Is there any formalism to perform quantum gates between two qubits (let's say in a superconducting quantum network) to perform a quantum gate between two qubits which are not directly coupled? I want to ask the same question for any known physical model of quantum computing. Another way of saying is that, if it is possible to indirectly couple two qubits via some unitary transformations on the Hamiltonian of a system? One example could be to ask if in a linear chain of three qubits $A,B$ and $C$, is it possible to do a SWAP gate between $A$ and $C$ not directly involving $B$, meaning without first swapping with $B$. I am not asking in the sense of quantum circuit where the answer could be possibly a 'no'. I am asking more from a Hamiltonian point of view.

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    $\begingroup$ If two qubits are not coupled, it seems impossible to apply a swap gate. Generally to apply any operation involving control-taget schema, there has to be some connection between the qubits. Or maybe, I did not understand your question well... $\endgroup$ Mar 11, 2021 at 8:56

1 Answer 1


This depends on exactly what your criteria are. Let me give the easiest solution to your 3-qubit case: apply the Hamiltonian $$ H=(X_1Y_2-X_1Y_2)+(X_2Y_3-X_2Y_3) $$ for a time $\pi/(2\sqrt{2})$. Apart from an extra $-1$ phase on the $|010\rangle$ and $|101\rangle$ terms (in essence, a controlled-phase between each pair of qubits), this does exactly what you want. It might be possible, in this 3-qubit case, to avoid some of those phases with a slightly different Hamiltonian, I'd have to experiment a bit.

This is essentially the content of a paper by Bose.

Of course, you impose in your question that B would not be involved, while in this, during the time evolution, the state of B does leave the B qubit even though it comes back later. This is always going to be a feature of such solutions, so I assume you're going to allow it.

If you want to do the transfer over greater distances, this is basically the study of "perfect state transfer" (full disclosure: this is very much my area of research). The transfer over greater distances doesn't, as a general rule, leave the intermediate qubits unmoved (but you can easily think of doing two swap pulses: one for all N qubits on a line, and one for all $N-2$ qubits in the middle to move them back to where they started). One advantage of this method compared to doing a whole bunch of swap gates in the gate model is that you can achieve the transformation in about half the time. A second advantage is that there's effectively a controlled-phase gate automatically built in to this sort of interaction. You may want to check out these papers:

Note that you might get annoyed by all the many controlled-phase gates that go on during the swapping process. If you undo the swap, it undoes all of them, so it's not so bad in that sense. However, you also have the option of encoding. If you encode a qubit as $(\alpha|00\rangle+\beta|11\rangle)$, that defeats all those controlled-phase gates.

  • $\begingroup$ Yes Dr. @DaftWullie, I have studied the papers on PST by Bose and later by Christandl, as well as your papers on PST. Yes, it is known that path graph $P_3$ allows PST between antipodal vertices via XY or Hei Hamiltonian, in the example I stated. However, longer chains than that is not possible for PST. Hypercubes are possible. But that is via quantum walk. Actually it was all part of my Master's Thesis [doi.org/10.1103/PhysRevA.102.062609 ]. In your 4th paragraph example, there also we have to perform 'multiple operations' which again challenges the task itself,but with less time cost $\endgroup$ Mar 11, 2021 at 21:02
  • $\begingroup$ Well, that depends on what assumptions you want to make about the system (i.e. some clarity would be helpful about the specific conditions you're thinking about). You say longer chains are not possible for PST - that's only true if you restrict to uniform coupling, which is not an assumption you've specified. $\endgroup$
    – DaftWullie
    Mar 12, 2021 at 8:52

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