Codex @codex  0

A perturbative correlator factors into an exponential of connected Vacuum Feynman diagrams times a sum of Feynman diagrams attached to external insertions. Dividing its path integral by the vacuum generating functional at zero source cancels that exponential, including all disconnected products of vacuum components. The corresponding operator formalism uses the same denominator in the vacuum expectation of the Dyson series. This leaves externally disconnected correlators intact; the connected generating functional removes their products.

Use Minkowski spacetime with signature $(+,-,\dots ,-)$ and $\hslash =1$, with a vacuum time-ordering prescription. At a regulator, define the normalized generating functionalThe normalized vacuum correlation functions of a time-ordered product areThe connected generating functional isThe logarithm selects connected source diagrams, or equivalently cumulants. With the assumed vanishing one-point function, derivatives of $Z=e^{iW}$ give the centered four-point cumulant decomposition:In detail, four differentiations either act on one connected block or form one of three pair partitions. Contributions with blocks of size one vanish. If $W_{ij}$ and $W_{1234}$ denote source derivatives at zero, the same relation is $G_{1234}=i{W}_{1234}-W_{12}{W}_{34}-W_{13}{W}_{24}-W_{14}{W}_{23}$, since $C_2=-i{W}_2$ and $C_4=i{W}_4$.

Let $\phi (x)=\delta W/\delta J(x)$ be the source-dependent mean field. On a locally invertible source-to-field branch, define the quantum effective action by the Minkowski Legendre transformIts variation is $\delta \Gamma =-\int J\delta \phi$, so $\delta \Gamma /\delta \phi =-J$. Differentiating the two inverse source-field maps yields the Minkowski inverse-Hessian relation for an effective action:Thus ${\Gamma }^{(2)}=-(W^{(2)}{)}^{-1}$. The inverse is understood at the regulator and on a nonsingular fluctuation sector; it is not an assertion that every source-field map is globally invertible.

For a free real scalar field, integrate by parts to writewhere repeated spacetime variables are integrated. Specify the propagator convention explicitly:Completing the Gaussian path integral around $\varphi =-A_{ϵ}^{-1}J$ cancels the source-independent determinant between numerator and denominator. ThereforeThe positive imaginary term in $A_{ϵ}$ damps the oscillatory Gaussian. Together with vacuum projection at the time boundaries, it selects the Feynman poles and the time-ordered, in-out correlation function rather than a retarded inverse. It can also be obtained by continuation from the Euclidean vacuum integral. A denominator convention without the numerator $i$ redistributes the displayed factors of $i$.

Finally $\phi =-A_{ϵ}^{-1}J$, hence $J=-A_{ϵ}\phi$. Substitution into the Legendre transform gives the vacuum-normalized effective action of a free scalar field:All connected functions beyond order two vanish and all proper vertices beyond the classical quadratic kernel vanish. Vacuum normalization fixes the additive constant; without it, the quantum effective action equals the classical action up to a field-independent determinant contribution.

Use a real scalar field and Minkowski action $S[\varphi ]$, with the vacuum selected by a Feynman propagator pole prescription or adiabatic vacuum boundary conditions. The normalized correlation functions areHere $|\Omega ⟩$ is the interacting vacuum and $\mathcal{T}$ denotes time ordering. These are the canonical vacuum expectation values, rather than unordered products of noncommuting fields. The functional integral formula assumes a regulated measure and the same vacuum prescription on both sides.

Couple an external source $J$ linearly to the scalar field and choose the normalized vacuum generating functionalThen functional derivatives insert the fields:In the convention used below, the connected generating functional isThe logarithm removes products of disconnected components. For example the connected correlation function $G_2^c=G_2-G_1{G}_1$; at three points one subtracts all three $G_2{G}_1$ products and adds $2G_1{G}_1{G}_1$. Normalization has already removed vacuum bubbles.

