Why is the superposition of all states an eigenvector, with eigenvalue 1, of the QFT?

Question

In many descriptions of order finding, but also in this answer here, it is shown that the superposition of all states is an eigenvector for eigenvalue 1.0. To cite:

Having found the eigenvalues, we now want to find the corresponding eigenvectors. We begin with the observation that 𝑈 permutes the states $|x^0\pmod N\rangle,\dots|x^{r-1}\pmod N\rangle$. Consequently, the uniform superposition

$$|v_0\rangle = \frac{1}{\sqrt{r}}\sum_{k=0}^{r-1}|x^k\pmod N\rangle\tag3$$

is an eigenvector associated with eigenvalue 1.

Sometimes this is being phrased as all eigenvectors adding up to $|1\rangle$.

I really don't get this conclusion. Any insights / pointers?

Thanks

Answer 1

Perhaps a simpler example will help. Let the unitary $U$ permute $\vert 0\rangle$ and $\vert 1 \rangle$. That means $U\vert 0\rangle = \vert 1\rangle$ and $U\vert 1\rangle = \vert 0\rangle$.

Then clearly, $$U\left(\frac{\vert 0\rangle +\vert 1 \rangle}{\sqrt{2}}\right) = \frac{U\vert 0\rangle + U\vert 1\rangle}{\sqrt{2}} = \frac{\vert 1\rangle + \vert 0\rangle}{\sqrt{2}} = \frac{\vert 0\rangle + \vert 1\rangle}{\sqrt{2}}.$$

Thus you have that the equal superposition is an eigenvector with eigenvalue 1. Now extend this argument to a permutation acting on $r$ states.