Is there a mistake in this paper when calculating the reduced density matrix (simple model)? This question regards this paper by Zurek, Cucchietti and Paz from 2008 (DOI: https://doi.org/10.48550/arXiv.quant-ph/0611200). There the following calculation is made:
$$|\psi_{SE}(t)\rangle=a|0\rangle|\epsilon_o(t)\rangle +b|1\rangle|\epsilon_1(t)\rangle$$
$$\rho_S=\text{Tr}_{\epsilon}|\psi_{SE}(t)\rangle\langle\psi_{SE}(t)|=|a|^2|0\rangle\langle0|+ ab^*r(t)|0\rangle\langle1|+a^*br^*(t)|1\rangle\langle0|+|b|^2|1\rangle\langle1|$$
with
$$r(t)=\langle\epsilon_1(t)|\epsilon_0(t)\rangle$$
With my own calculation, especially by using the following two "rules", I arrive at
$$r(t)=\langle\epsilon_0(t)|\epsilon_1(t)\rangle$$
Rules (1) and (2):
$\left(|a\rangle \otimes|b\rangle\right)\left(\langle c|\otimes \langle d|\right) =|a\rangle\langle c|\otimes|b\rangle\langle d|\tag{1}$
$\text{Tr}\left({\sum_{i,j=1}^2|\alpha_i\rangle\langle\alpha_j|\otimes|\beta_i\rangle\langle\beta_j|}\right)=\sum_{i,j=1}^2|\alpha_i\rangle\langle\alpha_j|\cdot \langle\beta_i|\beta_j\rangle\tag{2}$
Am I making a mistake or is the calculation in the paper "wrong" (maybe just $r(t)$ and $r^*(t)$ get mixed up)?
 A: 
Rules (1) and (2):
$\left(|a\rangle \otimes|b\rangle\right)\left(\langle c|\otimes \langle d|\right) =|a\rangle\langle c|\otimes|b\rangle\langle d|\tag{1}$
$\text{Tr}\left({\sum_{i,j=1}^2|\alpha_i\rangle\langle\alpha_j|\otimes|\beta_i\rangle\langle\beta_j|}\right)=\sum_{i,j=1}^2|\alpha_i\rangle\langle\alpha_j|\cdot \langle\beta_i|\beta_j\rangle\tag{2}$

Eq (2) is not correct. The $i$ and $j$ are flipped. You can see this because
$$
\text{Tr}_{\beta}\left({\sum_{i,j=1}^2|\alpha_i\rangle\langle\alpha_j|\otimes|\beta_i\rangle\langle\beta_j|}\right)
$$
$$
=
\sum_{i,j=1}^2|\alpha_i\rangle\langle\alpha_j|\cdot\text{Tr}_{\beta}\left({|\beta_i\rangle\langle\beta_j|}\right)
$$
$$
=\sum_{i,j=1}^2|\alpha_i\rangle\langle\alpha_j|\cdot \sum_n\langle n|\beta_i\rangle\langle\beta_j|n\rangle
$$
$$
=\sum_{i,j=1}^2|\alpha_i\rangle\langle\alpha_j|\cdot \sum_n\langle\beta_j|n\rangle \langle n|\beta_i\rangle
$$
$$
=\sum_{i,j=1}^2|\alpha_i\rangle\langle\alpha_j|\cdot \langle\beta_j|\sum_n|n\rangle \langle n||\beta_i\rangle
$$
$$
=\sum_{i,j=1}^2|\alpha_i\rangle\langle\alpha_j|\cdot \langle\beta_j|\beta_i\rangle
$$
