# Non-entangling two-qubit gates

The non-entangling gates in $$SU_4$$ contains the entire group of gates of the form $$SU_2 \otimes SU_2.$$ It also contains $$\zeta_8 SWAP= \zeta_8 \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$ where $$\zeta_8=e^{2\pi i/8}= e^{\pi i/4}$$ is a primitive eighth root of unity.

Are there any other non-entangling two-qubit gates? A related (perhaps equivalent?) question is what is the normalizer of $$SU_2 \otimes SU_2$$ in $$SU_4$$? Does the normalizer $$N(SU_2 \otimes SU_2)$$ just have two connected components (the component of the identity and the component of SWAP)? Does it have more connected components? Do these other components correspond to other non-entangling gates? Also, interesting to note that $$(\zeta_8 SWAP)^2=iI \not \in SU_2 \otimes SU_2$$ is not in $$SU_2 \otimes SU_2$$ even though we wouldn't think of it as an entangling gate since it is just a global phase and moreover it is in $$U_2 \otimes U_2$$.

• Nit: don't forget gates from $\text{SWAP} \cdot (SU_2 \otimes SU_2)$ such as "swap and bit flip". Your example gate at the end is a SWAP and an S gate. Sep 19, 2022 at 17:31
• Ah good point. The second gate I wrote down $$M= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0& i & 0 & 0 \\ 0 & 0 & 0 & i \\ \end{bmatrix}$$ is actually just $\zeta_8 (SWAP)$ times $\overline{\zeta_8}(S \otimes I) =(\overline{\zeta_8}S) \otimes I \in SU_2 \otimes SU_2$ (where $S$ is the phase gate) so it is a non-entangling gate from the same connected component of $N(SU_2 \otimes SU_2)$ as $SWAP$. Since $M$ isn't adding anything new to the question I'll just delete it. Sep 19, 2022 at 18:09
• What does $\zeta_8$ mean? Also, why swap is also non-entangling gate since I can't see that it can be written into $U_1\otimes U_2$ form. Is it just because when swap act on a quantum state it can never create entanglement? Oct 1, 2022 at 7:13

There are no other non-entangling gates in $$SU(d^2)$$ in any dimension $$d=2,3,\dots$$. Note that the global phase is irrelevant to the problem, so we lose no generality by considering non-entangling gates in $$U(d^2)$$ instead. We will prove that if $$U\in U(d^2)$$ is non-entangling then either $$U\in U(d)\otimes U(d)$$ or $$\text{SWAP}\circ U\in U(d)\otimes U(d)$$.

## Preliminaries

If $$A$$ is a linear subspace of $$\mathbb{C}^d$$ and $$|\psi\rangle\in\mathbb{C}^d$$ a pure state, then let $$|\psi\rangle\otimes A$$ denote the set $$\{|\psi\rangle\otimes|\phi\rangle:|\phi\rangle\in A\}$$ which is a linear subspace of $$\mathbb{C}^d\otimes\mathbb{C}^d$$. Similarly, for $$A\otimes|\phi\rangle$$. We say that a linear space $$B\subseteq\mathbb{C}^d\otimes\mathbb{C}^d$$ is entanglement-free if every element of $$B$$ is (a scalar multiple of) a product state.

Lemma If a linear subspace $$B\subseteq\mathbb{C}^d\otimes\mathbb{C}^d$$ is entanglement-free, then either $$B=A\otimes|\psi\rangle$$ or $$B=|\psi\rangle\otimes A$$ for some state $$|\psi\rangle\in\mathbb{C}^d$$ and some linear subspace $$A\subseteq\mathbb{C}^d$$.

Proof. Assume otherwise. Then we can find $$|a\rangle\otimes|b\rangle\in B$$ and $$|x\rangle\otimes|y\rangle\in B$$ such that $$|x\rangle$$ is not a scalar multiple of $$|a\rangle$$ and $$|y\rangle$$ is not a scalar multiple of $$|b\rangle$$. However, then $$|a\rangle\otimes|b\rangle + |x\rangle\otimes|y\rangle$$ is entangled$$^1$$.$$\square$$

## Two types of non-entangling gates

Now, suppose that $$U\in U(d^2)$$ is non-entangling. Then the image $$U[B]$$ of any entanglement-free subspace $$B=A\otimes|\psi\rangle$$ under $$U$$ is entanglement-free and the lemma above implies that either

$$U[A\otimes|\psi\rangle] = A'\otimes|\psi'\rangle\tag{1}$$

or

$$U[A\otimes|\psi\rangle] = |\psi'\rangle\otimes A'\tag{2}$$

for some subspace $$A'$$ of $$\mathbb{C}^d$$ and some state $$|\psi'\rangle\in\mathbb{C}^d$$. In the first case, $$U\in U(d)\otimes U(d)$$. In the latter case $$\text{SWAP}\circ U\in U(d)\otimes U(d)$$. Since $$(1)$$ and $$(2)$$ exhaust all possibilities, no other non-entangling gates exist.

