text:
\section{Introduction}

$S$-matrix is the central object for scattering physics in quantum field theories.
However, the conventional $S$-matrix is not well-defined for some theories involving massless particles because of  \textit{infrared (IR) divergences}.
A famous example of such theories is \textit{quantum electrodynamics (QED)} in 4-dimensional Minkowski spacetime.
When we compute the $S$-matrix perturbatively, the contributions from low-energy virtual photons cause divergences of loop diagrams.
Because the infrared divergences cannot be eliminated by any renormalization procedure, the $S$-matrix is  not a well-defined object at any order of perturbation (except for the tree level).
The resummation of all orders of the IR divergent terms gives exponentially suppressed factors to $S$-matrix elements for any nontrivial processes, and  makes the $S$-matrix trivial \cite{Yennie:1961ad, Weinberg:1965nx, Weinberg:1995mt, Carney:2017jut}.

The traditional prescription for avoiding the infrared problem is to calculate the \textit{inclusive} cross-sections \cite{Bloch:1937pw}.
In this prescription, we compute the sum of the cross-sections for all processes including possible emissions of real soft bosons (photons, gravitons)  that are physically indistinguishable.
As is well known, the inclusive cross-sections are IR finite.
The IR divergences caused by virtual soft bosons are canceled out by
those arising from emissions of real soft bosons.
However, the $S$-matrix itself remains ill-defined in this prescription.

 An alternative approach is to directly construct the well-defined $S$-matrix without IR divergences by using more appropriate asymptotic states instead of Fock states \cite{Chung:1965zza, Greco:1967zza, Kibble:1968sfb, Kibble:1969ip, Kibble:1969ep, Kibble:1969kd, Kulish:1970ut, Greco:1978te}.
In conventional computations of $S$-matrix, it is usually assumed that the asymptotic states for charged particles obey free dynamics although  massless bosons mediate infinitely long-range interactions.
The key idea of the alternative approach is to use appropriate asymptotic states incorporating the effect of long-range interactions to compute the $S$-matrix.

A candidate for such asymptotic states was first proposed in QED by Chung \cite{Chung:1965zza}.
The states are dressed by an infinite number of coherent soft photons, which are sometimes referred to as \textit{dress} or \textit{cloud} of soft photons.
It was shown that the naive $S$-matrix for the dressed states is infrared finite to all orders perturbatively \cite{Chung:1965zza}.
Kulish and Faddeev derived other similar dressed states by solving the asymptotic dynamics of QED \cite{Kulish:1970ut}.
Analogous dressed states for perturbative gravity were also obtained by \cite{Ware:2013zja}.
It is also known that the Kulish-Faddeev (KF) dressed states can be obtained by solving the gauge invariant (BRST) condition in asymptotic regions \cite{Hirai:2019gio}.\footnote{In \cite{Kulish:1970ut}, the dressed states derived by solving the infrared dynamics are modified by introducing  an artificial null vector $c_{\mu}$ in order to resolve the problem that the dressed states do not satisfy the conventional Gupta-Bleuler condition  ($k^{\mu}a_{\mu}(k)\ket{\psi}=0$).
 In \cite{Hirai:2019gio}, it is shown that the condition is not adequate for dressed states and  also shown that the unmodified KF dressed states are gauge invariant without introducing the vector $c_{\mu}$.
 In this paper, KF dressed states mean the unmodified states.
 }

Although the dressed states  were proposed many years ago, they have been recently reinvestigated in the connection to the asymptotic symmetry (see, e.g., \cite{Mirbabayi:2016axw, Gabai:2016kuf, Kapec:2017tkm, Choi:2017bna, Choi:2017ylo, Carney:2018ygh, Neuenfeld:2018fdw, Hirai:2019gio, Gonzo:2019fai, Choi:2019rlz,  Choi:2019sjs, Furugori:2020vdl}).
The asymptotic symmetry in QED is a part of large $U(1)$ gauge transformations.
It is a physical symmetry in the sense that the conservation law of the Noether charge associated with the symmetry leads to a nontrivial constraint.
It was pointed out in \cite{Kapec:2017tkm} that the vanishing of $S$-matrix elements in the conventional computations is consistent with the asymptotic symmetry of QED.
Initial and final Fock states used in the conventional computations generally belong to different sectors of the asymptotic symmetry.
Therefore, the amplitude between them should vanish since otherwise it breaks the conservation law.
Thus it was argued that dressed states are needed in order to obtain non-vanishing amplitudes \cite{Kapec:2017tkm}.

Let us mention a slight but important difference between Chung's dressed states that are a candidate for the  well-defined $S$-matrix and the KF dressed states that are naturally appeared by solving the dynamics of QED.
The difference is the existence of the oscillating phase factors taking the form of
$
\exp\left(\pm i\frac{p\cdot k}{E_{p}}t\right)
$
 in KF dresses (see \eqref{def:Rf}, \eqref{def:Ri}).
Here $p^{\mu}=(E_p,\vec{p})$ is a four momentum of an (anti)electron, $k^{\mu}$ is that of a photon, and $t$ is the time when the initial or final state of the scattering is defined. We will take the limit as $t\rightarrow \pm \infty$ at the end of the calculation.
The phase factors naturally appear in the dress by solving the asymptotic dynamics in QED \cite{Kulish:1970ut} or by requiring the gauge invariance \cite{Hirai:2019gio}.
In \cite{Kulish:1970ut} and the literature, the phase factors are set to one by adopting the approximation:
\begin{align}
\label{eq:phaseapp}
\exp\left(\pm i\frac{p\cdot k}{E_{p}}t\right)\sim 1
\end{align}
for``small" $k$.
If we use this approximation, the KF dressed states become almost identical to Chung's ones.
However, the approximation \eqref{eq:phaseapp} is valid only for the parameter region
\begin{align}
\left|\frac{p\cdot k}{E_{p}}\right|\ll \left|\frac{1}{t}\right|\,.
\end{align}
When we take the limit $t\rightarrow \pm \infty$, this parameter region disappear.
Therefore the approximation \eqref{eq:phaseapp} is not justified in this limit.

In this paper, we propose new gauge invariant dressed states which are consistent with the asymptotic symmetry in QED,
and show that the $S$-matrix for the dressed states is IR finite.
If we remember the derivation of the soft photon theorems \cite{Weinberg:1965nx} or evaluation of IR divergences in the loop diagrams in QED, the leading divergences are caused by soft photons interacting with on-shell charged particles.
This fact suggests that we can replace the charged current by a background current of point-particles if we focus just on the IR problem.
Thus we first consider QED with the background current.
In other words,
we consider the Maxwell theory with background Wilson lines,
which is exactly solvable but shares the same IR problems as the full QED.
The dynamics of this theory is actually identical to the ``asymptotic dynamics'' considered by Kulish and Faddeev  \cite{Kulish:1970ut}.
We see in the model that  usual Fock states lead to IR divergences in the $S$-matrix.
In addition, the original KF dressed states cannot remove the IR divergences,
and are also inconsistent with the asymptotic symmetry.
The consistency with the asymptotic symmetry implies that we should introduce an additional dressing factor.
We confirm that we can use Chung's dressing factor as the additional one.
After all, we have to use simultaneously both of Chung's dressing factor and KF's one.
We then show that the $S$-matrix for the new dressed states is IR finite.
We finally argue that we can apply the new dressed states to the full QED, and conclude that appropriate dressed asymptotic states define the IR finite $S$-matrix.

The rest of this paper is structured as follows.
In section \ref{sec:main}, we resolve the IR problem in QED with a fixed current of point-particles which we call the background current model.
We first explain the model in section \ref{subsec:model} and construct the $S$-matrix in the model in section \ref{subsec:Smatrix}.
In section \ref{subsec:IRdiv}, we see that the same IR divergences as QED appear in the model, and also that the IR divergences cannot be eliminated even if we add the KF dressing operators to initial and finial states.
In section \ref{sec:Gauss}, we obtain physically allowed dressed states by solving the gauge invariant condition.
In section \ref{sec:AS}, we express that the asymptotic symmetry requires an additional dressing factor other than KF's factor.
We find that the additional factor can be Chung's one.
In subsection \ref{sec:asyCL}, we present the expressions of charges associated with asymptotic symmetry at asymptotic regions and also comment on their conservation laws.
Section \ref{sec:IRfinite} shows that the $S$-matrix for new dressed states has no IR divergences.
In section \ref{sec:fullQED}, the implications for the full QED are discussed.
Section \ref{sec:conclusion} contains our conclusion  and discussions about future directions.
Appendix \ref{app:BRST} shows the basic results of  BRST quantization of our model.
Appendix \ref{app:as} contains the derivation of charges of asymptotic symmetry in our model.
Appendix \ref{app:falloff} shows that the Li\'enard-Wiechert potential does not contribute to the charge of asymptotic symmetry in the asymptotic limit.
Appendix \ref{app:soft} shows that the current can be approximated by the point-particles current at asymptotic region in QED.

\section{Resolution of IR problem in background current model of QED\label{sec:main}}

In this section, we consider QED with a fixed current of point-particles.
We call this theory the background current model.
This model shares  the same IR problem as the full QED, and we see how dressed states resolve it.

\subsection{QED with a fixed current}\label{subsec:model}

The IR problem in QED is universal in the sense that the detailed information of scatterings is not important.
More concretely, the IR divergences of $S$-matrix elements depend only on charges and momenta of the initial and the final states of charged particles.
It is expected that the detailed dynamics of charged particles are not so relevant to understand the IR structure.
It leads us to  consider QED with a fixed current as
\begin{align}
\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}
+j^\mu_{\mathrm{pp}}A_\mu,
\label{eq:Lag}
\end{align}
where
\begin{align}
\label{j_cl}
j^\mu_{\mathrm{pp}}(x)=\Theta(-t)\sum_{n\in I} \frac{e_n p_n^\mu}{E_{n}}\, \delta^{3}(\vec{x}-\vec{p}_n t/E_{n})
+\Theta(t)\sum_{n \in F} \frac{e_{n} p_{n}^\mu}{E_{n}}\, \delta^{3}(\vec{x}-\vec{p}_{n}t/E_{n}).
\end{align}
This current $j^\mu_{\mathrm{pp}}$ corresponds to uniformly moving charged point-particles. Initially the charged particles have momenta $\{p_n\}_{n\in I}$, then scatter instantaneously at $t=0$, and finally have momenta $\{p_{n}\}_{n\in F}$ (see Fig.~\ref{fig:current}).
Here, $I$ and $F$ denotes \textit{initial} and \textit{final}.
\begin{figure}[H]
\centering
\includegraphics[height=6cm]{scat.pdf}
\caption{The trajectories of charged particles corresponding to \eqref{j_cl}.}
\label{fig:current}
\end{figure}\vspace{5mm}
We represent the mass of each particle as $m_n$ where $p_n^2=-E_n^2+\vec{p}_n^{~2}=-m_n^2$.
We suppose that the total charge is conserved; $\sum_{n \in I}e_n=\sum_{n \in F}e_{n}$. It ensures the current conservation $\partial_\mu j^\mu_{\mathrm{pp}}=0$.
The trajectories of charged particles in eq.~\eqref{j_cl} mean that we have cusped Wilson lines with the cusp at the origin $x=0$.  It implies that this theory can be a good approximation of the full QED when the charged particles are very massive, and shares the same IR structures with the full QED.
In fact, we will see that the naive $S$-matrix in the theory \eqref{eq:Lag} has the same IR divergences as in  the full QED for the scatterings of charged particles $\{p_n\}_{n\in I} \to \{p_{n}\}_{n\in F }$ and hard photons. Therefore, this theory is a good toy model to understand the IR problems in QED. Note that the term $j^\mu_{\mathrm{pp}}A_\mu$ in eq.~\eqref{eq:Lag} is actually nothing but the ``asymptotic interaction" considered by Kulish and Faddeev in \cite{Kulish:1970ut}.

Before proceeding, we briefly comment on the classical counterpart of this theory.
The theory \eqref{eq:Lag} is the free Maxwell theory with a fixed source. In the  Lorenz gauge, the general solutions of the classical equation of motion, $\Box A^\mu=-j^\mu_{\mathrm{pp}}$ are given by
\begin{align}
   A^\mu(x)=A^\mu_{in}(x)+\int d^4x' G(x,x')j^\mu_{\mathrm{pp}}(x'),
   \label{eq:cl_sol}
\end{align}
where $A^\mu_{in}$ is an initial configuration satisfying $\Box A^\mu_{in}=0$, and $G(x,x')$ is the retarded Green's function. The second term does not depends on the initial field $A^\mu_{in}$. In this sense, the scattering of electromagnetic waves is trivial in this simplified situation.
We will see the same result in the quantum case. In the case, the effect of the second term is realized as  a dressing operator (which is a displacement operator creating a coherent state of photons). Except for this dressing effect, the scattering of photons should be trivial. It means that for appropriate dressed asymptotic states the $S$-matrix of the photon sector is trivial in this simple toy model \eqref{eq:Lag}.

