# 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 $$H_{2}$$. 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.

• In your first inequality, is $y$ supposed to be $x$? Jun 21, 2022 at 23:11
• When you say $\Lambda$ is a quantum operation, is it a unitary? Jun 22, 2022 at 0:05
• @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. Jun 22, 2022 at 1:57