# Is the square root of SWAP gate "maximally entangling"?

I'm not sure if this is a good question for the site, but here goes.

The $$\sqrt{\mathrm{SWAP}}$$ gate is not, however maximally entangling; more than one application of it is required to produce a Bell state from product states.

I'm a bit confused by this remark, its definition for the gate in the $$|00\rangle, |01\rangle, |10\rangle, |11\rangle$$ basis is $$\sqrt{\mathrm{SWAP}} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \frac{1}{2}(1+i) & \frac{1}{2}(1-i) & 0 \\ 0 & \frac{1}{2}(1-i) & \frac{1}{2}(1+i) & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \,.$$

It seems like if I apply the gate on the product state $$|01\rangle$$ I should end up with something proportional to $$\frac{1}{\sqrt2}|01\rangle - \frac{i}{\sqrt2}|10\rangle$$ which looks maximally entangled to me? Am I misunderstanding something?

• I agree with you. Of course, technically you're not producing a Bell state, but something that is locally equivalent to one (and therefore maximally entangled). That would seem to make it maximally entangling by the definition I'd usually work to. Sep 2 '20 at 14:50
• @DaftWullie I see, thank you for your comment. I'm not sure what to do with this question now, since the answer does seem quite obvious, the quote from wikipedia just kind of threw me off a bit (depends on what they mean by 'maximally entangling' I guess). I suppose I'll leave this up for as long as that stays in the wiki article. Sep 2 '20 at 15:51
• @DaftWullie the wikipedia page should probably be amended. On this note, Do you know if there is a universally accepted precise notion of "entangling gate"? Is it just about whether it generates an output entangled state for some input?
– glS
Sep 2 '20 at 18:08
• @glS Any definition which is not invariant under applying local unitaries seems to have little, if at all, to do with what we call entanglement. Sep 2 '20 at 20:21
• @NorbertSchuch for the output state, sure, but isn't that taken care of automatically requiring the gate to generate "an entangled state", entanglement as a notion being invariant under local unitaries? Or do you mean that for the gate itself to be entangling we should also require it to generate entanglement given, e.g., any separable input state (which would indeed be false in this case)?
– glS
Sep 2 '20 at 20:59

It seems like the article you're referencing is defining "maximally entangling" as "capable of producing Bell states from product states". However there are other ways to describe how much entanglement an operator $$U$$ introduces into a state.

One definition that I like but haven't had much success with is the "entangling power" introduced in [1], which describes the maximum entropy resulting from applying the operation to a bipartite input state (note that $$|\psi\rangle,|\phi\rangle$$ are each defined over two-qubit systems): $$K_E(U) \equiv \max_{|\psi\rangle,|\phi\rangle}E\left(U |\psi\rangle|\phi\rangle \right)$$

where $$E$$ is the Von Neumann entropy of the partial trace state. The article computes the entangling power of $$CNOT$$ as 1 ebit and entangling power of $$SWAP$$ as 2 ebits.

Another way to describe operator entanglement is the operator Schmidt rank of [2,3], which just finds the number of nonzero coefficients in the decomposition

$$V = \sum_{j=0}^3 c_j \sigma_j \otimes \sigma_j$$ where $$V$$ is related to $$U$$ by local rotations only. Then the Schmidt rank of $$CNOT$$ is 2 and the Schmidt rank of $$SWAP$$ is 4.

So to address your question, I believe its impossible to construct a Bell state starting from a separable two-qubit state and using only $$SWAP$$'s for entanglement, but both of the definitions above suggest that the $$SWAP$$ gate is capable of creating more entanglement than a $$CNOT$$ in at least some sense. This suggests that the definition of "maximally entangling" provided is either inconsistent or poorly defined or both. As for how much entanglement $$\sqrt{SWAP}$$ produces, you might look into computing either of the quantities defined above for a more rigorous (albeit less conceptual) answer.

[1] (Shen and Chen, 2018) Entangling power of two-qubit unitary operations. https://iopscience.iop.org/article/10.1088/1751-8121/aad7cb

[2] (Nielsen, 2002) Quantum dynamics as a physical resource. https://journals.aps.org/pra/abstract/10.1103/PhysRevA.67.052301

[3] (Nielsen, 2000) Quantum Information Theory. https://arxiv.org/abs/quant-ph/0011036

• Thanks for the answer. I've skimmed through [1], but I'm having a hard time understanding why the entangling power of $SWAP$ is 2 bits, I'm under the impression that the maximum value of $K_E(U)$ for a two qubit operation is 1 bit. Do you mind elaborating on this further? Sep 3 '20 at 1:11
• @eugenhu So the bipartite state they're computing entropy on is composed of two 2-qubit systems, which better reflects the idea that you can reconstruct either reduced system using bell states plus local operations within that system. I edited the response to make that more clear. Sep 3 '20 at 2:45
• I see, I don't completely understand but I'll have to read through [1] a bit more thoroughly at a later date. Just one more comment, you describe that the entangling power "describes the maximum possible single qubit entropy resulting from ...". Is the maximal entropy of a single qubit not also 1 bit though? Sep 3 '20 at 3:01
• That was an error on my part; I've edited that and left the definition otherwise vague so that interested readers will refer to [1] to get a more precise description Sep 3 '20 at 17:10

$$\sqrt{SWAP}|01\rangle\ = \frac{1+i}{2}|01\rangle + \frac{1-i}{2}|10\rangle$$

$$\sqrt{SWAP}(\frac{1+i}{2}|01\rangle + \frac{1-i}{2}|10\rangle)=\frac{1+i}{2}(\frac{1+i}{2}|01\rangle + \frac{1-i}{2}|10\rangle)+\frac{1-i}{2}(\frac{1-i}{2}|01\rangle + \frac{1+i}{2}|10\rangle)$$

$$=\frac{i}{2}|01\rangle+\frac{1}{2}|10\rangle-\frac{i}{2}|01\rangle+\frac{1}{2}|10\rangle=|10\rangle$$

completing the action of the SWAP gate. Repeated application is only going to transition between these two possibilities. I don't believe, given a product state like you have shown, you can generate entanglement via multiple applications, like the article states. As for why it is not being called maximally entangling, I think it's due to coefficients being complex, as opposed to real for the bell states. So this seems more like the editor of that sections personal definition.

• why would the result of two applications of the gate matter? You already get a maximally entangled state with one application: $\sqrt2 \sqrt{\operatorname{SWAP}}|01\rangle=|01\rangle-i|10\rangle$
– glS
Sep 2 '20 at 15:29
• Because their wikipedia article states multiple applications were required. I was simply showing that isn't the case. I am also not sure how the $\sqrt{2}$ creates the state you have shown, shouldn't it be $\frac{1+i}{\sqrt{2}}|01\rangle+\frac{1-i}{\sqrt{2}}|10\rangle$? Sep 2 '20 at 15:43
• This is completely wrong. You write an entangled state yourself. Sep 2 '20 at 15:45
• @NorbertSchuch I am aware the first state is entangled, that was why I mentioned it being a personal definition of the editor The second is product. The article state multiple applications would be required, which is not the case as these either give the $SWAP$ or $\sqrt{SWAP}$ gates, which is what I showed. Sep 2 '20 at 16:01
• @glS Yeah that's what I stated in my response, that the author/editor of the article's claim that the $\sqrt{SWAP}$ doesn't create an entangled state was down to their definition of entangled, which I assume they were basing of whether or not it created a state in the bell basis with no complex relative phase, ie, no $i$. As the state created after one application is an entangled state. Sep 3 '20 at 12:58