Pauli Hamiltonians

Every quantum system has a Hamiltonian — a Hermitian operator $H$ whose eigenvalues are the system's energy levels. To simulate $H$ on a quantum computer we need to express it as a sum of Pauli terms, because each individual Pauli exponential $e^{-i\theta P}$ translates directly into a short circuit (the subject of the next lesson).

The Pauli basis

The four single-qubit Pauli matrices

I = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}, \quad X = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}, \quad Y = \begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}, \quad Z = \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}

form an orthogonal basis for the space of all $2\times 2$ Hermitian matrices. Any single-qubit Hamiltonian can be written uniquely as

H = c_I\,I + c_X\,X + c_Y\,Y + c_Z\,Z, \qquad c_P \in \mathbb{R}.

For $n$ qubits the basis extends to all $4^n$ tensor products $P_1 \otimes P_2 \otimes \cdots \otimes P_n$ with $P_j \in \{I,X,Y,Z\}$ , often written as strings such as $ZXI$ or $ZZ$ . A general $n$ -qubit Hamiltonian is then

H = \sum_{P \in \{I,X,Y,Z\}^n} c_P\,P.

Extracting Pauli coefficients

The Pauli matrices are orthonormal under the Hilbert–Schmidt inner product $\langle A,B\rangle = \tfrac{1}{2}\operatorname{tr}(A^\dagger B)$ . Concretely, $\tfrac{1}{2}\operatorname{tr}(P_j^\dagger P_k) = \delta_{jk}$ for any two Pauli strings $P_j$ , $P_k$ . This orthonormality lets you read off each coefficient by a trace:

c_P = \frac{1}{2^n}\operatorname{tr}(P\,H).

A worked example: single-qubit Hamiltonian

Consider the 1-qubit Hamiltonian $H = 3Z + X$ . In matrix form:

H = 3\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix} + \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix} = \begin{pmatrix}3 & 1\\ 1 & -3\end{pmatrix}.

Let us verify the coefficient of $Z$ using the trace formula:

c_Z = \tfrac{1}{2}\operatorname{tr}(Z H) = \tfrac{1}{2}\operatorname{tr}\!\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}3&1\\1&-3\end{pmatrix} = \tfrac{1}{2}\operatorname{tr}\begin{pmatrix}3&1\\-1&3\end{pmatrix} = \tfrac{1}{2}(3+3) = 3.

We can confirm the other coefficients:

$c_I = \tfrac{1}{2}\operatorname{tr}(I H) = \tfrac{1}{2}(3 + (-3)) = 0$ .
$c_X = \tfrac{1}{2}\operatorname{tr}(X H) = \tfrac{1}{2}\operatorname{tr}\begin{pmatrix}1 & -3\\ 3 & 1\end{pmatrix} = \tfrac{1}{2}(1+1) = 1$ .
$c_Y = \tfrac{1}{2}\operatorname{tr}(Y H)$ . Since $Y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}$ , we get $YH = \begin{pmatrix}-i & 3i\\ 3i & i\end{pmatrix}$ , so $\operatorname{tr}(YH) = -i + i = 0$ .

The decomposition $H = 0\cdot I + 1\cdot X + 0\cdot Y + 3\cdot Z = X + 3Z$ matches what we started with.

Multi-qubit Hamiltonians

For two qubits the 16 Pauli strings $\{II, IX, IY, IZ, XI, XX, \ldots, ZZ\}$ span the space of $4\times 4$ Hermitian matrices. A paradigmatic example is the transverse-field Ising model on two sites:

H_{\text{Ising}} = -J\,(Z\otimes Z) - h\,(X\otimes I + I\otimes X),

where $J$ is the coupling strength and $h$ is the transverse field. This Hamiltonian has only three non-zero Pauli terms, which is typical: physically motivated Hamiltonians are sparse in the Pauli basis — they have far fewer than $4^n$ terms — and that sparsity is what makes Trotterisation efficient.

Why the Pauli representation matters for simulation

Each Pauli string $P$ satisfies $P^2 = I$ , so its eigenvalues are $\pm 1$ . The exponential of a single Pauli term is therefore a rotation:

e^{-i\theta P} = \cos(\theta)\,I - i\sin(\theta)\,P.

This closed-form expression means that any individual Pauli exponential can be realised as a small, exact quantum circuit. When the Hamiltonian is a sum $H = \sum_k c_k P_k$ , the Trotter–Suzuki formula decomposes $e^{-iHt}$ into a product of these simple rotations, each of which maps directly to a circuit. The Pauli decomposition is therefore not just a mathematical convenience — it is the bridge between the physical description of a system and its gate-level implementation.

Try it

The 1-qubit Hamiltonian $H = 3Z + X$ has the matrix $\begin{pmatrix}3 & 1\\ 1 & -3\end{pmatrix}$ . Use the trace formula $c_Z = \tfrac{1}{2}\operatorname{tr}(ZH)$ to compute the coefficient of $Z$ and return it.

Run your code to see the quantum state.