Is kronecker product identifiable?

Question

I have a unitary matrix $U$ and a quantum state $\vert \Psi \rangle$ such that $$ U \vert \Psi \rangle = e^{i \theta} \vert \Psi \rangle.$$ I also know that my unitary matrix and my quantum state can be written as $U = U_1\otimes U_2$ and $\vert \Psi \rangle = \vert \Psi_1\rangle\otimes\vert\Psi_2\rangle$ with matching dimensions (i.e. $U_1$ acts on the Hilbert state $\vert\Psi_1\rangle$ belongs to, same for $U_2$ and $\vert\Psi_2\rangle$).

My first equation then becomes $$U_1\vert\Psi_1\rangle \otimes U_2 \vert\Psi_2 \rangle = e^{i\theta} \vert \Psi_1 \rangle \otimes \vert \Psi_2 \rangle.$$

Can I deduce the 3 following assertions from my formula above?

1. $U_1 \vert\Psi_1\rangle = e^{i\theta_1}\vert\Psi_1\rangle$
2. $U_2 \vert\Psi_2\rangle = e^{i\theta_2}\vert\Psi_2\rangle$
3. with $\theta = \theta_1 + \theta_2 + 2k\pi$.

The implication seems "logical" as we just separate $2$ non-related quantum states from each other, but I do not have enough confidence in my reasoning to accept this result yet. I searched for a mathematical proof, but the closer I found is the following theorem from here:

Theorem: Let $A$ and $B$ be two complex square matrices. If $\lambda$ is an eigenvalue of $A$ with corresponding eigenvector $x$ and $\mu$ is an eigenvector of $B$ with corresponding eigenvector $y$, then $\lambda\mu$ is an eigenvalue of $A\otimes B$ with corresponding eigenvector $x\otimes y$. Moreover, every eigenvalue of $A\otimes B$ arises as such a product.

that does not fit with my problem (not the good implication and more oriented toward eigenvectors).

Answer 1

Let's say I know that $$ U_1|\Psi_1\rangle\otimes U_2|\Psi_2\rangle=e^{i\theta}|\Psi_1\rangle\otimes|\Psi_2\rangle. $$ Now, let's imagine that $U_1|\Psi_1\rangle=|\phi_1\rangle$ and $U_2|\Psi_2\rangle=|\phi_2\rangle$. So, $$ |\phi_1\rangle\otimes |\phi_2\rangle=e^{i\theta}|\Psi_1\rangle\otimes|\Psi_2\rangle. $$ Now, I would just read off your desired relations, but as a proof, it depends on how deep we need to go... Let's take the inner product of both sides with $|\phi_1\rangle\otimes |\phi_2\rangle$, and calculate the absolute value. You have $$ 1=|\langle\phi_1|\Psi_1\rangle|\ |\langle\phi_2|\Psi_2\rangle|. $$ Both terms on the right-hand side satisfy $|\langle\phi_1|\Psi_1\rangle|\leq 1$, and can only achieve equality if $|\phi_i\rangle=e^{i\theta_i}|\Psi_i\rangle$ (do you need a detailed proof of that? Think of real unit vectors $\underline{v},\underline{u}$: $\underline{u}\cdot\underline{v}=\cos\gamma\leq 1$, where $\gamma$ is the angle between them.)

Hence, we get overall equality if $$ e^{i\theta}=e^{i\theta_1}e^{i\theta_2}, $$ which is true provided $$ \theta\equiv\theta_1+\theta_2\text{ mod }2\pi. $$

Answer 2

Let $H_1,H_2$ be two Hilbert spaces and for unit vectors $u_1,v_1 \in H_1$ and $u_2,v_2 \in H_2$ we have $$ u_1 \otimes u_2 = v_1 \otimes v_2 \in H_1 \otimes H_2. $$ Then it must be $$ v_1 = \lambda u_1, ~~ v_2 = \frac{1}{\lambda} u_2, $$ for some $\lambda \in \mathbb{C}$, $|\lambda|=1$.

To see why consider a scalar product $$ 1 = (u_1\otimes u_2, u_1\otimes u_2) = (u_1\otimes u_2, v_1\otimes v_2) = (u_1,v_1)(u_2,v_2). $$ Since $|(u_i,v_i)|\leq 1$ it must be $|(u_i,v_i)|=1$ for $i=1,2$. Hence $v_i = \lambda_i u_i$, where $|\lambda_i|=1$. But $u_1 \otimes u_2 = v_1 \otimes v_2 = \lambda_1 u_1 \otimes \lambda_2 u_2 = \lambda_1 \lambda_2 (u_1 \otimes u_2)$, thus $\lambda_1\lambda_2=1$.

From this you can deduce your 3 assertions, though DaftWullie essentially used this argument directly in his answer.