Example conventions, energy units and constructed/formal data

From ${ϵ}_0{\mu }_0{c}^2=1$ and positivity, $\alpha =e^2{\mu }_{0}c\mathrm{/}(4\pi \mathrm{\hslash })>0$. Hence $g^2=e^2{\mu }_{0}c\mathrm{/}\mathrm{\hslash }=4\pi \alpha$, and the positive square root gives the stated equality $g=\sqrt{4\pi \alpha }$. The units of $g^2$ are ${\mathrm{C}}^2(\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{C}}^{-2})(\mathrm{m}{\mathrm{s}}^{-1})\mathrm{/}(\mathrm{J}\mathrm{s})=1$, so $g$ is dimensionless.

Since $[ct]=\mathrm{m}$, $x^0$ is a length; since $[c{\mathbf{p}}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}]=\mathrm{J}$, every component of $p$ and $M$ is an energy. Thus $p^2$ and $s$ have units ${\mathrm{J}}^2$, while division of energy by $\mathrm{\hslash }c$ has units $\mathrm{J}\mathrm{/}(\mathrm{J}\mathrm{m})={\mathrm{m}}^{-1}$. For the adopted dimensionless kinetic actions, $4+2[\phi ]-2=0$ and $4+2[\mathrm{\Psi }]-1=0$ in length degrees, yielding the assigned field dimensions. The stated $\mu$ has units $\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-1}\mathrm{/}(\mathrm{J}\mathrm{s})={\mathrm{m}}^{-1}$.

In SI, $[\mathrm{T}]=\mathrm{k}\mathrm{g}{\mathrm{s}}^{-1}{\mathrm{C}}^{-1}$, so $[q{A}_a\mathrm{/}\mathrm{\hslash }]={\mathrm{m}}^{-1}$ and $[q\chi \mathrm{/}\mathrm{\hslash }]=1$ because $\mathrm{W}\mathrm{b}=\mathrm{J}\mathrm{s}{\mathrm{C}}^{-1}$. Also $A_{\mathrm{n}\mathrm{a}\mathrm{t}}$ has units ${\mathrm{m}}^{-1}$ and $q_{\mathrm{n}\mathrm{a}\mathrm{t}}$ is dimensionless. Therefore the derivative and gauge phase terms in the Definition are dimensionally compatible. Finally $u^{†}v=\overline{(u,v{)}_1}$, so their norms and modulus squares agree.

The matrix exponential is defined by its norm-convergent power series; termwise differentiation on compact time intervals gives $W^{\mathrm{\prime }}(t)=i{H}_{0}W(t)\mathrm{/}\mathrm{\hslash }$ for $W(t)=e^{it{H}_0\mathrm{/}\mathrm{\hslash }}$. Hence $(W(t{)}^{†}W(t){)}^{\mathrm{\prime }}=0$ and $W(0{)}^{†}W(0)=I$, so $W$ is unitary (in finite dimension a left inverse is an inverse). Therefore $V_I(t)=W(t)V(t)W(t{)}^{†}$ is Hermitian and norm-continuous. Its norm is bounded on compact $[0,T]$, so $B<\mathrm{\infty }$.

Recursion expresses ${\lambda }^n{S}_n(T)$ as an $n$-fold ordered integral with integrand of norm at most $(\mathrm{\mid }\lambda \mathrm{\mid }B\mathrm{/}\mathrm{\hslash }{)}^n$. The volume of $0<t_n<\cdots <t_1<T$ is $T^n\mathrm{/}n!$: induction integrates the preceding volume $t^{n-1}\mathrm{/}(n-1)!$ from $0$ to $T$. Thus $\mathrm{\parallel }{\lambda }^n{S}_n(T)\mathrm{\parallel }\le z^n\mathrm{/}n!$ and the same bound with $T$ replaced by any $t\in [0,T]$ holds uniformly. The scalar exponential series proves uniform absolute convergence.

