# Time evolution of Hamiltonian

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.

## 1 Answer

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).