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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 $H_{2}$. 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.

$$ \|e^{-itX\otimes P} \rho\otimes\sigma e^{itX\otimes P} - \Lambda_{2}\big\{e^{-itX\otimes P} \rho\otimes\sigma e^{itX\otimes P}\big\}\|_{1} = $$ $$\big\|e^{-itX\otimes P}\bigg\{\int\int K_{\rho}(x,y)|x\rangle\langle y| dxdy\otimes \sigma\bigg\} e^{itX\otimes P} - \Lambda_{2}\Bigg\{ e^{-itX\otimes P}\bigg\{\int\int K_{\rho}(x,y)|x\rangle\langle y| dxdy\otimes \sigma\bigg\} e^{itX\otimes P}\Bigg\} \big\|_{1} = $$

$$ \big\|\int\int K_{\rho}(x,y)|x\rangle\langle y| \otimes e^{-itx P}\sigma e^{ity P}dxdy - \Lambda_{2}\Bigg\{\int\int K_{\rho}(x,y)|x\rangle\langle y| \otimes e^{-itx P}\sigma e^{ity P}dxdy\Bigg\} \big\|_{1} = $$ $$\big\|\int\int K_{\rho}(x,y)|x\rangle\langle y| \otimes e^{-itx P}\sigma e^{ityP}dxdy - \int\int K_{\rho}(x,y)|x\rangle\langle y| \otimes \Lambda_{2}\Big\{e^{-itx P}\sigma e^{ity P}\Big\}dxdy\big\|_{1} $$

$$\big\|\int\int K_{\rho}(x,y)|x\rangle\langle y| \otimes \Big(e^{-itx P}\sigma e^{ityP} - \Lambda_{2}\Big\{e^{-itx P}\sigma e^{ity P}\Big\}\Big)dxdy\big\|_{1} $$

Up to here I believe that all is in order, correct me if I am wrong. However, what baffels me is the following. This is my question to StackExchange. How do I proceed from $$\big\|\int\int K_{\rho}(x,y)|x\rangle\langle y| \otimes \Big(e^{-itx P}\sigma e^{ityP} - \Lambda_{2}\Big\{e^{-itx P}\sigma e^{ity P}\Big\}\Big)dxdy\big\|_{1} $$. without expanding the kernel $K_{\rho}$.

Is it ever the case that
$$\big\|\int\int K_{\rho}(x,y)|x\rangle\langle y| \otimes \Big(e^{-itx P}\sigma e^{ityP} - \Lambda_{2}\Big\{e^{-itx P}\sigma e^{ity P}\Big\}\Big)dxdy\big\|_{1} \leq $$ $$\int K_{\rho}(x,x) \big\|e^{-itx P}\sigma e^{ityP} - \Lambda_{2}\Big\{e^{-itx P}\sigma e^{ity P}\Big\}\big\|_{1}dx $$ I highly doubt this.

My other hypothesis is that $$\big\|\int\int K_{\rho}(x,y)|x\rangle\langle y| \otimes \Big(e^{-itx P}\sigma e^{ityP} - \Lambda_{2}\Big\{e^{-itx P}\sigma e^{ity P}\Big\}\Big)dxdy\big\|_{1} \leq$$ $$\int\int K_{\rho}(x,y) \big\|e^{-itx P}\sigma e^{ityP} - \Lambda_{2}\Big\{e^{-itx P}\sigma e^{ity P}\Big\}\big\|_{1}dxdy $$

But I am not having successproving or disproving these.

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  • $\begingroup$ In your first inequality, is $y$ supposed to be $x$? $\endgroup$ Jun 21, 2022 at 23:11
  • $\begingroup$ When you say $\Lambda$ is a quantum operation, is it a unitary? $\endgroup$ Jun 22, 2022 at 0:05
  • $\begingroup$ @JoshuaLin Yes, the $y$ you saw in the first inequality was indeed supposed to be an $x$. Now, the map $\Lambda$ could be a unitary map but in general it need not be. Such a map is characterized by its Krauss operator representation. let $\rho$ be a density operator. In general $\Lambda( \rho) = \sum_{i} B_{i} \rho B_{i}^{*}$ where $\{B_{i}\}_{i}$ are these so called Krauss operators satisfying $\sum_{i} B_{i}B^{*}_{i} = \mathbb{I}$. This is compactly presented in the wikipedia page en.wikipedia.org/wiki/Quantum_operation kindly scroll down to the Krauss Operator section. $\endgroup$
    – Hldngpk
    Jun 22, 2022 at 1:57

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