For ${\sigma }_x^2=I$, separate the even and odd powers in the absolutely convergent exponential series: $e^{-iz{\sigma }_x}={\sum }_{k\ge 0}\frac{(-1{)}^k{z}^{2k}}{(2k)!}I-i{\sum }_{k\ge 0}\frac{(-1{)}^k{z}^{2k+1}}{(2k+1)!}{\sigma }_x=\cos zI-i\sin z{\sigma }_x,$ where $z=g_{E}t\mathrm{/}\mathrm{\hslash }$. Its off-diagonal entry has modulus square ${\sin }^{2}z$, and the first-order term has modulus square $z^2$. Taking $\mathrm{\mid }z\mathrm{\mid }>1$ makes the latter exceed $1$. Also ${\cos }^{2}z$ has derivative $0$ at $t=0$, whereas $e^{-\mathrm{\Gamma }t}$ has derivative $-\mathrm{\Gamma }<0$ there; so they cannot agree for all $t$ when $\mathrm{\Gamma }>0$.

For the tail, write $n=N+1+k$. Then ${\sum }_{n=N+1}^{\mathrm{\infty }}\frac{z^n}{n!}=\frac{z^{N+1}}{(N+1)!}{\sum }_{k=0}^{\mathrm{\infty }}\frac{z^k}{(N+2)\cdots (N+1+k)}\le \frac{z^{N+1}}{(N+1)!}{\sum }_{k=0}^{\mathrm{\infty }}\frac{z^k}{k!}=e^z\frac{z^{N+1}}{(N+1)!},$ where an empty product is $1$ and $(N+2)\cdots (N+1+k)\ge k!$. The norm tail is bounded by this scalar tail.

The family $\mathcal{H}(t)=\lambda V_I(t)$ satisfies the exact hypotheses of [F2] and [F3] on the compact interval $[0,T]$: it is bounded, norm-continuous and self-adjoint, and the required ${\mathrm{A}\mathrm{C}}_{\omega }$ is [A1]. The recursion above is its Dyson recursion, so [F2] supplies the Volterra equation and uniqueness, [F3] supplies unitarity of $S(t)$, and step 1.2 supplies the displayed sharper bound with $B$.

From $S=I+iT$, $S^{†}S=I$ expands to $i(T-T^{†})+T^{†}T=0$. Its $ii$ entry is $-2\mathrm{Im}{T}_{ii}+{\sum }_f\mathrm{\mid }{T}_{fi}{\mathrm{\mid }}^2=0$, proving the exact identity. The norm-convergent power series makes $T(\lambda )$ analytic near $0$ (indeed entire), so coefficient comparison is valid: the $\lambda$ term gives $T_1=T_1^{†}$, and the ${\lambda }^2$ term gives $i(T_2-T_2^{†})+T_1^{†}{T}_1=0$, hence $2{\mathrm{Im}}_{\mathrm{H}}{T}_2=T_1^{†}{T}_1$.

For the linear truncation $S_{\mathrm{t}\mathrm{r}}=I+i\lambda T_1$ with $T_1=T_1^{†}$, multiplication gives $S_{\mathrm{t}\mathrm{r}}^{†}{S}_{\mathrm{t}\mathrm{r}}=I+{\lambda }^2{T}_1^2$. If $\lambda T_1\mathrm{\ne }0$, then $T_1^2$ is nonzero positive semidefinite because $⟨T_1^{2}v,v⟩=\mathrm{\parallel }{T}_{1}v{\mathrm{\parallel }}^2$; hence this truncation is not unitary. The exact coefficient relation in step 3.1 therefore requires the higher-order correction whenever its right side is nonzero.