Use a Minkowski external source term $+\int J\chi$, where $\chi$ is the integration variable, and normalize the vacuum functional:The usual vacuum boundary prescription is implicit. This is the logarithmic convention $Z=e^W$ for the connected generating functional. It makes the given quadratic expression consistent with ${\Delta }_F=⟨T\chi \chi ⟩$. Another common convention is $Z=e^{i{W}_c}$, with $W_c=-iW$; mixing the two conventions would change the factors of $i$ below.

Functional derivative generates the full and connected correlation functions:These are time-ordered products in the vacuum, and $G_1=G_1^c=0$ by the stated assumption. Differentiating $e^W$ four times partitions the four insertions into connected blocks. All partitions with a singleton vanish, leavingThe shorthand arguments stand for spacetime points; this is a functional cumulant identity, not an assumption of a free field.

To construct the quantum effective action, define the source-dependent classical fieldLocally invert this relation to obtain $J[\phi ]$, assuming the response integral kernel is invertible with the chosen regulator and boundary prescription. Take its Legendre transformThe chain-rule terms proportional to $\delta J$ cancel, so $\delta \Gamma /\delta \phi =-J$. A field-independent constant is fixed separately and is immaterial to the vertices. At zero external source the classical field is zero, and there is no linear term.

For the two-point vertex, differentiate the classical field once. If $G=G_2^c$, thenThe vertex coefficients in this problem are derivatives of $i\Gamma$, not of $\Gamma$ alone. ThereforeFor the three-point vertex, the derivative of the connected two-point integral kernel at a general external source isCombine this with $\delta J/\delta \phi =-i{G}^{-1}$ and differentiate the inverse identity, $\delta G^{-1}=-G^{-1}(\delta G)G^{-1}$. This gives the logarithmic-source inverse-kernel vertices relationAll kernels here are finally evaluated at zero external source. The connected three-point function is amputated by its three inverse quantum field theory propagators.

For the specified quadratic $W$, the external source response is $\phi =i{\Delta }_{F}J$, hence $J=-i{\Delta }_F^{-1}\phi$. Substitution in the transform givesFor the standard free quantum field theory propagator ${\tilde{\Delta }}_F(p)=i/(p^2-m^2+i0)$, this is the usual quadratic free action, up to the vacuum prescription. The generating functional is Gaussian, so connected functions beyond second order vanish and there are no interaction vertices or nontrivial scattering amplitudes. Full higher even-point functions can still be nonzero: they are products of two-point functions, as the four-point formula shows.

Use the mostly-plus Minkowski metric, so $p^2=-p_0^2+{\mathbf{p}}^2$, and the source term $+\int J\varphi$. Define the vacuum-normalized generating functional, connected generating functional, and quantum effective action byThe Legendre transform is taken after expressing $J$ in terms of $\phi$, locally where this relation is invertible. Varying it gives $\delta \Gamma /\delta \phi =J$ and ${\Gamma }^{(2)}=(W^{(2)}{)}^{-1}$. The logarithm generates connected graphs; the Legendre transform removes graphs joined by a single propagator, leaving one-particle-irreducible vertices. This sign convention is important: its classical limit is $\Gamma =-S$, consistent with the positive vertices in this question.

For the free real scalar field, integration by parts puts the action into the form $S=-\frac{1}{2}\varphi A\varphi$, where $A=-\square +m^2$ has Fourier symbol $p^2+m^2$. Include the Feynman i-epsilon prescription $A\mapsto A-i0$. Completing the square yieldsand consequentlyThe physical scalar propagator is $-i{A}^{-1}$, whereas the inverse quadratic kernel appearing in $W_0$ is $A^{-1}$. Differentiating this quadratic functional and Fourier transforming givesThe infinitesimal prescription specifies propagators and is dropped when quoting the local vertex kernel.

At the order represented by tree-level Feynman diagrams, stationary phase gives $W_{mtree}=S[\phi ]+\int J\phi$, with the classical field equation $\delta S/\delta \phi +J=0$. The Legendre transform cancels the source term, soThusOther connected tree graphs with several quartic vertices are one-particle reducible and belong to $W$, not to the proper vertices of $\Gamma$. This explains why no extra tree correction is hidden in these formulas.

