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For a composition free energy $F=\int [f(\varphi )+\kappa |\nabla \varphi {|}^2/2]$, the local chemical potential is the functional derivativeAt phase coexistence, exchanges between the bulk phases require equal chemical potentials. Mechanical equilibrium additionally requires equal pressure for a flat interface, or a Young–Laplace equation pressure jump for a curved interface.

A generating functional is a path integral with a source $J$ coupled to a field. Its functional derivatives with respect to $J$ generate time-ordered correlation functions.

For nonconserved order-parameter dynamics, the positive kinetic coefficient $\Gamma$ converts the functional derivative of the free energy into a relaxation rate through $\dot{p}=-\Gamma \delta F/\delta p$. Microscopic reversibility fixes the accompanying Gaussian white noise variance by the Model A fluctuation-dissipation relation.

For a vacuum generating functional $Z[J]$, normalization gives $\hat{Z}[J]=Z[J]/Z[0]$ and $\hat{Z}[0]=1$. Functional derivatives of $\hat{Z}$ give normalized vacuum expectation values. The linked-cluster theorem shows that this ratio cancels Vacuum Feynman diagrams, while $-i\log \hat{Z}[J]$ gives connected correlators. Components attached to external insertions are not removed merely by the normalization.

A functional derivative of the quantum effective action gives an amputated one-particle-irreducible correlation function. Its perturbative graphs stay connected after cutting any single internal propagator. A free quadratic effective action has no vertices of order greater than two; interactions generate higher vertices.

The positive order-parameter mobility $M$ relates the diffusive transport flux to the chemical potential gradient: $\mathbf{J}=-M\nabla \mu$. Combining this law with local conservation gives the Cahn--Hilliard equation when $\mu$ is the functional derivative of a composition free energy.

A linear operator on finite-dimensional vectors acts by $(Au{)}_i={\sum }_j{A}_{ij}{u}_j$. For a field the discrete index becomes a continuous coordinate, and a suitable operator is represented by an integral kernel, possibly a distribution:The identity operator has kernel ${\delta }^{(d)}(x-y)$. If an inverse exists on the chosen domain and with the chosen boundary prescription, its kernel satisfieswith the analogous reversed identity. A differential operator is a distributional kernel: for example, $A(x,y)=(-{\square }_x-m^2){\delta }^{(d)}(x-y)$. Its inverse is a Green function. Boundary conditions, a causal prescription and the treatment of zero modes are part of the inverse's definition, not consequences of formal matrix notation. An operator with an untreated zero mode does not have an inverse on the full field space.

The functional derivative is defined through the first variation: for arbitrary smooth test variation $h$,In particular,For a symmetric kernel, differentiating $F[\varphi ]=\frac{1}{2}\int dxdy\varphi (x)A(x,y)\varphi (y)$ gives $\delta F/\delta \varphi =A\varphi$. The two terms from varying the two fields coincide; for a nonsymmetric kernel the symmetric part $(A+A^T)/2$ appears instead. Integration by parts is understood when $A$ is differential and the boundary variations vanish. A finite-volume, finite-mode regulator makes these statements ordinary matrix identities before a continuum limit.

For a finite-dimensional integral, $u_i{e}^{b\cdot u}={\partial }_{b_i}{e}^{b\cdot u}$. Expanding a polynomial interaction $V(u)$ formally and interchanging each finite-order derivative with the regulated integral giveswhere $I_0$ is the free Gaussian integral. This is a perturbation-series identity; it does not assert convergence of the full expansion for an arbitrary potential.

For the Minkowski functional integral, each field insertion is generated byExpand $e^{i{S}_I[\varphi ]}$, replace its field factors by these functional derivatives, and take the derivatives outside the free integral. The result is polynomial interactions as source differential operators:If the field interaction term contains derivatives of the field, the corresponding spacetime derivatives act on the source-insertion operators as well. For the scalar polynomial interactions the operator in the exponent isThe free quadratic source exponential then supplies all Wick contractions, reproducing the Feynman rules. For normalized correlation functions, divide the resulting interacting functional by $Z[0]$; the interaction operator changes this denominator too, so it cannot simply be ignored.

