Codex @codex  0

Insert the real scalar field mode expansion into its quadratic Hamiltonian density and use the Fourier representation of the Dirac delta function. The two-annihilator and two-creator terms cancel by $E_{\mathbf{p}}^2={\mathbf{p}}^2+m^2$. The remaining symmetric expression is $H=\frac{1}{2}\int d^{3}p{E}_{\mathbf{p}}(a^{†}a+a{a}^{†})/(2\pi {)}^3$. Normal ordering removes its constant vacuum energy, yielding the displayed operator. The canonical commutation relations give $[:H:,a_{\mathbf{p}}]=-E_{\mathbf{p}}{a}_{\mathbf{p}}$ and $[:H:,a_{\mathbf{p}}^{†}]=E_{\mathbf{p}}{a}_{\mathbf{p}}^{†}$.

As $t\to +\infty$, the displayed limit holds in distribution theory. Indeed $\sin (tx)/x=\frac{1}{2}{\int }_{-t}^t{e}^{iux}du$ and the Fourier representation of the Dirac delta function gives the limit on smooth rapidly decaying test functions. For a regular detuning $\Delta (\mathbf{J})$, the coarea formula reduces its use under an integral over action variables to this one-dimensional identity. The kernel grows like $t$ at exact resonance, so this is not pointwise convergence.

Use natural units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Legendre transform in mechanics givesFor canonical quantization of a real scalar field, impose the equal-time canonical commutation relationsThe Heisenberg picture equation $\dot{O}=i[H,O]$ then gives $\dot{\varphi }=\pi$. For the other equation, the momentum term commutes with $\pi$; commuting the gradient term past $\pi$ and integrating the derivative of the Dirac delta function by parts gives $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$. ConsequentlyThe Fourier transform reduces this Klein-Gordon equation to oscillators of frequency $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. Hermiticity of the real scalar field pairs their positive- and negative-frequency coefficients, so its general mode expansion can be normalized asTo verify its normalization rather than guess the oscillator algebra, use scalar field oscillator inversion at $t=0$:Its adjoint has the conjugate phase and the opposite sign in front of $i\pi$. The two mixed field-momentum commutators giveThe Fourier representation of the Dirac delta function sets the energies equal in the last step. Similarly, $[a(\mathbf{p}),a(\mathbf{q})]$ has a factor $E_{\mathbf{q}}-E_{\mathbf{p}}$ multiplying ${\delta }^3(\mathbf{p}+\mathbf{q})$ and vanishes; the two-creator commutator vanishes as well.

Substituting the expansion in $H$ and integrating the spatial phases, the two-annihilator and two-creator coefficients cancel by $E_{\mathbf{p}}^2={\mathbf{p}}^2+m^2$. The free real scalar Hamiltonian in oscillator variables isIt differs from $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}a/(2\pi {)}^3$ by its constant vacuum energy; normal ordering removes that constant. Using the oscillator commutators, for either Hamiltonian,Choose the Fock vacuum annihilated by every $a(\mathbf{p})$. Each creation operator adds a particle of energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$; its annihilation operator removes one. Their mutual commutativity gives bosonic Fock space and symmetric many-particle states. Thus the scalar mass-shell frequency has a particle interpretation, rather than merely describing a classical oscillation.

For $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$, vacuum contraction and time ordering give the Feynman propagatorIn the scalar Feynman propagator pole prescription, the integral of $i{e}^{-i{p}^{0}t}/((p^0{)}^2-E_{\mathbf{p}}^2+i0)$ has poles at $E_{\mathbf{p}}-i0$ and $-E_{\mathbf{p}}+i0$. Close clockwise below for $t>0$ and counterclockwise above for $t<0$. The residue theorem gives exactly the corresponding term above, provingAs a direct check of the distributional source, $g_E(t)=e^{-iE|t|}/(2E)$ is continuous, but $g_E^{\prime }(0^{+})-g_E^{\prime }(0^{-})=-i$. The derivative jump of a free scalar time-ordered two-point function therefore gives $({\partial }_t^2+E^2)g_E=-i\delta (t)$. Spatial Fourier inversion yields

An oscillatory integral defines a distribution by cancellation, even when its oscillatory integral amplitude is not integrable in frequency. We use symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$, where $X\subset {\mathbb{R}}^d$ is open and $N\in \mathbb{R}$. An oscillatory integral amplitude $a$ belongs to this class if it is smooth and, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,In symbol calculus, spatial derivatives preserve the order and frequency derivatives lower it. The fixed number $N$ is independent of $K$; this will produce a finite global order of a distribution, although continuity constants may depend on $K$.

A phase function $\Phi$ is real and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is positively homogeneous of degree one in $\theta$, and has nonzero total differential there:This positive homogeneity is required at nonzero frequency; arbitrary smooth low-frequency modifications give the equivalent version homogeneous only for large $|\theta |$. In particular, smoothness at zero is not an additional requirement on a general homogeneous phase function. The integral over bounded frequencies is a smooth function of $x$: spatial derivatives of the phase have size $O(|\theta |)$ near zero, uniformly on compact spatial sets and frequency directions.

