We know the QFT gives us a new orthogonal basis from the original one, however, when I apply it on two qubits, I am not getting the output vectors orthogonal.
$|out(k)\rangle = \Sigma^{N-1}_{j=0} e^{\frac{2\pi ij.k}{N}}|j\rangle$
Where $j.k$ is the bitwise 'AND' and then summed up.
Applying this on the basis:
$|00\rangle , |01\rangle , |10\rangle , |11\rangle$
I get the following vectors (ignore normalization factor):
$|00\rangle + |01\rangle + |10\rangle + |11\rangle$
$|00\rangle + i|01\rangle + |10\rangle + i|11\rangle$
$|00\rangle + |01\rangle - |10\rangle - |11\rangle$
$|00\rangle + i|01\rangle + |10\rangle -i|11\rangle$
However these are not orthogonal, as you can see, the dot product does not equal zero for the first and second vector.
Why is this the case?