I'm trying to apply a time evolution algorithm for a physical system I'm trying to simulate on QISkit, however, in order to do that, I need to use the so-called Ising coupling gate:

$I=\begin{pmatrix} e^{ia} & 0 & 0 &0 \\ 0 & e^{-ia} & 0 & 0 \\ 0 & 0 & e^{-ia} & 0 \\ 0 & 0 & 0 & e^{ia} \end{pmatrix}$

I've tried performing rotations in the z-axis in both quits with the rz gate, also I've tried combining crz gates, as well as rzz and cu1 gates, but nothing seems to work.

The closest I could get was by implementing a zzz gate followed by a cu1 gate with oposite angle, however $[I]_{1,1}$ still remains at 1, no phase detected by the Aer unitary simulator.

How can I implement this gate?

Thank you very much in advance.


AHusain's answer is absolutely correct, but perhaps lacks some detail. The circuit that you want to implement is enter image description here Basically, the key is to realise that you want to apply phase $e^{i\alpha}$ to the basis elements $|00\rangle$ and $|11\rangle$, and $e^{-i\alpha}$ otherwise. In other words, you care about the parity of the two bits. If you can compute that parity of the two bits somewhere, you can perform a phase gate on that output, then undo the computation. Controlled-not computes the parity.

Here, I'm assuming that $$ R_z(\alpha)=\left(\begin{array}{cc} e^{i\alpha} & 0 \\ 0 & e^{-i\alpha} \end{array}\right). $$ This might be inconsistent with whatever definition you wish to use by a global phase or by a factor of 2 on the angle.

  • $\begingroup$ I followed the circuit you draw using: circ.cx(q[0],q[1]) circ.rz(a,q[1]) circ.cx(q[0],q[1]) with $a$ being a parameter y defined previously. However $[I]_{1,1}$ and $[I]_{4,4}$ still remained at 1, with no phase at all. $\endgroup$ – Jorge Rodríguez Peña Jul 15 at 13:53
  • $\begingroup$ Thanks. I was on mobile, so was hard to TeX. $\endgroup$ – AHusain Jul 15 at 14:06
  • $\begingroup$ @JorgeRodríguezPeña So you've got what you want up to a global phase. Global phases don't matter. $\endgroup$ – DaftWullie Jul 15 at 14:22
  • $\begingroup$ Did the others get the right phase or a factor of 2 off? $\endgroup$ – AHusain Jul 15 at 14:40

Conjugate by a CNOT, you'll see a controlled unitary of multiplying by $e^{\pm 2i a}$ depending on which CNOT you do and an overall phase of $e^{\mp i a}$.


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