3
$\begingroup$

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 \begin{equation} 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 \}. \end{equation}

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ ".....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? $\endgroup$
    – BlackHat18
    Sep 12 at 11:59
  • 1
    $\begingroup$ @BlackHat18, you are right. I meant the case when there is no antisymmetric subspace, i.e., when $d>n$. I modified the answer accordingly. $\endgroup$ Sep 12 at 12:09
  • $\begingroup$ A last nitpick: I think your latest edit of the answer has a typo and says that $n > d$ instead of $d > n$. $\endgroup$
    – BlackHat18
    Sep 12 at 12:52
  • 1
    $\begingroup$ It should be $n >d$, the typo was in my comment, not in my answer. $\endgroup$ Sep 12 at 12:56
  • $\begingroup$ Oh oops. Yes, you're right! Thanks a lot. $\endgroup$
    – BlackHat18
    Sep 12 at 12:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.