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.

For a bosonic quadratic action $S_2=\frac{1}{2}\int \varphi K\varphi$, completing the square in the generating functional gives $Z_0[J]/Z_0[0]=\exp (-iJ{K}^{-1}J/2)$. Two functional derivatives therefore give the Feynman propagator $\Delta =i{K}^{-1}$, with boundary conditions and a pole prescription included in the inverse. Zero modes must be removed or fixed before an inverse exists.

Take Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$, Fourier convention $e^{-ikx}$, and the Yang-Mills field strength $F_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a+g{f}^{abc}{A}_{\mu }^b{A}_{\nu }^c$. The interaction terms do not contribute to the quadratic action. Integration by parts in $-F_{\mu \nu }^a{F}^{a\mu \nu }/4-(\partial \cdot A^a{)}^2/(2\alpha )$ yieldsThe color kernel is diagonal. For $k^2\ne 0$, use the projectors $P_L^{\mu \nu }=k^{\mu }{k}^{\nu }/k^2$ and $P_T^{\mu \nu }={\eta }^{\mu \nu }-P_L^{\mu \nu }$. They obey $P_L^2=P_L$, $P_T^2=P_T$, $P_L{P}_T=0$, soMultiplication explicitly gives the identity on both transverse and longitudinal components. Continuing this inverse with the Feynman propagator prescription gives the gluon propagatorThe second term has the causal double-pole prescription. Away from the pole this is the usual $-i{\delta }^{ab}[{\eta }_{\mu \nu }-(1-\alpha )k_{\mu }{k}_{\nu }/k^2]/(k^2+i0)$. For $\alpha =1$ it becomes the Feynman-gauge adjoint propagator; $\alpha \to 0$ gives a transverse covariant Landau gauge propagator.

The limit $\alpha \to \infty$ removes gauge fixing and has no finite inverse on the full field space. The longitudinal eigenvalue $-k^2/\alpha$ tends to zero and its inverse diverges. A contraction with conserved external currents kills the longitudinal term, but that does not define a gauge-unfixed Gaussian integral. This massless limit is not the massive-vector unitary-gauge limit: here it exposes the unfixed gauge orbit degeneracy.

Because $⟨B⟩=0$, a periodic primitive $H$ exists. Fix its additive constant by $⟨H⟩=0$. Integrating the conserved-field equation once in space gives $H_T=\sigma (H_X+\delta A^2{)}_X$: the possible integration constant vanishes on taking the spatial mean. Treat the spatial average as the inner product defining functional derivatives. Integration by parts, with no boundary terms under periodic boundary conditions, givesThus the conserved-field real amplitude equation is a gradient flow ascending $V$, or descending $-V$, with positive diagonal mobilities. Along a classical solution,For the mixed term put $P=⟨A^4{⟩}^{1/2}$ and $Q=⟨H_X^2{⟩}^{1/2}$. The Cauchy-Schwarz inequality and a completed square giveConsequently, for $\beta =1-\delta >0$,The negative leading quartic coefficient proves the upper bound. The stronger pointwise square completionis also useful: a superlevel set $V\ge V(0)$ bounds the amplitude in $L^4$, its spatial derivative in $L^2$, and $B=H_X$ in $L^2$, because a quartic polynomial with negative leading coefficient dominates its lower-order terms.

The dynamics dissipates toward the steady-state set rather than sustaining nonconstant periodic motion. Indeed $V$ increases to a finite limit, and the time integral of $⟨A_T^2+H_T^2/(2\delta \sigma )⟩$ is finite. For global regular trajectories with the usual parabolic precompactness, every limit point is stationary, by the LaSalle invariance principle. A nonconstant periodic or recurrent orbit would force a strictly increasing $V$ to return to its original value and is impossible. Boundedness of a monotone functional alone does not prove convergence to one particular equilibrium; isolation of the limiting equilibrium or an additional convergence theorem supplies that stronger claim.

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.

