Merge branch 'main' of github.com:project-ida/notes

Author: mklilley

latex/Pion Dressing and Magnetic Transitions.tex

@@ -1435,6 +1435,347 @@ \subsubsection{9.8 Symmetric limit for identical
 symmetric identical-system limit, that is precisely the condition for
 destructive interference between the \(bb\) and \(dd\) virtual pathways.
 
+\subsubsection{9.9 Solving for the actual low-energy
+roots}\label{solving-for-the-actual-low-energy-roots}
+
+We now keep the symmetric identical-system assumptions of Section 9.8
+and go one step further: instead of treating \(E\) as a common external
+parameter, we solve the Bloch-Horowitz self-consistency equations for
+the actual low-energy roots.
+
+For this subsection, take
+
+\[
+E_0=0,
+\qquad
+E_\chi=E_0=0,
+\label{eq:E0andEchiZero}
+\]
+
+and define
+
+\[
+E_\pi \equiv \omega_\pi.
+\label{eq:EpiDef}
+\]
+
+Then Eq. \(\ref{eq:commonDeltaEdef}\) reduces to
+
+\[
+\Delta(E)=E_\pi-E.
+\label{eq:DeltaSimple}
+\]
+
+It is also convenient to absorb the symmetric coupling into
+
+\[
+\Omega^2 \equiv 2W^2.
+\label{eq:OmegaDefLast}
+\]
+
+With this notation, the diagonal entries of Eq.
+\(\ref{eq:HABdiagIdentical}\) become
+
+\[
+\lambda_{bb}(E)=-\frac{2\Omega^2}{E_\pi-E},
+\label{eq:lambdabbSimple}
+\]
+
+\[
+\lambda_{bd}(E)=\lambda_{db}(E)=-\frac{\Omega^2}{E_\pi-E},
+\label{eq:lambdabdSimple}
+\]
+
+\[
+\lambda_{dd}(E)=0.
+\label{eq:lambdaddSimple}
+\]
+
+The physical dressed energies are found from the self-consistency
+conditions
+
+\[
+E_{bb}=\lambda_{bb}(E_{bb}),
+\qquad
+E_{bd}=E_{db}=\lambda_{bd}(E_{bd}),
+\qquad
+E_{dd}=0.
+\label{eq:selfConsLast}
+\]
+
+\paragraph{Exact quadratic equations}\label{exact-quadratic-equations}
+
+For the singly-bright states \(bd\) and \(db\),
+
+\[
+E_{bd}=-\frac{\Omega^2}{E_\pi-E_{bd}},
+\label{eq:EbdImplicit}
+\]
+
+so
+
+\[
+E_{bd}(E_\pi-E_{bd})=-\Omega^2,
+\label{eq:EbdQuadraticStep}
+\]
+
+and therefore
+
+\[
+E_{bd}^2-E_\pi E_{bd}-\Omega^2=0.
+\label{eq:EbdQuadratic}
+\]
+
+Thus
+
+\[
+E_{bd}^{(\pm)}=E_{db}^{(\pm)}=\frac{E_\pi\pm\sqrt{E_\pi^2+4\Omega^2}}{2}.
+\label{eq:EbdRoots}
+\]
+
+For the doubly-bright state \(bb\),
+
+\[
+E_{bb}=-\frac{2\Omega^2}{E_\pi-E_{bb}},
+\label{eq:EbbImplicit}
+\]
+
+so
+
+\[
+E_{bb}(E_\pi-E_{bb})=-2\Omega^2,
+\label{eq:EbbQuadraticStep}
+\]
+
+and hence
+
+\[
+E_{bb}^2-E_\pi E_{bb}-2\Omega^2=0.
+\label{eq:EbbQuadratic}
+\]
+
+Therefore
+
+\[
+E_{bb}^{(\pm)}=\frac{E_\pi\pm\sqrt{E_\pi^2+8\Omega^2}}{2}.
+\label{eq:EbbRoots}
+\]
+
+Finally,
+
+\[
+E_{dd}=0.
+\label{eq:EddExactLast}
+\]
+
+The plus roots in Eqs. \(\ref{eq:EbdRoots}\) and \(\ref{eq:EbbRoots}\)
+sit near the pion scale and are not the low-energy branches we want. The
+relevant low-energy solutions are therefore
+
+\[
+E_{bd}^{\rm low}=E_{db}^{\rm low}=\frac{E_\pi-\sqrt{E_\pi^2+4\Omega^2}}{2},
+\label{eq:EbdLow}
+\]
+
+\[
+E_{bb}^{\rm low}=\frac{E_\pi-\sqrt{E_\pi^2+8\Omega^2}}{2},
+\label{eq:EbbLow}
+\]
+
+\[
+E_{dd}=0.
+\label{eq:EddLow}
+\]
+
+\paragraph{Small-parameter expansion}\label{small-parameter-expansion}
+
+Define the small parameter
+
+\[
+x\equiv \frac{\Omega^2}{E_\pi^2}\ll 1.
+\label{eq:xSmall}
+\]
+
+Using
+
+\[
+\sqrt{1+u}=1+\frac{u}{2}-\frac{u^2}{8}+O(u^3),
+\label{eq:sqrtExpand}
+\]
+
+we expand Eq. \(\ref{eq:EbdLow}\) as
+
+\[
