# eigenstates not eigenvectors of spin operator

Ive been reading Griffiths's introduction to Quantum mechanics and just reached the chapter about spin. I really dont understand Griffiths's derivation of the eigenspinors of for instance $$S_x$$. This is supposed to be a very simple calculation, basically just calculating the eigenvectors to the following matrix:

$$S_x = \begin{pmatrix} 0 & \frac{h}{4\pi} \\ \frac{h}{4\pi} & 0 \\ \end{pmatrix}$$

When calculating the eigenvectors for this very simple matrix i get: $$v_1 = \begin{pmatrix} 1\\ 1 \end{pmatrix}$$ and $$v_2 = \begin{pmatrix} -1\\ 1 \end{pmatrix}$$

This, however, seems to be the incorrect result. because Griffiths gets the same $$v_1$$ but $$v_2 = \begin{pmatrix} 1\\ -1 \end{pmatrix}$$

I really don't understand this, because $$v_2$$ is not even an eigenvector to $$S_x$$. Letting Griffiths $$v_2$$ transform with $$S_x$$ should give a vector that is proportional to his $$v_2$$, but that ends up not happening. It's instead proportional to my $$v_2$$, and therefore $$v_2$$ can't be an eigenvector of $$S_x$$. I've obviously messed up somewhere, but I really don't see what it might be. I'd really appreciate some help. Thanks.

• Note that if $\Lambda \vec v=\lambda \vec v$ for some operator $\Lambda$ and some eigenvector $\vec v$, then $\alpha \vec v$ is also an eigenvector $\forall \alpha\in \Bbb C-\{0\}$ (assuming your vector space is defined over $\Bbb C$ as is the case here) Commented Jul 30, 2022 at 19:51

His $$v_2$$ and your $$v_2$$ simply differ by a $$-1$$ factor, so they're colinear. If one is an eigenvector of $$S_x$$, the other is too, for the same eigenvalue.