Show that any two product states of the same dimension are LU-equivalent

Question

States $|Ψ \rangle$, $|Φ\rangle$ on $C_d⊗C_{d′}$ are said to be equivalent up to Local Unitarities (LU-equivalent) if there exist unitaries $U : C_d → C_d$ and $V : C_{d′} → C_{d′}$ such that:

$|Ψ \rangle = (U ⊗ V )|Φ \rangle$

I'm trying to prove the following:

a) Show that any two product states of the same dimension are LU-equivalent. That is, for any $|ψ \rangle$, $|φ \rangle$ $∈ C_d$ and $|ψ'\rangle$, $|φ'\rangle ∈ C_{d′}$, we have $|ψ\rangle ⊗ |ψ′\rangle$ is LU-equivalent to $|φ\rangle ⊗ |φ′\rangle$

b) Show that a product state is never LU-equivalent to an entangled state

For part a I thought of maybe manually constructing a $d$ x $d$ matrix $U$ and a $d'$ x $d'$ matrix $V$ s.t. $U|\phi \rangle = |ψ \rangle$ and $V|\phi ' \rangle = |ψ' \rangle$. While constructing such matrices isn't hard, I'm not sure about whether or not they'd be unitary.

For part b, since LU equivalences can be shown to be reflexive, we have:

$|Ψ⟩=(U⊗V)(|\alpha⟩⊗|\beta⟩)$

$= (U|\alpha⟩⊗V|\beta⟩)$

which seems to be a product state(hence a contradiction) as $U|\alpha⟩$ and $V|\beta⟩$ are vectors but I'm not sure if this is correct?

Answer 1

For part (a), you can certainly manually construct a unitary that for which $U|\phi\rangle = |\psi\rangle$ (though there may be a more elegant proof of this fact). For example, suppose $|\phi\rangle = W|0\rangle$ for some unitary $W$ and define $U' = UW$ then we have $$ U'|0\rangle = |\psi\rangle \tag{1} $$

This means that $U'$ can be any unitary whose first column is $|\psi\rangle$. You can always find one such unitary by completing the remaining columns of $U'$ using a Gram-Schmidt procedure.

For part (b), this can be shown using the Schmidt decomposition: Given a bipartite $|\Psi\rangle$ defined over two $d$-dimensional systems, you can always write $$ |\Psi\rangle = \sum_{i=1}^d c_i |u_i\rangle \otimes |v_i\rangle \tag{2} $$

where $\{|u_i\rangle\}_{i=1}^d$ and $\{|v_i\rangle\}_{i=1}^d$ are orthonormal sets. A separable state can only have one term in this sum ("Schmidt rank 1"), e.g.

$$ |\Psi_{sep}\rangle = |\alpha\rangle \otimes |\beta \rangle \tag{3} $$ while entangled states are defined as those which have more than one term, e.g.

$$ |\Psi_{ent}\rangle = c_1|u_1\rangle \otimes |v_1\rangle + c_2|u_2\rangle \otimes |v_2\rangle \tag{4} $$

where $\langle u_1 | u_2\rangle = \langle v_1| v_2\rangle = 0$ and $c_1>0, c_2 > 0$. Since orthogonality is preserved under unitary transformations, there is no way to reduce the number of terms in Eq. (4) (you can't rotate $|u_1\rangle$ and $|u_2\rangle$ to combine them somehow). As a result, there is no way to show equality between Eqs. (3) and (4).