dropping the density-gradient term only in the long-wave limit and completing the density Gaussian square gives the Bose-gas phase-only actionThis has the same sound pole. A term $i\hslash n_0{\partial }_{\tau }\theta$ is also present before restricting to the smooth zero-winding sector; although locally a total derivative, it cannot in general be ignored in topologically nontrivial sectors. The density and phase variables describe the same coupled Bogoliubov branch, not a second independent massive particle simply because the density has a static restoring cost.^[ref]

A phase gradient carries superflow ${\mathbf{v}}_s=(\hslash /m_b)\nabla \theta$. Its energy cost defines the superfluid number density. At leading mean-field order the stiffness uses $n_0$, but condensate density and superfluid number density are different observables: for a Galilean-invariant continuum gas at zero temperature the latter is the full density even when interactions deplete the condensate. Bogoliubov coefficients give $v_k^2=[({\epsilon }_k+g{n}_0)/E_k-1]/2$ and a small zero-temperature depletion of relative order $\sqrt{n{a}_s^3}$ in three dimensions. Large depletion or strong long-wavelength phase fluctuations would invalidate the small-fluctuation saddle approximation.

The Landau criterion for creating excitations is $v_L={\inf }_{k>0}{E}_k/(\hslash k)=c_s$ for this homogeneous dispersion. Thus repulsion produces a linear sound mode and a positive critical velocity from the Landau criterion, whereas an ideal quadratic gas has zero value of this criterion. Vortices, obstacles and nucleation can reduce an actual flow threshold below $c_s$; the criterion alone is not a complete theory of metastable flow. At finite temperature thermal quasiparticles reduce the superfluid number density. Dimensionality and finite size also matter: the uniform three-dimensional weak-depletion condensate expansion must not be transferred unchanged to a one- or two-dimensional thermodynamic system. The field integral provides systematic corrections only within its dilute, stable and sufficiently weak-fluctuation regime.

Continue the free-particle propagator to $t=-i\mathcal{T}$ and replace its coordinate mass by the angular inertia $I$. The Euclidean free-particle bridge amplitude at coincident endpoints isThis normalization includes the time-sliced Gaussian measure, not just a formal determinant with an unspecified multiplicative constant. Set $A=2{\pi }^{2}I/({\hslash }^2\beta )>0$. The correct Poisson summation formula is ${\sum }_{m}F(m)={\sum }_n{\int }_{-\infty }^{\infty }dxF(x)e^{-2\pi inx}$; the source PDF omits the integral sign before its $d\theta$. For $F(x)=e^{-A{x}^2+i2\pi fx}$ the Gaussian Fourier integral givesThe overall prefactor cancels exactly: $2\pi Z_0\sqrt{\pi /A}=1$, and ${\pi }^2/A=\beta {\hslash }^2/(2I)$. Hencein agreement with the angular-momentum trace. Pairing $m$ and $-m$ makes the winding expression real, and the momentum expression shows positivity. Both are periodic in flux and invariant under $f\mapsto -f$. At high temperature the zero-winding term gives $Z\simeq \sqrt{2\pi I/(\beta {\hslash }^2)}$, with exponentially small flux dependence; at low temperature the nearest integer angular-momentum state dominates, with two equal lowest terms at half-integer flux. These independent limits also check the normalization and the flux phase.

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.

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 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 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 ${\mathcal{V}}_4=i{\Gamma }^{(4)}$ denote the amputated Feynman vertex, with tree term $v_4=-i\lambda$. A one-loop one-particle-irreducible Feynman diagram has $V_3+2V_4=4$, allowing three families: two quartic vertices, two cubic vertices and one quartic vertex, or four cubic vertices. DefineUsing $I_2,I_3$ from part c and writing $P_{ij}=p_i+p_j$, the complete one-loop result isIn each term of the middle sum, $k,l$ are the two complementary indices. Their order does not change the integral after a loop-momentum redefinition.

The first line contains three bubble channels, one for each partition of the four labelled external legs into two pairs. Each has $S=2$ from interchanging its two identical internal lines. In the second line, the quartic vertex carries a chosen pair of external legs, giving $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ mixed triangles. Each has $S=1$. In the last line, cyclic orderings of four labelled external legs are identified under rotation and reversal, leaving $(4-1)!/2=3$ boxes, again with $S=1$ each. The figure in part b shows one representative of each family and its multiplicity. External-leg corrections and tadpoles connected by bridges are excluded because ${\Gamma }^{(4)}$ is a one-particle-irreducible vertex. All integrals are left unevaluated, as appropriate here.

