4
$\begingroup$

Consider a $2$-qubit system with Hilbert space $\mathcal{H} \cong \mathcal{H}_1 \otimes \mathcal{H}_2$. A pure separable state in $\mathcal{H}$ is of the form $\lvert \psi_1 \rangle \otimes \lvert \psi_2 \rangle$ where $\lvert \psi_i \rangle \in \mathcal{H}_i$. The associated density matrix is $\rho \equiv \lvert \psi_1 \rangle \langle \psi_1 \lvert \otimes \lvert \psi_2 \rangle \langle \psi_2 \lvert$.

Consider a unitary operator $U: \mathcal{H} \to \mathcal{H}$ that is not of the form $U_1 \otimes U_2$ where $U_i: \mathcal{H}_i \to \mathcal{H}_i$.

Does there exist such a $U$ such that $U\rho U^\dagger = \lvert \psi'_1 \rangle\langle \psi'_1 \lvert \otimes \lvert \psi'_2 \rangle \langle \psi'_2\lvert \equiv \rho'$ where $\lvert \psi'_i \rangle \in \mathcal{H}_i$ and $\rho \neq \rho'$? In words, does there exist a non-local unitary operator that maps a particular pure product density matrix $\rho$ to a distinct pure product density matrix $\rho'$?

I throw out the case in which $\rho = \rho'$ above because it is conceivable that if $U$ stabilizes $\rho$ $$U\rho U^\dagger = \rho,$$ then $U$ could be non-local.

$\endgroup$
0

2 Answers 2

6
$\begingroup$

The swap operator preserves separability. It is a non-separable unitary, mapping $\rho$ to $\rho'$, with $\rho \neq \rho'$ and if $\rho$ is separable, then it is $\rho'$ as well.

$\endgroup$
5
  • $\begingroup$ Is it possible to provide a reference that proves that the SWAP operator is a non-separable unitary? (or prove in your answer if convenient) $\endgroup$ Commented Apr 17 at 0:49
  • 1
    $\begingroup$ An easy way to convince yourself (in a non-rigorous manner) is to think that $\text{SWAP}$ moves information from $\mathcal{H}_2$ into $\mathcal{H}_1$ (and vice versa) for which you need to access information within $\mathcal{H}_2$, which you definitely cannot do if $\text{SWAP}$ is of the form $U_1 \otimes U_2\,.$ $\endgroup$
    – FDGod
    Commented Apr 17 at 1:40
  • $\begingroup$ As for the reference, this paper paper cites this paper saying " the SWAP operation on a bipartite system preserves separability of the bipartite state, but the SWAP is not separable". I think you should be able to find what you are looking for in second link (although I haven't checked). $\endgroup$
    – FDGod
    Commented Apr 17 at 1:46
  • $\begingroup$ @SillyGoose Just compute the rank of SWAP across the A vs B partition. $\endgroup$ Commented Apr 17 at 6:45
  • $\begingroup$ It’s somewhat interesting that the SWAP gate, which preserves separability of the subsystems, can be decomposed into three CNOT gates, each of which most certainly do not. $\endgroup$ Commented Apr 20 at 13:09
0
$\begingroup$

Just think about $\text{CNOT}$ gate

$$\text{CNOT} =\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{bmatrix}\,.$$

let's say that $\text{CNOT}$ gate could be represented as tensor product $A \otimes B\,.$ Take $$A= \begin{bmatrix} a&b\\ c&d \end{bmatrix}\,\,\,,\,\,\, B=\begin{bmatrix} e&f\\ g&h \end{bmatrix}\,.$$ Then $A\otimes B$ is $$A\otimes B=\begin{bmatrix} ae & af & be & bf\\ ag & ah & bg & bh\\ ce & cf & de & df\\ cg & ch & dg & dh \end{bmatrix}\,.$$

Now, considering this is equal to the $\text{CNOT}$, try to compute the value of $a,b,c,d,e,f,g,h$. It turns out there is no solution to this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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