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: $$ ||\rho_{AE} - \rho_{UE} || \le \epsilon. $$

  • 1
    $\begingroup$ 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. $\endgroup$
    – rnva
    Commented Oct 27, 2020 at 17:27
  • $\begingroup$ Thanks for pointing that out @rnva . You are right. $\endgroup$ Commented Oct 27, 2020 at 17:31

1 Answer 1


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 $-I_A \leq M_A \leq I_A$.

Actually you can even prove a stronger result: the trace distance decreases under CPTP maps of which the partial trace is a special case.


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