# What are the Pauli-Y eigenvectors?

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?

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

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 $$Y$$.

• Of course, you'and @Ohad are both right, shame on me for my stupid mistake. I'll leave my question here as a reminder :) Aug 10, 2022 at 14:17
• @tigerjack you can still accept the answer that seems more adequate to you ;-) Aug 10, 2022 at 16:28

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.