For $n<N$, direct application gives $a_N{a}_N^{†}\mathrm{\mid }n⟩=(n+1)\mathrm{\mid }n⟩$ and $a_N^{†}{a}_N\mathrm{\mid }n⟩=n\mathrm{\mid }n⟩$, so the commutator equals $\mathrm{\mid }n⟩$. At $n=N$, the two terms are $0$ and $N\mathrm{\mid }N⟩$, so their difference is $-N\mathrm{\mid }N⟩$. These are exactly the values of $I-(N+1)\mathrm{\mid }N⟩⟨N\mathrm{\mid }$ on each basis vector. For arbitrary finite matrices $A,B$, $\mathrm{tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}=\mathrm{tr}(BA)$ by swapping dummy indices, so $\mathrm{tr}[A,B]=0$; but $\mathrm{tr}I=d>0$. Thus the unmodified canonical commutator cannot hold on a finite-dimensional space.

For $\omega \mathrm{\ne }0$, integration of the exponential gives ${\int }_0^T{e}^{i\omega t}dt=(e^{i\omega T}-1)\mathrm{/}(i\omega )$. Fact [F2] gives its squared modulus $4{\sin }^2(\omega T\mathrm{/}2)\mathrm{/}{\omega }^2$. At $\omega =0$ the integral is $T$, so the continuation is $K_T(0)=T$. For any integrable complex $q$, put $I=\int q$. If $I=0$, $\mathrm{\mid }I\mathrm{\mid }\le \int \mathrm{\mid }q\mathrm{\mid }$ is immediate; otherwise $\alpha =\bar{I}\mathrm{/}\mathrm{\mid }I\mathrm{\mid }$ gives $\mathrm{\mid }I\mathrm{\mid }=\int \mathrm{Re}(\alpha q)\le \int \mathrm{\mid }q\mathrm{\mid }$. Applied to $q(t)=e^{i\omega t}$ on $[0,T]$, with $\mathrm{\mid }q\mathrm{\mid }=1$ by [F2], this gives $K_T\le T$. Also $K_T\le 4\mathrm{/}(T{\omega }^2)$ for $\omega \mathrm{\ne }0$. These bounds show $K_T$ is integrable and establish the tail estimate by integrating ${\omega }^{-2}$ on both rays.

Let $g=1_{[0,T]}$. It belongs to $L^1(\mathbb{R})\cap L^2(\mathbb{R})$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=T$. For $\nu =-\omega \mathrm{/}(2\pi )$, its integral transform is $\widehat{g}(\nu )={\int }_0^T{e}^{i\omega t}dt$. By [F1], this integral represents the $L^2$ transform almost everywhere and Plancherel gives ${\int }_{\mathbb{R}}\mathrm{\mid }\widehat{g}(\nu ){\mathrm{\mid }}^{2}d\nu =T$. The linear substitution $\omega =-2\pi \nu$ [F3] gives ${\int }_{\mathbb{R}}{K}_T(\omega )d\omega =\frac{2\pi }{T}{\int }_{\mathbb{R}}\mathrm{\mid }\widehat{g}(\nu ){\mathrm{\mid }}^{2}d\nu =2\pi \mathrm{.}$ The affine substitution is the one-dimensional case of [F3].

Since $\int K_T\mathrm{/}(2\pi )=1$, subtracting $f(0)$ writes the error as $\int (K_T\mathrm{/}(2\pi ))(f-f(0))$. On $\mathrm{\mid }\omega \mathrm{\mid }\le r$ the absolute difference is at most ${\omega }_f(r)$ and the kernel mass there is at most one. On $\mathrm{\mid }\omega \mathrm{\mid }>r$ it is at most $2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, and the mass is at most $4\mathrm{/}(\pi Tr)$ by step 1.1. Adding these bounds proves the estimate.

Given $\epsilon >0$, continuity at zero gives $r>0$ with ${\omega }_f(r)<\epsilon \mathrm{/}2$. For this fixed $r$, choose $T$ large enough that $8\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}(\pi Tr)<\epsilon \mathrm{/}2$. Step 2.1 then gives error below $\epsilon$. Thus the approximate-identity limit is $f(0)$.

