Is the square-root-of-SWAP for a pair of 4-dimensional qudits isomorphic to two square-root-of-SWAPS for two pairs of qubits?

Question

This may be a very naïve question indicative of a lot of confusion, but I am trying to understand more about Hamiltonian simulation. I'm starting to intuit that the $n^{th}$-root-of-SWAP acting on a single pair of qubits somehow corresponds to what's meant by Hamiltonian simulation of a SWAP gate (much as a Lie algebra is to a Lie group). But what about the $n^{th}$-root-of-SWAP qutrits or qudits, with $d=4?$

For example consider a pair of SWAP gates acting on four qubits; the first SWAP gate swaps the first two qubits, and the second SWAP gate swaps the second two qubits. That is, consider a two-qubit gate such as $\mathsf{SWAP}\otimes\mathsf{SWAP}$.

The $16\times 16$ matrix $\mathsf{SWAP}\otimes\mathsf{SWAP}$ of such a gate may be as below:

$$\mathsf{SWAP}\otimes\mathsf{SWAP}=\begin{pmatrix} 1 & & & & & & & & & & & & & & & &\\ & & 1 & & & & & & & & & & & & & &\\ & 1 & & & & & & & & & & & & & & &\\ & & & 1 & & & & & & & & & & & & &\\ & & & & & & & & 1 & & & & & & & &\\ & & & & & & & & & & 1 & & & & & &\\ & & & & & & & & & 1 & & & & & & &\\ & & & & & & & & & & & 1 & & & & &\\ & & & & 1 & & & & & & & & & & & &\\ & & & & & & 1 & & & & & & & & & &\\ & & & & & 1 & & & & & & & & & & &\\ & & & & & & & 1 & & & & & & & & &\\ & & & & & & & & & & & & 1 & & & &\\ & & & & & & & & & & & & & & 1 & &\\ & & & & & & & & & & & & & 1 & & &\\ & & & & & & & & & & & & & & & 1 &\\ \end{pmatrix}.$$

Notice that $\mathsf{SWAP}\otimes\mathsf{SWAP}$ is unitary (by virtue of it being a permutation matrix) and also hermitian (by virtue of it being symmetric around the diagonal). This I believe is isomorphic to a SWAP gate acting to swap a pair of qudits ($d=4$).

I'd like to see if I can somehow do a local Hamiltonian simulation to simulate such a gate, which may be part of a larger simulation. For example, I'd like to act locally on one of the pairs of qubits, and also act locally on the other of the pair of qubits; but I'm not sure if I'm missing something. The matrix $\mathsf{SWAP}\otimes\mathsf{SWAP}$ does not seem to be composed of a sum of two separate hermitian matrices.

Nonetheless, it might make sense to simulate such a matrix with repeated applications of an "$n^{th}$-root-of-SWAP" on the first pair of qubits and an "$n^{th}$-root-of-SWAP" on the second pair of qubits?

Is $\sqrt {\mathsf{SWAP}}$ acting on a pair of 4-dimensional qudits isomorphic to $\sqrt {\mathsf{SWAP}}\otimes\sqrt{\mathsf{SWAP}}$ acting on two pairs of qubits?

Answer 1

Swapping 4-level qudits is equivalent to swapping pairs of qubits. Because you can encode a 4-level qudit into a pair of qubits. Similarly, swapping 8-level qudits will be equivalent to swapping triplets of qubits. The swap gate is convenient enough that this should hold regardless of how you map the 4-level qudit into the qubits (e.g. whether you map qudit |2> to big endian qubits |10> or little endian qubits |01>).

That being said, in general it is not the case that $\sqrt{U \otimes U}$ is defined to be $\sqrt{U} \otimes \sqrt{U}$ even though $(\sqrt{U} \otimes \sqrt{U})^2 = U \otimes U$, so you can't go from "swapping qubit pairs is the same as individual qubit swaps" to "the square root of swapping qubit pairs is the same as individual square roots of swapping qubits".

Let the principle square root of $U$ be what you get by computing its eigendecomposition, halving the angles of the eigenvalues in polar coordinates, then putting the matrix back together. Under this definition $\sqrt{SWAP \otimes SWAP}$ looks like this:

Whereas $\sqrt{SWAP} \otimes \sqrt{SWAP}$ looks like this:

You can see they're not the same. In particular, the former entangles qubits that the latter does not.