+E_{bd}^{\rm low}
+=
+\frac{E_\pi-E_\pi\sqrt{1+4x}}{2}
+=
+-E_\pi x+E_\pi x^2+O(E_\pi x^3),
+\label{eq:EbdExpandStep}
+\]
+
+so
+
+\[
+E_{bd}^{\rm low}=E_{db}^{\rm low}
+=
+-\frac{\Omega^2}{E_\pi}
++
+\frac{\Omega^4}{E_\pi^3}
++
+O\!\left(\frac{\Omega^6}{E_\pi^5}\right).
+\label{eq:EbdExpand}
+\]
+
+Similarly, Eq. \(\ref{eq:EbbLow}\) gives
+
+\[
+E_{bb}^{\rm low}
+=
+\frac{E_\pi-E_\pi\sqrt{1+8x}}{2}
+=
+-2E_\pi x+4E_\pi x^2+O(E_\pi x^3),
+\label{eq:EbbExpandStep}
+\]
+
+so
+
+\[
+E_{bb}^{\rm low}
+=
+-\frac{2\Omega^2}{E_\pi}
++
+\frac{4\Omega^4}{E_\pi^3}
++
+O\!\left(\frac{\Omega^6}{E_\pi^5}\right).
+\label{eq:EbbExpand}
+\]
+
+And of course
+
+\[
+E_{dd}=0.
+\label{eq:EddExpand}
+\]
+
+\paragraph{Energy differences}\label{energy-differences}
+
+Now compare the singly-bright state to its neighbours above and below.
+
+The exact spacings are
+
+\[
+E_{dd}-E_{bd}^{\rm low}
+=
+\frac{\sqrt{E_\pi^2+4\Omega^2}-E_\pi}{2},
+\label{eq:ddMinusBdExact}
+\]
+
+and
+
+\[
+E_{bd}^{\rm low}-E_{bb}^{\rm low}
+=
+\frac{\sqrt{E_\pi^2+8\Omega^2}-\sqrt{E_\pi^2+4\Omega^2}}{2}.
+\label{eq:bdMinusBbExact}
+\]
+
+Expanding these to the same order gives
+
+\[
+E_{dd}-E_{bd}^{\rm low}
+=
+\frac{\Omega^2}{E_\pi}
+-
+\frac{\Omega^4}{E_\pi^3}
++
+O\!\left(\frac{\Omega^6}{E_\pi^5}\right),
+\label{eq:ddMinusBdExpand}
+\]
+
+\[
+E_{bd}^{\rm low}-E_{bb}^{\rm low}
+=
+\frac{\Omega^2}{E_\pi}
+-
+3\frac{\Omega^4}{E_\pi^3}
++
+O\!\left(\frac{\Omega^6}{E_\pi^5}\right).
+\label{eq:bdMinusBbExpand}
+\]
+
+So the two spacings are equal at leading order but differ at the next
+order:
+
+\[
+\big(E_{dd}-E_{bd}^{\rm low}\big)-\big(E_{bd}^{\rm low}-E_{bb}^{\rm low}\big)
+=
+\frac{2\Omega^4}{E_\pi^3}
++
+O\!\left(\frac{\Omega^6}{E_\pi^5}\right).
+\label{eq:asymmetryExpand}
+\]
+
+Equivalently, the midpoint shift is
+
+\[
+E_{bd}^{\rm low}-\frac{E_{bb}^{\rm low}+E_{dd}}{2}
+=
+-\frac{\Omega^4}{E_\pi^3}
++
+O\!\left(\frac{\Omega^6}{E_\pi^5}\right).
+\label{eq:midpointShift}
+\]
+
+So the \(bd/db\) pair lies slightly \textbf{below} the exact midpoint
+between \(bb\) and \(dd\) once the self-consistent \(E\)-dependence is
+kept.
+
+\paragraph{\texorpdfstring{Writing the asymmetry in terms of
+\(E_{bd}\)}{Writing the asymmetry in terms of E\_\{bd\}}}\label{writing-the-asymmetry-in-terms-of-e_bd}
+
+Introduce the leading-order singly-bright energy
+
+\[
+E_{bd}^{(0)}\equiv -\frac{\Omega^2}{E_\pi}.
+\label{eq:Ebd0}
+\]
+
+Then
+
+\[
+\frac{\Omega^2}{E_\pi}\frac{E_{bd}^{(0)}}{E_\pi}
+=
+-\frac{\Omega^4}{E_\pi^3}.
+\label{eq:Ebd0Identity}
+\]
+
+So Eq. \(\ref{eq:midpointShift}\) can be written as
+
+\[
+E_{bd}^{\rm low}-\frac{E_{bb}^{\rm low}+E_{dd}}{2}
+=
+\frac{\Omega^2}{E_\pi}\frac{E_{bd}^{(0)}}{E_\pi}
++
+O\!\left(\frac{\Omega^6}{E_\pi^5}\right)
+\label{eq:midpointShiftEbd0}
+\]
+
+Likewise Eq. \(\ref{eq:asymmetryExpand}\) becomes
+
+\[
+\big(E_{dd}-E_{bd}^{\rm low}\big)-\big(E_{bd}^{\rm low}-E_{bb}^{\rm low}\big)
+=
+-2\,\frac{\Omega^2}{E_\pi}\frac{E_{bd}^{(0)}}{E_\pi}
++
+O\!\left(\frac{\Omega^6}{E_\pi^5}\right).
+\label{eq:asymmetryEbd0}
+\]
+
+To the order kept here, one may replace \(E_{bd}^{(0)}\) on the
+right-hand sides by the full low-energy \(E_{bd}^{\rm low}\), because
+the difference between them only feeds in at \(O(\Omega^6/E_\pi^5)\).
+
 \printbibliography