As requested through the comment by the OP.
Given a Hermitian matrix $H$, we can always write it as linear combination of Pauli strings. That is,
$$ H = \sum_i \alpha_i P_i \hspace{1 cm} P_i \in \{I,X,Y,Z\}^{\otimes n} $$
note this linear combination can have up to $4^n$ terms. Thus, for a general Hamiltonian splitting it into linear combination of Pauli terms can be hard and inefficient. However, many interesting Hamiltonian have very efficient decomposition (the linear combination scales polynomial). For instance, the electronic structure Hamiltonian for molecule.
Now, the coefficients $\alpha_i$ can be determined through the formula
$$ \alpha_i = \dfrac{1}{2^n}Tr(P_i H)$$
So here, we are given that $H = A =\dfrac{1}{4} \begin{pmatrix} 15 & 9 & 5 & -3\\ 9 & 15 & 3 & -5\\ 5 & 3 & 15 & -9 \\ -3 & -5 & -9 & 15 \end{pmatrix} $.
so for instance,
$$ \dfrac{1}{2^2}Tr(II \cdot H)=\dfrac{1}{2^2} Tr\bigg(\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \dfrac{1}{4} \begin{pmatrix} 15 & 9 & 5 & -3\\ 9 & 15 & 3 & -5\\ 5 & 3 & 15 & -9 \\ -3 & -5 & -9 & 15 \end{pmatrix} \bigg)= \dfrac{1}{2^2} Tr\bigg(\begin{pmatrix} 3.75 & 2.25 & 1.25 & -0.75\\ 2.25 & 3.75 & 0.75 & -1.25\\ 1.25 & 0.75 & 3.75 & -2.25 \\ -0.75 & -1.25 & -2.25 & 3.75\end{pmatrix} \bigg) = 3.75 $$
Similarly, $\dfrac{1}{4}Tr( XZ \cdot H) = 1.25$, $\dfrac{1}{4}Tr( ZX \cdot H) = 2.25$, and $\dfrac{1}{4}Tr( YY \cdot H ) = 0.75$.
The rest of the coefficients are $0$. For instance, $\dfrac{1}{4}Tr( IX) = 0$, and $\dfrac{1}{4}Tr( YX) = 0$, etc.
Once you have the Pauli string decomposition, you can input this into quiskit to create a circuit that approximate (if the pauli terms don't commute with each other) the operation $e^{iAt}$ through Trotterization (look at page 207 here to know more about this technique) pretty quickly using their WeightedPauliOperator as follow:
from qiskit.aqua.operators import WeightedPauliOperator
pauli_dict = {'paulis': [{"coeff": {"imag": 0, "real": 3.75 }, "label": "II"},
{"coeff": {"imag": 0, "real": 1.25 }, "label": "XZ"},
{"coeff": {"imag": 0, "real": 0.75 }, "label": "YY"},
{"coeff": {"imag": 0, "real": 2.25 }, "label": "ZX"},
]}
operator = WeightedPauliOperator.from_dict(pauli_dict)
approx_circuit = operator.evolve(evo_time= 1, num_time_slices=1).decompose()
Note that you have option to change the evolution time and number of time slices in the evolve method. But note that all the Pauli terms here commute with each other so this is not an approximation but exact representation of the circuit. The reason is if $A$ and $B$ are two commuting matrix, $[A,B] = AB - BA = 0$, then $e^{A+B} = e^Ae^B $. Here $II$ commute with everything, and $[XZ,ZX] = 0$, $[ZX,YY] = 0$, and $[XZ,YY] = 0$ .
Which would output something like:
and if you want to transpiled to a specific gate set that you want to use, you can use the transpile function:
from qiskit.compiler import transpile
transpiled_circuit = transpile(approx_circuit, basis_gates=['cx', 'u3'], optimization_level = 3)
Then now, if you want to create the controlled version of $Controlled-e^{iAt}$ then you can create a controlled gate to all the gates you have in the circuit. But then you would still need to decompose those controlled operations to native gates that matches the device's native gate set. You can also do this automatically with qiskit too, actually. That is, you can just create a controlled version of the entire circuit above. Let's suppose the circuit to simulate $e^{iAt}$ is described as the transpiled_circuit we have above, then what I can do is:
from qiskit import QuantumRegister, QuantumCircuit
xs_gate = transpiled_circuit.to_gate()
cxs_gate = xs_gate.control()
qreg_q = QuantumRegister(3, 'q')
circuit = QuantumCircuit(qreg_q)
circuit.append(cxs_gate, [0,1,2])
This will create the circuit:
where the circuit231 represents the entire transpiled_circuit we had above. You can then again* decompose this circuit to see what it would look like in term of more elementary gates:
decomposed_circuit = circuit.decompose().decompose()
which can give you something ridiculously long:
What I described is for a general case. For specific problem, you can use different tricks to make this neater. For instance, see this amazing answer by Adam.