\subsection{Dyson's $S$-matrix}\label{subsec:Smatrix}
We consider the time-evolution for the theory \eqref{eq:Lag}.
In the Feynman gauge, the Hamiltonian in the Schr\"{o}dinger picture\footnote{The superscript $``\mathrm{s}"$ denotes the Schr\"{o}dinger picture. We also use the superscript $``I"$ for the interaction picture.} is given by\footnote{
See section 2 in \cite{Hirai:2019gio} for more details.
}
\begin{align}
    H^\mathrm{s}(t)=H_0+V^\mathrm{s}(t)
\end{align}
where
\begin{align}
 & H_0= \int \!\! d^3 x \left[\frac12 \Pi^{\mathrm{s}}_\mu \Pi^{\mu \mathrm{s}} +(\partial_i\Pi_0^\mathrm{s})   A^{i\mathrm{s}} +(\partial_i \Pi^{i\mathrm{s}})  A^{0\mathrm{s}} +\frac14 F^\mathrm{s}_{ij} F^{ij \mathrm{s}}\right],\quad\\[3mm]
   &V^\mathrm{s}(t)=-\int \!\! d^3 x j^\mu_{\mathrm{pp}}(t,\vec{x})A^{\mathrm{s}}_{\mu }(\vec{x}),
    \label{H0V}
\end{align}
and $\Pi_{\mu}$ are the conjugate momenta of $A^\mu$ satisfying the commutation relation
\begin{align}
    [A_\mu^{\mathrm{s}}(\vec{x}), \Pi_\nu^{\mathrm{s}}(\vec{x}')]=i\eta_{\mu\nu}\delta^3(\vec{x}-\vec{x}').
\end{align}
Note that the Hamiltonian has an explicit time-dependence, even in the Schr\"{o}dinger picture, through the classical background current.

Introducing the annihilation and creation operators of photons with the commutation relation
\begin{align}
    [a_\mu(\vec{k}),a_\nu^{\dagger}(\vec{k'})]=(2\omega)\eta_{\mu\nu}(2\pi)^3  \delta^3(\vec{k}-\vec{k'}),
\end{align}
we can write the fields $A^\mu, \Pi^\mu$ as
\begin{align}
\label{a-A}
&A_\mu^\mathrm{s}(\vec{x})=\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \left[a_\mu(\vec{k})e^{-i\omega t_s +i \vec{k}\cdot \vec{x}} +a_\mu^{\dagger}(\vec{k}) e^{i\omega t_s -i \vec{k}\cdot \vec{x}}\right],
\\
\label{a-P0}
&\Pi_0^\mathrm{s}(\vec{x})=-i \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \left[k^\mu a_\mu(\vec{k})e^{-i\omega t_s +i \vec{k}\cdot \vec{x}} -k^\mu a_\mu^{\dagger}(\vec{k}) e^{i\omega t_s -i \vec{k}\cdot \vec{x}}\right],
\\
\label{a-Pi}
&\Pi_i^\mathrm{s}(\vec{x})=-i \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \left[(k_i a_0(\vec{k})+\omega a_i(\vec{k}))e^{-i\omega t_s +i \vec{k}\cdot \vec{x}} -(k_i a_0^{\dagger}(\vec{k})+\omega a_i^{\dagger}(\vec{k}))e^{i\omega t_s -i \vec{k}\cdot \vec{x}}\right],
\end{align}
where
\begin{align}
    \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)}:=\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)}
\end{align}
is the Lorentz invariant measure and $t_s$ is an arbitrary time that defines operators in the Schr\"{o}dinger picture.
The free Hamiltonian is given by (up to the zero point energy)
\begin{align}
    H_0=\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \omega\,a_\mu^{\dagger}(\vec{k})a^{\mu}(\vec{k}),

\end{align}
and the interaction is
\begin{align}
   V^\mathrm{s}(t)= -\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \left[a_{\mu}(\vec{k})j^\mu_{\mathrm{pp}}(t,-\vec{k})e^{-i\omega t_s} +a^{\dagger}_{\mu}(\vec{k})j^\mu_{\mathrm{pp}}(t,\vec{k}) e^{i\omega t_s}\right],
\end{align}
where
\begin{align}
    j^\mu_{\mathrm{pp}}(t,\vec{k}):=\int \!\! d^3 x \,e^{-i \vec{k}\cdot \vec{x}} j^\mu_{\mathrm{pp}}(t,\vec{x}).
\end{align}
The time-evolution operator in the Schr\"{o}dinger picture is given by
\begin{align}
\label{eq:U}
    U(t_1,t_2)=\mathrm{T}\exp\left(-i\int^{t_1}_{t_2} dt H^\mathrm{s}(t)\right),
\end{align}
where $\mathrm{T}$ represents the time-ordering.
What we want to compute is the amplitude defined as
\begin{align}
\label{eq:defSmatrix}
S_{\alpha,\beta}:=\tensor[_s]{\bra{\alpha(t_f)}}{}U(t_f,t_i)\ket{\beta(t_i)}_s
\end{align}
for scattering states $\ket{\alpha(t_f)}_s$ and $\ket{\beta(t_i)}_s$.

We also introduce the free time-evolution operator as
\begin{align}
    U_0(t_1,t_2)=e^{-iH_0(t_1-t_2)}.
\end{align}
The time-evolution operator \eqref{eq:U} is the same as that for the ``asymptotic interaction" in  \cite{Kulish:1970ut} and can be computed easily as follows (see also \cite{Hirai:2019gio}).
Using the M\o ller operator defined by
\begin{align}
\label{eq:Moller}
    \Omega(t):=U(t_s,t)U_0(t,t_s),
\end{align}
we can write the operator \eqref{eq:U} as
\begin{align}
   U(t_f,t_i)=U_0(t_f,t_s)\Omega^\dagger(t_f)\Omega(t_i)U_0(t_s,t_i):=U_0(t_f,t_s)S_0(t_f,t_i)U_0(t_s,t_i),
\end{align}
where $S_0(t_f,t_i):=\Omega^\dagger(t_f)\Omega(t_i)$ is  the usual (finite time) $S$-matrix operator in the interaction picture.
It can be expressed by the Dyson series as
\begin{align}
\label{eq:S0}
    S_0(t_f,t_i)&=\mathrm{T}\exp\left(-i\int^{t_f}_{t_i} dt\, V^I(t)\right),
\end{align}
where $V^I$ is the interaction term in the interaction picture
\begin{align}
      V^I(t)&:= U_0(t,t_s)^{-1} V^\mathrm{s}(t)U_0(t,t_s)=-\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \left[a_{\mu}(\vec{k})j^\mu_{\mathrm{pp}}(t,-\vec{k})e^{-i\omega t} +a^{\dagger}_{\mu}(\vec{k})j^\mu_{\mathrm{pp}}(t,\vec{k}) e^{i\omega t}\right].
\end{align}
Since $V^I$ is linear in $a^{\mu}$ or $a^{\mu\dagger}$, the commutator $[V^I(t_1),V^I(t_2)]$ is a c-number function as
\begin{align}
\label{commu_VI}
    [V^I(t_1),V^I(t_2)]=-2i\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)}
   \eta_{\mu\nu} j^{\mu}_{\mathrm{pp}}(t_1,-\vec{k})j^{ \nu}_{\mathrm{pp}}(t_2,\vec{k})
    \sin [\omega(t_1-t_2)].
\end{align}
We can simplify the time-ordering in \eqref{eq:S0} as
\begin{align}
\label{eq:S0-2}
    S_0(t_f,t_i)=e^{i\Phi(t_f,t_i)}e^{-i\int^{t_f}_{t_i} dtV^I(t)},
\end{align}
where
\begin{align}\label{eq:defphase}
    \Phi(t_f,t_i):=\frac{i}{2}\int^{t_f}_{t_i}dt_1 \int^{t_1}_{t_i} dt_2 [V^I(t_1),V^I(t_2)]
\end{align}
which takes a real value because $[V^I(t_1),V^I(t_2)]$  is pure imaginary as \eqref{commu_VI}.
Thus, $e^{i\Phi(t_f,t_i)}$ is just an oscillating factor with phase $\Phi(t_f,t_i)$.
See \eqref{eq:phase-int} and \eqref{eq:FKphase} in appendix \ref{app:phase} for the concrete expression of the phase $\Phi(t_f,t_i)$.
The expression \eqref{eq:S0-2} is a general formula which holds for any background current.
Using the explicit form of $j^\mu_{\mathrm{pp}}$ in \eqref{j_cl}, we obtain\footnote{In this paper,  we assume $t_f>0$ and $t_i<0$.}
\begin{align}
    -i\int^{t_f}_{t_i} dt\, V^I(t)
    =R_{out}(t_f)-R_{out}(0)
    +R_{in}(0)-R_{in}(t_i)
\end{align}
with
\begin{align}
\label{def:Rf}
    R_{out}(t)&:=\sum_{n\in F}\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \frac{ e_{n} p_{n}^\mu}{p_{n}\cdot k} \left[a_{\mu}(\vec{k})e^{i\frac{p_{n} \cdot k}{E_n}t}
    -a_\mu^{\dagger}(\vec{k})
    e^{-i\frac{p_{n} \cdot k}{E_n}t}
    \right],
    \\
    \label{def:Ri}
    R_{in}(t)&:=\sum_{n\in I}\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \frac{ e_{n} p_{n}^\mu}{p_{n}\cdot k} \left[a_{\mu}(\vec{k})e^{i\frac{p_{n} \cdot k}{E_{n}}t}
    -a_\mu^{\dagger}(\vec{k})
    e^{-i\frac{p_{n} \cdot k}{E_n}t}
    \right].
\end{align}
Note that $R_{out}$ and $R_{in}$ are anti-Hermitian operators, and thus $e^{R_{out}}$ and $e^{R_{in}}$ are unitary operators.

In short,  the time-evolution operator \eqref{eq:U} can be written as
\begin{align}
\label{teo}
   U(t_f,t_i)=
   U_0(t_f,t_s)e^{i\Phi(t_f,t_i)}
   e^{R_{out}(t_f)+\Delta R-R_{in}(t_i)}

   U_0(t_s,t_i),
\end{align}
where
\begin{align}
\Delta R:=R_{in}(0)-R_{out}(0)\,.
\end{align}
Using the expression \eqref{teo}, we can write the $S$-matrix in \eqref{eq:defSmatrix} as
\begin{align}
\label{tr-amp}
S_{\alpha,\beta}

   =e^{i\Phi(t_f,t_i)}\,\tensor[_I]{\bra{\alpha(t_f)}}{}
   e^{R_{out}(t_f)+\Delta R-R_{in}(t_i)}\ket{\beta(t_i)}_I\,,
\end{align}
where we have written the states in the interaction picture by using  $\ket{\alpha(t)}_I:=U_0(t_s,t)\ket{\alpha(t)}_s$.
\subsection{Naive IR divergences}\label{subsec:IRdiv}
Here we see that IR divergences generally appear in the transition amplitude \eqref{tr-amp}  unless we prepare appropriate states $\ket{\alpha}, \ket{\beta}$.

First, the phase $\Phi$ in \eqref{eq:defphase} has terms which diverge in the limit $t_i\rightarrow -\infty, t_f\rightarrow \infty$ as explained in appendix \ref{app:phase}.
The divergent part is given by \eqref{eq:divphase}.
The divergence does not so matter because such a divergent phase can be eliminated by simultaneously redefining the basis states. (See appendix \ref{app:phase} for more concrete discussion.)