The bounds in step 1.1 make $K_T(\omega )w(E_0+\mathrm{\hslash }\omega )$ integrable because $w$ is bounded. In the definition of $P_2$, substitute $E=E_0+\mathrm{\hslash }\omega$; this is an affine $C^1$ diffeomorphism with Jacobian $\mathrm{\hslash }>0$, so [F3] applies to the nonnegative integrand. Since ${\mid {\int }_0^T{e}^{i\omega t}dt\mid }^2=T{K}_T(\omega ),$ we obtain $\frac{P_2(T)}{T}=\frac{{\lambda }^2}{\mathrm{\hslash }}{\int }_{\mathbb{R}}{K}_T(\omega )w(E_0+\mathrm{\hslash }\omega )d\omega =\frac{2\pi {\lambda }^2}{\mathrm{\hslash }}{\int }_{\mathbb{R}}\frac{K_T(\omega )}{2\pi }f(\omega )d\omega ,$ where $f(\omega )=w(E_0+\mathrm{\hslash }\omega )$ is bounded and continuous at zero. Step 3.1 gives the claimed limit. The units are ${\mathrm{s}}^{-1}$ because $w(E_0)\mathrm{/}\mathrm{\hslash }$ has units $\mathrm{J}\mathrm{/}(\mathrm{J}\mathrm{s})$. No exact-dynamics or probability claim follows from this coefficient calculation.

Let $\mathrm{\ell }=\sqrt{\mathrm{\hslash }\mathrm{/}(m_o\omega )}$ and ${\psi }_n(x)={\mathrm{\ell }}^{-1\mathrm{/}2}{g}_n(x\mathrm{/}\mathrm{\ell })$ as in [F2] and [F11]. The exact hypotheses include the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ [A1]. Rescaling [F3] gives $X{\psi }_n=\sqrt{\frac{\mathrm{\hslash }}{2m_o\omega }}(\sqrt{n}{\psi }_{n-1}+\sqrt{n+1}{\psi }_{n+1}),$ with ${\psi }_{-1}=0$. Thus $X$ maps the finite span $\mathcal{C}$ into itself, so all products in the statement are defined on ${\psi }_0$. On a finite vector ${\sum }_{n=0}^N{c}_n{\psi }_n$, the phase rule preserves norm by orthonormality [F11]; its inverse is the opposite phase rule. Density of $\mathcal{C}$ follows from completeness in [F11], and Hilbert completeness [F12] gives unique continuous extensions of both inverse isometries to the whole oscillator Hilbert space. Hence $U(t)$ is unitary and agrees on $\mathcal{C}$ with the energy evolution [F2].