A free contraction must connect one $\varphi$ and one ${\varphi }^{\ast }$ field. In momentum space,Represent the nonzero propagator by a directed line so the two conjugate ends remain distinguishable; an undirected line alone would incorrectly allow same-charge contractions. Choose one consistent arrow convention, for example pointing from the starred end to the unstarred end.

The interaction contains two identical $\varphi$ fields and two identical ${\varphi }^{\ast }$ fields. Their $2!2!$ assignments cancel the coefficient $1/4$, givingThe vertex has two incoming and two outgoing arrows. Include $(2\pi {)}^d{\delta }^d(\sum p_i)$ at every vertex, integrate $d^{d}k/(2\pi {)}^d$ for each independent loop, and divide by the oriented graph's symmetry factor. There is no fermion-loop sign.

The propagator carries a real-species label rather than a charge-flow arrow:Differentiating the interaction with respect to its four external real fields gives the compact tensor ruleThus four ${\varphi }_1$ legs or four ${\varphi }_2$ legs give $-3i{\lambda }_0/2$, while two of each give $-i{\lambda }_0/2$. Vertices with an odd number of either species vanish. For example, $4!({\lambda }_0/16)=3{\lambda }_0/2$, whereas $2!2!(2{\lambda }_0/16)={\lambda }_0/2$. These factors already account for assignments at one vertex; graph symmetry factors must still be included separately. Momentum conservation and loop measures are the same as in part a.

Use renormalized propagators of mass $m$ and tree vertex $-ig{\mu }^{ϵ}$. The interaction counterterms have Feynman factors $-i{\delta }_{m^2}$ for a two-leg mass insertion and $-i{\mu }^{ϵ}g{\delta }_g$ for a four-leg insertion. At one loop the two-point one-particle-irreducible vertex isThe directed tadpole has $S=1$: interchanging its two ends would reverse the field labels, so the real-scalar tadpole's factor $1/2$ does not apply. It is independent of external momentum, so no wave-function counterterm is required at this order.

For the four-point vertex, label the two unstarred incoming fields by $1,2$ and the starred ones by $3,4$, with all momenta incoming. The tree contact, the three oriented bubbles and the contact counterterm giveThe first bubble has two equivalent arrows along the same direction between its vertices, giving $S=2$. In the other two channels the arrows oppose one another, and interchanging the lines does not preserve their oriented attachments, giving $S=1$. This is the oriented complex-scalar quartic graph factors rule.

The mass tadpole is order $g{m}^2$, while the four-point loop is order $g^2$. In the multiplicative coupling convention of the source,The distinction between ${\delta }_g$ and the whole additive four-leg coefficient $g{\delta }_g$ matters for the requested order counting.

The canonical momenta are ${\pi }_k={\dot{\varphi }}_k$, soDifferent species commute at equal times, so these products have no ordering ambiguity and $Q$ is Hermitian. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. Substituting the mode expansion of a free field and using the spatial Fourier transform identity givesSubtract the expression with species interchanged. The two-annihilator terms cancel after $\mathbf{p}\mapsto -\mathbf{p}$, as do the two-creator terms. Commuting different species in the remaining terms yieldsThere is no infinite vacuum contribution to the Noether charge to subtract: the contractions that would generate a species-diagonal constant are absent.

The bosonic canonical commutation relations giveDefine the creation operatorsTheir commutators with $Q$ are $[Q,c_{\pm }^{†}]=\pm c_{\pm }^{†}$. Since $Q|0⟩=0$, the one-particle states haveThe fixed-momentum states are delta-function normalized; a square-integrable wave packet of either species is a normalizable state with the same Noether charge eigenvalue. In this basis $Q={\int }_{\mathbf{p}}(c_{+}^{†}{c}_{+}-c_{-}^{†}{c}_{-})$, so it measures a difference of occupation numbers. The equal-time identity $i[Q,{\varphi }_1]={\varphi }_2$, $i[Q,{\varphi }_2]=-{\varphi }_1$ independently checks the sign against a.

