# Given $|\psi\rangle=(U_A\otimes U_B)|0,0\rangle$, is $|\psi\rangle\!\langle\psi|$ always a product state?

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

• Have you tried writing out $|\psi \rangle \langle \psi |$? Commented Apr 21, 2021 at 18:14
• Yes, and also $\rho_{A,B}$ using the definitions, but I couldn't show its identical Commented Apr 21, 2021 at 18:16
• 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. Commented Apr 21, 2021 at 18:20
• Suppose $U_A |0\rangle = |\psi_A \rangle$ and $U_B |0\rangle = |\psi_B\rangle$ then this is essentially showing $\big( |\psi_A \rangle \otimes |\psi_B \rangle \big) \big( \langle \psi_A | \otimes \langle \psi_B| = |\psi_A\rangle \langle \psi_A| \otimes |\psi_B\rangle \langle \psi_B |$ Commented Apr 21, 2021 at 18:29

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

• Thank you very much! Commented Apr 22, 2021 at 17:46