# If $\langle\psi_{AB}\vert\rho_A\otimes\rho_B\vert\psi_{AB}\rangle = 0$, then $\langle\psi_{AB}\vert\rho_{AB}\vert\psi_{AB}\rangle = 0$

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?

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

## 2 Answers

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.

• Very clean analysis! Apr 9, 2020 at 13:26

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 $$\langle\rho\rangle=\sum_k q_k \langle\rho_k\rangle=0$$