Can someone please explain to me why when we calculate the Kraus operators (the action of a unitary $U$ on a restricted system we have $$M_{j} = \langle j|U|0\rangle\,,$$ where $j \in \text{Span(Environment)}$.
However, I don't understand what the $|0\rangle$ is there for. Perhaps it is not only ever $|0\rangle$, but in all the examples I looked at, it was. I can't figure out where this comes from.
Starting from the system-environment form of a $d$-dimensional channel $$ \Phi(\rho):={\rm tr}_E(U(\rho\otimes|\psi\rangle\langle\psi|)U^\dagger) $$ where the environment is in the pure state $|\psi\rangle$, the Kraus operators of $\Phi$ are precisely $M_j=\langle j|_EU|\psi\rangle_E\in\mathbb C^{d\times d}$ (where $j$ runs from $0$ to $d_E-1$, with $d_E$ the environment dimension). The point is that the Kraus operators depend not only on $U$, but also on the state the environment is initialized in. And if $|\psi\rangle=|0\rangle$, i.e., if the environment is initially in the basis state $|0\rangle$, then you recover the formula you asked about.
As an aside, if the environment starts out in a mixed state $\omega=\sum_{i=1}^mr_i|\psi_i\rangle\langle\psi_i|$, then you have even more Kraus operators: $\sqrt r_i\langle j|_EU|\psi_i\rangle_E$, $i=1,\ldots,m$, $j=0,\ldots,d_E-1$. In other words operators of the form $\langle j|_EU|0\rangle_E$ could—up to rescaling—also appear for mixed environment, but then there would necessarily be more Kraus operators than just the ones you listed in your question.
