Unitary transformation from pure unentangled state to pure entangled state

Question

I believe that for any state $|\Psi\rangle$, that is unentangled, and for any state $|\Phi\rangle$, of the same dimension that is entangled, there is a unitary operator $\hat{U}$ such that $\hat{U}|\Psi\rangle=|\Phi\rangle$ and $\hat{U^{\dagger}}|\Phi\rangle=|\Psi\rangle$. Is that true? I can't seem to prove it or find such an operator even for simple cases such as $(a^2 + b^2 + c^2 + d^2)^{-1/2}\hat{U}\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}2^{-1/2}$ where $(a^2 + b^2 + c^2 + d^2)^{-1/2}\begin{bmatrix} a \\ b \\ c \\ d\end{bmatrix} = \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix} \otimes \begin{bmatrix} \beta_1 \\ \beta_2 \end{bmatrix}$, the tensor product of two systems.

Thanks.

Answer 1

A pure state $|\psi\rangle$ that is an element of a Hilbert space $\mathcal{H}$ is not entangled if it can be written as a product state of subsystems, for example $$|\psi\rangle = |\phi\rangle\otimes |\varphi\rangle \in \mathcal{H}_1\otimes\mathcal{H}_2$$ Acting on the total system will keep the system itself pure (in order to maintain unitarity), however we can generate entanglement between the subsystems. This requires a unitary transformation that forces the subsytsems to interact, for example $$ U = e^{it(H_1 + H_1 + H_{12})} $$ Where $H_1$ acts only on Hilbert space 1 and so forth. Only if $H_{12}$ is non-zero will you generate entanglement between the two systems, of the form $$|\psi\rangle \neq |\phi\rangle\otimes |\varphi\rangle$$ If $H_{12} = 0$, then the unitary operator will itself be factorisible and each subsystem will remain pure and unentangled.

Answer 2

I'm not sure I understand the connection with entanglement. If two states are in the same vector space, there is always a unitary transformation from one to the other, irrespective of their entanglement.

Any state $$ \vert\psi_1\rangle =\left(\begin{array}{c} \alpha_1\\ \alpha_2 \\ \vdots \\ \alpha_n\end{array}\right) =U_1\left(\begin{array}{c} 1\\ 0 \\ \vdots \\0\end{array}\right) $$ where $U_1$ is unitary with the first column $$ U_1=\left(\begin{array}{cccc} \alpha_1&* & \ldots &* \\ \alpha_2 &* & \ldots &* \\ \vdots & \vdots &\vdots &* \\ \alpha_n &* & \ldots &* \end{array}\right) $$ where $*$ is some entry that need not be specified beyond the condition that will make $U_1$ unitary. This is true wether $\vert\psi_1\rangle$ is entangled or not.

Write then $$ \vert\psi_2\rangle = U_2\left(\begin{array}{c} 1\\ 0 \\ \vdots \\0\end{array}\right) = U_2\cdot U_1^{-1} \vert\psi_1\rangle\, . $$ Since $U_2$ and $U_1$ are both unitary, the product $U_2\cdot U_1^{-1}$ is also unitary and is the transformation needed to go from $\vert\psi_1\rangle $ to $\vert\psi_2\rangle$, irrespective of the entanglement of either states.