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The question should be pretty simple, but it turns out that there's more to it with respect to what I initially expected.

Starting from the definition of the gate $Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$, Wikipedia states that the eigenvectors are $$\lambda_{+1} = \frac{1}{\sqrt{2}}\begin{bmatrix}1\\i\end{bmatrix}=: |i\rangle, \lambda_{-1} = \frac{1}{\sqrt{2}}\begin{bmatrix}1\\-i\end{bmatrix} =: |-i\rangle$$

So, I should be able to derive the matrix Pauli-Y as

$$Y = (+1) |i\rangle \langle i| + (-1) |-i\rangle \langle -i| = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$$

which is clearly different from the first matrix.

numpy, instead, gives as eigenvectors $$\lambda_{+1} = \frac{1}{\sqrt{2}}\begin{bmatrix}-i\\1\end{bmatrix}=: |i\rangle, \lambda_{-1} = \frac{1}{\sqrt{2}}\begin{bmatrix}1\\-i\end{bmatrix} =: |-i\rangle$$

which, by using the previous formula, returns a value for the Pauli Y gate equal to

$$Y = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$$

which is even stranger.

So, what am I doing wrong? Is it possible to uniquely define the eigenvectors?

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    $\begingroup$ Can you write out what $\langle i|$ and $\langle -i|$ explicitly? Check it, you will know. $\endgroup$
    – narip
    Aug 10, 2022 at 13:39

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Your first equation is wrong.

If $|i\rangle=\frac1{\sqrt2} (|0\rangle + i|1\rangle)$,

then its dual is the tranposed-conjugated version

$\langle i|=\frac1{\sqrt2} (\langle 0| - i\langle 1|)$,

which yields

$|i\rangle\langle i|=\frac12(|0\rangle\langle 0|-i|0\rangle\langle 1|+i|1\rangle\langle 0|+|1\rangle\langle 1|)$.

Similarly $|-i\rangle\langle -i|=\frac12(|0\rangle\langle 0|+i|0\rangle\langle 1|-i|1\rangle\langle 0|+|1\rangle\langle 1|)$.

So we get $|i\rangle\langle i|-|-i\rangle\langle -i|=-i|0\rangle\langle 1|+i|1\rangle\langle 0|=Y$ as you know it.

Note that I can modify one of the the eigenvectors as such $|i'\rangle=e^{i\alpha}|i\rangle$, with any real $\alpha$, and the relation still holds. So no, the eigenvector is not unique, it is defined up to a phase factor. numpy is providing valid eigenvectors, but be sure you are using the right complex functions to build $Y$.

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  • $\begingroup$ Of course, you'and @Ohad are both right, shame on me for my stupid mistake. I'll leave my question here as a reminder :) $\endgroup$
    – tigerjack
    Aug 10, 2022 at 14:17
  • $\begingroup$ @tigerjack you can still accept the answer that seems more adequate to you ;-) $\endgroup$
    – Mauricio
    Aug 10, 2022 at 16:28
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The eigenvectors that you have written above (both from Wikipedia and those plotted with Numpy) are valid eigenstates of the $pauli-Y$ operator.

I have done the same calculations that you have presented and the result in both cases is the $Y$ operator matrix.

I think that your mistake is in the computation of each outer product - it seems like that you forgot to complex-conjugate the entries of the bras $⟨i|$ and $⟨-i|$.

If $|i⟩ = \frac{1}{\sqrt2} \begin{bmatrix} 1 \\ i \end{bmatrix}$ then $⟨i| = \frac{1}{\sqrt2} \begin{bmatrix} 1 & -i \end{bmatrix}$.

Fix this for both $⟨i|$ and $⟨-i|$, perform the calculations again and you should get the right answers.

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