# Relation between symmetric subspaces and $n$-exchangeable density matrices

Let us consider $$n$$ elements, each taken from the set $$\{1, 2, \ldots, d\}$$ and let $$S_n$$ be the set of all permutations on these $$n$$ elements.

Define a permutation operator on the set of $$n$$ qudits as

$$P_d(\pi) = \sum_{i_1, i_2, \ldots, i_n \in [d]} |i_{\pi(1)}, \ldots, i_{\pi(n)}\rangle \langle i_1, \ldots i_n|.$$

Define the symmetric subspace on $$n$$ qudits as $$$$V^{n}(\mathbb{C}^{d}) = \text{span}\{|\phi\rangle \in (\mathbb{C}^{d})^{\otimes n} : P_d(\pi)|\phi\rangle = |\phi\rangle~~\text{for all}~~\pi \in S_n \}.$$$$

Define an $$n$$ exchangeable density matrix $$\rho \in \text{Density}\bigg(\big(\mathbb{C}^{d}\big)^{\otimes n}\bigg)$$ as

$$P_d(\pi) \rho P_d(\pi) = \rho, ~\text{for all} ~\pi \in S_n.$$

The notations are borrowed from here.

It is easy to see that any linear combination like

$$\rho = \sum_i \alpha_i |\phi_i\rangle\langle \phi_i|,$$ where each $$|\phi_i\rangle \in V^{n}(\mathbb{C}^{d})$$ and the $$\alpha_i$$s are arbitrary complex numbers is an $$n$$ exchangeable state.

I could not prove the converse, however. Is it true that any $$n$$ exchangeable state is a linear combination of density matrices of states in the symmetric subspace? If not, what is an example to the contrary?

The answer is: no, it is not true that any $$n$$ exchangeable state is a linear combination of density matrices of states in the symmetric subspace (that is supported on the symmetric subspace). Actually, there are even pure state counterexamples when $$n=2$$. Consider the state

$$\rho = |\phi\rangle\langle \phi|,$$

where

$$|\phi\rangle = \frac{1}{\sqrt{2}}(|0 \rangle \otimes |1 \rangle - |1 \rangle \otimes |0 \rangle).$$

In this case the permutation group is simply $$S_2$$. The action of the trivial permutation obviously maps $$|\phi\rangle$$ to $$|\phi\rangle$$; while for the nontrivial permutation $$\pi=(12)$$ (which is simply the swap operation), we have $$P_{\pi} |\phi\rangle=-|\phi\rangle$$. This means that $$P_{\pi}|\phi\rangle \langle \phi|P_{\pi} = (-1)^2 |\phi\rangle \langle \phi|= |\phi\rangle \langle \phi|$$, which implies that (according to your definition) $$\rho$$ is exchangeable. However, $$\rho$$ is clearly not supported on the symmetric subspace (it is actually supported on the antisymmetric subspace), thus it cannot be written as a linear combination of density matrices of states in the symmetric subspace.

Also in the $$n > 2$$ case there are states that are not supported on the symmetric subspace (i.e., their density matrices cannot be written as the requested linear combination in the question). However, when $$n >d$$ (i.e., when there is no totally antisymmetric subspace) these states must be mixed, and in this case (when $$n>d$$) every $$n$$-exchangeable pure state is supported on the symmetric subspace.

• ".....and in this case every 𝑛-exchangeable pure state is supported on the symmetric subspace." Why is this? Can't it also be supported on the anti-symmetric subspace? It is still true that for a state in the anti-symmetric subspace, $P_{\pi} |\psi\rangle = - |\psi\rangle$ or $P_{\pi} |\psi\rangle = |\psi\rangle$ (depending on the sign of the particular permutation in consideration) and that $P_{\pi} |\psi\rangle \langle \psi| P_{\pi} = |\psi\rangle \langle \psi|$ for every $\pi$ --- is it not? Sep 12 at 11:59
• @BlackHat18, you are right. I meant the case when there is no antisymmetric subspace, i.e., when $d>n$. I modified the answer accordingly. Sep 12 at 12:09
• A last nitpick: I think your latest edit of the answer has a typo and says that $n > d$ instead of $d > n$. Sep 12 at 12:52
• It should be $n >d$, the typo was in my comment, not in my answer. Sep 12 at 12:56
• Oh oops. Yes, you're right! Thanks a lot. Sep 12 at 12:58