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I know for a separable state $\rho_{AB}$ we have

$$P_{AB}= P_{A}=P_{B}=1$$

Where P is purity. What if the relation between the purities when the bipartite state is entangled? Is there any inequality relation?

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  • $\begingroup$ I don't understand how you came up with the first equation. If you have a pure product state, say $|\psi\rangle_A |\phi\rangle_B$, Then $$P_{AB} = P_A = P_B= 1$$ $\endgroup$
    – FDGod
    Commented Mar 5 at 20:26
  • $\begingroup$ Thanks you i fixed that. $\endgroup$
    – reza
    Commented Mar 5 at 20:52

2 Answers 2

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This question seems to presume a pure bipartite state. For any such state, there exists a Schmidt decomposition that effectively finds a local basis in which to expand each of the subsystems such that $$|\Psi\rangle=\sum_{i=1}^n \lambda_i |\psi_i\rangle\otimes |\phi_i\rangle$$ for some integer $n$, some coefficients satisfying $\sum_{i=1}^n |\lambda_i|^2=1$, some set of orthonormal states $\{|\psi_i\rangle\}$ for system A, and another set of orthonormal states $\{|\phi_i\rangle\}$ for system B. Any bipartite pure state can be expressed in this form; it is separable if and only if $n=1$, equivalent to a single coefficient $|\lambda_1|=1$ and the rest of the coefficients vanishing.

Because any state can be written like this, we know that the reduced state for system A is $$\rho_A=\sum_{i=1}^n|\lambda_i|^2 |\psi_i\rangle\langle\psi_i|$$ with purity $$P_A=\mathrm{Tr}(\rho_A^2)=\sum_{i=1}^n|\lambda_i|^4$$ and that the reduced state for system B is $$\rho_B=\sum_{i=1}^n|\lambda_i|^2 |\phi_i\rangle\langle\phi_i|$$ with purity $$P_B=\mathrm{Tr}(\rho_B^2)=\sum_{i=1}^n|\lambda_i|^4=P_A.$$

This directly tells us all equalities and inequalities that we would like. First, if the bipartite state is pure, then $P_A=P_B$. Second, if it is separable, then $|\lambda_1|=1$, so $P_A=P_B=1$, and if it is not separable, then there is more than one nonzero coefficient $\lambda_i$, and we always have that $$\sum_i |\lambda_i|^4\leq \sum_i |\lambda_i|^2=1,$$ so the reduced states are always mixed. The more coefficients $\lambda_i$ there are, the more mixed the reduced state will be, in general (maximally mixed comes from maximal entanglement which comes from $n$ as large as possible and equal-magnitude $|\lambda_i|=\mathrm{constant}$.


What happens if the initial state is not pure; then what can we say? Take a generic state expressed in some orthonormal basis: $$\rho=\sum_{ij;kl}\rho_{ij;kl}|i\rangle\langle j|\otimes|k\rangle\langle l|$$ subject to positivity and normalization ($\sum_{ik}\rho_{ii;kk}=1$) constraints. The purity is $$P_{AB}=\mathrm{Tr}(\rho^2)=\sum_{ij;kl}\rho_{ij;kl}\rho_{ji;lk}=\sum_{ij;kl}|\rho_{ij;kl}|^2.$$ Then, tracing out one system yields something like $$\rho_A=\sum_{ij}(\sum_k \rho_{ij;kk})|i\rangle\langle j|$$ with purity $$P_A=\sum_{ij}|\sum_k \rho_{ij;kk}|^2;$$ similarly, $$P_B=\sum_{kl}|\sum_i \rho_{ii;kl}|^2.$$ The first thing to note is that, in general $P_A\neq P_B$ for mixed states. Next, we can note some inequalities with complex numbers like $$|\sum_k \rho_{ij;kk}|^2\leq \sum_k |\rho_{ij;kk}|^2 \leq \sum_{kl} |\rho_{ij;kl}|^2,$$ which directly show that $$P_A\leq \sum_{ij}\sum_{kl} |\rho_{ij;kl}|^2=P_{AB}$$ and the same inequality can be done to show that $P_B\leq P_{AB}$.

