One-mode oscillator, Feynman boundary prescription, and retarded contrast

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 $\mathrm{\mid }\phi (0)\mathrm{\mid }\le p_{00}(\phi )$.

Put $d_o=\sqrt{\mathrm{\hslash }\mathrm{/}(2m_o\omega )}$. From step 1.1, $X{\psi }_0=d_o{\psi }_1$ and $X{\psi }_1=d_o({\psi }_0+\sqrt{2}{\psi }_2)$. For $t\ge 0$, applying $X(t)=U(t{)}^{†}XU(t)$ to $X{\psi }_0$ and taking the ground-state coefficient gives $(X(t)X{\psi }_0,{\psi }_0{)}_1=d_o^2{e}^{-i\omega t}\mathrm{.}$ For $t<0$, time ordering gives $(XX(t){\psi }_0,{\psi }_0{)}_1$. Since $U(t){\psi }_0=e^{-i\omega t\mathrm{/}2}{\psi }_0$ and $U(t{)}^{†}{\psi }_1=e^{3i\omega t\mathrm{/}2}{\psi }_1$, step 1.1 gives $X(t){\psi }_0=d_o{e}^{i\omega t}{\psi }_1$; applying $X$ and taking the ground-state coefficient gives $d_o^2{e}^{i\omega t}$. Multiplying by $m_o\mathrm{/}\mathrm{\hslash }$ and using $m_o{d}_o^2\mathrm{/}\mathrm{\hslash }=1\mathrm{/}(2\omega )$ proves $C_F(t)=e^{-i\omega \mathrm{\mid }t\mathrm{\mid }}\mathrm{/}(2\omega )$.

For $ϵ>0$, the damped kernel is integrable, and its positive-sign Fourier transform is the sum of two ordinary half-line integrals: ${\int }_{\mathbb{R}}{e}^{i\nu t}{e}^{-ϵ\mathrm{\mid }t\mathrm{\mid }}{C}_F(t)dt=\frac{1}{2\omega }(\frac{1}{ϵ+i(\omega -\nu )}+\frac{1}{ϵ+i(\omega +\nu )})=\frac{i}{2\omega }(\frac{1}{\nu -\omega +iϵ}-\frac{1}{\nu +\omega -iϵ})\mathrm{.}$ The first equality follows by integrating $e^{-(ϵ+i(\omega -\nu ))t}$ on $t\ge 0$ and $e^{-(ϵ+i(\omega +\nu ))\mathrm{\mid }t\mathrm{\mid }}$ on $t<0$; both real decay rates are $ϵ>0$.

All bounded kernels used here are tempered: $\mathrm{\mid }\int f\phi \mathrm{\mid }\le 2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}(p_{00}(\phi )+p_{20}(\phi ))$. Derivative actions are continuous because $p_{jk}({\phi }^{(r)})=p_{j,k+r}(\phi )$. The function $C_F$ is continuous at $0$, solves $C_F^{\mathrm{\prime }\mathrm{\prime }}+{\omega }^2{C}_F=0$ classically on each open half-line, and has one-sided derivatives $C_F^{\mathrm{\prime }}(0+)=-i\mathrm{/}2$ and $C_F^{\mathrm{\prime }}(0-)=i\mathrm{/}2$. For a test $\phi \in \mathcal{S}(\mathbb{R})$, integrate ${\int }_{-\mathrm{\infty }}^0{C}_F{\phi }^{\mathrm{\prime }\mathrm{\prime }}+{\int }_0^{\mathrm{\infty }}{C}_F{\phi }^{\mathrm{\prime }\mathrm{\prime }}$ by parts twice; rapid decay of the test and its derivative removes the boundary terms at infinity. The boundary terms in ${\phi }^{\mathrm{\prime }}(0)$ cancel because $C_F$ is continuous; the remaining coefficient of $\phi (0)$ is $C_F^{\mathrm{\prime }}(0+)-C_F^{\mathrm{\prime }}(0-)=-i$. Thus $({\mathrm{\partial }}_t^2+{\omega }^2)C_F=-i{\delta }_0$.

The function $G_R$ is zero on $t<0$, solves $G_R^{\mathrm{\prime }\mathrm{\prime }}+{\omega }^2{G}_R=0$ on $t>0$, is continuous at zero with $G_R(0)=0$, and has derivative jump $G_R^{\mathrm{\prime }}(0+)-G_R^{\mathrm{\prime }}(0-)=1$. The same two integrations by parts show $({\mathrm{\partial }}_t^2+{\omega }^2)G_R={\delta }_0$. Since $C_F(t)$ has modulus $1\mathrm{/}(2\omega )>0$ for $t<0$ but $G_R(t)=0$ there, the Feynman kernel is not the retarded response.

To justify the distribution limit, the Fourier transform in [F6] and the rescaling in [F5] preserve the Schwartz test space. The time-side kernels $e^{-ϵ\mathrm{\mid }t\mathrm{\mid }}{C}_F(t)$ converge to $C_F$ against every such transformed test: on $\mathrm{\mid }t\mathrm{\mid }\le R$, $\mathrm{\mid }{e}^{-ϵ\mathrm{\mid }t\mathrm{\mid }}-1\mathrm{\mid }\le ϵR$, and the tail is bounded by $(1\mathrm{/}\omega ){\int }_{\mathrm{\mid }t\mathrm{\mid }>R}\mathrm{\mid }{\mathcal{T}}_{+}\phi (t)\mathrm{\mid }dt$, which tends to zero as $R\to \mathrm{\infty }$. For $ϵ>0$, Fubini [F7] identifies the transform function in step 3.1 with this dual action, since the absolute double integral is at most $\mathrm{\parallel }{e}^{-ϵ\mathrm{\mid }\cdot \mathrm{\mid }}{C}_F{\mathrm{\parallel }}_1\mathrm{\parallel }\phi {\mathrm{\parallel }}_1$. Applying the Plemelj limits in steps 1.2 and 2.1 to the two shifted poles gives the displayed boundary distribution. Combining the rational terms at finite $ϵ$ gives $(i+ϵ\mathrm{/}\omega )\mathrm{/}({\nu }^2-{\omega }^2+2iϵ\omega +{ϵ}^2)$, so the numerator tends to $i$ and the two poles approach from opposite sides. In the $i0$ shorthand, the exact partial-fraction definition is $\frac{i}{{\nu }^2-{\omega }^2+i0}:=\frac{i}{2\omega }(\frac{1}{\nu -\omega +i0}-\frac{1}{\nu +\omega -i0}),$ where each $i0$ is the boundary proved above; the opposite sign at the second pole follows from ${\nu }^2-{\omega }^2=(\nu -\omega )(\nu +\omega )$ and $\omega >0$. No contour integration or unproved pole rule is used.