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say I have some state in the combined space $\psi$$H_A\otimes H_B$, where $\psi=U_A \otimes U_B|0,0\rangle$ (operators from respective spaces), and $\rho_A, \rho_B$ the respective density matrices.

Is the following statement true: $|\psi\rangle \langle\psi|$ = $\rho_A \otimes \rho_B$ ?

If yes, how can I show it? Thanks in advance

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  • $\begingroup$ Have you tried writing out $|\psi \rangle \langle \psi |$? $\endgroup$
    – Rammus
    Apr 21 at 18:14
  • $\begingroup$ Yes, and also $\rho_{A,B}$ using the definitions, but I couldn't show its identical $\endgroup$
    – Frogfire
    Apr 21 at 18:16
  • $\begingroup$ Maybe you should add what you found to the question body (by using the edit button). Note also that $|\psi\rangle = (U_A |0\rangle) \otimes (U_B |0 \rangle)$, maybe this helps you. $\endgroup$
    – Rammus
    Apr 21 at 18:20
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    $\begingroup$ Suppose $U_A |0\rangle = |\psi_A \rangle$ and $U_B |0\rangle = |\psi_B\rangle$ then this is essentially showing $ \big( |\psi_A \rangle \otimes |\psi_B \rangle \big) \big( \langle \psi_A | \otimes \langle \psi_B| = |\psi_A\rangle \langle \psi_A| \otimes |\psi_B\rangle \langle \psi_B |$ $\endgroup$
    – KAJ226
    Apr 21 at 18:29
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Suppose that $U_A |0\rangle = |\psi_A \rangle = \begin{pmatrix}A_1 \\ A_2 \end{pmatrix}$ and $U_B |0\rangle = |\psi_B\rangle = \begin{pmatrix} B_1 \\ B_2 \end{pmatrix}$.

Then we have $|\psi \rangle = (U_A \otimes U_B) |00\rangle =(U_A \otimes U_B) (|0\rangle \otimes 0\rangle ) = U_A|0\rangle \otimes U_B|0\rangle = |\psi_A\rangle \otimes |\psi_B\rangle$

We can also write $|\psi \rangle = \begin{pmatrix}A_1 \\ A_2 \end{pmatrix} \otimes \begin{pmatrix}B_1 \\ B_2 \end{pmatrix} = \begin{pmatrix}A_1B_1 \\ A_1B_2 \\ A_2B_1 \\ A_2B_2\end{pmatrix}$

and hence $|\psi \rangle \langle \psi|$ is

$$ \begin{pmatrix}A_1B_1 \\ A_1B_2 \\ A_2B_1 \\ A_2B_2\end{pmatrix} \begin{pmatrix} A_1^*B_1^* & A_1^*B_2^* & A_2^*B_1^* & A_2^*B_2^* \end{pmatrix} = \begin{pmatrix} |A_1|^2 |B_1|^2 & |A_1|^2B_1B_2^* & A_1A_2^*|B_1|^2 & A_1A_2^*B_1B_2^* \\ |A_1|^2 B_2B_1^* & |A_1|^2|B_2|^2 & A_1A_2^*B_2B_1^* & |A_2|^2|B_2|^2 \\ A_2A_1^* |B_1|^2 & A_2A_1^*B_1B_2^* & |A_2|^2|B_1|^2 & |A_2|^2B_1B_2^* \\ A_1^*A_2 B_1^*B_2 & A_2A_1^* |B_2|^2 & |A_2|^2B_2B_1^* & |A_2|^2 |B_2|^2 \\ \end{pmatrix}$$

Now, we also have that $\rho_A = |\psi_A \rangle \langle \psi_A | $ and $\rho_B = |\psi_B \rangle \langle \psi_B| $ so therefore

$$\rho_A = \begin{pmatrix} |A_1|^2 & A_1 A_2^*\\ A_2 A_1^* & |A_2|^2 \end{pmatrix} \hspace{1 cm} \rho_B = \begin{pmatrix} |B_1|^2 & B_1 B_2^*\\ B_2 B_1^* & |B_2|^2 \end{pmatrix} $$

and therefore,

$$\rho_A \otimes \rho_B = \begin{pmatrix} |A_1|^2 & A_1 A_2^*\\ A_2 A_1^* & |A_2|^2 \end{pmatrix} \otimes \begin{pmatrix} |B_1|^2 & B_1 B_2^*\\ B_2 B_1^* & |B_2|^2 \end{pmatrix} = \begin{pmatrix} |A_1|^2 |B_1|^2 & |A_1|^2B_1B_2^* & A_1A_2^*|B_1|^2 & A_1A_2^*B_1B_2^* \\ |A_1|^2 B_2B_1^* & |A_1|^2|B_2|^2 & A_1A_2^*B_2B_1^* & |A_2|^2|B_2|^2 \\ A_2A_1^* |B_1|^2 & A_2A_1^*B_1B_2^* & |A_2|^2|B_1|^2 & |A_2|^2B_1B_2^* \\ A_1^*A_2 B_1^*B_2 & A_2A_1^* |B_2|^2 & |A_2|^2B_2B_1^* & |A_2|^2 |B_2|^2 \\ \end{pmatrix}$$

And therefore, $|\psi \rangle \langle \psi |= \rho_A \otimes \rho_B$

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    $\begingroup$ Thank you very much! $\endgroup$
    – Frogfire
    Apr 22 at 17:46

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