# Why are these two QFT circuits equivalent?

I am new to quantum computing and have been trying to understand the Quantum Fourier Transform (QFT). Through my research using both the Qiskit textbook and other sources, I see differences in how the circuit is actually implemented. My question is: Why are the two circuits below equivalent? Does the order of the wires in each circuit matter? To me they look significantly different as even the operations are being done on different qubits.

For example;

From the Qiskit Textbook: From other sources: Qiskit uses little endian convention, meaning that in the state $$|ABC\rangle$$, $$q_0$$ corresponds to $$|C\rangle$$, $$q_1$$ to $$|B\rangle$$, and $$q_2$$ to $$|A\rangle$$. In literature, it is common to see big endian convention which is the other way around.

It can also be because of qubit ordering on the circuit. Qiskit uses the top qubit as $$q_0$$. But you can find literature in which the bottom qubit is $$q_0$$.

If you reverse the wires, they are the same circuit.

The circuits use opposite qubit order and different gate name conventions. You can see the change in qubit order by noting that the reflection of the first circuit in a horizontal line produces a circuit that overlaps with the second.

The gate name conventions are

$$S = P\left(\frac{\pi}{2}\right) = \begin{pmatrix}1&\\&e^{i\pi/2}\end{pmatrix} \\ T = P\left(\frac{\pi}{4}\right) = \begin{pmatrix}1&\\&e^{i\pi/4}\end{pmatrix}.$$

The qubit order in the second circuit is from top to bottom: $$q_2, q_1, q_0$$.