- #1
FatPhysicsBoy
- 62
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Homework Statement
Hi so I have the eigenvalue equation [itex]S\vec x = λ\vec x[/itex] where [itex]S = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}[/itex] I have correctly calculated the eigenvalues to be [itex]λ=±\frac{\hbar}{2}[/itex] and the corresponding normalised eigenvectors to be: [itex]\hat{e}_{1}= \frac{1}{\sqrt2}\begin{pmatrix} 1 \\ 1 \end{pmatrix} [/itex] and [itex]\hat{e}_{2}= \frac{1}{\sqrt2}\begin{pmatrix} -1 \\ 1 \end{pmatrix} [/itex]
2. My questions
So I have confirmed the eigenvalues and they are correct since this was a 'show that' question, however, I have a few questions about the eigenvectors.
1) The second eigenvector [itex]\hat{e}_{2}[/itex] was supposed to be shown to be: [itex]\hat{e}_{2}= \frac{1}{\sqrt2}\begin{pmatrix} 1 \\ -1 \end{pmatrix} [/itex]. Now I understand that if I think about the line corresponding to [itex]\hat{e}_{2}[/itex] then both of these eigenvectors refer to the same line, so they are both correct. However, is there one which is 'more correct'? I noticed that it's totally arbitrary where the negative sign is since I can just divide the ratio equation by -1 and get the other one and vice versa.
2) I distinctly rememeber writing eigenvectors with a k multiplied by the eigenvector to show that any multiple of it is also an eigenvector. What happens to this k when you normalise? How come we don't still write the k?
3) Is the way you normalise all eigenvectors the same? No matter what they have in them or what size they are is it always the 'Pythagoras' equivalent? I.e. dividing by the norm? [itex]||\vec e_{n}||=\sqrt{a^{2} + b^{2} + ...}[/itex]
Thank you