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Let $\rho_{AB}$ be some bipartite quantum state. Let $\rho_{A}$ and $\rho_{B}$ be the marginal states. I am reading some notes where the following statement is made.

The support of $\rho_{AB}$ is always contained in the support of $\rho_{A}\otimes\rho_B$

I believe this is equivalent to the expression below (please correct me if this is wrong) for any $\psi_{AB}$

$$\langle\psi_{AB}\vert\rho_A\otimes\rho_B\vert\psi_{AB}\rangle = 0 \implies \langle\psi_{AB}\vert\rho_{AB}\vert\psi_{AB}\rangle = 0$$

How can one prove this statement?

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  • $\begingroup$ Are you sure they were talking about matrix elements instead of $Tr$? $\endgroup$ Apr 9, 2020 at 0:28

2 Answers 2

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  1. The statement is true for pure states: For $\rho_{AB}=|\psi\rangle\langle\psi|$, write $|\psi\rangle=\sum\lambda_i|i\rangle_A|i\rangle_B$ in the Schmidt basis. Then the claim is immediate, since the support of $\rho_A\otimes \rho_B$ is $$ \mathrm{span}\{|i\rangle_A\}\otimes \mathrm{span}\{|i\rangle_B\}\ , $$ which clearly contains $|\psi\rangle$.

  2. For a state $\rho_{AB}=\sum p_i|\psi_i\rangle\langle\psi_i|$, the claim follows by observing that

    • the support of $\rho_{AB}$ is $\mathrm{span}\{|\psi_i\rangle\}$,

    • $\rho_A = \sum p_i \rho_A^i$, and thus the support of $\rho_A$ contains the support of the $\rho_A^i$ (using positivity), and thus, the support of $\rho_A\otimes \rho_B$ contains the support of $\rho_A^i\otimes \rho_B^i$, and

    • and combining this with the statement 1. above.

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  • $\begingroup$ Very clean analysis! $\endgroup$ Apr 9, 2020 at 13:26
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Assume $\rho\equiv\rho_{AB}$ is pure, $\rho=|\Phi\rangle\!\langle\Phi|$, and write its Schmidt decomposition as $|\Phi\rangle=\sum_k \sqrt{p_k} |u_k\rangle|v_k\rangle$. Notice that the reduced states then have the form $$\rho_A = \sum_k p_k |u_k\rangle\!\langle u_k|, \qquad \rho_B = \sum_k p_k |v_k\rangle\!\langle v_k|.$$ It follows that $\langle\psi|\rho_A\otimes\rho_B|\psi\rangle=\sum_{jk} p_j p_k |\langle u_j,v_k|\psi\rangle|^2 = 0$, which implies $\langle u_j,v_k|\psi\rangle=0$ for all $j,k$. The conclusion then follows from $$\langle\psi|\rho|\psi\rangle=|\langle\Phi|\psi\rangle|^2 =\left|\sum_k \sqrt{p_k}\langle u_k, v_k|\psi\rangle\right|^2 = 0. $$ To generalise to mixed states, write $\rho=\sum_k q_k \rho_k$ with $\rho_k$ pure, and observe that

  1. $\rho_A\otimes\rho_B=\sum_{jk} q_j q_k (\rho_j)_A\otimes(\rho_k)_B$
  2. thus $\langle \rho_A\otimes\rho_B\rangle=0$ implies $\langle(\rho_j)_A\otimes(\rho_k)_B\rangle=0$ for all $j,k$
  3. thus in particular $\langle(\rho_j)_A\otimes(\rho_j)_B\rangle=0$, implying $\langle \rho_j\rangle=0$ (because each $\rho_j$ is pure)
  4. thus $\langle\rho\rangle=\sum_k q_k \langle\rho_k\rangle=0$
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