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

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?

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).