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

Question

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

On the "Quantum logic gate" Wikipedia page, it is said that:

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?

Answer 1

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.

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[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

Answer 2

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

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

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

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.