Next, we consider the operator $e^{R_{out}(t_f)+\Delta R-R_{in}(t_i)}$ in \eqref{tr-amp}.
This is  a displacement operator of photons because   $R_{out}$ and $R_{in}$ are anti-Hermitian and  linear in $a_{\mu}$ or $a_{\mu}^{\dagger}$  as \eqref{def:Rf} and \eqref{def:Ri}.
In general a displacement operator
\begin{align}
    D:=\exp\left(
    \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \left[C^\mu(\vec{k})a_\mu(\vec{k})
    -C^{\mu\ast}(\vec{k})
    a_\mu^{\dagger}(\vec{k})
    \right]
    \right)
\end{align}
    can be rewritten into the normal ordering as
\begin{align}
\label{D}
    D
    &=e^{
    -\frac{1}{2}
    \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} C^\mu(\vec{k})
    C_\mu^{\ast}(\vec{k})}
    e^{-
    \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} C^{\mu\ast}(\vec{k})
    a_\mu^{\dagger}(\vec{k})}
    e^{
    \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} C^\mu(\vec{k})a_\mu(\vec{k})}.
\end{align}
The first factor may have IR divergences. These divergences are real numbers unlike the divergence in the phase $i\Phi$.

For example,  the transition amplitude between the Fock vacuum for the operator $e^{\Delta R}$ is given by
\begin{align}
\bra{0} e^{\Delta R}\ket{0}
 &= \bra{0}e^{-R_{out}(0)+R_{in}(0)}\ket{0}\nonumber\\[3mm]
    &=
    \exp\left(
    -\frac{1}{2}\sum_{n,n'} \eta_n \eta_{n'} e_n e_{n'}
    \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \frac{p_n \cdot p_{n'}}{(p_n\cdot k)(p_{n'}\cdot k)}
    \right),
    \label{ampIRdiv}
\end{align}
where $\eta_n=-1$ for incoming particles ($n\in I$) and  $\eta_{n}=1$ for outgoing particles ($n\in F$).
This integral is IR divergent and in fact  is the same as that appearing in the  $S$-matrix element  for the process $\{p_n\}_{n\in I} \to \{p_{n}\}_{n\in  F}$ in the full QED
(see, \textit{e.g.}, \cite{Weinberg:1995mt}).
As explained in \cite{Weinberg:1995mt}, if we introduce the cutoff in the $k$-integral as $\lambda<|\vec{k}|<\Lambda$,
\eqref{ampIRdiv} depends on the cutoff as
\begin{align}
    \exp\left(
    -\frac{1}{2}\sum_{n,n'} \eta_n \eta_{n'} e_n e_{n'}
    \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \frac{p_n \cdot p_{n'}}{(p_n\cdot k)(p_{n'}\cdot k)}
    \right)
    \propto   \left(\frac{\lambda}{\Lambda}\right)^A
\end{align}
where $A$ is a positive constant for any nontrivial scattering process \cite{Weinberg:1995mt, Carney:2017jut}.
The limit $\lambda\to 0$ makes the amplitude vanish.

Because the first factor in \eqref{D} is independent of photon states, we always have the vanishing factor
$\left(\frac{\lambda}{\Lambda}\right)^A$ in any transition amplitudes  unless states have something cancelling it.
Thus, transition amplitudes
$\bra{\gamma_1}e^{\Delta R}\ket{\gamma_2}$ between arbitrary Fock states of hard photons $\ket{\gamma_{1,2}}$ vanish in the limit $\lambda\to 0$.
This is not the case for dressed states including soft photons as we will see in section~\ref{sec:IRfinite}.

The result that there is no transition between Fock states is actually the conservation law of the asymptotic symmetry in QED as pointed out in \cite{Kapec:2017tkm}.
In a scattering process, the asymptotic symmetry requires that  the initial and final states belong to different soft-charge sectors.\footnote{More precisely, the initial and final states must have the same asymptotic charge $Q_{as}=Q_{soft}+Q_{hard}$. The hard charge $Q_{hard}$ generally changes in the scattering, and thus the soft charge $Q_{soft}$ also has to change. }
Since the Fock states belong to the same sector, the transition between them are not allowed.
It means that we have to dress the initial or final states so that it is consistent with the asymptotic symmetry.
We will show this in section~\ref{sec:AS} for the background current model \eqref{eq:Lag}.

In the above discussions, we have ignored $R_{out}(t_f)$ and  $R_{in}(t_i)$ in $e^{R_{out}(t_f)+\Delta R-R_{in}(t_i)}$ for simplicity.
One might think that these extra  operators cancel the IR divergences in \eqref{ampIRdiv}.
Such cancellation does not occur.
In \cite{Kulish:1970ut} and the literature, it is discussed that the oscillating factors $e^{\pm i\frac{p_{n} \cdot k}{E_{p_n}}t}$ in \eqref{def:Rf} and \eqref{def:Ri} can be set to 1 because we consider the IR region $|\vec{k}|\ll 1$.
If this is correct, we have $R_{out}(t_f)\sim R_{out}(0)$ and $R_{out}(t_i)\sim R_{in}(0)$, and there is no IR divergence in $e^{R_{out}(t_f))+\Delta R-R_{in}(t_i)}$ because
\begin{align}
e^{R_{out}(t_f)+\Delta R-R_{in}(t_i)}=e^{R_{out}(t_f)-R_{out}(0)+R_{in}(0)-R_{in}(t_i)}\sim e^{0}\,.
\end{align}
However, we cannot use this approximation because it is only valid for the parameter region,
$
\left|\frac{p\cdot k}{E_{p}}\right|\ll \left|\frac{1}{t}\right|\,,
$
and this region vanishes when we take the limit $t\rightarrow \pm \infty$.
In general, $e^{R_{out}(t_f)-R_{in}(t_i)}$ produces IR divergences oscillating  with $t_f, t_i$ which are not the same as \eqref{ampIRdiv}.
Nevertheless, we do not need to worry much about these divergences from $R_{out}(t_f)$ and $R_{in}(t_i)$.
As we will see in next section, the gauge invariance requires the asymptotic states have the dressing factors $e^{-R_{out}(t_f)}$, $e^{R_{in}(t_i)}$ which almost cancel $R_{out}(t_f)$ and $R_{in}(t_i)$ coming from the time-evolution operator \eqref{teo}.
After this cancellation, we still have the IR divergence given by \eqref{ampIRdiv}.\footnote{Some problems on the Kulish-Faddeev dressed sates are also discussed for a massless Yukawa model in  \cite{Dybalski:2017mip}.}
This is the reason  why we need other dressings in addition to $e^{-R_{out}(t_f)}$, $e^{R_{in}(t_i)}$.

\subsection{Gauge invariant states are dressed states \label{sec:Gauss}}
The initial and final states should be physical states.
In the BRST formalism, the physical state condition is given by the BRST closed condition \cite{Kugo:1977zq},
\begin{align}
\label{BRST-cond}
    Q_{BRST}\ket{phys}=0.
\end{align}
It was shown in \cite{Hirai:2019gio} that the physical states cannot be the Fock states if there are interactions.
This fact just represents that Coulomb-like fields must exist around charged particles.
This fact also holds in the background current model \eqref{eq:Lag}.
The discussion is slightly different from the full QED because of the explicit time-dependence of the background current.
Thus, we briefly review the analysis of the physical state condition \eqref{BRST-cond} in \cite{Hirai:2019gio} for the background current model \eqref{eq:Lag}.

The BRST charge in the Schr\"{o}dinger picture is given by
\begin{align}
\label{eq:BRSTop}
Q_{BRST}^\mathrm{s}(t)=-\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \left[
c(\vec{k}) \{k^\mu a_\mu^{\dagger}(\vec{k}) +e^{-i\omega t_s}j^{0}_{\mathrm{pp}}(t,-\vec{k})\}
+c^\dagger(\vec{k})  \{k^\mu a_\mu(\vec{k}) +e^{i\omega t_s}j^{0}_{\mathrm{pp}}(t,\vec{k})\}
\right],
\end{align}
where $c(\vec{k}), c^\dagger(\vec{k})$ are annihilation and creation operators of the ghost field. We summarize the BRST formalism in appendix~\ref{app:BRST}.

The physical Hilbert space is given by the BRST cohomology of
this BRST charge.
However,
there is a tricky point because
the BRST charge \eqref{eq:BRSTop} does not commute with the Hamiltonian unlike the full QED case (see appendix~\ref{app:BRST}).
Thus, one might worry that the BRST closed condition is not consistent with the time-evolution.
This is not the case.
The BRST charge \eqref{eq:BRSTop} has an explicit time-dependence through the background current $j^{\mu}_{\mathrm{pp}}$,  and it helps the consistency.
One can show (see appendix~\ref{app:BRST} for the proof) that if a state $\ket{\psi(t)}_s$ is a BRST closed state at time $t$ as  $Q^\mathrm{s}_{BRST}(t)\ket{\psi(t)}_s =0$, the time-evolved state
$U(t',t)\ket{\psi(t)}_s$ is also a BRST closed state at time $t'$ as $Q^\mathrm{s}_{BRST}(t')U(t',t)\ket{\psi(t)}_s =0$.

We now solve the BRST closed condition $Q^\mathrm{s}_{BRST}(t)\ket{\psi(t)}_s =0$.
We restrict states to the ghost vacuum as $c(\vec{k})\ket{\psi(t)}_s=0$ since the ghost is decoupled from the dynamics of gauge fields.
The condition $Q^\mathrm{s}_{BRST}(t)\ket{\psi(t)}_s =0$ then becomes
\begin{align}
\label{Gauss-law}
     \left[k^\mu a_\mu(\vec{k}) +e^{i\omega t_s}j^{0}_{\mathrm{pp}}(t,\vec{k})\right] \ket{\psi(t)}_s=0
\end{align}
for any $\vec{k}$.
The key point is that this condition is different from the conventional Gupta-Bleuler condition $k^\mu a_\mu(\vec{k}) \ket{\psi(t)}_s=0$.
The form of \eqref{Gauss-law} means that physical states are coherent states.
Non-zero charges must be surrounded by an infinite number of (longitudinal) photons, namely  Li\'enard-Wiechert potential.
In this sense, the states are not elements of the Fock space where the number of photons is finite.

In the interaction picture, the condition \eqref{Gauss-law} is written as
\begin{align}
\label{Gauss-law_int}
     \left[k^\mu a_\mu(\vec{k}) +e^{i\omega t}j^{0}_{\mathrm{pp}}(t,\vec{k})\right] \ket{\psi(t)}_I=0\,.
\end{align}
The explicit form of the background current \eqref{j_cl} means
\begin{align}
    j^{0}_{\mathrm{pp}}(t,\vec{k})=\Theta(-t)\sum_{n \in I} e_n\,e^{-i \frac{\vec{p}_n\cdot \vec{k} }{E_n}t}
+\Theta(t)\sum_{n \in  F} e_{n}\,e^{-i \frac{\vec{p}_{n}\cdot \vec{k} }{E_n}t}.
\end{align}
Using this, we can solve \eqref{Gauss-law_int} as\footnote{We suppose $\Theta(0)=1/2$. It implies  that $\ket{\psi(t=0)}_I=e^{\frac{1}{2}\left(R_{in}(0)+R_{out}(0)\right)}\ket{\psi}_0$.}
\begin{align}
\label{FK-dress-st}
   \ket{\psi(t)}_I=  \left\{ \begin{array}{ll}
e^{R_{in}(t)}\ket{\psi}_0 & (t<0) \\
e^{R_{out}(t)}\ket{\psi}_0 & (t>0)
\end{array} \right.,
\end{align}
where $\ket{\psi}_0$ are arbitrary states satisfying the conventional  Gupta-Bleuler condition $k^\mu a_\mu(\vec{k}) \ket{\psi}_0=0$, and $R_{out}$ and $R_{in}$ are given in \eqref{def:Rf}, \eqref{def:Ri}.
These $R_{out}$ and $R_{in}$ are nothing but the Kulish-Faddeev
dressing operators \cite{Kulish:1970ut}.
As mentioned above,
the condition \eqref{Gauss-law} is Gauss's law.
The dressing factors in \eqref{FK-dress-st} actually correspond to the Li\'enard-Wiechert potential
\cite{Bagan:1999jf, Hirai:2019gio}.

An important remark is that the states $\ket{\psi}_0$ in \eqref{FK-dress-st} are not necessarily the Fock states.
$\ket{\psi}_0$ also can be dressed states if they satisfy the condition $k^\mu a_\mu(\vec{k}) \ket{\psi}_0=0$. In this sense, the dressing factors are not uniquely fixed by Gauss's law condition \eqref{Gauss-law_int}.
It was argued in \cite{Hirai:2019gio} that this freedom of the dressing factors is important in the conservation law of asymptotic symmetry.
We will see that we need dressing factors other than $R_{out}(t), R_{in}(t)$ to have IR finite $S$-matrix elements.

