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A Hermitian operator $A$ satisfies $⟨\phi ,A\psi ⟩=⟨A\phi ,\psi ⟩$ on its operator domain. HereTwo integration by parts operations, together with the boundary conditions in the Hamiltonian domain, show that the kinetic term is Hermitian; multiplication by the real function $V$ is Hermitian as well. Thus $H$ is Hermitian.

For a time-independent operator $A$, the Schrodinger equation and its adjoint give the Ehrenfest theoremUsing the canonical commutation relation $[x,p]=i\hslash$,Consequently

Let $T:H^1(U)\to L^2(\partial U)$ be the trace map from the Sobolev trace theorem. The weak Robin problem for a uniformly elliptic operator uses all of $H^1(U)$ as the test space, since a Robin boundary condition does not require a zero trace. DefineThe weak formulation isFor a classical solution, integration by parts produces the boundary term $-{\int }_{\partial U}(\sum a^{ij}{u}_{x_i}{\nu }_j)v$. Substituting the Robin boundary condition replaces it by ${\int }_{\partial U}\beta uv$, giving the displayed identity.

Conversely, if $u\in C^2(\overline{U})$ satisfies the identity, tests $v\in C_c^{\infty }(U)$ first show $(L+\lambda )u=f$ in the sense of distributions. Subtracting this interior equality after integration by parts givesEvery smooth boundary function is the trace of a smooth function on $\overline{U}$. The boundary residual is continuous, so this proves the boundary equation pointwise. Thus the formulations agree for $C^2$ solutions. If $f$ is continuous the interior equation holds pointwise; for merely $f\in L^2$, equality with that given representative of $f$ is understood almost everywhere.

Take the difference of two weak solutions, so $f={\psi }_0={\psi }_1=0$. At the stated $H^1(U_T)$ regularity, it is not justified simply to test by $u_t$, which need not belong to the spatial test space $H_0^1$. Instead use the antiderivative test underlying uniqueness of Sobolev weak solutions of a wave equation.

Fix $s\in (0,T]$ and defineThese are Bochner integrals in $H_0^1$. The function $v$ is an admissible space-time $H^1$ test, $v_t=-u$ before $s$, and $v(t)=z(s)-z(t)$. The time term in the weak formulation isbecause $u(0)=0$. Symmetry of $a^{ij}$ and integration by parts in time giveFor the lower-order part, integration by parts in space givesBounded coefficients, the Poincare inequality for $v$, and the Young inequality now implySince $Dv(t)=Dz(s)-Dz(t)$,For $s$ in a sufficiently short fixed interval, absorb the last term into the left-hand side. The Gronwall inequality applied to $\Vert u(s){\Vert }_2^2+(\theta /2)\Vert Dz(s){\Vert }_2^2$ proves $u=0$ on that interval. The equation gives continuity of $u_t$ in $H^{-1}$, so both initial traces at its endpoint are again zero. Repeating with the same uniform coefficient bounds covers $[0,T]$. Therefore

For each $t$, letA weak energy solution of a variable-coefficient wave equation is a function $u\in H^1(U_T)$ with zero lateral trace, equivalently $u\in L^2(0,T;H_0^1(U))$, whose time trace satisfies $u(0)={\psi }_0$ in $L^2(U)$ and for whichfor every $v\in H^1(U_T)$ with zero lateral trace and $v(T)=0$. Here $u\in H^1(0,T;L^2(U))$ has a continuous $L^2$ representative, so the displacement trace is well defined.

The velocity condition is encoded by the boundary term in time; an arbitrary space-time $H^1$ function need not have an $L^2$ trace of $u_t$. The equation implies $u_{tt}=f-L(t)u\in L^2(0,T;H^{-1}(U))$, hence $u_t$ has a continuous $H^{-1}$ representative and $u_t(0)={\psi }_1$ in that sense. This is the weak formulation obtained by integration by parts once in time and once in space.

The expression is a Test-function pairing with a Gaussian free field, rather than the pointwise integral of an ordinary random function. For a real test function $\varphi$, take the Dirichlet solutionusing the Green-kernel identity allowed in the original PDF. In particular, $-\Delta f_{\varphi }=2\pi \varphi$. Integration by parts gives, initially for smooth compactly supported $u$ and then by continuity in the Dirichlet energy space,The integral functional is continuous in the energy norm: pull back to the unit disc, where the transformed test function has compact support, and use the Poincare inequality. Hence $f_{\varphi }$ is its energy-space representer. DefineBy the isonormal Gaussian process definition, it has mean zero and varianceThe logarithmic singularity is locally integrable, so this variance is finite for smooth compactly supported test functions. ThusThe local TeX corrupted the double integral and the bracketed Green-kernel identity; the PDF supplies the identity above and does not assert an infinite variance.

