# Trace distance bound after partial trace

Let's say I have a pair of states among three parties Alice(A), Bob(B) and Eve(E), $$\rho_{ABE}$$ and $$\rho_{UUE}$$ where the first two parties hold uniform values U.}

I know that the trace distance between them is upper bounded by some small quantity $$\epsilon$$. Like:

$$||\rho_{ABE} - \rho_{UUE} ||\le \epsilon .$$ Now, I take the partial trace on B's subsystem, leaving only operators $$\rho_{AE}, \rho_{UE}$$. What can I say about the trace distance of these two operators? Would the same bound be followed? I.e. is it the case that: $$||\rho_{AE} - \rho_{UE} || \le \epsilon.$$

• A comment on notation: I think what you want to say is that there is another state $\sigma_{ABE}$ where $\sigma_A$ and $\sigma_B$ are maximallly mixed. The subscripts refer to registers (such as Alice's register or Bob's register) and using $U$ means there is another party involved.
– rnva
Oct 27, 2020 at 17:27
• Thanks for pointing that out @rnva . You are right. Oct 27, 2020 at 17:31

Yes, the trace distance can only decrease under partial trace. One can see this via the variational characterization of the trace norm $$\|\rho\|_1 = \max_{-I \leq M \leq I} \mathrm{Tr}[M\rho]$$ where $$M$$ is some hermitian operator satisfying the two operator inequalities $$M \leq I$$ and $$M \geq - I$$. This is sometimes also known as the duality between conjugate norms as $$\| M \|_{\infty} \leq 1$$.
Now if we have a bipartite system $$AB$$ we find \begin{aligned} \|\rho_{A} - \sigma_A\| &= \max_{-I_A \leq M_A \leq I_A} \mathrm{Tr}[M_A(\rho_A - \sigma_A)] \\ &= \max_{-I_A \leq M_A \leq I_A} \mathrm{Tr}[(M_A \otimes I_B)(\rho_{AB} - \sigma_{AB})] \\ &\leq \max_{-I_{AB} \leq M_{AB} \leq I_{AB}} \mathrm{Tr}[M_{AB}(\rho_{AB} - \sigma_{AB})] \\ &= \| \rho_{AB} - \sigma_{AB} \|_{1}. \end{aligned} The second line follows from the identity $$\mathrm{Tr}[\mathrm{Tr}_A[X_{AB}]] = \mathrm{Tr}[X_{AB}]$$ and the third line from the fact that $$-I_{AB} \leq M_{A} \otimes I_B \leq I_{AB}$$ is satisfied for all $$-I_A \leq M_A \leq I_A$$.