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?


  • 1
    $\begingroup$ An eigenvector with eigenvalue $1$ of what? $\endgroup$ Commented Jan 19, 2021 at 15:07
  • $\begingroup$ given the linked thread, and the context, I'm assuming you meant eigenvector of the QFT. Feel free to revert the edit if that is not so (possibly clarifying what you meant) $\endgroup$
    – glS
    Commented Jan 20, 2021 at 10:48

1 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.

  • $\begingroup$ Nice answer @rnva , for those who are new, this U is the Pauli X-gate. $\endgroup$ Commented Jun 5, 2021 at 15:02

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