Fix $\phi \in \mathcal{S}(\mathbb{R})$. Set $\chi =1_{[-1,1]}$ and $\psi =\phi -\phi (0)\chi$. The function $\chi$ is even and integrable, $\psi (0)=0$, and $\psi (x)\mathrm{/}x$ is bounded near zero because $\chi =1$ there and $\phi$ is differentiable. Outside $[-1,1]$, $\psi =\phi$, so $\psi (x)\mathrm{/}x$ is integrable at infinity; on compact sets away from zero it is also integrable. For any integrable complex $q$, put $I=\int q$. If $I=0$, $\mathrm{\mid }I\mathrm{\mid }\le \int \mathrm{\mid }q\mathrm{\mid }$ is immediate; otherwise $\alpha =\bar{I}\mathrm{/}\mathrm{\mid }I\mathrm{\mid }$ gives $\mathrm{\mid }I\mathrm{\mid }=\int \mathrm{Re}(\alpha q)\le \int \mathrm{\mid }q\mathrm{\mid }$. Decompose the regularized integral as $R_{ϵ}-i{J}_{ϵ}$, where $R_{ϵ}=\int x\phi (x)\mathrm{/}(x^2+{ϵ}^2)dx$ and $J_{ϵ}=\int ϵ\phi (x)\mathrm{/}(x^2+{ϵ}^2)dx$. These are complex kernel contributions, not scalar real and imaginary parts. The first contribution is ${\int }_{\mathbb{R}}\frac{x\psi (x)}{x^2+{ϵ}^2}dx={\int }_{\mathbb{R}}\frac{\psi (x)}{x}\frac{x^2}{x^2+{ϵ}^2}dx,$ because the omitted $\phi (0)\chi$ term is odd. For $0<r<1$, the error from replacing $x^2\mathrm{/}(x^2+{ϵ}^2)$ by $1$ on $\mathrm{\mid }x\mathrm{\mid }<r$ is at most $2r{p}_{01}(\phi )$, since $\mathrm{\mid }\psi (x)\mathrm{/}x\mathrm{\mid }\le p_{01}(\phi )$ there (assign its derivative limit at $x=0$). On $\mathrm{\mid }x\mathrm{\mid }\ge r$, the error is at most $({ϵ}^2\mathrm{/}{r}^2)\mathrm{\parallel }\psi \mathrm{/}x{\mathrm{\parallel }}_1$. Given $\eta >0$, choose a fixed $r\in (0,1)$ such that $2r{p}_{01}(\phi )<\eta \mathrm{/}2$; then choose ${ϵ}_0>0$ such that $({ϵ}_0^2\mathrm{/}{r}^2)\mathrm{\parallel }\psi \mathrm{/}x{\mathrm{\parallel }}_1<\eta \mathrm{/}2$. The total error is below $\eta$ for every $0<ϵ<{ϵ}_0$, proving convergence to $\int \psi (x)\mathrm{/}xdx$. Since $\mathrm{PV}\int \chi (x)\mathrm{/}xdx=0$, this is $\mathrm{PV}\int \phi (x)\mathrm{/}xdx$. Moreover its limiting action is continuous: $\mathrm{\mid }\int \psi \mathrm{/}x\mathrm{\mid }\le 2p_{01}(\phi )+p_{20}(\phi )$, by the derivative bound on $\mathrm{\mid }x\mathrm{\mid }<1$ and the tail bound $\mathrm{\mid }\phi (x)\mathrm{/}x\mathrm{\mid }\le p_{20}(\phi )\mathrm{\mid }x{\mathrm{\mid }}^{-3}$ for $\mathrm{\mid }x\mathrm{\mid }>1$. Thus the principal value is a tempered distribution.

The second kernel contribution is $-i{J}_{ϵ}=-i\int \phi (x)ϵ\mathrm{/}(x^2+{ϵ}^2)dx$. The nonnegative kernel has integral $\pi$, by the antiderivative $\arctan (x\mathrm{/}ϵ)$. For any $r>0$, continuity bounds the error from replacing $\phi (x)$ by $\phi (0)$ on $\mathrm{\mid }x\mathrm{\mid }\le r$ by $\pi {\sup }_{\mathrm{\mid }x\mathrm{\mid }\le r}\mathrm{\mid }\phi (x)-\phi (0)\mathrm{\mid }$. On $\mathrm{\mid }x\mathrm{\mid }>r$, the kernel mass is at most $2ϵ\mathrm{/}r$, so the error is at most $4\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}ϵ\mathrm{/}r$. First choose $r$ small and then $ϵ$ small. This proves $J_{ϵ}\to \pi \phi (0)$ and hence $R_{ϵ}-i{J}_{ϵ}\to \mathrm{PV}\int \phi (x)\mathrm{/}xdx-i\pi \phi (0)$ for complex tests. Replacing $+iϵ$ by $-iϵ$ changes the second contribution to $+i{J}_{ϵ}$ and reverses the delta sign. Evaluation is continuous since $\mathrm{\mid }\phi (0)\mathrm{\mid }\le p_{00}(\phi )$.