How do you compute the compiled unitary of a quantum circuit comprised of different $n$-input gates?

Question

Given a quantum circuit consisting of two qubits, how is the compiled unitary of the circuit computed when we have different input type gates? (X-gate, H-gate are single-input gates, CNOT is a 2-input gate).

qc = QuantumCircuit(2)
qc.x(1)
qc.h(0)
qc.cnot(0,1)
usim = Aer.get_backend('unitary_simulator')
qobj = assemble(qc)
unitary = usim.run(qobj).result().get_unitary()
array_to_latex(unitary, pretext="\\text{Circuit = } ")

Section 2.2 of Qiskit shows how the simultaneous operations of two single-input gates are represented by computing the tensor product of the two gates, and that when we apply a gate to one bit at a time, we represent the unitary as the tensor product with the identity matrix. For the circuit above, how do we compile the circuit's unitary for the $4\times 4$ CNOT matrix,

$$ \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ \end{pmatrix}, $$

and the $2\times 2$ $X$ and $H$ matrices?

Answer 1

Your circuit is looking like this:

Thus, your unitary matrix is:

$$ U = CNOT \cdot \big( H \otimes X \big) $$

where

\begin{align} H \otimes X &= \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \cdot \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & 1 \cdot \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ 1 \cdot \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & -1 \cdot \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{pmatrix} \\ &= \dfrac{1}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & -1\\ 1 & 0 & -1 & 0 \\ \end{pmatrix} \end{align}

and CNOT (textbook convention) is

$$ CNOT = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{pmatrix} $$

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If you us little endian (like qiskit does) then instead of $H \otimes X$, you have to do $X \otimes H$. You have to read the circuit in reverse.

$$ U = CNOT \cdot \big( X \otimes H \big) $$

so here you have

\begin{align} X \otimes H &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 0 \cdot \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} & 1 \cdot \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \\ 1 \cdot \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} & 0 \cdot \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \end{pmatrix} \\ &= \dfrac{1}{\sqrt{2}} \begin{pmatrix} 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \\ 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ \end{pmatrix} \end{align}

and in this convention, we have

$$ \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ \end{pmatrix}, $$

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