The minimum exceeds $t$ exactly when both times exceed $t$. By independence,Differentiating its survivor function givesThe tail integral formula for moments then givesEquivalently, integration by parts in ${\int }_0^{R}t{f}_X(t)dt$ gives $-R{F}_A(R)F_B(R)+{\int }_0^R{F}_A(t)F_B(t)dt$. The assumed $t{F}_B(t)\to 0$ makes the boundary term vanish, since $0\le F_A\le 1$.

The stated tail condition does not itself guarantee a finite mean. For example, the proper survivor functionssatisfy $t{F}_B(t)\to 0$, but their product behaves as $1/[t\log t\log \log t]$, whose integral diverges. Thus the displayed expectation is valid as an extended nonnegative value; finiteness needs integrability of the product. The exponential distribution in the next part guarantees that finiteness.

For independent and identically distributed random variables with distribution function $F$, let $M_n={\max }_{1\le i\le n}{X}_i$ be the sample maximum. In extreme value theory, $F$ belongs to the maximum domain of attraction $D(G)$ of a nondegenerate distribution function $G$ when there exist $a_n>0$ and $b_n\in \mathbb{R}$ such thatat every continuity point of $G$. Thus the normalized sample maxima have convergence in distribution to $G$.

A positive measurable function $q$ has regular variation at infinity with index $\rho$, written $q\in {\mathrm{RV}}_{\rho }$, ifThe case $\rho =0$ defines a slowly varying function; equivalently a regularly varying function is $q(t)=t^{\rho }\ell (t)$ with $\ell$ a slowly varying function.

Write $\overline{F}=1-F$ for the survival function and $x_F=\sup \{x:F(x)<1\}$ for the right endpoint of a distribution. For $\alpha >0$, use the following standard extreme value theory normalization conventions:They are respectively the Fréchet distribution, the negative Weibull distribution, and the Gumbel distribution for maxima. The negative Weibull distribution is supported to the left of its finite endpoint; it is not the usual positive Weibull distribution.

The necessary and sufficient maximum domain of attraction criteria are:

For the corresponding sufficient hazard function conditions, assume $F$ is absolutely continuous near $x_F$, its survival function tends to zero there, and its probability density function $f$ is positive there. Write $\eta (t)=f(t)/\overline{F}(t)$. The Von Mises conditions for extreme values giveThe last condition additionally assumes that the reciprocal hazard function is continuously differentiable near the endpoint. It concerns the derivative of the reciprocal hazard function, not the derivative of the hazard function itself. Equivalently, when $f$ is differentiable, it is $\overline{F}(t)f^{\prime }(t)/f(t{)}^2\to -1$.

To see the connection, taking an integral of the hazard function givesThe first two limits make the logarithmic tail ratios over multiplicative or endpoint-distance intervals tend to $-\alpha \log u$ or $\alpha \log u$, giving the required regular variation. For the Gumbel distribution, set $a(t)=1/\eta (t)$. The derivative condition makes $a(t+a(t)v)/a(t)\to 1$ uniformly on bounded $v$-intervals, soThe endpoint assumptions ensure these shifts are admissible: at an infinite endpoint $a(t)/t\to 0$, and at a finite endpoint $a(t)/(x_F-t)\to 0$. These are sufficient smooth-tail conditions; the tail-ratio characterizations above are necessary and sufficient without differentiability assumptions. For a source using the same reciprocal-hazard convention, see Smith and Weissman's university lecture notes, equations (2.18)–(2.24).

For the Gamma distribution with integer shape $m$ and rate one, let ${\overline{F}}_m(t)={\int }_t^{\infty }{u}^{m-1}{e}^{-u}du/(m-1)!$. The base case is ${\overline{F}}_1(t)=e^{-t}$, and integration by parts givesInduction provesThis survival function has the asymptotic form ${\overline{F}}_m(t)\sim e^{-t}{t}^{m-1}/(m-1)!$; for $m=1$ the relation is exact. Put $t_n={\beta }_n+x$. For every fixed real $x$, $t_n\to \infty$ and $t_n/\log n\to 1$. The proposed centering satisfiessoOnly the highest-degree term contributes to the limit, with the same reasoning covering $m=1$. The Taylor expansion $\log (1-u)=-u+O(u^2)$ now givesbecause $n{\overline{F}}_m({\beta }_n+x{)}^2\to 0$. Therefore the Gumbel limit for gamma maxima isEquivalently, $M_n-{\beta }_n$ has convergence in distribution to the standard Gumbel distribution, with scale $a_n=1$.