\subsection{Asymptotic symmetry requires a new dress}\label{sec:AS}
Gauge theories on manifolds with asymptotic boundaries generally have asymptotic symmetries.
For QED on 4d Minkowski space, the asymptotic symmetry consists of large U(1) gauge transformations \cite{He:2014cra, Campiglia:2015qka, Kapec:2015ena}.
Although the asymptotic symmetry is a part of local U(1) symmetry, it is not a gauge redundancy.\footnote{See \cite{Hirai:2018ijc} for an explanation in the BRST formalism.}
In general, a physical symmetry leads to a conservation law.
For example, total electric charges must conserve in any scattering process.
This is the global U(1) charge conservation.
Similarly, we have an infinite number of conservation laws associated with a part of large U(1) gauge transformations.
The relation between the asymptotic symmetry and dressed states was discussed, \textit{e.g.}, in \cite{Mirbabayi:2016axw, Gabai:2016kuf, Kapec:2017tkm,  Carney:2018ygh, Hirai:2019gio}.
In \cite{Hirai:2019gio}, it was argued that the Kulish-Faddeev dressing operator is not enough to realize the conservation laws of the asymptotic symmetry.
In this section we will see that the asymptotic symmetry requires additional dressing factors in order to have non-vanishing $S$-matrix elements.

The background current model \eqref{eq:Lag} with the Feynman gauge is invariant under the residual gauge symmetry $A_\mu \to A_\mu+\partial_\mu \epsilon$ where the gauge parameter $\epsilon$ satisfies $\Box \epsilon=0$.
The asymptotic symmetry is then given by these transformations such that the parameters $\epsilon(x)$ are $\mathcal{O}(1)$ at the asymptotic boundary.
We can reach the future null infinity $\mathscr{I}^+$ by taking $t, r \to +\infty$ with the retarded time $u=t-r$ fixed.
$\mathscr{I}^+$ is parameterized by $u$ and angular coordinates $\Omega^A\, (A=1,2)$.
By solving $\Box \epsilon=0$ near the future null infinity $\mathscr{I}^+$,  we can find that the asymptotic behavior of $\epsilon(x)$ at $\mathscr{I}^+$ is given by an arbitrary function of the angular coordinates $\epsilon^{(0)}(\Omega)$.
If we specify this boundary behavior $\epsilon^{(0)}(\Omega)$, the bulk parameter $\epsilon(x)$ is fixed up to an addition of ``small" gauge parameter which is a gauge redundancy.
Thus, the asymptotic symmetry is an infinite dimensional symmetry parameterized by  functions $\epsilon^{(0)}(\Omega)$ on the unit two-sphere.
The expression of $\epsilon(x)$ as a functional of $\epsilon^{(0)}(\Omega)$ is given by \eqref{LGP} in appendix~\ref{app:as}.

In the Schr\"{o}dinger picture, the Noether charge for the gauge transformation is given by
\begin{align}
Q_{as}^\mathrm{s}[\epsilon]=\int \!\! d^3 x \left[
-\Pi^{0s} \partial_0 \epsilon -\Pi^{is}\partial_i \epsilon
+j^0_{\mathrm{pp}}\epsilon
\right].
\label{neother_as}
\end{align}
Note that, in order to match the convention with that in the full QED, we have added the last term $j^0_{\mathrm{pp}}\epsilon$, which is a trivial operator (c-number) because $j^\mu_{\mathrm{pp}}$ is a classical background current.
If $\epsilon$ vanishes at the asymptotic boundary, the charge \eqref{neother_as} is BRST exact\footnote{This charge can be written as $Q_{as}^\mathrm{s}[\epsilon]=-\int \!\! d^3 x\, \partial_i (\Pi^{is}\epsilon)
+ \left\{Q^\mathrm{s}_{BRST}, \int \!\! d^3 x(-\bar{c}^\mathrm{s} \partial_0 \epsilon+i \pi_{(c)}^\mathrm{s} \epsilon)
\right\}$ where $\bar{c}^\mathrm{s}, \pi_{(c)}^\mathrm{s}$ are ghost operators (see appendix~\ref{app:BRST}).} and does not play any role on the physical Hilbert space.
It just means that $Q_{as}^\mathrm{s}[\epsilon]$ for  a ``small" gauge parameter $\epsilon$ generates a gauge redundancy which is not  physical symmetry.
If $\epsilon$ is a  ``large" gauge parameter as discussed above,  $Q_{as}^\mathrm{s}[\epsilon]$ is a charge for the asymptotic symmetry, which we call the asymptotic charge.
It can act nontrivially on the physical Hilbert space unlike small gauge charges.
We can classify physical states by the values of the asymptotic charges because
  $Q_{as}^\mathrm{s}[\epsilon]$ are commutative $[Q_{as}^\mathrm{s}[\epsilon], Q_{as}^\mathrm{s}[\epsilon']]=0$ and also commute with the BRST charge $[Q_{as}^\mathrm{s}[\epsilon],Q^\mathrm{s}_{BRST}(t)]=0$.

We now consider a time-evolution of an initial state $\ket{\beta(t_i)}_s$, which is converted to the interaction picture as $\ket{\beta(t_i)}_s=U_0(t_i,t_s)\ket{\beta(t_i)}_I$ where $\ket{\beta(t_i)}_I$ should be dressed as $e^{R_{in}(t_i)}\ket{\beta}_0$ (see $\eqref{FK-dress-st}$).
For simplicity, we assume that $\ket{\beta(t_i)}_I$ is an eigenstate of $Q_{as}^I[\epsilon(t_i)]$.\footnote{General states can be obtained by the superposition of the eigenstates.}
It means that $\ket{\beta}_0$ is also an eigenstate of $Q_{as}^I[\epsilon(t_i)]$ because the commutator of $Q_{as}^I[\epsilon(t_i)]$ and the dressing factor $R_{in}(t_i)$ is a c-number.
Supposing that the eigenvalue for $\ket{\beta}_0$ is $Q_{\beta}[\epsilon(t_i)]$, we have
\begin{align}
\label{Q^Ii-ei}
  Q_{as}^I[\epsilon(t_i)]\ket{\beta(t_i)}_I
  =\left([Q_{as}^I[\epsilon(t_i)],R_{in}(t_i)]+Q_\beta[\epsilon(t_i)]\right)\ket{\beta(t_i)}_I .
\end{align}
We can also find that a time-evolved state $\ket{\beta(t_f)}_I$ is an eigenstate of $Q_{as}^I[\epsilon(t_i)]$ (not $Q_{as}^I[\epsilon(t_f)]$) as follows.
From \eqref{teo}, $\ket{\beta(t_f)}_I$ can be written as
\begin{align}
\ket{\beta(t_f)}_I
=U^{-1}_{0}(t_f,t_s)U(t_f,t_i)\ket{\beta(t_i)}_s
=e^{i\Phi(t_f,t_i)}e^{\tilde{R}}  \ket{\beta(t_i)}_I,

\end{align}
where
\begin{align}\label{eq:Rtilde}
\tilde{R}:=R_{out}(t_f)+\Delta R-R_{in}(t_i).
\end{align}
Using the expression, we obtain
\begin{align}
 Q_{as}^I[\epsilon(t_i)]\ket{\beta(t_f)}_I
 &=e^{i\Phi(t_f,t_i)}Q_{as}^I[\epsilon(t_i)]e^{\tilde{R}}  \ket{\beta(t_i)}_I \nonumber\\[3mm]
 &=e^{i\Phi(t_f,t_i)}\left([Q_{as}^I[\epsilon(t_i)], e^{\tilde{R}}]+  e^{\tilde{R}}Q_{as}^I[\epsilon(t_i)]\right)\ket{\beta(t_i)}_I\nonumber\\[3mm]
 &=\left([Q_{as}^I[\epsilon(t_i)], \tilde{R}]+[Q_{as}^I[\epsilon(t_i)], R_{in}(t_i)]+Q_\beta[\epsilon(t_i)]\right)\ket{\beta(t_f)}_I,
 \label{eq:Qieigen}
\end{align}
where we have used the fact that
$[Q_{as}^I[\epsilon(t_i)], \tilde{R}]$ is a c-number and eq.\eqref{Q^Ii-ei} in the third equality.
Putting \eqref{eq:Rtilde} into \eqref{eq:Qieigen}, we obtain
\begin{align}
Q_{as}^I[\epsilon(t_i)]\ket{\beta(t_f)}_I=\left(\,[Q_{as}^I[\epsilon(t_i)], R_{out}(t_f)]+\Delta Q+Q_\beta[\epsilon(t_i)]\,\right)\ket{\beta(t_f)}_I ,
\end{align}
where
\begin{align}
\Delta Q:=[Q_{as}^I[\epsilon(t_i)], \Delta R].
\end{align}
Therefore, $\ket{\beta(t_f)}_I$ is the eigenstate of $Q_{as}^I[\epsilon(t_i)]$ with the eigenvalue $[Q_{as}^I[\epsilon(t_i)], R_{out}(t_f)]+\Delta Q+Q_\beta[\epsilon(t_i)]$.
This is an important result for finding the appropriate asymptotic state for the following reason.

The amplitude in \eqref{eq:defSmatrix}
\begin{align}
S_{\alpha,\beta}=\tensor[_s]{\bra{\alpha(t_f)}}{}U(t_f,t_i)\ket{\beta(t_i)}_s
=\tensor[_I]{\bra{\alpha(t_f)}\ket{\beta(t_f)}}{_I}
\end{align}
 vanishes unless $\ket{\alpha(t_f)}_I$ has the same eigenvalue $[Q_{as}^I[\epsilon(t_i)], R_{out}(t_f)]+\Delta Q+Q_\beta[\epsilon(t_i)]$ of $Q_{as}^I[\epsilon(t_i)]$ as $\ket{\beta(t_f)}_I$.
On the other hand, the gauge invariance requires that  $\ket{\alpha(t_f)}_I$ be dressed as $\ket{\alpha(t_f)}_I=e^{R_{out}(t_f)}\ket{\alpha}_0$.
It means that, in order to have a non-zero amplitude $S_{\alpha,\beta}$, the state $\ket{\alpha}_0$ must satisfy the following eigenstate equation:
\begin{align}
    Q_{as}^I[\epsilon(t_i)] \ket{\alpha}_0
    =\left(\Delta Q+Q_\beta[\epsilon(t_i)]\right)\ket{\alpha}_0\,.
    \label{change_Qas}
\end{align}
Here let us represent the Hilbert space satisfying the conventional Gupta-Bleuler condition by $\mathcal{H}^0$,
\begin{align}
\ket{\phi}\in \mathcal{H}^0\ \Leftrightarrow \ k^\mu a_{\mu}(k)\ket{\phi}=0\quad \text{for any $k^\mu$},
\end{align}
and the subspace by $\mathcal{H}^0_{Q}$  where the eigenvalue of $Q_{as}^I[\epsilon(t_i)]$ is $Q[\epsilon(t_i)]$.
Then the above  says that  $\ket{\alpha}_0$ and  $\ket{\beta}_0$ need to belong to different sectors with respect to the asymptotic charges $Q_{as}^I[\epsilon(t_i)]$ as
\ba{
\label{eq:shift_ab}
\ket{\beta}_0 \in\mathcal{H}^0_{Q}\ \Rightarrow \ \ket{\alpha}_0\in\mathcal{H}^0_{Q+\Delta Q}\,,
}
in order to have the non-zero matrix element $S_{\alpha,\beta}$.
We need a dressing factor associated to the shift $\Delta Q$ in $\ket{\alpha}_0$.
This transition can be realized by adding the transverse part of the dressing operator $\Delta R$ as follows.

