Worked mathematical example B06 — a cutoff coherent soft state

Put the incoming leg at $p_{\mathrm{i}\mathrm{n}}=E(1,\mathbf{0})$ with ${\eta }_{\mathrm{i}\mathrm{n}}=-1,Q_{\mathrm{i}\mathrm{n}}=1$, and the outgoing leg at $p_{\mathrm{o}\mathrm{u}\mathrm{t}}=E^{\mathrm{\prime }}(1,v{\mathbf{e}}_z)$ with ${\eta }_{\mathrm{o}\mathrm{u}\mathrm{t}}=+1,Q_{\mathrm{o}\mathrm{u}\mathrm{t}}=1$. The charge balance is $-1+1=0$. For $x=n_z$, substitution into the definition of $K$ gives $K(n)=(\frac{vx}{1-vx},0,0,\frac{v}{1-vx}),K^0=n\cdot \mathbf{K}\mathrm{.}$ Therefore $\mathrm{\mid }{K}_{\perp }(n){\mathrm{\mid }}^2=\mathrm{\mid }\mathbf{K}{\mathrm{\mid }}^2-(n\cdot \mathbf{K}{)}^2=\frac{v^2(1-x^2)}{(1-vx{)}^2}\mathrm{.}$ Every denominator is positive because $1-vx\ge 1-v>0$.

Use the spherical chart $n=(\sqrt{1-x^2}\cos \varphi ,\sqrt{1-x^2}\sin \varphi ,x)$ for $-1<x<1$, $0<\varphi <2\pi$. Its two tangent vectors are orthogonal and have squared lengths $(1-x^2{)}^{-1}$ and $1-x^2$, so the induced area density is $1$. Away from the poles, the azimuthal seam is the coordinate line $x_2=0$ in the graph chart over $(x_2,x_3)$; at each pole, its graph chart over $(x_1,x_2)$ maps the pole to a singleton contained in a coordinate line. Under [A1], these parameter sets are null by [F4]; integrating the measurable chart area density over a null parameter set gives zero, so the omitted seam and poles have surface measure zero. Hence $d\mathrm{\Omega }=dxd\varphi$ in this chart. It follows from step 1.1 that ${\int }_{S^2}\mathrm{\mid }{K}_{\perp }{\mathrm{\mid }}^{2}d\mathrm{\Omega }=2\pi I(v),I(v)={\int }_{-1}^1\frac{v^2(1-x^2)}{(1-vx{)}^2}dx\mathrm{.}$ With $u=1-vx$, reversing the limits gives $I(v)=\frac{1}{v}{\int }_{1-v}^{1+v}(\frac{v^2-1}{u^2}+\frac{2}{u}-1)du=\frac{2}{v}\log \frac{1+v}{1-v}-4.$ The antiderivative is $v^{-1}((1-v^2)\mathrm{/}u+2\log u-u)$, using the derivative and integral formula for $\log$ from [F5]; evaluation at both endpoints gives the displayed result. Positivity also follows directly from the integrand, which is strictly positive for $-1<x<1$.

Substituting the angular integral into [F3] gives $C(v)=\frac{g^2}{2(2\pi {)}^3}2\pi I(v)=\frac{g^2}{4{\pi }^2}(\frac{1}{v}\log \frac{1+v}{1-v}-2)=\frac{\alpha }{\pi }(\frac{1}{v}\log \frac{1+v}{1-v}-2),$ where [F1] gives $g^2=4\pi \alpha$. At $v=1\mathrm{/}2$, the bracket is $2\log 3-2>0$ by the positive integral in step 2.1. For $\lambda =\mathrm{\Lambda }\mathrm{/}\mathrm{e}$, [F3] and [F6] give $N_{\lambda }=C\log \mathrm{e}=C$. The sector formula then yields $s_0=\exp (-C)$ and ${\sum }_{m\ge 0}m{s}_m=C$ by the exponential series.

The free-energy quadratic expectation from [F3] at this cutoff is $C(\mathrm{\Lambda }-\lambda )=C\mathrm{\Lambda }(1-{\mathrm{e}}^{-1})$. If the lower cutoff is halved, then $N_{\lambda \mathrm{/}2}-N_{\lambda }=C\log 2>0$. The substitution $\omega =(\lambda \mathrm{/}2)u$ and [F5] give the shell norm $C{\int }_{\lambda \mathrm{/}2}^{\lambda }d\omega \mathrm{/}\omega =C\log 2$. For each integer $j\ge 0$ set ${\lambda }_j=\lambda 2^{-j}$; these are all admissible cutoffs. The exact coherent overlap identity of [F3] gives $\mathrm{\mid }⟨{\mathrm{\Psi }}_{{\lambda }_j},{\mathrm{\Psi }}_{{\lambda }_{j+1}}⟩\mathrm{\mid }=\exp (-C\log 2\mathrm{/}2)$. For every $\mathrm{\mid }z\mathrm{\mid }=1$, expanding the norm gives $\mathrm{\parallel }{\mathrm{\Psi }}_{{\lambda }_{j+1}}-z{\mathrm{\Psi }}_{{\lambda }_j}{\mathrm{\parallel }}^2\ge 2(1-\exp (-C\log 2\mathrm{/}2))>0$. By the projective distance defined in [F3], adjacent rays are therefore separated by at least $\sqrt{2(1-\exp (-C\log 2\mathrm{/}2))}>0$. If the ray sequence converged, the triangle inequality for that metric would force the adjacent distances to tend to zero, a contradiction. The example does not use the supplier's weak-zero or no-strong-limit conclusions. No Born rule, actual hard-scattering dynamics, detector law, or empirical claim is part of this calculation.