Use the free Hamiltonian operator and the canonical commutation relations. The kinetic momentum term givesThe Heisenberg picture equation therefore yields ${\dot{\varphi }}_1={\pi }_1$.

For the momentum commutator, the mass term gives $i{m}_1^2{\varphi }_1(\mathbf{x})$. The gradient term giveswhere integration by parts differentiates the field instead of the Dirac delta distribution. ThusCombining the two first-order operator equations gives the Klein-Gordon equation:Suitable decay or spatial smearing removes the boundary term in the integration by parts.

Let the charge parameter depend on position temporarily. Using the gauge-covariant derivative of a charged scalar field, the variation of the matter action is $\delta S=\int d^{4}x{j}^{\mu }{\partial }_{\mu }\alpha$. The resulting Noether current isHere ${\nabla }_{\mu }\bar{\varphi }={\partial }_{\mu }\bar{\varphi }+i{A}_{\mu }\bar{\varphi }$. The potential does not contribute, and $j^{\mu }=-\delta S_{\mathrm{matter}}/\delta A_{\mu }$. Because $A$ has not been rescaled to a canonical kinetic term, there is no factor $e$ in this Noether current or in the matter interaction vertices.

A local change of variables in the path integral gives $\delta ⟨\varphi (y)\bar{\varphi }(z)⟩=⟨\delta (\varphi (y)\bar{\varphi }(z))-(\delta S)\varphi (y)\bar{\varphi }(z)⟩=0$. Integration by parts in the term containing $\partial \alpha$ produces the two opposite contact terms of the Ward identity.

For clarity about phases, transform the current correlator with $e^{i(p-p^{\prime })\cdot x-ip\cdot y+i{p}^{\prime }\cdot z}$ and suppress its overall momentum conservation Dirac delta function. Denote the result by $C^{\mu }(p,p^{\prime })$. The coordinate Ward identity becomesAmputation of the two external quantum field theory propagators then fixes the longitudinal part of the current interaction vertex. Since the action couples $A_{\mu }$ to $-j^{\mu }$, the gauge interaction vertex has the opposite sign. The phase convention for the paper's gauge interaction vertices is ${\Gamma }^{\mu }=i{G}^{\mu }$, where $G^{\mu }$ is the corresponding derivative of the Euclidean quantum effective action.

One can derive the exact one-particle-irreducible correlation function directly, without assuming the entire connected current correlator is one-particle irreducible. Local gauge invariance of the quantum effective action impliesA linear covariant gauge fixing adds a breaking term independent of the scalars, which disappears after the scalar differentiations below. The identity requires a regularization in quantum field theory and counterterms preserving the Abelian gauge theory Ward identity; scalar quantum electrodynamics has no gauge anomaly.

Differentiate with respect to $\varphi (y)$ and $\bar{\varphi }(z)$, then set all background fields to zero. The two-point derivative is the inverse exact quantum field theory propagator. Use the Fourier transform convention $\varphi (x)={\int }_p{e}^{ip\cdot x}\varphi (p)$, $\bar{\varphi }(x)={\int }_{p^{\prime }}{e}^{-i{p}^{\prime }\cdot x}\bar{\varphi }(p^{\prime })$, with the photon momentum $p^{\prime }-p$ incoming. This givesAt tree level $G^{\mu }=-(p+p^{\prime }{)}^{\mu }$, so ${\Gamma }^{\mu }=-i(p+p^{\prime }{)}^{\mu }$, and the identity reduces to $-i(p^2-p^{\prime 2})=i(p^{\prime 2}-p^2)$.

To obtain the two-photon scalar Ward identity, differentiate the same functional Ward identity also with respect to $A_{\nu }(w)$. In terms of Euclidean action derivatives it readsThis construction automatically includes the seagull vertex: the Noether current depends on $A$, with $\delta j^{\mu }(x)/\delta A_{\nu }(w)=-2{\delta }^{\mu \nu }{\delta }^4(x-w)|\varphi (x){|}^2$. It cannot be discarded when differentiating a current insertion.

