# Is kronecker product identifiable?

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).

• I am not satisfied with my title but could not find anything more descriptive... Any edit is welcomed. Mar 11 '20 at 15:56
• I suppose, technically, you might have $\theta=\theta_1+\theta_2+2k\pi$ for some integer $k$. Otherwise, yes. Mar 11 '20 at 16:31
• You are right, I edited my question, thank you. About the "Otherwise, yes", does it mean I can deduce the 3 assertions? If so, is there any "proof"? Mar 11 '20 at 16:32

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 $$\theta\equiv\theta_1+\theta_2\text{ mod }2\pi.$$
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 $$\lambda_1\lambda_2=1$$.