2
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.