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\begin{document}
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\title{EPAPS}

\date{\today}
\author{}

\begin{abstract}
\end{abstract}

\maketitle


In this supplementary material, we detail the proof of Observation 4 of the main text.
We again label in the Stinespring dilation
$T(\rho) = \text{tr}_E(
    U \rho U^\dagger
    )$ of the quantum
    channel $T$ the input by $A$, the output by $B$ and the environment by $E$, but also keep a system $R$
holding a purification of the input. We have to show that if $Q(T) = S_{\max}(T)$, then also
\begin{eqnarray}
    Q_1(T)= S_{\max}(T).
\end{eqnarray}
If $Q(T) = S_{\max}(T)$ indeed holds true, then
we find for the classical and quantum entanglement-assisted capacities $C_E$ and $Q_E$,
respectively,
\begin{eqnarray}\label{ineq}
    C_E(T) = 2Q_E(T) \geq 2Q(T) = 2S_{\max}(T).
\end{eqnarray}
We also know that
\begin{eqnarray}
    C_E(T) &=& \max_\Psi I(\Psi,  R : B),
\end{eqnarray}
where the maximum is taken over pure states $\Psi$.
One way of expressing the right hand side, so the mutual information,
in our notation of input $A$, output $B$, environment $E$, and
purification $R$ is
\begin{eqnarray}\label{b}
    C_E(T) = \max_\Psi \left(
    S(\rho_R) + S(\rho_B)  - S(\rho)\right),
\end{eqnarray}
where $\rho$ is a mixed state that is shared between $B$ and $R$
that is obtained as $\rho = (\id_R\otimes T)(\Psi)$. From the
subaddivity of the von-Neumann entropy, it follows that for the
input state that achieves the maximum in Eq.\ (\ref{b}),
\begin{equation}
    2S(\rho_B) \geq S(\rho_R)+S(\rho_B) -S(\rho) \geq 2S_{\max}(T).
\end{equation}
Since at the same time, by definition $S(\rho_B)\leq S_{\max}(T)$, we find that there exists an
input pure state $\Psi = |\psi\rangle\langle \psi| $
shared between $R$, $B$, and $E$, with $|\psi\rangle = U|\phi\rangle$,
$|\phi\rangle$ being shared between $R$ and $A$, such that
\begin{eqnarray}
    S(\rho_B) &= & S _{\max}(T),\\
    S(\rho_R)+S(\rho_B) -S(\rho) &=& 2 S_{\max}(T).
\end{eqnarray}
The latter equality also implies that
\begin{eqnarray}
    S(\rho_{B,E})  = S(\rho_B) + S(\rho_E),
\end{eqnarray}
which in turn means
$\rho_{B,E}= \rho_B\otimes \rho_E$.

In the final step, it is the aim to construct an input to the channel that certifies that the single shot
quantum capacity $Q_1(T) = S_{\max}(T)$. Based on the above properties of the channel, this can easily be done.
By virtue of the Schmidt decomposition,
there exists a unitary $V$ supported on $R$ such that
\begin{eqnarray}\label{out}
    V|\psi\rangle &=& |\xi\rangle_{R_1, B} |\eta\rangle_{R_2,E}\nonumber\\
    &=&
    \sum_{i,j}
    \alpha_i \beta_j |i\rangle_{R_1} |e_i\rangle_B |j\rangle_{R_2} |e_{j}\rangle_E,
\end{eqnarray}
where for convenience the tensor factors have been indicated. So the input
\begin{equation}
    V|\phi\rangle = \sum_{i,j}
    \alpha_i \beta_j |i\rangle_{R_1} |j\rangle_{R_2} |d_{i,j}\rangle_A,
\end{equation}
will give rise to the output as in Eq.\ (\ref{out}). From this input to the original problem we can construct a new input which
achieves the desired bound: Take
\begin{eqnarray}
    |\nu\rangle = \sum_i\alpha_i  |i\rangle_{R_1}  |0\rangle_{R_2} |d_{i,0}\rangle_A,
\end{eqnarray}
yielding the output
\begin{eqnarray}
     U|\nu\rangle =  |\xi\rangle_{R_1, B} |0\rangle_{R_2}|e_0\rangle_E,
\end{eqnarray}
which is a tensor product between $R$ and $E$. This means that $Q_1(T) = \max_\Psi ( S(\rho_R) - S(\rho) )= S_{\max}(T)$,
which is that was to be shown.

\end{document}