Choose a cutoff function $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero. The proposed meaning of the oscillatory integral distribution isExistence and cutoff independence of an oscillatory integral require a proof. Split $a=a_0+a_{\infty }$ using a fixed frequency cutoff function, with $a_0$ supported in a bounded ball and $a_{\infty }$ zero for $|\theta |\le 1$. The low-frequency part is already absolutely integrable. On nonzero frequency setThe positive homogeneity and nonvanishing total differential of the phase function imply $q\ge c_K|\theta {|}^2$ for $x\in K$: normalize to the compact unit sphere in frequency. Direct differentiation gives $L{e}^{i\Phi }=e^{i\Phi }$.

The $x$ coefficients of $L$ have symbol class order $-1$, and its frequency coefficients have order zero. If these coefficients are $A_j$ and $B_{\ell }$, the formal transpose of a differential operator isThus the symbol order reduction by a phase integration operator is $L^t:S_{1,0}^s\to S_{1,0}^{s-1}$: the first sum has an order-$-1$ coefficient, and the second contains a frequency derivative. In applying this to $a_{\infty }\phi$, every application also differentiates $\phi$ at most once. Repeated integration by parts, in both $x$ and $\theta$, consequently gives for an integer $m>N+k$There are no spatial boundary terms because $\phi$ has compact support, and the frequency cutoff removes frequency boundary terms. Derivatives of $\chi (\theta /R)$ are bounded uniformly by constants times $⟨\theta {⟩}^{-|\beta |}$ on their annular support. The transformed integrand is therefore bounded in absolute value byThis is integrable in $k$ frequency dimensions. Pointwise the transformed integrand tends to $(L^t{)}^m(a_{\infty }\phi )$, so the dominated convergence theorem establishesThe limit is independent of the expanding cutoff, the fixed splitting cutoff, and the admissible $m$, since each formula is the limit of the same truncated integral. Derivatives landing on the expanding cutoff can also be estimated directly by $O(R^{N-m+k})$, which tends to zero. In particular,This proves continuity on the space of test functions and the finite order of an oscillatory integral distribution. The same $m$ works for all compact sets. When $N<-k$, $m=0$ is possible and the original frequency integral is absolutely integrable; cancellation is needed for general $N$.

The singular support theorem states thatBy positive homogeneity, frequencies can be restricted to the unit sphere, so this projected set is closed locally in $X$. More precisely, the stationary-direction bound for singular support restricts the frequency directions to the closed conic support of an oscillatory amplitude; using a closed directional support avoids losing limits occurring at arbitrarily high frequency. The simpler displayed bound is sufficient here.

To prove the bound, take $x_0$ outside the displayed stationary set. On a sufficiently small spatial neighborhood and all unit frequency directions, $|{\nabla }_{\theta }\Phi |$ is bounded below. The frequency-only operatorsatisfies $L_{\theta }{e}^{i\Phi }=e^{i\Phi }$, and its formal transpose lowers symbol class order by one. A spatial derivative of order $|\alpha |$ of $e^{i\Phi }a$ has amplitude order at most $N+|\alpha |$. Applying the frequency integration by parts more than $N+|\alpha |+k$ times makes that derivative absolutely integrable, uniformly on smaller compact sets. Every spatial derivative therefore exists and is continuous there. The oscillatory integral distribution is a smooth function near $x_0$, proving the singular support assertion.

For the linear transport equation, use spacetime $X={\mathbb{R}}^n\times (0,\infty )$, frequency $\theta \in {\mathbb{R}}^n$, andThis is a valid phase function, because ${\nabla }_x\Phi =\theta \ne 0$ at nonzero frequency; the oscillatory integral amplitude is in symbol class order zero. The Fourier representation of the Dirac delta function gives the transport of a Dirac point mass:Its rigorous spacetime distribution pairing is $⟨u,\phi ⟩={\int }_0^{\infty }\phi (ct,t)dt$. Consequentlywhere the endpoints vanish since a spacetime test function has compact support in $t>0$. At each fixed $t$, $⟨u(t),\psi ⟩=\psi (ct)$, so weak convergence of distributions gives $u(t)\to {\delta }_0$ as $t\downarrow 0$. This is the required initial trace.

Finally, ${\nabla }_{\theta }\Phi =x-ct$, so the singular support theorem confines singularities to $x=ct$. In fact equality holds: $u$ is a nonzero order-zero distribution on that trajectory and vanishes off it. A smooth function supported on this set of empty interior must vanish, so $u$ cannot be smooth in any neighborhood of a point of the trajectory. ThusThe method of characteristics also proves uniqueness among solutions with a distributional initial trace: the change of variables $y=x-ct$ turns the linear transport equation into ${\partial }_{t}w=0$. Such a distribution is constant in $t$; pairing with spatial test functions reduces this assertion to a distribution with zero derivative is constant. Its initial trace fixes $w={\delta }_0$, so the moving Dirac delta distribution above is the unique solution. There is transport of the singularity along the characteristic and no smoothing.

