The circuit to simulate the term $e^{i Z \otimes Z t}$ can be construct as

and the circuit to simulate the term $e^{i X \otimes Y t}$ can be construct as

Now to simulate $H = X \otimes Y + Z \otimes Z$, we can use Trotter approx with one time slice to get the following circuit to approximate $e^{i (X \otimes Y + Z \otimes Z) t}$ :

Now as commented by @tsgeorgios, in fact, $X \otimes Y$ and $Z \otimes Z$ are commute. That is,

$$ [X\otimes Y, Z \otimes Z] = X\otimes Y \cdot Z \otimes Z - Z \otimes Z \cdot X\otimes Y = \boldsymbol{0}$$

you can see this explicitly as follows:
$$ X \otimes Y =
\begin{pmatrix}
0 & 0 & 0 & -i\\
0 & 0 & i & 0\\
0 & -i & 0 & 0\\
i & 0 & 0 & 0\\
\end{pmatrix} \ \ \ \ \ \ \textrm{and} \ \ \ \
Z \otimes Z =
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & 1\\
\end{pmatrix}
$$
hence you can see that $$ X \otimes Y \cdot Z \otimes Z - Z \otimes Z \cdot X \otimes Y = \begin{pmatrix}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix} $$

that is they are commute with one another. This is **important** because if two matrix $A$ and $B$ are commute then we have that (See here. )
$$e^{A + B} = e^{A}e^{B}$$

What this tells us is that the above circuit is the **exact representation** of the term $e^{i (X \otimes Y + Z \otimes Z) t}$ and not just an approximation!

Now, the circuit to simulate $e^{i X \otimes I t}$ is:

and the circuit to simulate $e^{i I \otimes Y t}$ is:

As you can see, the **Identity operator** doesn't do anything! so you can ignore them out of the circuit.

Thus, the trotter approx circuit for $e^{i (X \otimes I + I \otimes Y + Z \otimes Z) t}$ is:

I would like to linked a very good and detail answer by @Davit Khachatryan to a similar question here.

1

A related answer

– Davit Khachatryan – 2020-11-06T11:44:56.1301Another point is that a tensor product in the Hamiltonian couples qubits together while the tensor product for gates does not. This means $H = X_{1}\otimes Y_{2}$ requires a two qubit gate to implement the associated evolution, while $H = X_{1} + Y_{2}$ can be implemented by two single qubit gates. In terms of circuit diagrams this means the tensor product $X_{1}\otimes Y_{2}$ should have a gate that connects $q_0$ and $q_1$, while the sum $X_{1} + Y_{2}$ would have separate gates for $q_0$ and $q_1$ (as you have in your figure above). – N A McMahon – 2020-11-06T13:26:48.577

1Note that what kind of two qubit gate(s) is used (and any other single qubit gates that are also used) depends on the hardware and how you implement the unitary. In the answer below these are the CNOT gates. – N A McMahon – 2020-11-06T13:29:35.337

1@Davit Khachatryan Thanks!! – Zhengrong – 2020-11-06T14:06:59.707

@N A McMahon Thank you so much, that's really helpful:) – Zhengrong – 2020-11-06T14:15:38.180