Take the first photon momentum to be incoming $k$, and the second incoming $p^{\prime }-p-k$. On the first scalar leg the contact term shifts $p$ to $p+k$; on the other it shifts $p^{\prime }$ to $p^{\prime }-k$. ConsequentlyWith ${\Gamma }^{\mu \nu }=i^2{G}^{\mu \nu }=-G^{\mu \nu }$, the result isThere is a momentum-routing sign error in the printed second identity. With the scalar momentum convention of the first identity, its first shifted argument must be $p^{\prime }-k$, not $k-p^{\prime }$. This is already forced at tree level: the seagull vertex has $G^{\mu \nu }=2{\delta }^{\mu \nu }$ and hence ${\Gamma }^{\mu \nu }=-2{\delta }^{\mu \nu }$. The corrected right-hand side is $-2k^{\nu }$, matching the left-hand side, whereas the printed right-hand side is $-2p^{\prime \nu }$. Changing to an all-incoming scalar convention would also change the first identity, so it does not fix both printed formulas simultaneously.

The Grassmann field $c$ is odd and the adjoint scalar field $\varphi$ is even. The two printed signs are consistent with a right-acting BRST differential, whose graded Leibniz rule isThe bracket between two odd fields is graded, so $[c,c]=2c^2$. Applying this rule to the ordinary Lie bracket $[c,\varphi ]=c\varphi -\varphi c$ givesThe second bracket here is graded because both its entries are odd. The graded Jacobi identity gives $[c,[c,\varphi ]]=\frac{1}{2}[[c,c],\varphi ]$. ThereforeThe side of the odd derivation is essential. With the usual left graded Leibniz rule and the same two printed signs, the result would be $-[[c,c],\varphi ]$, which is generally nonzero. A left-acting convention must reverse one of those signs. All subsequent BRST symmetry formulas here use the right-acting convention.

Choose an anti-Hermitian basis $T^a$ with $[T^b,T^c]=f^{abc}{T}^a$, an invariant positive invariant bilinear form on a Lie algebra, and the adjoint covariant derivative $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$. Couplings are absorbed into this convention; restoring $D_{\mu }={\partial }_{\mu }+g[A_{\mu },\cdot ]$ multiplies each ghost interaction vertex below by $g$. The associated BRST charge acts as $Q{A}_{\mu }=-D_{\mu }c$, $Q\bar{c}=ih$, and $Qh=0$.

Write $F^a=\partial \cdot A^a+(n\cdot \partial {\varphi }^a)/\sqrt{2}$. For the gauge-fixing fermion $\Psi =\int {\bar{c}}^a(F^a-i{h}^a/2)$, the right graded Leibniz rule givesThe Gaussian functional integral over the Nakanishi-Lautrup field produces the positive gauge fixing term $F^a{F}^a/2$. The ghost operator is $-\partial \cdot D-(n\cdot \partial ){\mathrm{ad}}_{\varphi }/\sqrt{2}$.

Now use the canonical free kinetic terms $F_{\mu \nu }^a{F}_{\mu \nu }^a/4+(\partial {\varphi }^a{)}^2/2$. At nonzero momentum in Euclidean space $p$, put $\eta =n\cdot p$. The quadratic kernel for $(A_{\mu }^a,{\varphi }^a)$, per color, isThe transverse gauge-field kernel plus the gauge-fixing longitudinal term has become $p^2{\delta }_{\mu \nu }$. The scalar quantum field theory propagator is the inverse Schur complement, not merely the inverse of the scalar diagonal entry:Hence the free adjoint-scalar propagator in scalar-dependent gauge fixing isFor completeness the mixed quantum field theory propagator is $⟨A_{\mu }^a{\varphi }^b{⟩}_0=-{\delta }^{ab}\eta p_{\mu }/(\sqrt{2}{p}^4)$; ignoring this mixing would give an incorrect scalar answer.

The scalar propagator is independent of the gauge vector $n$, not of the momentum component parallel to $n$. For a unit $n$, $p^2=p_{\perp }^2+p_{\parallel }^2$, and the free scalar quantum field theory propagator still depends on $p_{\parallel }$. Thus the literal momentum-independence clause in the PDF is false for the standard minimally coupled massless scalar action; the cancellation above establishes the natural gauge-vector-independence statement. The usual massless zero mode in field theory at $p=0$ needs a separate infrared prescription.

Finally, integration by parts gives the ghost action in an unambiguous convention:Use $e^{ip\cdot x}$ for the Fourier transform of every field, with all momenta incoming. Let the antighost carry color $a$ and momentum $q$, the boson color $b$ and momentum $k$, and the ghost color $c$ and momentum $p$, so $q+k+p=0$. Expansion of $e^{-S_{\mathrm{gh}}}$ gives the ghost vertices in scalar-dependent gauge fixingThe free Faddeev-Popov ghost field propagator is ${\delta }^{ab}/p^2$ and every closed ghost loop contributes a minus sign. There are no further ghost interaction vertices in this gauge. Factors of $i$ and an overall ghost-vertex sign depend on the Fourier transform and ghost-ordering conventions; the displayed ghost action fixes both here.