We first introduce our notation of the polarization vectors.
We represent two transverse polarization vectors by $\epsilon_A^\mu(\hat{k})=(0,\vec{\epsilon}_A(\hat{k}))$ ($A=1,2$) where $\vec{\epsilon}_A$ depends only on the direction of photon momentum $\vec{k}$ and  satisfies $\vec{k} \cdot \vec{\epsilon}_A(\hat{k})=0$.
Here, we take a real basis such that $\epsilon^{\mu}_A=\epsilon^{\mu\ast}_A$.
We also represent the inner product of two polarization vectors by $\eta^{T}_{AB}$
as $\eta^{T}_{AB}(\hat{k})=\eta_{\mu\nu}\epsilon_A^{\mu}(\hat{k})\epsilon_B^\nu(\hat{k})$.
This matrix $\eta^{T}_{AB}$ must be  invertible so that the two polarization vectors are independent.
We have the completeness relation
\begin{align}
    \eta^{\mu\nu}=\eta^{AB}_{T}\epsilon_A^\mu(\hat{k})\epsilon^{\nu}_B(\hat{k})
    -\frac{k^\mu \tilde{k}^\nu+\tilde{k}^\mu k^\nu}{2\omega^2},
\end{align}
where $\eta^{AB}_{T}$ is the inverse matrix of $\eta^{T}_{AB}$ as $\eta^{AB}_{T}\eta^{T}_{BC}=\delta^A_C$, and  $\tilde{k}^\mu$ is the spatially reflected vector of $k^\mu$ as  $\tilde{k}^\mu=(\omega, -\vec{k})$ for $k^\mu=(\omega, \vec{k})$.
Then, we can decompose the dressing operator $\Delta R$ into the transverse and longitudinal parts as
\begin{align}
\Delta R=  -R_{out}(0)+R_{in}(0)=\Delta R_{T}  +\Delta R_{L}
\end{align}
with
\begin{align}
\label{def:RT}
    \Delta R_{T}&=-\sum_{n}\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \eta^{AB}_{T}(\hat{k})\frac{\eta_n e_{n} p_{n}\cdot \epsilon_A(\hat{k})}{p_{n}\cdot k} \left[\epsilon_B(\hat{k}) \cdot a(\vec{k})
    -\epsilon_B(\hat{k}) \cdot a^{\dagger}(\vec{k})
    \right],
    \\
    \label{def:RL}
    \Delta R_{L}&=\sum_{n}\int \frac{\widetilde{d^3 k}}{2\omega^2
    \eta_n e_{n} \left[\tilde{k} \cdot a(\vec{k})
    -\tilde{k} \cdot a^{\dagger}(\vec{k})
    \right]
    +
    \sum_{n}\int  \frac{\widetilde{d^3 k}}{2\omega^2} \frac{\eta_n e_{n} p_{n}\cdot \tilde{k}}{p_{n}\cdot k} \left[k \cdot a(\vec{k})
    -k \cdot a^{\dagger}(\vec{k})
    \right]
    \nonumber\\
    &= \sum_{n}\int \frac{\widetilde{d^3 k}}{2\omega^2} \frac{\eta_n e_{n} p_{n}\cdot \tilde{k}}{p_{n}\cdot k} \left[k \cdot a(\vec{k})
    -k \cdot a^{\dagger}(\vec{k})
    \right],
\end{align}
where we have used the total electric charge conservation $\sum_n \eta_n e_n=0$.
One can easily show that the longitudinal part $\Delta R_{L}$ commutes with $\Pi^I_\mu$.
It means that $\Delta R_{L}$ commutes with the asymptotic charges $Q_{as}^I[\epsilon]=\int \!\! d^3 x \left(
-\Pi^{0I} \partial_0 \epsilon -\Pi^{iI}\partial_i \epsilon
+j^0_\mathrm{pp}\epsilon
\right)$.
Thus, the shift of the asymptotic charges $\Delta Q=[ Q_{as}^I[\epsilon(t_i)], \Delta R]$, which appeared in \eqref{change_Qas}, can be written as
\begin{align}
    \Delta Q=[ Q_{as}^I[\epsilon(t_i)], \Delta R_{T}].
\end{align}
It shows that the dressing factor $e^{\Delta R_{T}}$ shifts the asymptotic charges by $\Delta Q$.
If a state $\ket{\psi}_0$ satisfies the conventional Gupta-Bleuler condition, the dressed state $e^{\Delta R_{T}}\ket{\psi}_0$ also does, because $\Delta R_{T}$ consists only of transverse components.
Therefore, the transition from $\mathcal{H}^0_{Q}$ to $\mathcal{H}^0_{Q+\Delta Q}$ can be implemented by the dressing factor $e^{\Delta R_{T}}$ as
\begin{align}
\mathcal{H}^0_{Q+\Delta Q}=e^{\Delta R_{T}}\mathcal{H}^0_{Q}.
\end{align}
Note that each term in the sum in $\Delta R_{T}$ is the same as  Chung's dressing operator which is acted on each charged particle to cancel IR divergences in \cite{Chung:1965zza}.
We will see in section \ref{sec:IRfinite} that this additional factor $e^{\Delta R_{T}}$ makes the $S$-matrix IR finite.

\subsubsection{Asymptotic charge conservation\label{sec:asyCL}}
In this subsection, we comment on the conservation law of the asymptotic symmetry.\footnote{You may skip this part if you would like to see soon the IR finiteness of the $S$-matrix.}

We have shown that $\ket{\alpha}_0$ and  $\ket{\beta}_0$ belong to different sectors with respect to the asymptotic charges $Q_{as}^I[\epsilon(t_i)]$ as \eqref{eq:shift_ab}.
The point is that we considered  $Q_{as}^I[\epsilon(t_i)]$, not $Q_{as}^I[\epsilon(t_f)]$, even for the final state $\ket{\alpha}_0$.
It is difficult to find the relation between $Q_{as}^I[\epsilon(t_i)]$ and $Q_{as}^I[\epsilon(t_f)]$ for finite $t_i$ and  $t_f$.
However, a simplification occurs in the asymptotic limit
$t_f\to \infty, t_i\to -\infty$.
We will show that, if the initial state $\ket{\beta}_0$ is an eigenstate of the past charge $\lim_{t_i\to -\infty}Q_{as}^I[\epsilon(t_i)]$ with the eigenvalue $Q[\epsilon^{(0)}]$, the final state $\ket{\alpha}_0$ is an eigenstate of the future charge $\lim_{t_f\to \infty}Q_{as}^I[\epsilon(t_f)]$ with the same eigenvalue $Q[\epsilon^{(0)}]$ in order to have the nonzero amplitude $\tensor[_s]{\bra{\alpha(t_f)}}{}U(t_f,t_i)\ket{\beta(t_i)}_s$.

We first consider the asymptotic limits of the asymptotic charge, $\lim_{t\to \pm\infty} Q_{as}^I[\epsilon(t)]$.
The gauge parameter $\epsilon(x)$ approaches a function on the celestial sphere, $\epsilon^{(0)}(\Omega)$, near the future null infinity $\mathscr{I}^+$.
As will be shown in appendix~\ref{app:as}, the limits of the asymptotic charge can be computed as
\begin{align}
\label{eq:as_consv}
&Q_{as}^{-I}[\epsilon^{(0)}]
:=\lim_{t_i\to -\infty} Q_{as}^I[\epsilon(t_i)]=Q_{hard}^{in}[\epsilon^{(0)}]+Q_{soft}[\epsilon^{(0)}]\,,\nonumber\\[3mm]
&Q_{as}^{+I}[\epsilon^{(0)}]
:=\lim_{t_f\to +\infty} Q_{as}^I[\epsilon(t_f)]=Q_{hard}^{out}[\epsilon^{(0)}]+Q_{soft}[\epsilon^{(0)}],
\end{align}
where
\begin{align}
 Q_{hard}^{out\,/\,in}[\epsilon^{(0)}]
 &:=\sum_{n \in F\, /\,I}
     \int d^2 \Omega  \sqrt{\gamma}
   \frac{e_{n} m_{n}^2  \epsilon^{(0)}(\Omega)}{4\pi(-E_{n}+\vec{p}_{n} \cdot \hat{x}(\Omega))^2},\\
Q_{soft}[\epsilon^{(0)}]&:=\frac{1}{8\pi}\lim_{\omega \to 0} \omega
  \int d^2\Omega \sqrt{\gamma}\gamma^{AB}\partial_A  \hat{x}^i\partial_B \epsilon^{(0)}
\left[ a_i(\omega\hat{x}) +  a_i^{\dagger}(\omega\hat{x})\right].
\end{align}
The hard charges $Q_{hard}^{out/in}$ are c-numbers which agree with those computed in the classical case in \cite{Hirai:2018ijc}.
$Q_{soft}$ is an operator acting on the soft photon sector.

In the asymptotic limit, $\Delta Q$ appeared in \eqref{change_Qas} is also simplified as
\begin{align}
    \lim_{t_i \to -\infty}\Delta Q&=[Q_{soft}, \Delta R]
   =-Q_{hard}^{out}+Q_{hard}^{in}.
\end{align}

We suppose that the initial state $\ket{\beta}_0$ is an eigenstate of $Q_{as}^{-I}[\epsilon^{(0)}]$ with the eigenvalue $Q[\epsilon^{(0)}]$.
Then, as shown by eq.\eqref{change_Qas},
in order to have the nonzero amplitude $\tensor[_s]{\bra{\alpha(\infty)}}{}U(\infty,-\infty)\ket{\beta(-\infty)}_s$,
the final state $\ket{\alpha}_0$ must be the eigenstate of the past charge $Q_{as}^{-I}[\epsilon^{(0)}]$ with the eigenvalue
\begin{align}
    Q[\epsilon^{(0)}]+\lim_{t_i\to -\infty}\Delta Q
    = Q[\epsilon^{(0)}]-Q_{hard}^{out}+Q_{hard}^{in}.
\end{align}
Combining this equation and eq.\eqref{eq:as_consv},  we find that
$\ket{\alpha}_0$ is the eigenstate of the future charge
$Q_{as}^{+I}[\epsilon^{(0)}]$ with the eigenvalue $Q[\epsilon^{(0)}]$.

The above properties of the eigenvalues of the asymptotic charges mean the asymptotic charge conservation, that is, the equation
\begin{align}
\label{as_cnsv_I}
\tensor[_I]{\bra{\alpha(\infty)}}{}
Q_{as}^{+I}[\epsilon^{(0)}]
S_0(\infty,-\infty)\ket{\beta(-\infty)}_I
=\tensor[_I]{\bra{\alpha(\infty)}}{}
S_0(\infty,-\infty)
Q_{as}^{-I}[\epsilon^{(0)}]
\ket{\beta(-\infty)}_I
\end{align}
holds if the amplitude $\tensor[_I]{\bra{\alpha(\infty)}}{}S_{0}(\infty,-\infty)\ket{\beta(-\infty)}_I$ is nonzero.
Since $\ket{\beta(t_i)}_I=e^{R_{in}(t_i)}\ket{\beta}_0$,
the initial state
$\ket{\beta(t_i)}_I$ is the eigenstate of $Q_{as}^I[\epsilon(t_i)]$ with the eigenvalue
\begin{align}
    Q[\epsilon(t_i)]+[Q_{as}^I[\epsilon(t_i)],R_{in}(t_i)]
\end{align}
if $\ket{\beta}_0$ has the eigenvalue $Q[\epsilon(t_i)]$.
The commutator $[Q_{as}^I[\epsilon(t_i)],R_{in}(t_i)]$ is computed as
\begin{align}\label{eq:chargeofLW}
    [Q_{as}^I[\epsilon(t_i)],R_{in}(t_i)]=\sum_{n\in I}\int\!\! d^3 x F_{0i}^{LW}(t_i,\vec{x};p_n)\partial^i \epsilon(t_i,\vec{x}),
\end{align}
where $F_{0i}^{LW}(x;p_n)$ represents the classical Coulomb field for the Li\'enard-Wiechert potential of a uniformly moving charged particle with momentum $p_n$,
\begin{align}
  F_{0i}^{LW}(x;p_n)
  :=i e_n\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \frac{ E_{n}k_i-p_{ni}\omega }{p_{n}\cdot k} \left(e^{-i\vec{k}\cdot\left(\vec{x}-\frac{\vec{p}_{n} t_i}{E_{n}}\right)}-e^{i\vec{k}\cdot\left(\vec{x}-\frac{\vec{p}_{n} t_i}{E_{n}}\right)}\right).
\end{align}
The charge \eqref{eq:chargeofLW}  represent the contribution of the Coulomb field to  the asymptotic charge. It vanishes at asymptotic infinities because the Coulomb field  falls off at asymptotic infinity (see appendix \ref{app:falloff} for more
detailed analysis.).
Thus, $\ket{\beta(-\infty)}_I$ has the same eigenvalue of $Q_{as}^{-I}[\epsilon^{(0)}]$ as $\ket{\beta}_0$.
Similarly,
$\ket{\alpha(\infty)}_I$ has the same eigenvalue of $Q_{as}^{+I}[\epsilon^{(0)}]$ as $\ket{\alpha}_0$.
Since the eigenvalue of $Q_{as}^{+I}[\epsilon^{(0)}]$ for $\ket{\alpha}_0$ agrees with that of $Q_{as}^{-I}[\epsilon^{(0)}]$ for $\ket{\beta}_0$ as shown above, we obtain the asymptotic conservation law \eqref{as_cnsv_I}.
This conservation law can be symbolically written as
\begin{align}
    Q_{soft}+Q_{hard}^{out}=Q_{soft}+Q_{hard}^{in}.
\end{align}
This is the quantum analog of the electromagnetic memory effect \cite{Bieri:2013hqa, Susskind:2015hpa, Hirai:2018ijc} (see also section~4 in \cite{Hirai:2019gio}).