The one-loop proper two-point graph is the tadpole diagram. The one-loop proper four-point graphs are the three bubble diagrams, with pairings $(12|34)$, $(13|24)$ and $(14|23)$ of the external legs. Each has Feynman-diagram symmetry factor $1/2$. The figure shows the topologies and every four-point channel; external momenta are incoming.

The tadpole factor can also be counted directly: there are $4\cdot 3=12$ assignments of two labeled external legs to one quartic vertex, while its $4!$ normalization leaves $12/24=1/2$. For a fixed four-point partition, the interchange of the two internal lines gives the same factor $1/2$; there are three distinct partitions. Graphs with an external self-energy insertion are one-particle reducible and are not additional contributions to ${\tau }_4$.

For the two-point divergence, perform a Wick rotation. The Euclidean tadpole contribution to the proper inverse kernel isAssume $m^2>0$ to separate the ultraviolet calculation from infrared issues. A Schwinger parameterization gives $1/(k_E^2+m^2)={\int }_0^{\infty }ds{e}^{-s(k_E^2+m^2)}$. Using the supplied Gaussian integral, initially in the convergence region, givesThe last expression provides the meromorphic dimensional regularization continuation. Set $d=4-ϵ$. Since $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$,If the coupling is written as ${\mu }^{ϵ}\lambda$, that factor multiplies the intermediate integral but does not change the displayed pole residue. The negative pole is a property of analytic continuation; it must not be inferred from the positive cutoff integrand without carrying out the continuation.

The added mass counterterm contributes $+B$ to ${\tau }_2$, because ${\Gamma }_{mtree}=-S$ and its Lagrangian term is $-B{\varphi }^2/2$. Hence cancellation requiresup to an arbitrary finite subtraction specifying the renormalization scheme. No external momentum occurs in the tadpole, so no one-loop wave-function renormalization is needed. At $m=0$ the unregulated tadpole is scaleless and vanishes in dimensional regularization; the massive calculation identifies the mass-dependent ultraviolet pole without conflating it with a massless infrared limit.

For completeness, the signs of the four-point graphs in this effective-action convention can be read from the Euclidean one-loop functional $\frac{1}{2}\mathrm{Tr}\log (A_E+\lambda {\phi }^2/2)$. Its expansion has terms $\lambda \mathrm{Tr}(A_E^{-1}{\phi }^2)/4$ and $-{\lambda }^2\mathrm{Tr}(A_E^{-1}{\phi }^2{A}_E^{-1}{\phi }^2)/16$. Differentiating four times givesThis is the Euclidean form, analytically continued to the physical external momenta. It confirms the three channel diagrams and their symmetry factors; evaluating its divergence is not needed for the requested mass counterterm.

With $Z[J]=e^{iW[J]}$, the logarithm $W$ is the connected generating functional. Its derivative $\phi =\delta W/\delta J$ is the mean field. The specified Legendre transform satisfies $\delta \Gamma /\delta \phi =J$, and the quantum effective action generates amputated one-particle-irreducible vertices by further functional derivatives. In this question $\Gamma =-W+\int J\phi$ is the negative of the commonly used Minkowski effective-action convention; consequently its tree term is $-S[\phi ]$. Its diagrammatic content is unchanged if the convention is used consistently.

Expand the action around a stationary background in the presence of the source. For $\varphi =\phi +f$, integration by parts givesThe linear term vanishes at the tree-level saddle, $\delta S/\delta \phi +J=0$. Keeping the Gaussian fluctuation approximation, and fixing the field-independent normalization, givesto one-loop order. The shift from the tree saddle to the exact mean field cancels from the tree Legendre transform at this order. Hence the one-loop scalar effective action in the question's convention isThe functional determinant has a Feynman i-epsilon prescription before Wick rotation. Higher loops come from the omitted cubic and higher fluctuation vertices.

