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.


2 Answers 2


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.

  • $\begingroup$ 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

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.


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