# Simulating $e^{iX_1\otimes X_2t}$

I'm trying to figure out the quantum circuit to simulate the time-evolution of a 2-qubit Hamiltonian $$e^{iX_1\otimes X_2t}$$, where $$X$$ is a Pauli gate. From this answer, the quantum circuit performs $$e^{iZ_1\otimes Z_2t}$$ could be constructed as an $$R_z$$ gate sandwiched by 2 controlled x gates. Can I replace $$R_z$$ by $$R_x$$ (and same goes for $$R_y$$) to simulate $$e^{iX_1\otimes X_2t}$$ and $$e^{iY_1\otimes Y_2t}$$? Thanks!

You can't just replace $$R_z$$ with $$R_x$$ - the $$R_x$$ would commute with the controlled-not, so the two controlled-nots cancel and all you get out is just the $$R_x$$.

The trick that you want to use instead is that $$Ue^{-iHt}U^\dagger=e^{-iUHU^\dagger t}$$. Now, we know that you can transform $$Z\otimes Z$$ into $$X\otimes X$$ using hadamards. So set $$U$$ to be the two-fold tensor product of Hadamards. Having said that, there are further circuit identities that you can use to simplify the circuit. You ultimately find that to get the $$X\otimes X$$ term, you need a circuit where you (i) replace the $$R_z$$ with $$R_x$$ and (ii) reverse the direction of the two controlled-nots.

You can follow a similar process to get $$Y\otimes Y$$ terms using $$U=\frac12(Z+Y)\otimes(Z+Y),$$ but I'm not sure the circuit simplifies to the same extent.

Since you mention both $$X\otimes X$$ and $$Y\otimes Y$$ terms, I should point out that if the Hamiltonian that you actually want to simulate is $$X\otimes X+Y\otimes Y$$, you might be better off doing that directly rather than breaking it down into a pair of terms. What you need to think about is what is the unitary that diagonalises the term $$X\otimes X+Y\otimes Y$$, which is basically the inverse of the unitary that produces the Bell basis.

• You can treat the two terms separately but if this is literally all you're doing, you'd be better treating them together as you'll be able to get away with fewer two-qubit gates. Aug 26 '21 at 13:23
• You can append the two circuits, but a pair of controlled-nots is not a swap (unless you know that one of the input qubits is definitely 0). It's three controlled-nots that give swap. Aug 26 '21 at 13:24
• I think it's sufficient to do the sequence cNOT - $R_x(\theta)\otimes R_z(\theta)$ - cNOT where the control is on the first qubit. Aug 26 '21 at 14:11
• I just checked the two circuits and they are the same! Could you explain a bit about how to come up with the circuit?
– IGY
Aug 26 '21 at 16:37
• You set $U$ to be the unitary that diagonalises your $H$. Then all you need is to implement $e^{-i D t}$ where $D$ is a diagonal matrix, i.e. it's just going to require phase gates on the two qubits (and possibly a controlled-phase gate, but that turns out to not be necessary in this case) Aug 26 '21 at 17:54