Unitary transformation from pure unentangled state to pure entangled state 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.
 A: 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.
A: 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.