Expand the normalized Euclidean path integral with weight $e^{-S_E}$ and apply Wick's theorem. Use labelled external momenta and a fixed Fourier transform convention throughout. A line carrying momentum $q$ contributes $(q^2+m_0^2{)}^{-1}$. Each $2k$-valent interaction vertex contributes $-{\lambda }_{2k}=-{\mu }^{d_k}{g}_{2k}$: its $(2k)!$ possible contractions cancel the factorial in the Lagrangian density. A vertex also carries $(2\pi {)}^D{\delta }^{(D)}(\sum q)$ with all incident momenta incoming. Integrate each independent loop momentum with $d^{D}q/(2\pi {)}^D$ and divide by the Feynman-diagram symmetry factor for fixed external labels. There are no Minkowski factors of $i$.

For the full $N$-point Green's function, include every graph in which all vertices belong to components attached to external insertions. In particular, disconnected components carrying external insertions must be included, and the zeroth-order result is the sum of free pairings. Pure vacuum bubbles cancel against the denominator. The connected Green's function restricts to connected graphs, while a one-particle-irreducible vertex imposes the stronger restriction that removing one internal line cannot disconnect the graph. These are different objects. Odd $N$ vanishes in the symmetric vacuum because every interaction has even valence. At fixed perturbative order, multiply by the appropriate power of the coupling and sum all allowed graph topologies and external attachments.

If the overall momentum-conservation delta function is removed by convention, report the remaining function as its coefficient. The ordinary unamputated Green's function includes external scalar propagators; dividing them out defines an amputated function. The definition in c performs this amputation but does not, by itself, impose the usual one-particle-irreducibility condition.

A phi-fourth theory interaction has mass dimension $4-D$, so its bare coefficient must contain a renormalization scale factor ${\mu }^{4-D}=({\mu }^2{)}^{2-D/2}$ when expressed through a dimensionless renormalized coupling. The tree-level four-leg interaction vertex is linear in that coupling. At the next order, two quartic vertices form the three one-loop bubble diagram graphs shown in c, with external pairings $12|34$, $13|24$ and $14|23$.

For each pairing there are two identical internal lines and Feynman-diagram symmetry factor 2. The loop integral has the large-momentum behaviourIt is logarithmically divergent in four dimensions. Dimensional regularization represents that local divergence by a simple pole proportional to $1/(D-4)$. Its residue is independent of external momenta, so a local quartic counterterm cancels it. The three graphs sum to one constant coefficient; this argument establishes its structure without requiring its value.

The relation consequently has the formThe sign of $C$ incorporates the chosen pole denominator and vertex convention. In particular, a negative $C$ here corresponds to a positive coefficient of $1/(4-D)$ in the bare coupling. Finite pieces depend on the renormalization scheme; a pure-pole relation is natural in minimal subtraction scheme. One-loop two-point tadpole diagrams contribute mass renormalization, but not the momentum-dependent field renormalization that could change this coupling relation at order $g^2$. The four-point object relevant here is the amputated one-particle-irreducible vertex; external-leg self-energy insertions are dealt with separately.

The displayed oscillator expansion and the displayed action require two separate corrections to describe one conventional positive-energy theory. First compute what the oscillator algebra literally implies. Assume the two oscillator species commute with each other and annihilate the vacuum. Same-charge contractions vanish, soFor the two mixed contractions the printed phases give the same kernel, with $z=x-y$:Thus $⟨T\varphi (x){\varphi }^{\ast }(y){⟩}_0=⟨T{\varphi }^{\ast }(x)\varphi (y){⟩}_0=C_{\mathrm{lit}}(z)$. In $D$ dimensions replace $d^{3}p$ and $d^{4}p$ by $d^{D-1}p$ and $d^{D}p$; the source's explicit three-dimensional spatial measure is the $D=4$ specialization. The covariant integral is manifestly Lorentz invariant. Withwe have $C_{\mathrm{lit}}(z)=D_F(z{)}^{\ast }$. This reversed-frequency scalar contraction has the opposite pole prescription to the usual positive-energy Feynman propagator. For a single zero-momentum mode at positive time separation, the printed contraction is $e^{+imt}/(2m)$, while $D_F$ is $e^{-imt}/(2m)$: this is a concrete phase counterexample, rather than an immaterial change of integration variable.