# Perfect correlations in bipartite (impure) density operators

Consider a bipartite system, defined on a Hilbert space $$\mathcal{H}= \mathcal{H}_A \otimes \mathcal{H}_B$$. Consider a basis $$\{|A_i\rangle \otimes |B_j\rangle\}$$. What is the general form of a density operator such that it predicts a perfect correlations between pairs $$(A_i, B_i)$$ upon measurement? (i.e. null probability of a joint measurement outcome $$(A_i, B_j)$$ for $$i \neq j$$).

If one had a pure state for the bipartite system, then one could appeal to the following Schmidt decomposition: $$|\psi\rangle= \Sigma_i \alpha_i |A_i\rangle |B_i\rangle$$ But if the quantum state is not pure, no such Schmidt decomposition is avaiable.

In other words, what is the equivalent of a Schmidt decomposition for an impure density operator?

• Well, assuming the basis is a complete ONB associated to a non-deg. observable for the sake of simplicity, the probability to measure the outcome associated to the event $a_i,b_j$ is $\mathrm{Tr}\rho P_{ij}$ with $P_{ij}:=|A_iB_j\rangle\langle A_i B_j|$. Now if the said probability should vanish for $i\neq j$ (this is how I interpret your question), then it follows that $\rho P_{ij}=0$ and hence $\rho |A_iB_j\rangle=0$ for all $i\neq j$. This basically gives you constraints of the matrix representation of $\rho$ in the said basis. Nov 15, 2023 at 18:08
• I.e. the only non-vanishing matrix elements are the diagonal elements of the form $\langle A_jB_j| \rho |A_jB_j\rangle$ and the off-diagonal elements of the form $\langle A_k B_k|\rho |A_jB_j\rangle$. You should double check everything if this is correct etc. I also don't know if I understand the question correctly... Anyways, it might help to consider a $2\times 2$-dimensional example, too... Nov 15, 2023 at 18:15
• Thank you very much for your comment. I think you are right. I could not quite formulate the question precisely, but you made it precise and offered the answer. What I am asking for is a density operator $\rho$ such that $Tr(\rho |A_i, A_j\rangle \langle A_i, A_j|)=0$ for all $i\neq j$. This implies $\langle A_i, A_j| \rho |A_i, B_j\rangle =0$. However, I can’t see why it implies $\rho |A_i, B_j\rangle=0$.
– Pol
Nov 17, 2023 at 11:24
• I cannot follow your comment. I've used: For any $A,B\geq 0$, it holds that $\mathrm{Tr}AB\geq 0$ with equality iff $AB=BA=0$. To prove this, make use of the fact that for a non-negative operator a unique hermitian square root exists with $\sqrt A\sqrt A=A$, and hence $\mathrm{Tr}AB=\mathrm{Tr}\sqrt A B \sqrt A$ and notice that the operator $\sqrt A B \sqrt A$ is non-negative. To proceed, if $A|\psi\rangle\langle\psi|=0$, then $A|\psi\rangle=0$. To see this, just apply the operator to the state $|\psi\rangle$: $A|\psi\rangle\langle \psi|\psi\rangle=||\psi||^2 A|\psi\rangle\overset{!}{=}0$. Nov 17, 2023 at 11:41
• Lovely, thank you very much. If would like to write it up in an answer I will accept it. THanks!
– Pol
Nov 17, 2023 at 11:44

The bipartite state $$\sigma_{AB}$$ is said to have Schmidt rank at most D if it can be written as $$\sigma_{AB} = \sum_ip_i|\psi_i\rangle\langle\psi_i|$$
where each $$|\psi_i\rangle$$ has at most D Schmidt coefficients different from zero. These states define a set, $$S_D$$. If a state $$\rho_{AB}$$ is not in this set, is said to be D-dimensional entangled.