\subsection{IR finiteness of dressed $S$-matrix}
\label{sec:IRfinite}
We now reconsider the transition amplitude $S_{\alpha,\beta}$ in \eqref{eq:defSmatrix} and see that there is no IR divergences for appropriate dressed states.
As shown in section~\ref{sec:Gauss}, the gauge invariance requires that
the initial and final states be dressed states.
In the interaction picture, the initial states $\ket{\beta(t_i)}_I$ are dressed as $\ket{\beta(t_i)}_I=e^{R_{in}(t_i)}\ket{\beta}_0$.
The state  $\ket{\beta}_0$ is an element of $\mathcal{H}^0$ subject to the conventional Gupta-Bleuler condition $k^\mu a_\mu(\vec{k})\ket{\beta}_0=0$.
The space $\mathcal{H}^0$ can be classified by the eigenvalues of the asymptotic charges $Q_{as}^I[\epsilon(t_i)]$ as $\mathcal{H}^0=\oplus_{Q} \mathcal{H}^0_Q$.
We focus on the case where the state $\ket{\beta}_0$ has a definite eigenvalue $Q$, \textit{i.e.} $\ket{\beta}_0 \in \mathcal{H}^0_Q$, since arbitrary states in $\mathcal{H}^0$ can be written as superposition of eigenstates of $Q_{as}^I[\epsilon(t_i)]$.
As we showed in section~\ref{sec:AS},
the final state has to have an additional dressing factor to realize the shift of the asymptotic charges.
We dress the final state as
\begin{align}
    \ket{\alpha(t_f); \Delta R_T}_I=e^{R_{out}(t_f)}e^{\Delta R_T}\ket{\alpha}_0 \qquad \text{with} \quad \ket{\alpha}_0 \in \mathcal{H}^0_Q,
\end{align}
where we write $\Delta R_T$ in the ket to stress that the dressing is different from $\ket{\beta(t_i)}_I$ by the additional dressing factor $e^{\Delta R_T}$.\footnote{Note that $\ket{\alpha}_0$ is different from that in section~\ref{sec:AS} where $\ket{\alpha}_0 \in \mathcal{H}^0_{Q+\Delta Q}$.}
Then,
the amplitude $S_{\alpha,\beta}$ turns out to be diagonal as
\begin{align}
S_{\alpha,\beta}
&=\tensor[_s]{\bra{\alpha(t_f);\Delta R_{T} }}{}U(t_f,t_i)\ket{\beta(t_i) }_s\nonumber\\
&=e^{i\Phi(t_f,t_i)}\,\tensor[_0]{\bra{\alpha}}{}
e^{-\Delta R_T}e^{-R_{out}(t_f)}
e^{R_{out}(t_f)+\Delta R-R_{in}(t_i)}
e^{R_{in}(t_i)}\ket{\beta}_0
\nonumber\\
&=
e^{i\Phi(t_f,t_i)}e^{-\frac{1}{2}[R_{out}(t_f)-R_{in}(t_i),\, \Delta R ]+\frac{1}{2}[R_{out}(t_f), R_{in}(t_i)]}
\tensor[_0]{\bra{\alpha}}{}
e^{\Delta R_L}\ket{\beta}_0
\nonumber\\
&=
e^{i\tilde{\Phi}(t_f,t_i)}\delta_{\alpha,\beta},
\end{align}
where the phase $\tilde{\Phi}$ is a c-number defined by
\begin{align}
  i\tilde{\Phi}(t_f,t_i):= i\Phi(t_f,t_i)-\frac{1}{2}[R_{out}(t_f)-R_{in}(t_i), \Delta R]+\frac{1}{2}[R_{out}(t_f), R_{in}(t_i)],
\end{align}
which is independent of states $\alpha, \beta$.
Note that $\Delta R_L$ given by \eqref{def:RL} contains only longitudinal photons, and  thus we have $\tensor[_0]{\bra{\alpha}}{}e^{\Delta R_L}\ket{\beta}_0=\tensor[_0]{\braket{\alpha}{\beta}}{_0}=\delta_{\alpha,\beta}$.

Combining the expression of $\Phi$ given by \eqref{eq:phase-int} with \eqref{eq:FKphase},
we can find that
the total phase $\tilde{\Phi}$ is given by
\begin{align}
\label{eq:totalphase}
  \tilde{\Phi}(t_f,t_i)  = -\theta_{out}(t_f)+\theta_{in}(t_i)
\end{align}
where
\begin{align}
&\theta_{out}(t_f):=
\sum_{n\in F} \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \frac{e_{n}^2 m_n^2 }{(k \cdot p_n) E_n}t_f
+ \sum_{\substack{n,m \in F\\  n\neq m}}
\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)}\frac{e_{n}e_{m}  (v_{n}\cdot v_{m})}{(k \cdot v_{m})\, [k \cdot \pa{ v_{n}- v_{m}}]}\sin\pa{k \cdot \pa{ v_{n}- v_{m}}t_f},
\\
&\theta_{in}(t_i):=
\sum_{n\in I} \int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)} \frac{e_{n}^2 m_n^2 }{(k \cdot p_n) E_n}t_I
+\sum_{\substack{n,m \in I\\  n\neq m}}
\int \widetilde{d^3k}\,}%{\int \!\! \frac{d^3 k}{(2\pi)^3(2\omega)}\frac{e_{n}e_{m}  (v_{n}\cdot v_{m})}{(k \cdot v_{m})\, [k \cdot \pa{ v_{n}- v_{m}}]}\sin\pa{k \cdot \pa{ v_{n}- v_{m}}t_i}.
\end{align}

The first terms in $\theta_{out}(t_f), \theta_{in}(t_i)$ are linear of $t_f, t_i$ respectively, and diverge in the limit $t_f\to \infty, t_i \to -\infty$.
Because the phases are independent of photon states, we may also absorb the phases into the dressing factor as
\begin{align}
&\ket{\alpha(t_f);\Delta R_{T}}_I:=e^{R_{out}(t_f)}e^{\Delta R_T}e^{-i\theta_{out}(t_f)}\ket{\alpha}_0\,,\\ &\ket{\beta(t_i)}_I:=e^{R_{in}(t_i)}e^{-i\theta_{in}(t_i)}\ket{\beta}_0.
\end{align}
Using these redefined dressed states, the $S$-matrix is trivial in the sense as
\begin{align}
S_{\alpha, \beta}
=\tensor[_s]{\bra{\alpha(t_f);\Delta R_{T} }}{}U(t_f,t_i)\ket{\beta(t_i) }_s
=\tensor[_0]{\braket{\alpha}{\beta}}{_0}
=\delta_{\alpha,\beta}\,.
\end{align}
Thus we have confirmed that the $S$-matrix for the dressed states is IR finite.
This result shows that the $S$-matrix in this model is trivial except for the dressing operators $R_{out}(t_f), R_{in}(t_i),  \Delta R_T$ and $e^{-i\theta_{out}(t_f)}, e^{-i\theta_{in}(t_i)}$.
The first two operators $R_{out}(t_f), R_{in}(t_i)$ are necessary for Gauss's law constraint.
$\Delta R_T$ is an operator realising the shift of the asymptotic charges, and $e^{-i\theta_{out}(t_f)}, e^{-i\theta_{in}(t_i)}$ are state-independent overall phases.
This is consistent with the classical situation where the scattering of electromagnetic waves is trivial as discussed around eq.~\eqref{eq:cl_sol}.

\section{Discussion on full QED}\label{sec:fullQED}

We have seen that there are no IR divergences in the $S$-matrix elements in the background current model if we use the appropriate dressed states.
We will discuss that this is true even for the full QED.

As we saw, the naive IR divergences take  the same expressions \eqref{ampIRdiv} in the background current model and the full QED.
In fact, this result is roughly the soft photon theorem as follows.
We define the classical current operator $j^\mu_{\mathrm{pp}}$ as
\begin{align}
\label{eq:j_cl_op}
  j^\mu_{\mathrm{pp}}(t,\vec{x})&:=  \sum_n e_n \int \!\! \frac{d^3 p}{(2\pi)^3(2 E_p)} \frac{p^\mu}{E_p}\, \delta^3(\vec{x} -\vec{p} t/E_p) \rho_n(\vec{p}),
\\
\rho_n(\vec{p})&:=b_n^\dagger(\vec{p})b_n(\vec{p})-d_n^{\dagger}(\vec{p})d_n(\vec{p}),
\end{align}
where $b_n^\dagger$ (and $d_n^{\dagger}$) are creation operators of the charged particles with charge $e_n$ (and anti-particles).\footnote{Here we omit labels for spins to simplify the expression.
The normalization of the ladder operators is the same as \cite{Hirai:2018ijc, Hirai:2019gio}.
$\rho_n$ represents  the number density of charge $e_n$.  }
The operator $j^\mu_{\mathrm{pp}}$ acts on the charged particle sectors as the classical current of point particles like \eqref{j_cl}.
In the Feynman diagrams for the full QED,
the IR divergences arise when virtual soft photons interact with the external lines.
This interaction among soft photons and the on-shell charged particles is classical in the sense that we can replace the current operator $j^\mu$ with the above classical current operator $j^\mu_{\mathrm{pp}}$ as shown in the appendix~\ref{app:soft}.
The proof is almost the same as that of soft photon theorem \cite{Weinberg:1965nx}.
Thus, when we evaluate IR divergences in the full QED, we can approximate the interaction $A_\mu j^\mu$ by $A
_\mu j^\mu_{\mathrm{pp}}$.
This is the reason why we have the same IR divergences in the background current model and the full QED.

Since the structure of IR divergences are the same in the both model,
the divergences in the full QED are canceled if we can use the same dressed states used in section~\ref{sec:IRfinite}.
We need a clarification of the usage of these dressed states  because the physical state condition is now given by
\begin{align}
\label{full-Gauss-law}
     \left[k^\mu a_\mu(\vec{k}) +e^{i\omega t_s}j^{0}(t,\vec{k})\right] \ket{\psi(t)}_s=0,
\end{align}
which is different from \eqref{Gauss-law} in the current term.
Nevertheless, we can replace $j^{0}(t,\vec{k})$ by the classical current operator $j^{0}_{\mathrm{pp}}(t,\vec{k})$ in the asymptotic regions $t \sim \pm\infty$ using the saddle point approximation which becomes exact at $t = \pm\infty$ \cite{Hirai:2019gio}.
Thus, we can use as the asymptotic states in the full QED the same dressed states used in the background current model.
Then, there are no IR divergences in the $S$-matrix elements for these dressed states as we saw in the last section.
Of course, the $S$-matrix in the full QED is non-trivial unlike the background current model because the interaction involving hard photons is not the same.
It would be an interesting future work to compute $S$-matrix elements for the full QED in our  dressed state formalism.

\section{Conclusion and outlook}\label{sec:conclusion}
We have shown that the Kulish-Faddeev dressed states are not enough to remove IR divergences in the $S$-matrix of QED.
Although the Kulish-Faddeev dressed states, which were derived by solving the asymptotic dynamics in \cite{Kulish:1970ut}, are a solution of the physical state condition \eqref{Gauss-law_int} as eq.~\eqref{FK-dress-st},
it is able to add another dressing operator.
The asymptotic symmetry actually requires us to add such another dressing operator at least to the initial or final state as we saw in section~\ref{sec:AS}.
Reflecting this fact, if we use the original Kulish-Faddeev dressed states in both of the initial and final states, we encounter the IR divergences.
In this paper, we suggest putting the additional dressing operator
$e^{\Delta R_T}$ in addition to the original Kulish-Faddeev dressing operator $e^{R_{out}(t_f)}$ to the final state.\footnote{We also need to put some time-dependent but photon-state-independent phase factors to the initial and final states if we want to avoid an overall phase factor which infinitely oscillates in the asymptotic limit $t_i \to -\infty$, $t_f \to \infty$.}
For the new dressed states, IR divergences completely disappear, and the $S$-matrix is IR finite.