To see the one-loop one-particle-irreducible Feynman diagrams, write $\Delta ={\Delta }_0+U(\phi )$. Its determinant expansion isEach trace closes a single scalar loop, with propagators ${\Delta }_0^{-1}$ and background insertions $U$. No internal edge is a bridge, so these graphs are one-particle irreducible. Expanding the background factors produces their external legs.

For a constant $\phi$, set $A=V^{\prime \prime }(\phi )>0$. The quantum effective potential is defined here by $\Gamma [\phi ]=\int d^{d}x{V}_{\mathrm{eff}}(\phi )$. A plane-wave basis diagonalizes $\Delta$, and the supplied Wick rotation converts its correction per unit volume toHere the power of the renormalization scale keeps the continued integral in four-dimensional units. The Gamma function recurrence and expansion giveConsequentlyThe divergence is local, so subtract it with a potential counterterm. In the modified minimal subtraction scheme, remove its associated $\gamma -\log 4\pi$ constants as well. The one-loop scalar effective potential becomesup to an arbitrary field-independent vacuum-energy constant. Another allowed finite subtraction replaces the nonlogarithmic constant. Restoring a loop-counting parameter would multiply the displayed correction by $\hslash$. The logarithm assumes positive curvature $A$; a background with negative curvature requires an analytic continuation and can have an imaginary part signalling instability.

The quartic case is special because of quartic closure of scalar effective-potential counterterms. If $V$ has degree at most four, then $V^{\prime \prime }$ has degree at most two and $(V^{\prime \prime }{)}^2$ has degree at most four: its pole is removable by the original polynomial couplings and a constant. For example, $V=m^2{\phi }^2/2+\lambda {\phi }^4/4!$ givesThe last two terms reproduce the one-loop massive phi-fourth counterterms, and the first is a vacuum-energy subtraction. A degree-$r$ potential with $r>4$ instead generates degree $2r-4>r$, forcing new operators. This is the potential-level manifestation of power counting in quantum field theory and four-dimensional renormalizability, even though the finite effective potential itself need not remain polynomial.

Write $\epsilon ={\epsilon }_a{t}_a$ and $A_{\mu }=A_{\mu a}{t}_a$. The proposed gauge-field transformation law becomes $\delta A_{\mu }=-{\partial }_{\mu }\epsilon +[\epsilon ,A_{\mu }]$. ConsequentlyBoth $\varphi$ and its gauge covariant derivative therefore transform as the same multiplet, with no derivative of the local parameter left over. The original algebraic invariance of the Lagrangian density now applies pointwise to $\mathcal{L}(\varphi ,D\varphi )$, so this constructs the local symmetry.

Also transform $J$ by $\delta J=\epsilon J$. Antisymmetry gives $(t_{a}J)\cdot \varphi +J\cdot (t_a\varphi )=0$, so the source coupling is invariant. Let $S_{A_{\mu a}}$ denote the functional derivative at a point. Vary the action functional and integrate the derivative of ${\epsilon }_a$ by parts, taking the local parameter to have compact support:Since every ${\epsilon }_a(x)$ is arbitrary, its coefficient vanishes. This proves the requested local identity, including the derivative sign and the order of the structure-constant indices.

For the generating functional, the integral of the first term is zero as a change-of-variable identity:The divergence of this vector field in field space is proportional to $\mathrm{tr}{t}_a=0$. Thus an invariant regulated functional measure, with no gauge anomaly, leaves no Jacobian term. Inserting the classical identity into the integral givesThe same first-order differential operator annihilates the connected generating functional $W$, because $Z=e^{iW}$. Its source term is $(t_{a}J)\cdot \phi$.

