tain a teleported state identical to the original input state. We call the channel $f_q$ reversible, if it is injective, that is, for different input state $|\Phi\rangle_A$ ($\||\Phi\rangle_A\|=1$) the corresponding output state $f_q(|\Phi\rangle_A)$ is different. We remark, that the reversibility of teleportation channels has also been investigated in Ref. [@pra55_2547]. We adopt a more general definition here. Reversibility means that every input state can be recovered (theoretically) from the output state. One can easily verify that this condition is equivalent to that the linear operator $LL_q^\dag \colon {{\mathcal H}}_A \to {{\mathcal H}}_C$ is injective.

It may be the case, however, that the channel $f_q$ is not linear. This way, the input state can be recovered from the output only using some sophisticated nonlinear transformations, which may not be realistic. Therefore, it is a natural requirement for the channel to be linear.

We show that if the teleportation channel is reversible, then its linearity is equivalent to that the probability (\[eq:p\_q\]) of the outcome $q$ is independent of the input state $|\Phi\rangle_A$. Suppose that $|\Phi\rangle_1$ and $|\Phi\rangle_2$ are linearly independent, and let $(\alpha_1|\Phi\rangle_1 + \alpha_2|\Phi\rangle_2)$ be such that $\|\alpha_1|\Phi\rangle_1 + \alpha_2|\Phi\rangle_2\|=1$. From the linearity condition $f_q(\alpha_1 |\Phi\rangle_1 + \alpha_2 |\Phi\rangle_2) = \alpha_1
f_q(|\Phi\rangle_1) + \alpha_2 f_q(|\Phi\rangle_2)$, one can obtain: $$\begin{gathered}
  \alpha_1 \left( \frac1{\big\| LL_q^\dag (\alpha_1|\Phi\rangle_1 +
  \alpha_2|\Phi\rangle_2) \big\|} - \frac1{\big\| LL_q^\dag |\Phi\rangle_1
  \big\|} \right) LL_q^\dag |\Phi\rangle_1 
\\
  + \alpha_2 \left( \frac1{\big\|
  LL_q^\dag (\alpha_1|\Phi\rangle_1 + \alpha_2|\Phi\rangle_2) \big\|} -
  \frac1{\big\| LL_q^\dag |\Phi\rangle_2 \big\|} \right) LL_q^\dag
  |\Phi\rangle_2  =0.
\label{eq:linfq2}\end{gathered}$$ Since $f_q$ is injective, $LL_q^\dag |\Phi\rangle_1$ and $LL_q^\dag
|\Phi\rangle_2$ a