Differentiate the product of Dirac delta functions defining $\tilde{P}$, regarding $B$ and $\sigma$ as independent density arguments. The random continuity equation isFor $0<s<t$, the causal functional derivatives follow from the explicit strain solution and $\tilde{B}(t)=B_0\exp [{\int }_0^t\tilde{\sigma }(u)du]$:Consequently,In the Furutsu–Novikov formula, the covariance $\kappa \delta (t-s)$ samples the upper endpoint of the causal time integral. Symmetric regularization of the white noise gives half the mass there. The $B$ response vanishes as $s\uparrow t$, while the strain response tends to one, so $\mathbb{E}[f(t)\tilde{P}(t)]=-(\kappa /2){\partial }_{\sigma }P$. Averaging the continuity equation yields the Fokker-Planck equationThere is no direct diffusion term in $B$: its equation contains the time integral of colored noise, whereas the strain equation receives Gaussian white noise. The same result follows from the Itô diffusion with drift $(\sigma B,-\sigma /\tau )$ and diffusion vector $(0,\sqrt{\kappa })$.

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.

Treat $\psi$ and its complex conjugate as independent variables in the functional derivative. Integrating the gradient variation by parts, with vanishing boundary terms, givesThe factors follow from ${\partial }_{{\psi }^{\ast }}|\psi {|}^4=2|\psi {|}^2\psi$ and ${\partial }_{{\psi }^{\ast }}|\psi {|}^6=3|\psi {|}^4\psi$. Thus the cubic-quintic Gross–Pitaevskii equation is

Use decaying variations with the same complex bulk field one, so that integration by parts produces no variation boundary term. The functional derivatives of energy and renormalized momentum of a condensate areFor the second identity, variation of the two momentum terms and integration of the ${\partial }_z\delta {\psi }^{\ast }$ term each supply $(2i{)}^{-1}{\psi }_z$, so together they give $-i{\psi }_z$. It follows thatis exactly the stationary Gross–Pitaevskii solitary wave equation. Thus each solitary wave is a constrained energy critical point with multiplier $U$.

Let $s$ parametrize a differentiable family of these waves. Apply the critical-point identity to the variation ${\partial }_s\psi$, keeping the multiplier fixed at its value for that member of the family. Including the complex conjugate variation givesHence the energy–momentum slope of a solitary wave iswherever $dp/ds\ne 0$. At a momentum turning point the parametrized identity $dE/ds=Udp/ds$ remains the correct statement; a globally single-valued energy as a function of momentum is not needed. Differentiating $E-U(s)p$ as a composite function would include an extra $-p{U}^{\prime }$ term, so that composite derivative must not be set to zero.

Replace $\phi$ by $\phi +ϵ\eta$, with $\eta$ vanishing on the boundary, and differentiate the action at zero. Integrate the derivative of $\eta$ by parts:The coefficient of $\eta$ is the functional derivative $\delta S/\delta \phi$.

For the electromagnetic action, $\delta F_{ab}={\partial }_a\delta A_b-{\partial }_b\delta A_a$. Antisymmetry combines these terms and integration by parts givesThus the inhomogeneous Maxwell equations are ${\partial }_a{F}^{ab}={\mu }_0{J}^b$. The definition of $F$ also gives the homogeneous equations ${\partial }_{[a}{F}_{bc]}=0$, equivalently ${\partial }_a{\tilde{F}}^{ab}=0$. These are the two four-vector Maxwell systems.

Unrestricted variation of the alternative component-gradient action instead gives $\square A^b={\mu }_0{J}^b$. This is equivalent to the first Maxwell system only when supplemented by the Lorenz gauge ${\partial }_a{A}^a=0$, becauseThe alternative action is not by itself gauge-invariant; the additional condition must be supplied consistently, not silently omitted.

Expand the tensor square: $F^{ab}{F}_{ab}=2[{\partial }^b{A}^a{\partial }_b{A}_a-{\partial }^b{A}^a{\partial }_a{A}_b]$. HenceIntegrating this cross term by parts twice expresses its integral as a boundary term plus $\int ({\partial }_a{A}^a{)}^2{d}^{4}x$. Under Lorenz gauge it is therefore purely a boundary term. With fixed or vanishing boundary contributions, the two actions have the same variational equations once the additional gauge condition is included. This is the Lorenz-gauge reduction of the electromagnetic action.

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.