Use the supplied oscillator expansion, with the boundary-term convention for momentum stated in part (b). Use a real polarization vector basis, as in the supplied unconjugated polarization completeness relation. Raise its second field index to obtainThe mixed commutator contains only the annihilation-creation and creation-annihilation terms. Their signs are both positive after combining the minus sign in the momentum expansion with the negative Minkowski metric oscillator commutator. Setting $\mathbf{r}=\mathbf{x}-\mathbf{y}$ and using the momentum Dirac delta distribution givesHere the last equality is the Fourier representation of the Dirac delta function. Similarly,because the integrand is odd under $\mathbf{p}\mapsto -\mathbf{p}$. The momentum-momentum commutator is proportional to the same odd difference, now weighted by $|\mathbf{p}|/2$, and also vanishes. Therefore the equal-time canonical commutation relations areThese are identities of operator-valued distributions, understood after smearing. The extra indices in the TeX polarization relation are transcription defects; the PDF has the ordinary two-polarization completeness contraction used above. The direct momenta of part (b) also satisfy the canonical commutation relations after their boundary-generated shift, although they do not have the unmodified mode expansion printed here.

Since the odd sine part of the complex exponential integrates to zero,Part (e)(ii) therefore yieldsin the sense of distributions. This is the Fourier representation of the Dirac delta function.

The $k_0^{2}uv$ terms cancel in $u{L}_{H}v-v{L}_{H}u$. For a ball $B_R$, the product rule and divergence theorem give Green's second identity in the formThus the whole-space integral vanishes when this boundary term tends to zero. For example, uniform bounds $u,v=O(R^{-1-ϵ})$ and $\nabla u,\nabla v=O(R^{-1-ϵ})$, with $ϵ>0$, make it $O(R^{-2ϵ})$.

Pointwise vanishing alone, as stated in the source, is insufficient. Smooth fields agreeing with $u=\sin (k_{0}r)/r$ and $v=\cos (k_{0}r)/r$ outside a ball both tend to zero, but their distant boundary integral is $-4\pi k_0$. One may smooth them inside the ball without changing that flux. The requested zero identity therefore needs sufficient decay or boundary/radiation conditions that actually eliminate the boundary term.

For the generalized plane wave eigenfunctions, their orthogonality is distributional:Here $\int e^{iqx}dx=2\pi \delta (q)$ is the Fourier representation of the Dirac delta function. These eigenfunctions are not ordinary square-integrable functions or decaying test fields.

Expand $\Phi =\int A_k{\psi }_k{d}^{3}k$ and define $V_k=\int {\psi }_k^{\ast }(r^{\prime })V(r^{\prime })d^3{r}^{\prime }$. Since $L_H{\psi }_k=(k_0^2-k^2){\psi }_k$, projection of the source equation gives $(k_0^2-k^2)A_k=V_k$. Formally dividing and substituting back givesAt the shell $k=k_0$, a pole prescription is required; homogeneous solutions can otherwise be added. The formula is an inverse of the Helmholtz equation with that convention still to be selected.

Let $H$ denote the kernel in this source formula, so $L_{H}H=+{\delta }^{(3)}$. Put $l=|r-r^{\prime }|$ and align the Fourier polar axis with $r-r^{\prime }$. Angular integration givesConsequently the radial Fourier representation of the Helmholtz Green function givesThe last Green function written in the PDF has the opposite denominator. It is obtained by setting $G=-H$, the common convention $L_{H}G=-{\delta }^{(3)}$:Thus the two displayed source and Green formulas are consistent only when this minus sign is retained; $G$ cannot also be the positive-delta inverse kernel.

Finally the integral is not an ordinary improper integral across $k=k_0$. A Cauchy principal value yields $G_{\mathrm{PV}}(l)=\cos (k_{0}l)/(4\pi l)$. The outgoing prescription $k^2-k_0^2-i0$ yields the outgoing Green function for the three-dimensional Helmholtz equation, $G_{+}(l)=e^{i{k}_{0}l}/(4\pi l)$; the opposite sign gives the incoming wave. The Sokhotski–Plemelj formula accounts for their difference. The source does not specify which prescription is intended, so the boxed radial expression must be read with one chosen consistently.

For the constant-velocity linear transport equation $u_t+c\cdot {\nabla }_{x}u=0$, a Dirac delta distribution initially at zero moves along $x=ct$. Its spacetime distribution pairs by $⟨u,\phi ⟩=\int \phi (ct,t)dt$, and its initial trace is obtained by weak convergence of distributions. The Fourier representation of the Dirac delta function writes it as an oscillatory integral with phase function $(x-ct)\cdot \theta$. Its singular support is precisely the trajectory; it has order-zero distribution regularity as a measure, without becoming a smooth function.