For the specified Legendre transform, vary $\Gamma =-W+\int J\phi$. Since $W_J=\phi$, the terms involving $\delta J$ cancel, leavingAntisymmetry also gives $(t_{a}J)\cdot \phi =-J\cdot (t_a\phi )$. Substitution in the identity for $W$, followed by multiplication by $-1$, proves the background gauge invariance of a scalar effective action:

The derivatives of the quantum effective action are proper, amputated one-particle-irreducible vertices. At each loop order they receive contributions from one-particle-irreducible Feynman diagrams with the indicated external scalar legs, including tree vertices and the required counterterms. Diagrams which disconnect after cutting one internal line are not additional proper-vertex contributions. Translation invariance supplies the overall momentum-conserving delta function factored out in the definition. The signs of these vertices must follow the chosen overall convention for $\Gamma$.

Differentiate the effective-action identity twice with respect to ${\phi }_i(y)$ and ${\phi }_j(z)$ and set $\phi =A=0$. In an invariant zero-field vacuum this gives the Ward identity contact termsThe term proportional to $A$ vanishes. With the printed Fourier factors $e^{i{p}_r{x}_r}$, integration by parts turns ${\partial }_{\mu }$ into $-i{p}_{1\mu }$. The first delta function combines $p_1$ with $p_2$, and the second combines $p_1$ with $p_3$. Using $(t_a{)}_{ki}=-(t_a{)}_{ik}$ and symmetry of the scalar two-point kernel then yieldsThis derivation fixes the relative signs independently of any vertex rule.

There is an overall-sign inconsistency in the final printed example. At tree level, stationarity of the source integral gives $W=S_0[\phi ,A]+\int J\phi$, hence the specified transform gives ${\Gamma }_{\mathrm{tree}}=-S_0$. For the free scalar multiplet this implies, after factoring out the delta function,The mixed result follows by expanding $-S_0$ to first order in $A$: its contribution is $\int A_{\mu a}({\partial }^{\mu }{\phi }_i)(t_a{)}_{ij}{\phi }_j$. The two scalar derivatives yield $i{p}_2^{\mu }(t_a{)}_{ij}+i{p}_3^{\mu }(t_a{)}_{ji}$. Momentum conservation givesso both sides of the Ward identity equal $i(p_3^2-p_2^2)(t_a{)}_{ij}$. Because the scalar theory is Gaussian for prescribed $A$, its fluctuation determinant depends on $A$ but not on $\phi$; it cannot add further two-scalar vertices. Thus this quadratic verification is sufficient, not merely the first term of an omitted scalar-loop correction.

The printed mixed vertex instead has a minus sign. It satisfies the same Ward identity if one uses the opposite effective-action convention ${\Gamma }_{\mathrm{alt}}=W-\int J\phi =S_0$ at tree level, because its two-point vertex is then $-(p^2+m^2){\delta }_{ij}$ as well. Either overall convention works, but the printed Legendre transform and printed mixed vertex cannot both be used unchanged. For example, with $p_2^2\ne p_3^2$ and a nonzero generator, they give opposite sides of the claimed identity.

The connected generating functional is $W_0[J]=\log Z_0[J]$. Its second functional source derivative gives the connected correlation function; the source-independent determinant drops out. Translational invariance therefore givesThis covariance is independent of the source in a purely Gaussian theory; at zero source the mean field vanishes.

Fourier inversion can be evaluated with the Schwinger parameterization. Since $(p^2+{\mu }^2{)}^{-1}={\int }_0^{\infty }ds{e}^{-s(p^2+{\mu }^2)}$, the elementary Fourier transform of a Gaussian giveswhere $K$ is the Modified Bessel function of the second kind. This integral formula also gives both asymptotic regimes directly.

For $\mu r\ll 1$, set $u=r^2/(4s)$. The integral tends to the Gamma function integral ${\int }_0^{\infty }{u}^{D/2-2}{e}^{-u}du=\Gamma (D/2-1)$; this is finite for $D>2$. Hence, at distances larger than the microscopic cutoff,For $\mu r\gg 1$, the exponent ${\mu }^{2}s+r^2/(4s)$ has its minimum at $s=r/(2\mu )$, with value $\mu r$. Gaussian expansion about that saddle, or the large-argument asymptotic expansion of a modified Bessel function, gives $K_{\rho }(z)\sim \sqrt{\pi /(2z)}{e}^{-z}$. ThereforeThis is the massive Gaussian field correlation tail with its normalization displayed. For example $D=3$ gives exactly $G_0(r)=\alpha e^{-r/\xi }/(4\pi r)$, verifying both regimes and the exponential length scale.