The dressing operator $e^{\Delta R_T}$ is almost the same as that in Chung's paper \cite{Chung:1965zza}.
Chung's dressing is necessary to cancel IR divergences, but it is not compatible with the gauge invariance unless we add another dressing involving longitudinal photons.
Kulish and Faddeev argued that their dressing is the same as Chung's dressing because the difference is just a factor like  $e^{i \frac{p\cdot k}{E}t}$ which becomes 1 for soft momenta $k \sim 0$ if $t$ is finite.
However, we cannot use this approximation because  we take the limit $|t| \to \infty$.
Therefore, Kulish-Faddeev dressing is different from Chung's one.
After all, we need both of them to cure IR divergences in a gauge invariant way, that is, our new dressed states are a mixture of the Chung and Kulish-Faddeev dressed states.

Here we would like to stress that the conservation law of the asymptotic symmetry is a necessary condition to obtain the non-vanishing $S$-matrix elements, but it is not the sufficient condition.
We introduced $\Delta R_T$ to realize the shift of the soft  charges like \eqref{change_Qas}.
Just to realize this shift, we have many other possibilities.
However, such dressings might not cancel the IR divergences.

We also have a comment on the movability of the photon clouds discussed in \cite{Kapec:2017tkm, Choi:2017ylo}.
The dressing factor $e^{\Delta R_T}$ which we put to the final state in this paper can be moved to the initial state.
Furthermore, we can decompose $\Delta R_T$ into two parts,  and put a part of $\Delta R_T$ to the final states and the rest to the initial state.
Any decomposition is compatible with the gauge invariance and the asymptotic symmetry.
The IR finiteness still holds,
since the decomposition changes only an overall finite phase factor of the $S$-matrix element.
However, we cannot move the Kulish-Faddeev dressing operators because of the gauge invariance.
The KF dressing is a realization of Gauss's law, and we always need a cloud of photons associated with a charged particle.
Thus, a charged particle without a photon cloud is not allowed.

We have considered uniformly moving charges as \eqref{j_cl} in the classical background model in this paper.
Although this is enough to look at the leading IR behaviors of QED,
the true trajectories of particles deviate because of the interaction of them.
The deviation is proportional to $\log |t|$ in the asymptotic region \cite{Laddha:2018rle, Laddha:2018myi, Sahoo:2018lxl}.
It is discussed in \cite{Laddha:2018myi} that this log behavior is related to the subleading soft factor proportional to $\log \omega$ where $\omega$ is the energy of a soft photon.
It would be interesting to incorporate the log deviation into the classical current \eqref{j_cl}, and see how the obtained dressed states are related to the subleading soft factor and also sub-subleading factor \cite{Sahoo:2020ryf}.

In addition, we have assumed that the linear trajectory go through $\vec{x}=0$ at $t=0$.
We can change it so that $\vec{x} \neq 0$ at $t=0$.
In \cite{Hamada:2018cjj}, it is shown for the classical case that such a change affects the subleading behavior of the Li\'enard-Wiechert potential and the subleading soft factor in Low's subleading soft theorem \cite{Low:1954kd, Low:1958sn, GellMann:1954kc, Burnett:1967km} is obtained.\footnote{Low's subleading soft factor is $\mathcal{O}(\omega^0)$ and different from the subleading $\log \omega$ term discussed above.}
Thus, if we consider the point-particle current of the shifted trajectories, the dressed states probably include the subleading soft factor.
Low's subleading soft factor contains the total angular momentum of the charged particles.
On the other hand, the analysis in \cite{Hamada:2018cjj} is classical, and reproduces only the orbital angular momentum.
It is unclear whether we can obtain the spin angular momentum in our background current model.
Dressed states containing the information of this subleading factor is considered in \cite{Choi:2019rlz}.\footnote{Dressed states containing the subleading information are also investigated recently for general Non-abelian gauge theories by using the worldline formalism \cite{Bonocore:2020xuj}.}
It is worth investigating the dressed states from the shifted trajectories mentioned above.

It is also important to extend our dressed state formalism to non-abelian gauge theories and gravity.

\section*{Acknowledgement}
The work of H.H. was supported in part by JSPS KAKENHI Grant Number 19J10588.

redpajama_set_name: RedPajamaArXiv


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Michele Chagnon Holbrook of Casabella Interiors knows what it takes to make a house a home. From her design studio on Cape Cod, the principal designer hand selects every piece of furniture, accessories, lighting, rugs, fabrics, drapery and art sourced from vendors around the world. What makes her business so unique is Holbrook's eye for combining classic elegance with modern comfort to create layered interiors using subtle textures and muted hues. Whether it's a whole home renovation or creating a new look for an individual room, the experienced team at Casabella Interiors will create your perfect look.
We caught up recently with Michele to discuss trends, what every room needs and creating the perfect escape.
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I just got back from Show and try to go three or four times a year to see trends in the future of the home. There continue to be new neutrals like pale blues and seafoam. Soothing, timeless colors are being brought out of the bedroom to other rooms like the living room and dining room. It's creating a place to snuggle up to. We are seeing a lot of staining or colored wood. For example, you could have a beautiful turquoise sofa and stained turquoise dresser. We are also seeing textured fronts that are not just smooth, which is so natural to the living element. A lot of our inspiration comes from the high seas with nautical tones and crisp blue and white, which remains classically stylish. The trends with the home are all about family, creating open floor plans that have a more relaxed design with performance fabrics, beautiful chenilles, large kitchens with ample seating, big stoves, home theaters and master bedrooms that are now hotel and spa-like. It's all about the work/life balance. We are also seeing a lot of changes in furniture with dual pieces, pieces hiding docking stations and even wifi enabled window treatments.
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What inspires you to come up with new ideas?
When someone tells me it's a big seller and I don't love it, I won't get it. We don't design around the trends, which is why classic pieces are timeless. My style is more classic, more collected. When someone comes to us, I find out what inspires them. What inspires me could be something I see in a magazine or it could be a medallion on a piece of fabric. I'm a hunter, always looking for the right piece. We can do any style, traditional or transitional, but my love is in the mix of everything. I like taking something ordinary and making it into an extraordinary piece by adding something interesting that makes it yours. Our challenge is recreating colors, but we want each space to look different. You don't want the same lamp or the same artwork in each room. It's constant hunting.
Something whimsy. I showed a fabric once for a kitchen drape of a loosely sketched rooster. When it was pleated, you couldn't really see it. That's how we design. Rooms shouldn't be too serious. For each room, we start with something we love and go from there.
Where should people splurge and where can they be cost conscious?
That's a good question. Always make sure the quality of pieces is your splurge. For example, in the kitchen, the counter and bar stools are one of the most used pieces. Invest in those pieces of furniture. From there, you can cut back on the pillow fabric or drape fabric, but don't cut corners on things that never change. I am always amazed when I see people invest $100,000 to $200,000 in a kitchen, but don't want to invest in lighting. How often do you change the hard lighting in your house? Never. If you are building a mirror into a bathroom, you can cut back on the vanity, but put some fun knobs on it. We are all about the details.
What is it about Casbella Interiors that makes you so unique?
I think one of the things that has helped our success is that a lot of small mom and pop shops are going away. We employ local people. We have a staff of five year round and do turnkey services from soup to nuts. We did a house last year and ended up gutting it and they just made one call to us. We did the soft furnishings, stonework and exterior work. We can stock your refrigerator, have the bed made, lamps on, candles lit. We can accessorize your house in a day. We are a collaboration of designers with a 2,500 square foot shop and 4,000 square foot warehouse. We charge one hourly rate and you get a whole team of us. One might do CAD drawings while another is pulling fabrics. Everyone knows what is going on with the project.
What is the key to making a traditional space transform in a beautiful one?
The devil is in the details. We get so many people coming to us who used a designer, but it's not finished. It's the layering that turns a house into a home. We want to make it look like you have lived there forever. People need guidance and our job is to cut it down on a smaller level so people don't get overwhelmed. We have seen so many people spend a lot of money, but don't end up with the finished product they were looking for.


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Розташоване в Полтавській області на відстані 10 км від м. Полтава.

Знаходиться в центральній частині приосьової зони Дніпровсько-Донецької западини.

Підняття виявлене в 1959 р. У відкладах нижнього карбону структура є асиметричною брахіантикліналлю півд.-зах. простягання розмірами по ізогіпсі — 4420 м 8,0х6,0 м, амплітуда 130 м. Перший промисл. приплив газу отримано з відкладів середнього карбону з інт. 4810-4860 м у 1974 р.

Поклади пластові, склепінчасті, тектонічно екрановані та літологічно обмежені. Режим покладів газовий. Запаси початкові видобувні категорій А+В+С1: газу — 10660 млн. м³; конденсату — 560 тис. т.

Джерела

Газоконденсатні родовища України
Газоконденсатні родовища Глинсько-Солохівського газонафтоносного району
Газові родовища України із запасами від 10 до 100 млрд.м3

redpajama_set_name: RedPajamaWikipedia


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La seconda edizione di Eurovision France, c'est vous qui décidez è stata organizzata dall'emittente televisiva francese France Télévisions per selezionare il rappresentante nazionale all'Eurovision Song Contest 2022 a Torino.

I vincitori sono stati Alvan & Ahez con Fulenn.

Organizzazione
Il 3 maggio 2021 l'emittente televisiva pubblica France Télévisions ha confermato la partecipazione della Francia all'Eurovision Song Contest 2022 insieme all'organizzazione di una selezione pubblica per il proprio rappresentante a Torino. In seguito al secondo posto di Barbara Pravi al contest, il format Eurovision France, c'est vous qui décidez è stato riconfermato e dal 21 giugno al 24 ottobre 2021 sono stati aperti i canali per permettere agli artisti interessati di inviare le proprie proposte all'emittente. All'inizio di gennaio 2022 si sono svolte le audizioni dal vivo degli artisti per selezionare i finalisti.

L'evento si è tenuto in un'unica serata il 5 marzo 2022. I conduttori dell'edizione precedente, Stéphane Bern e Laurence Boccolini, sono stati riconfermati. 12 artisti si sono contesi la possibilità di rappresentare il proprio paese al festival europeo. Il vincitore è stato decretato decretato attraverso due fasi di votazione: dal primo round si sono qualificate 6 canzoni, di cui 5 via televoto e un'ultima in base al voto della giuria, mentre nella seconda fase il voto combinato di giuria e pubblico ha decretato i risultati.

Giuria
La giuria internazionale è stata composta da:

  Jenifer, cantante
  Gjon's Tears, rappresentante della Svizzera all'Eurovision Song Contest 2021
  Nicoletta, cantante
  Cyril Féraud, conduttore televisivo
  Yseult, cantante
  Agustín Galiana, cantante e attore
  Élodie Gossuin, modella, conduttrice televisiva e radiofonica
  André Manoukian, cantante e musicista
  Joyce Jonathan, cantante
  Sundy Jules, youtuber

Partecipanti
Come accaduto nell'anno precedente, gli artisti interessati hanno partecipato a un'audizione all'Apollo Théâtre di Parigi dove una giuria ha scelto i 12 finalisti. I loro nomi e i relativi brani sono stati resi noti il 16 febbraio 2022.

Finale
La finale si è tenuta il 5 marzo 2022 presso gli France TV Studio, ed è stata trasmessa su France 2 e TV5 Monde.

Prima fase – Les Qualifications
Nella prima fase della serata, denominata Les Qualifications, il televoto ha promosso cinque artisti da far accedere alla fase finale.

Seconda fase – L'Euro-ticket
Nella seconda fase della serata, denominata L'Euro-ticket, la giuria internazionale ha selezionato l'ultimo finalista da far accedere alla finale.

Fase finale – Le Vote ultime
Nell'ultima fase, nota anche come Le Vote ultime, il voto combinato della giuria con quello del televoto ha decretato il vincitore tra i sei finalisti.  telespettatori hanno seguito l'evento dal vivo trasmesso su France 2, rappresentando uno share pari all'8,8%.

Alvan & Ahez sono stati proclamati vincitori trionfando sia nel televoto che nel voto della giuria.

