Trace distance of two infinite rank tensor product states
Let $\sigma$ and $\rho$ be both density operators acting on different Hilbert spaces, $H_{1}$ and $H_{2}$ respectively. Also, let said operators have infinite rank. In the infinite dimensional case these may be treated as integral operators. Let $K_{\rho}(x,y)$ and $K_{\sigma}(u,v)$ the respective kernels. Now, let $H_{int} = X\otimes P$ be the intereaction Hamiltonian governing the total unitary dynamics. X is the position operator and $P$ is the momentum operator. Where $X$ acts on $H_{1}$ and $P$ acts on $H_{2}$. 956
Finally, let $\Lambda_{2}$ be a quantum operation that only acts non-trivially on the subspace $B(H_{2})$. i.e. $$ \Lambda_{\sigma}(\rho\otimes \sigma) = \rho \otimes \Lambda_{2}(\sigma).$$
I have been trying to compute the following trace distance.
$$ \|e^{-itX\otimes P} \rho\otimes\sigma e^{itX\otimes P} - \Lambda_{2}\big{e^{-itX\otimes P} \rho\otimes\sigma e^{itX\otimes P}\big}\|{1} $$ Where the trace distance is defined as follows. $$ \|A-B\|{1} = \frac{1}{2}Tr\sqrt{\big(A-B\big)^{\dagger}\big( A-B\big) } $$ My question is independent of the specific structure of the quantum operation that I have laid out, i.e. $\Lambda_{2}$, so we will need not select something concrete.
Now, there are two approaches that I have taken. Here is the one that I am the most dubious about. Please let me know if any step is falacious. I will rewrite the state $\rho$ using the dirac notation. 916