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I have given the following Hamiltonian

$$\tilde H = - J_x (X_0 X_1 + X_2 X_3) - J_z (Z_0 Z_2 + Z_1 Z_3) - h\sum_j X_j + Z_j$$

Where, for example, $X_0 = \sigma^x \otimes I \otimes I \otimes I$ and I should Implement the time evolution of this Hamiltonian with Trotter Suzuki Simulations with the following term:

$$e^{i \tilde H t} \approx \left( e^{-i X_0 X_1 \, J_x t/n} e^{-i X_2 X_3 \, J_x t/n} e^{-i Z_0 Z_2 \, J_z t/n} e^{-i Z_1 Z_3 J_z \, t/n}\prod_j e^{-i(X_j +Z_j)\, ht/n} \right)^n$$

So I'm still at the first part $e^{-i X_0 X_1 \, J_x t/n}$ because I can't find a way to express this with single qubit Gates and CNOTs. At first thought, I could use an Rx Rotation around the X-axis for the qubit 0 and a CNot with the first qubit as the control and the second as the target qubit. But Now I noticed (or at least I guess) that I need to multiply the Qubit 2 and 3 also with this value. By now we have just seen in the lecture that $e^{i\frac{\theta}{2}\sigma^x} = R_x(\theta)$ and the same for $\sigma^z$. I would be very thankful to get an idea how I can implement this so I can adapt it to the Rest of the Hamiltonian.

FDGod
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Ruebli
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1 Answers1

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I found out, that one can do that using the $Z$-rotations instead of $X$-rotations because the $Z$-rotation is already diagonal. that makes it way easier to see what $e^{\sigma^{z}}$ should be. So one can implement $e^{-i\theta X_1X_2}$ with $$ \big(H \otimes H \big) \cdot CNOT_{1\to2} \cdot \big(I \otimes R_z(2\theta) \big)\cdot CNOT_{1\to2} \cdot \big(H \otimes H\big)\,.$$If anything is wrong feel free to correct me. (I just didn't want to leave the question open when I think I got the answer).

FDGod
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Ruebli
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