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  • $\begingroup$ Thank you for answer. I was looking for a relation between purities for a state ( mixed or pure), i guess $P_{AB}<=P_A.P_B$ holds. But im not sure about it because could not prove it. $\endgroup$
    – reza
    Commented Mar 6 at 7:51
  • $\begingroup$ Seems for pure states, where you described in your answer, the relation i mentioned in the above comment, holds. $\endgroup$
    – reza
    Commented Mar 6 at 7:58
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    $\begingroup$ @reza I updated with some generic results for mixed states to show that $P_{AB}\geq P_A$ and $P_{AB}\geq P_B$ (so, if this is what you meant, $P_{AB}\geq P_A\times P_B$. $\endgroup$ Commented Mar 6 at 22:44
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    $\begingroup$ @Quantum Mechanic Thanks, trying to summarize things in a final conclusion always has this danger of oversimplifying! I glanced through it too quickly, of course bipartite doesn't imply qubits (and the $1/2$ should be $1/n$). $\endgroup$ Commented Mar 7 at 7:30
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    $\begingroup$ @reza the right side includes all of the terms on the left side when $k=l$, then adds the additional terms when $k\neq l$, and each term is positive because of the absolute value $\endgroup$ Commented Mar 22 at 5:00
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Special case I: Product states

A bipartite quantum state$^1$ $\rho_{AB}$ is said to be a product state if it can be written as $\rho_{AB}=\rho_A\otimes\rho_B$ for some quantum states $\rho_A$ and $\rho_B$ of the subsystems $A$ and $B$. If $\rho_{AB}$ is a product state, then $$ P_{AB}=\mathrm{tr}(\rho_{AB}^2)=\mathrm{tr}(\rho_A^2\otimes\rho_B^2)=P_A\cdot P_B.\tag1 $$

Special case II: Pure product states

If a product state $\rho_{AB}$ is pure, then $\rho_A$ and $\rho_B$ are pure, too$^2$. Therefore, if $\rho_{AB}$ is a pure product state, then $$ P_{AB}=P_A=P_B=1.\tag2 $$

General case: Separable states

A bipartite product state $\rho_{AB}$ is said to be separable if it is a convex combination of product states, i.e. if $\rho_{AB}=\sum_k c_k\rho_{A,k}\otimes\rho_{B,k}$ for some real numbers $c_k\in(0,1]$ that add up to one and quantum states $\rho_{A,k}$ and $\rho_{B,k}$ of the subsystems. Consequently, if $\rho_{AB}$ is separable, then $$ \begin{align} P_A&=\sum_{ij}c_ic_j\cdot\mathrm{tr}(\rho_{A,i}\rho_{A,j})\tag3\\ P_B&=\sum_{ij}c_ic_j\cdot\mathrm{tr}(\rho_{B,i}\rho_{B,j})\tag4\\ P_{AB}&=\sum_{ij}c_ic_j\cdot\mathrm{tr}(\rho_{A,i}\rho_{A,j})\cdot\mathrm{tr}(\rho_{B,i}\rho_{B,j})\tag5 \end{align} $$ but $0\leqslant \mathrm{tr}(\rho_{A,i}\rho_{A,j})\leqslant 1$ by Cauchy-Schwarz inequality and we obtain $$ P_{AB}\leqslant\min(P_A,P_B).\tag6 $$


$^1$ Here, a quantum state is a positive semidefinite operator with unit trace.
$^2$ This follows by considering the rank of the tensor product of two matrices. However, it also follows from $(1)$.

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  • $\begingroup$ Thank you adam for the answer. Im wondering about a general case ( a mixed entangled state) $\endgroup$
    – reza
    Commented Mar 6 at 9:47

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