## Intuition

The argument above attempts to capture the intuitive observation that if we vary$$^2$$ the state of the first qudit in a product state that is fed into a two-qudit unitary gate then that variation affects either the first qudit, the second qudit or both qudits at the output. However, if the variation affects both qudits then they become entangled$$^3$$. Therefore, since the gate is non-entangling, the variation can only feed through either to the first qudit or to the second qudit. These two cases correspond to the two possibilities $$(1)$$ and $$(2)$$ above.

## Normalizer

The normalizer $$N:=N(U(d)\otimes U(d))$$ of $$U(d)\otimes U(d)$$ in $$U(d^2)$$ does indeed have two connected components which correspond to the identity and the SWAP gate. First, note that every non-entangling gate belongs to $$N$$. Conversely$$^4$$, no entangling gate belongs to $$N$$.

Now, $$N$$ inherits its topology from $$U(d^2)$$ which inherits its topology from $$\mathbb{C}^{d^4}$$. Moreover, $$N$$ is closed, so connectedness and path-connectedness are equivalent in $$N$$. Thus, if $$N$$ was connected, then there would be a continuous path from a gate of the form $$U_1\otimes V_1$$ to a gate of the form $$\text{SWAP}\circ(U_2\otimes V_2)$$. However, this would mean that we can approximate the SWAP gate by product gates arbitrarily well, which is impossible. Therefore, $$N$$ has at least two connected components.

Finally, $$U(d)$$ is path-connected, so we can form a continuous path between any two gates. Taking the product of such paths, we see that any two gates of the form $$U_1\otimes V_1$$ live in the same connected component. Similarly for gates of the form $$\text{SWAP}\circ(U_2\otimes V_2)$$. Therefore, $$N(U(d)\otimes U(d))$$ has exactly two connected components.

$$^1$$ This can be proved rigorously by extending $$\{|a\rangle\otimes|b\rangle\}$$ to a basis and writing the coefficients of $$|a\rangle\otimes|b\rangle + |x\rangle\otimes|y\rangle$$ in that basis as a $$d\times d$$ matrix. Since $$|x\rangle$$ is not a scalar multiple of $$|a\rangle$$ and $$|y\rangle$$ is not a scalar multiple of $$|b\rangle$$, the matrix is full rank and therefore $$|a\rangle\otimes|b\rangle + |x\rangle\otimes|y\rangle$$ is not a product state.
$$^2$$ For example, we could imagine varying the state of the first qudit with time as in $$|\psi(t)\rangle\otimes|\phi\rangle$$.
$$^3$$ More generally, the qudits could become correlated classically. However, this possibility is ruled out by unitarity. It would be relevant if we considered two-qudit quantum channels instead of two-qudit unitary gates.
$$^4$$ We can arrange for the conjugation of a product unitary with non-degenerate spectrum by an entangling unitary to result in an operator with entangled eigenstates. Such an operator is not a product unitary.

• There's no need to restrict yourself to qubits, this is true for all dimensions. Oct 2, 2022 at 8:53
• @MateusAraújo Good point. Made the changes. Thanks! Oct 2, 2022 at 16:21

There's no other interesting non-entangling gates. The normalizer is just generated by $$SU_2 \otimes SU_2$$ together with $$\zeta_8 SWAP$$. Indeed in quantum mechanics we really want $$PU_4$$ rather than $$U_4$$. In $$PU_4$$ the group of non-entangling gates is exactly $$(PU_2 \times PU_2) \rtimes SWAP \subset PU_4$$ where $$SWAP$$ just swaps the two terms in the direct product.

For more details on some of the math here see

https://math.stackexchange.com/questions/4397313/so-4-mathbbr-and-su-2-otimes-su-2-subgroups-of-su-4?rq=1