# Qiskit Inverse of a quantum fourier transformation

In the photo provided the Quantum Fourier Transform is depicted in Qiskit before the barrier. I don't understand the result of inverse. Conceptually, should the inverse of the QFT be the same resultant as the original implementation? And is the inverse correctly depicted in Qiskit for this 3 bit QFT algorithm show below?

From linear algebra we know that $$(AB)^{-1} = B^{-1} A^{-1}$$. This is because $$(AB)*(AB)^{-1} = ABB^{-1}A^{-1} = AIA^{-1} = I$$.

Hence, if you have the circuit to generate the Bell state from the state $$|00\rangle$$ which takes the form of:

then its inverse would be:

Putting them together gives you:

This makes sense as the inverse of $$CNOT$$ is $$CNOT$$, so they would formed Identity operator. Similarly, the inverse of Hadamard gate is itself... so altogether they will cancel each other out completely.

Thus, if I generate the QFT circuit for 4 qubit without the SWAP at the end, I would have:

then similarly to the above explanation with respect to the bell circuit and its inverse, the inverse QFT would be:

This is the reason why you see that you see in your circuit. hope that helps...