# Is the trace distance between multipartite states invariant under permutations?

1. Consider two multipartite states $$\rho_{A_1A_2..A_L}$$ and $$\sigma_{A_1A_2..A_L}$$ in $$\mathcal{H}_{A_1} \otimes\mathcal{H}_{A_2} \otimes...\mathcal{H}_{A_L}$$. For an arbitrary permutation $$\pi$$ over $$\{ 1, \ldots ,L\}$$, is it true that

$$\lVert \rho_{A_1A_2..A_L} - \sigma_{A_1A_2..A_L} \lVert_1 = \lVert \rho_{A_{\pi(1)}A_{\pi(2)}..A_{\pi(L)}} - \sigma_{A_{\pi(1)}A_{\pi(2)}..A_{\pi(L)}} \lVert_1?$$

1. If 1. is not true, is the following true: If $$\rho_{A_1A_2..A_LB}$$ and $$\sigma_{A_1A_2..A_LB}$$ are two calssical-quantum states in $$\mathcal{H}_{A_1} \otimes\mathcal{H}_{A_2} \otimes...\mathcal{H}_{A_L} \otimes \mathcal{H}_{B}$$, i.e., one can write $$\rho_{A_1A_2..A_LB} = \sum_{a_1} \ldots \sum_{a_L} p(a_1,a_2,\ldots,a_L) |a_1\rangle \langle a_1| \otimes \ldots \otimes |a_L\rangle \langle a_L| \otimes \rho_B^{a_1,\ldots,a_L}$$, then $$\lVert \rho_{A_1A_2..A_LB} - \sigma_{A_1A_2..A_LB} \lVert_1 = \lVert \rho_{A_{\pi(1)}A_{\pi(2)}... B...A_{\pi(L)}} - \sigma_{A_{\pi(1)}A_{\pi(2)}... B...A_{\pi(L)}} \lVert_1,$$ where the position of the subscript $$B$$ is arbitrary on the righthand side.
• Interesting question. It may not be so. Because you are not imposing any restriction over the structure of the multipartite state. Are they i.i.d? If not, this may not be true in general. So I'm hoping to see expert answers for this question too. Nov 26 '20 at 1:56

## 2 Answers

A permutation of the qubits is a unitary operation. The trace distance is invariant under unitaries (https://en.wikipedia.org/wiki/Trace_distance#Properties). Thus, statement 1 is true.

I'd like to add a small addition to the answer of @DaftWullie about why you would expect this operationally to be true -- without knowing permutations correspond to unitary matrices.

It boils down to the Holevo-Helstrom Theorem (HHT) that says the trace distance between two states characterizes operationally the probability that we can distinguish the two states. Suppose some referee has two multipartite states and places each of their subsystems in different boxes. We are then given a collection of boxes $$A_1, \dots, A_n$$ which we are told contain the parts of either the state $$\rho$$ or the state $$\sigma$$ with $$50\%$$ probability of each. Tasked with guessing whether we got $$\rho$$ or $$\sigma$$ we know by the HHT that our probability of guessing correctly using the optimal strategy is a function of $$\lVert \rho_{A_1A_2..A_n} - \sigma_{A_1A_2..A_n} \lVert_1.$$ Moreover, from the perspective of a guessing game it is clear that our probability of guessing correctly shouldn't depend on which order we decide to arrange our boxes in the lab -- as part of the guessing protocol we could anyway rearrange the boxes to maximize the guessing probability. So for any permutation $$\pi$$ the probability of guessing correctly when we're given boxes $$A_{\pi(1)},\dots,A_{\pi(n)}$$ should be the same and hence, $$\lVert \rho_{A_1A_2..A_n} - \sigma_{A_1A_2..A_n} \lVert_1 =_{\mathrm{operationally}} \lVert \rho_{A_{\pi(1)}A_{\pi(2)}..A_{\pi(n)}} - \sigma_{A_{\pi(1)}A_{\pi(2)}..A_{\pi(n)}} \lVert_1.$$

Please note that this is in no way a proof and $$=_{\mathrm{operationally}}$$ is not a well-defined concept. This is merely some intuition to maybe complement the accepted answer. On reflection, thinking of the permutation as a rearranging the order of the boxes in the lab tells us that the permutation operation should be a unitary one -- we could always just reverse the arrangement if we please which should not affect the state.

• This is a nice extra to the accepted answer. A question though: "Moreover, from the perspective of a guessing game it is clear that our probability of guessing correctly shouldn't depend on which order the boxes came in". Why would this be the case? Doesn't not knowing which box corresponds to which part of the state introduce an additional difficulty in the guessing game? Nov 26 '20 at 12:43
• @user1936752 Sorry, maybe it was not clear. When we write $\rho_{A_{\pi(1)}A_{\pi(2)}..A_{\pi(n)}}$ we still know where the system $A_1$ has gone to. That is, we know what permutation was applied to our boxes. If the referee was to apply a random permutation and give us the box without knowing what permutation it was then our state of knowledge would be $\sum_{\pi \in S_n} p(\pi) \rho_{A_{\pi(1)}A_{\pi(2)}..A_{\pi(n)}}$ where $p$ is some probability distribution that the referee gives us. I'll edit my answer accordingly, thanks for the question. :) Nov 26 '20 at 15:20