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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?

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  • $\begingroup$ where does that definition of the QFT come from? $\endgroup$ – DaftWullie Sep 17 '18 at 13:54
  • $\begingroup$ @DaftWullie, I found it in Nielsen and Chuang, page 217 of the 2002 edition. $\endgroup$ – Mahathi Vempati Sep 17 '18 at 13:55
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You are mis-quoting the definition of the QFT. You simply take the product of the decimal values of $j$ and $k$, and don't use their binary representations.

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