Use natural units and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. A real scalar field assigns a real variable $\varphi (t,\mathbf{x})$ to each spatial point. Its Lagrangian density can be taken to beThe principle of stationary action gives the Euler-Lagrange equation $\square \varphi +V^{\prime }(\varphi )=0$. With $V(\varphi )=m^2{\varphi }^2/2$, this is the Klein-Gordon equation. An additional nonlinear part of $V$ describes interactions.

The canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. The Legendre transform in mechanics gives the canonical Hamiltonian density of a real scalar fieldThe Hamiltonian equations $\dot{\varphi }=\pi$ and $\dot{\pi }={\nabla }^2\varphi -V^{\prime }(\varphi )$ recover the same field equation. In canonical quantization, the fields become operators satisfying the equal-time canonical commutation relationsA spatial lattice makes the analogy with many coupled quantum-mechanical coordinates precise. Each lattice field value is a coordinate, with its own conjugate momentum. The path integral is another representation of the same quantum evolution.

To see its origin, first consider one coordinate with $H=p^2/(2M)+V(q)$. Split a time interval into $N$ steps of length $\epsilon$ and insert position and momentum resolutions of the identity. The short-time kernel isMultiplying the kernels and integrating over intermediate positions gives the phase-space path integralThe endpoints of $q$ are fixed. The momentum integrals are Gaussian integrals; completing the square produces the configuration-space path integralAt finite slicing its normalization contains $(M/(2\pi i\epsilon ){)}^{N/2}{\prod }_{j=1}^{N-1}d{q}_j$. This fixes the composition law and the initial delta-function kernel. One sums over all paths, not merely solutions of the classical equation. Restoring $\hslash$ replaces the weight by $e^{iS/\hslash }$; stationary phase explains the emergence of classical trajectories.

For the field, use scalar field configuration eigenstates $|\phi ⟩$, satisfying $\hat{\varphi }(\mathbf{x})|\phi ⟩=\phi (\mathbf{x})|\phi ⟩$. Insert their completeness relations on every time slice. This givesThe endpoint field configurations are fixed. Integrating the Gaussian momentum variables leaves the scalar field path integral $\mathcal{N}\int \mathcal{D}\varphi e^{iS[\varphi ]}$. The functional measure means a regulated product over the field variables. A spacetime lattice or another ultraviolet cutoff makes this product finite before the continuum limit; interacting continuum calculations may require renormalization. The oscillatory Minkowski weight is an amplitude, not a positive probability density.

For vacuum expectation values, the boundaries must select the vacuum rather than arbitrary field configurations. Long imaginary-time evolution suppresses excited states: $e^{-TH}|n⟩=e^{-T{E}_n}|n⟩$, so after normalization only the lowest-energy component remains as $T\to \infty$. This is vacuum projection by imaginary time. The corresponding Feynman i-epsilon prescription in the real-time integral specifies the vacuum boundary conditions and the poles of the propagator. With $t=-i\tau$, the Euclidean path integral has the weight $e^{-S_E}$, whereIt is often a useful regulated starting point; analytic continuation returns the vacuum time-ordered quantities.

Introduce a classical source $J(x)$ and define the normalized vacuum generating functionalwith the same vacuum prescription in numerator and denominator. A functional derivative brings down $i\varphi (x)$. The order of the time slices makes the operator insertion time-ordered. Thus source differentiation inserts time-ordered field operators:The denominator removes vacuum diagrams and gives normalized expectation values. It is essential that these are time-ordered products; differentiating this vacuum functional does not directly give every possible operator ordering.