Note

Voci correlate
 Francia all'Eurovision Song Contest
 Eurovision Song Contest 2022

Altri progetti

Collegamenti esterni

Festival musicali del 2022


text:
hackathons like this one View the winners
Best Overall Hack for LA App - 1st Prize
Judges will select the best 3 overall apps from all those submitted that have a civic, community, or public interest component that will benefit the Los Angeles community. Overall winners also receive automatic enrollment in SoftLayer's Catalyst Startup program, a 12-month program which includes free hosting credits ($1k/month; total of $12k for the year), mentorship (business and technical) and marketing support. Plus $500 in Amazon Web Services credit!
Best Overall Hack for LA App - 2nd Prize
Overall winners also receive automatic enrollment in SoftLayer's Catalyst Startup program, a 12-month program which includes free hosting credits ($1k/month; total of $12k for the year), mentorship (business and technical) and marketing support.
Best Overall Hack for LA App - 3rd Prize
Sierra Wireless Unite Device with 1 month 1GB data plan for each team member. Overall winners also receive automatic enrollment in SoftLayer's Catalyst Startup program, a 12-month program which includes free hosting credits ($1k/month; total of $12k for the year), mentorship (business and technical) and marketing support.
i.am.angel Award for Best Boyle Heights App
Prize for the team who create the app that best serves the needs of Boyle Heights, with potential to make a real-life impact on the community. Criteria: The app must be functional, incorporate local data sets or enable the creation of personal data, and have a tangible impact on the Boyle Heights community.
CGI Challenge for best use of Los Angeles data
CGI is sponsoring this prize for the best use of Los Angeles-specific data in an app or visualization.
Google Challenge: Best Environmental/ Health app - 1st prize
With so much public concern about the environment and health issues, but an overload of conflicting information in the public domain, Google is supporting a challenge to focused on using the wealth of public environmental and health data to clear up the confusion and help people make better choices.
Google Challenge: Best Environmental/ Health app - 2nd prize
Scopely Challenge: Best Civic App with a gaming or social element - 1st Prize
Scopely is redefining the way social games are developed! Scopely is offering prizes for the best civic apps with a gaming or social element.
Scopely Challenge: Best Civic App with a gaming or social element - 2nd Prize
Chase Bank Challenge: Best app related to jobs, business, and economic development
Chase Bank wants you to create the best app for Los Angeles economic development. Examples include apps that help people look for jobs, job training, apps that serve local small businesses, help foster local entrepreneurship etc.
memBRAIN Challenge: Best arts app
Create the best app related to LA arts and culture! Examples include apps that help preserve public art, chronicle the history of a community, creates or curates street art.
ESRI Map Challenge
Esri's GIS (geographic information systems) mapping software helps you understand and visualize data to make decisions. This challenge awards $1,000 for the best app that uses an ESRI map layer for a civic purpose.
AT&T API Challenge
Gift cards worth $250 for the best app using AT&T APIs - more details here: http://developer.att.com/developer/apiLandingPage.jsp?passedItemId=9700218
Amazon WS Cloud Challenge - First Prize
This award will be given to the team that made the best use of the AWS credits to improve LA through their app. The prizes (AWS credits) will allow teams to continue running their applications in the cloud.
Amazon WS Cloud Challenge - 2nd Prize (2)
Amazon WS Cloud Challenge - 3rd Prize (2)
Create innovative apps to solve LA's community challenges!
Following Los Angeles' election of our new mayor, the hackathon offers government agencies, businesses, nonprofits and civically engaged individuals the opportunity to work together to move into a new era of solving complex social problems through technology innovation. We are beginning the process of building a civic hacking movement here in Los Angeles, and this is our first event. We have multiple categories and challenges that you can respond to at the local and federal level. We are looking for innovative apps and ideas that address community issues and make use of open data.
View full rules
Any participating team registered through the Hack for LA challengepost site with a leader who is over 18 can enter the challenge. You must have an idea, functional overview, and at minimum technical specifications and some coding completed to enter for prizes. Any participating team can post and idea and overview to this page. All prizing is up to the judges and at the discretion of organizers and sponsors. No submission can win more than 2 prizes!
You must present an overview of your Hack for LA civic application, which should include the goals for the application, how it relates to the community needs of Los Angeles, which specific prize (if any) you are applying for, and the code you've created during the weekend, as well as any visual materials or other backup documents. We encourage mobile app submissions but any platform or software is acceptable.
All individuals and teams participating on-site during the Hack for LA event must reigster on Challengepost to be considering for prizes and judging.
Here are some datasets that can be used as part of you Hack for LA submission:
Metro Developer: Los Angeles Transit Data
City of Los Angeles Open Data Pilot Maps
Portal for Los Angeles County Data
Los Angeles Family and Neighborhood Survey (L.A.FANS)
Hack for LA Judges Team
Originality of the idea and potential impact on Los Angeles
Apps will be judged based on their originality and creativity as well as their potential impact on the community and the challenges we face here in Los Angeles.
Clear articulation of the idea
Teams will be judged on their ability to clearly articulate the details, goals, data, and process they used in development of their submission.
Technology used in the submission
Judges will rank submissions based on the technologies and platforms used as well as the extent and quality of coding completed during the weekend.
Feasibility and appeal
Judges will consider the feasibility of implementation and likelihood of adaption and use by the Los Angeles community.

redpajama_set_name: RedPajamaCommonCrawl


The Youth Award recognizes a youth leader who has undertaken habitat or conservation projects or activities and is a student in the elementary or secondary schools of our region.
The Education Award for excellence in conservation education programs or projects in schools, organizations, companies, youth groups, etc.
The Environmental Achievement Award for programs or projects which have enhanced the region's environment in a significant way. In addition to individuals and organizations, municipalities can also be nominated for environmental achievements.
Nominations can be submitted online, or nominators can simply write a one page summary of the individual or organizational accomplishments that merit the receipt of the award. Please be sure to include the category, name, address and phone number of the nominee, in addition to your own name, address and telephone number. Nominations close on Friday, December 14, 2018. The awards, presented since 1992, will be announced at ERCA's Annual General Meeting in January 2019.


Nowadays Group Discussion is being extensively used along with personal interviews for the selection of candidates. It plays a main role in selecting the best among the best and in choosing the socially suitable candidate among the academically superior achievers. It is one of the best tools to study the behavioral and attitudinal responses of the participants.
At Vidya Knowledge Park extra attention is paid to develop the GD skills by including it as an academic activity for assessment. This helps students to perform well in business meetings and emerge as leaders. Corporate experts help students to mould there GD skills to a level suitable for corporate discussions and business meets.
Simulated Interview is a process which helps the student to do their own competitive analysis and find their strengths and weaknesses. In Simulated Interview the University provides virtual corporate environment for the interviews. Vidya Knowledge Park invites industrial professional to be a part of every interview panel and judge the students.
Presentation is the most important part for a working professional and every company expects its employees to be a good presenter. Presentations are used to explain products to the client, results to management, future plans, processes to subordinates etc.Credit based system at Vidya Knowledge Park has made presentation a part of course curriculum. Thus a student is polished throughout the course duration till he reaches the perfection level, so that when he passes out of this university he shall be having all the qualities of a high performer employee.
Constant mentoring is a part of the LMS methodology. The student gets every piece of advice whenever he/she faces a problem with any aspect of the curricula.


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Area Employers Have Pick of Top IT Talent at "We Can Code IT" Recruiting Day
Columbus Student Spotlight: Brent Milan
Published by Coding Bootcamp at June 27, 2017
John Zitzner, President of Friends of Breakthrough Schools, congratulates Matthew Fothergill (right) after graduating from We Can Code IT with help from a $2,000 Breakthrough scholarship for parents with children in the charter school network.
Cleveland Dad Transforms Life and Career with Breakthrough Schools & We Can Code IT Partnership
CLEVELAND, Ohio – June 26, 2017 – Matthew Fothergill sends his 13-year-old son to Breakthrough School's E-Prep charter school to give him more opportunities to succeed in life. Now he wants the same kind of success for himself.
At age 36 with a high school diploma and a full-time job in building maintenance, Fothergill started taking courses to learn computer coding and software development. It all came about because of a partnership between the Breakthrough School network of high-performing free, public charter schools and Cleveland's We Can Code IT coding bootcamp.
"Our son's school sent home a flier about We Can Code IT courses for parents and my wife gave it to me," recalls Fothergill, the father of three sons. "I wanted to earn more money and do more with my life, so I decided to choose a path in coding and software development."
With an interest in computers but no technical background, Fothergill took the assessment and was accepted into the 20-week part-time evening program at We Can Code IT. Last Friday, Fothergill graduated during Career Day, where local employers came to see teams present new apps developed during the hands-on course. "It was a very big challenge for me, but I loved it, and I refused to quit! By the end of the course, I began to see myself as a developer," says Fothergill with a smile.
Katie Grootegoed is Director of Technology Enhanced Learning with Breakthrough Schools. "By partnering with We Can Code IT, we can reach more Cleveland parents with better opportunities and we plant seeds for our students that they can do this, too. There's no better role model than a parent."
The program provides training for those who can't afford college – with hopes of saturating the Cleveland area with competent programmers who can meet growing employer demands. Breakthrough already provides coding and robotics instruction to their K-8 students. "We're fostering a love for programming at an early age and setting students up for success," says Grootegoed.
The We Can Code IT boot camp focuses on the in-demand C# programming language. Graduates receive extensive job search support and earn an average annual starting salary of $55,000 for entry level software developer positions.
Fothergill is encouraging his middle son's dream to design video games and tells other parents that the We Can Code IT course is fast-paced and challenging, but very possible. "You need to be bold, have a can-do attitude and stick with it. It can be done part time in the evenings. It's a time commitment, but I've been blessed to have the support from my work and my family."
Breakthrough Schools' partnership with We Can Code IT is their most direct effort to serve students through their families, some of them with parents working 3 or 4 jobs to make ends meet. "If we can make an impact on just one family, it will spread," believes Grootegoed. "We will see career success but also the positive impact of parent role models."
The program is open to any member of a Breakthrough Schools' family who has a high school diploma, GED or equivalent, and is at least 18 years old. No previous coding or programming experience is necessary. Scholarships and financial aid for the coding bootcamp tuition is available. More information can be found at WeCanCodeIT.org and BreakthroughSchools.org
We Can Code IT teaches people how to code, with a special emphasis on inclusion and diversity in tech. With locations in Cleveland and Columbus, We Can Code IT is bridging the gap for those underrepresented in tech by graduating 80% women and people of color. Interested students can take a free online assessment and register for full time or part time courses at WeCanCodeIT.org.
Breakthrough Schools is a non-profit organization supporting the highest-performing network of free, public charter schools in Cleveland, Ohio. The mission of Breakthrough Schools is to develop a growing network of distinctive school options that prepare Cleveland area students for success in life. Learn more at BreakthroughSchools.org
This Bootcamp Graduate Is Prepared for the Real World
Coding Was the Answer for this Bootcamp Grad
Introducing New Coding Pre-Bootcamp in Cleveland


text:
News & Views 5/8/2019
Research Evaluates Disinfectant Wipe Effectiveness Against Candida Auris
The US Centers for Disease Control and Prevention (CDC) recommends the use of sporicidal disinfectants in hospital rooms of patients with Candida auris (C auris) colonization or infection. Specifically, daily and post-discharge cleaning of surfaces is recommended. But research says that non-sporicidal disinfectants can also be effective.
According to a Contagion Today article, a team of investigators from Northeast Ohio Veterans Affairs Healthcare System, Case Western Reserve University, University Hospital Cleveland Medical Center, and Louis Stokes Veterans Affairs Medical Center presented research on the effectiveness of disinfectant wipes in reducing C auris on surfaces.
That research included testing two distinct methods. The first evaluated the efficacy of different disinfectant wipes that were made up of various active ingredients. Specifically, the focus was on reducing C auris on steel, glass and equipment in the room.
The second method was conducted in a simulated patient room and examined efficacy of wipes and their potential for transferring C auris from surface-to-surface. Testing also included benefits of UV-C, use of quaternary ammonium and quaternary ammonium-alcohol wipes in a "real world" setting.
As reported in the article, "the study found that the use of all disinfectant wipes led to a >5 log10 colony forming units (CFU) reduction of C auris on steel and glass carriers, while non-disinfectant wipes led to a reduction of ~2 log."
Health experts have issued a warning regarding C auris, which has been quietly spreading around the world. The CDC says almost half of those infected die in 90 days, according to reports on CleanLink. Due to the growing and global spread, the CDC has just added C auris to its list of "urgent threats."
In recent years, superbugs have been a growing threat. As fungus and bacterias become immune to medications, the focus shifts to facility cleanliness, personal hygiene, infection control and a cross-contamination efforts.
The fungus can live on surfaces for weeks if not properly cleaned, CDC officials said.
As cleaning experts are aware, eradicating superbugs is no easy task — it requires specific products and processes.
With supervision from an infectious disease expert, "hospital workers used a special device to spray aerosolized hydrogen peroxide around a room used for a patient with C auris, the theory being that the vapor would scour each nook and cranny. They left the device going for a week. Then they put a 'settle plate' in the middle of the room with a gel at the bottom that would serve as a place for any surviving microbes to grow. Only one organism grew back. C auris."
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