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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.

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  • $\begingroup$ 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) $\endgroup$
    – Charlie
    Commented Jul 30, 2022 at 19:51

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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.

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  • $\begingroup$ Ohh i see, that makes sense. Thanks for your answer, but why is that the normal convention for defining the eigenstate? Feels so unnatural $\endgroup$
    – lumpludde
    Commented Jul 30, 2022 at 19:58
  • $\begingroup$ As far as I know, there is no convention. For a eigenvalue with multiplicity 1, the eigenvectors form a dimension-1 vector space, in other words they're all colinear. In quantum mechanics, the only important thing is that eigenvectors are usually normalized, which means that there are a infinite number of them, differing by a phase. $\endgroup$
    – Miyase
    Commented Jul 30, 2022 at 20:01

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