A Bogomolny bound expresses a static field energy as nonnegative squares plus a term fixed by the topological charge or boundary data. Setting the squares to zero gives the first-order Bogomolny equations. Their solutions minimize the energy in that sector and satisfy the second-order Euler-Lagrange field equations, although a general stationary solution need not attain the bound.

For one real scalar in one spatial dimension, take the Lagrangian densityA finite-energy field configuration in this static sector must approach scalar-field vacua ${\varphi }_{\pm }$ with $W^{\prime }({\varphi }_{\pm })=0$ at the two ends. Completing the square gives the square completion for a one-dimensional kink:Choose the sign for which the boundary term is nonnegative. ThusTaking the derivative of the equality equation gives ${\varphi }_{xx}=W^{\prime }{W}^{\prime \prime }=d([W^{\prime }{]}^2/2)/d\varphi$, the static Euler-Lagrange field equation. The boundary term is invariant under deformations keeping the asymptotic scalar-field vacua fixed; it is not a contribution from the local shape of the kink.

For a phi-four kink, let $c>0$, $W(\varphi )=c^2\varphi -{\varphi }^3/3$, and select the sector ${\varphi }_{-}=-c$, ${\varphi }_{+}=c$. Then $W^{\prime }=c^2-{\varphi }^2$, and the increasing Bogomolny equation is ${\varphi }_x=c^2-{\varphi }^2$. Separating variables yieldsThe integration constant $X$ is the translational collective coordinate. Directly, ${\varphi }_K^{\prime }=c^2{\mathrm{sech}}^2(c(x-X))$ and its energy density is $c^4{\mathrm{sech}}^4(c(x-X))$, whose integral is $4c^3/3$. The decreasing antikink has the reversed boundary values, profile $-{\varphi }_K$, and the same energy. A Lorentz boost produces the exact uniformly moving kink $c\tanh [c\gamma (x-X_0-vt)]$, with Lorentz factor $\gamma =(1-v^2{)}^{-1/2}$ and energy $\gamma M$.

In two spatial dimensions, choose the Abelian Higgs model at critical coupling, with $D_i={\partial }_i-i{A}_i$, $B={\partial }_1{A}_2-{\partial }_2{A}_1$, and energyThe gauge covariant derivative is used throughout this normalization. Integration by parts, using $[D_1,D_2]=-iB$, gives $\int |D_i\varphi {|}^2=\int |(D_1+i{D}_2)\varphi {|}^2+\int B|\varphi {|}^2$, with a vanishing boundary divergence for the decaying vortex fields. Combining this identity with the magnetic and potential terms gives the Bogomolny square completion for an Abelian Higgs vortex:Finite-energy field configurations have $|\varphi |\to 1$ at infinity, and $D_i\varphi \to 0$ makes the magnetic flux equal to the phase winding $2\pi N$. For $N>0$,These are the Bogomolny vortex equations for an Abelian Higgs vortex. Opposite signs give antivortices and the bound $E\ge \pi |N|$. The coefficient $\pi$ follows from the explicit energy normalization above; other conventions can give $2\pi$. Static solutions of fixed positive vortex number have equal energy independent of their positions, giving the Abelian Higgs vortex moduli space used for slow dynamics.

Use geometrized units $G=c=1$ and signature $(-,+,+,+)$. The original PDF starts with the d'Alembert operator $\square {\bar{h}}_{\mu \nu }=-16\pi T_{\mu \nu }$; the local TeX incorrectly transcribes its derivative indices. The harmonic condition is the Lorenz gauge in linearized gravity.

Choose the retarded solution, excluding an incoming homogeneous wave. For a spatially localized source, the Linearized Einstein equations giveLet the source size be $D$ and the characteristic angular frequency be $\omega$. Assume the weak-field approximation, nonrelativistic source velocities, and $D\ll {\omega }^{-1}\ll r$ for a radiative far-zone measurement. Then $|\mathbf{x}-\mathbf{y}|=r-\mathbf{n}\cdot \mathbf{y}+O(D^2/r)$. At leading order in $D/r$ and $\omega D$, one may replace the denominator by $r$ and the retarded time throughout the source by $t-r$, obtaining