How the single qubit unitary (U) calculates when apply a gate to only one qubit at a time?

Question

Qiskit Textbook, Chapter 2, Section 2.2. Single Qubit Gates on Multi-Qubit Statevectors (here).
In here, they have described that: If we want to apply a gate to only one qubit at a time (such as in the circuit below), we describe this using tensor product with the identity matrix, e.g.: $Xotimes I$.
By executing the given code, the tensor product matrix generated corresponds to when X-gate is applied to qubit 1 in circuit.
qc = QuantumCircuit(2)
qc.x(1)
qc.draw()
# Simulate the unitary
usim = Aer.get_backend('unitary_simulator')
qobj = assemble(qc)
unitary = usim.run(qobj).result().get_unitary()
# Display the results:
array_to_latex(unitary, pretext="text{Circuit = } ")

The output is:

When I tried to change the $X$ gate position to qubit 0 (qc.x(0)) then the tensor product matrix is changed (refer to image attached below). My doubt is: in this case how is the tensor product calculated, shouldn't it be the same?

epelaaez · Accepted Answer

Short answer, tensor product is not commutative. Therefore, in the general case, $A otimes B not = B otimes A$. By calculating the matrices $I otimes X$ and $X otimes I$, you'll see the difference.
Applying $X$ to qubit 0 corresponds to the following matrix:
$$
I otimes X = begin{pmatrix} 1 & 0  0 & 1 end{pmatrix} otimes begin{pmatrix} 0 & 1  1 & 0 end{pmatrix} = begin{pmatrix} 1 begin{pmatrix} 0 & 1  1 & 0 end{pmatrix} & 0 begin{pmatrix} 0 & 1  1 & 0 end{pmatrix}  0 begin{pmatrix} 0 & 1  1 & 0 end{pmatrix} & 1 begin{pmatrix} 0 & 1  1 & 0 end{pmatrix}  end{pmatrix} = begin{pmatrix} 0 & 1 & 0 & 0  1 & 0 & 0 & 0  0 & 0 & 0 & 1  0 & 0 & 1 & 0 end{pmatrix}
$$
And applying $X$ to qubit 1 corresponds to:
$$
X otimes I = begin{pmatrix} 0 & 1  1 & 0 end{pmatrix} otimes begin{pmatrix} 1 & 0  0 & 1 end{pmatrix} = begin{pmatrix} 0 begin{pmatrix} 1 & 0  0 & 1 end{pmatrix} & 1 begin{pmatrix} 1 & 0  0 & 1 end{pmatrix}  1 begin{pmatrix} 1 & 0  0 & 1 end{pmatrix} & 0 begin{pmatrix} 1 & 0  0 & 1 end{pmatrix} end{pmatrix}  = begin{pmatrix} 0 & 0 & 1 & 0  0 & 0 & 0 & 1  1 & 0 & 0 & 0  & 1 & 0 & 0 end{pmatrix}
$$
So, you write the matrices in the same order as the qubits. For example, if you have the register $|crangleotimes|brangleotimes|arangle$ and we apply the matrix $Cotimes Botimes A$, $A$ will be applied to $|arangle$, $B$ to $|brangle$, and $C$ to $|crangle$. Furthermore, $|arangle$ would correspond to qubit 0, $|brangle$ to qubit 1, and $|crangle$ to qubit 2.

How the single qubit unitary (U) calculates when apply a gate to only one qubit at a time?

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