Timeline for Negative sign in rotation operator again
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 17, 2015 at 15:18 | comment | added | a00 | Huh. Very interesting. I will have to think on it further. Thank you for your additional comments. | |
| Jul 14, 2015 at 14:00 | comment | added | CR Drost | Well, I mean, among spin-$\frac 12$ particles there are only the 2 degrees of freedom of what's called the "Bloch sphere". But yes, that's exactly what I was alluding to: the length of $\vec j$ is $\sqrt 3$ longer than its component in any direction, even the one that it is most "pointed along", meaning that it must be "pointed along" a circle of size $\hbar \sqrt{1/2}$ which is necessarily "averaged out", and $\vec j$ refuses to be located on anything other than such a circle. | |
| Jul 14, 2015 at 13:21 | comment | added | a00 | Your spatial spread statement is intriguing. I believe the fact that $\langle j^2 \rangle > \hbar^2/4$ is related to the fact that, for example, a supposedly "purely up spin" somehow has some character of "horizontal spin" that cannot be removed from it. (i.e., why we can measure x direction spin even when we are sure the spin is perfectly polarized in the positive z direction). It all ties into the bizarre nature of there being 6 dimensions of spin (+ and - z, + and - y, + and - x) rather than 3. | |
| Jul 10, 2015 at 14:18 | comment | added | CR Drost | The Pauli matrices give a great description of a spin-$\frac 1 2$ system where the ang. momentum in the $p$-direction of the state $|\psi\rangle$ is $\langle j_p\rangle=\frac\hbar 2\langle\psi|\sigma_p|\psi\rangle$. A nice check from orbital mechanics is that the $j^2$ "total spin" should be $\hbar^2s(s+1)$ which for $s=\frac 12$ is $\frac 34\hbar^2$, which is indeed what you get doing $\langle j_x^2+j_y^2+j_z^2\rangle$. Eigenvectors of those operators give us $x$, $y$, and $z$ states, but those eigenvectors aren't orthogonal because $\langle j^2\rangle>\hbar^2/4$, giving spatial "spread". | |
| Jul 9, 2015 at 15:36 | vote | accept | a00 | ||
| Jul 9, 2015 at 15:36 | comment | added | a00 | I think you're saying that to figure out which is the right way, try both formulations with the example that we know $\hat x $ rotated $\pi/2$ produces $\hat y$. I don't understand yet certain parts such as how you know $\sigma_z$ is involved, but I do see that the matrix you produced does rotate $\hat x$ correctly, assuming the assumption that these phase factors don't matter. I'm sure I will understand these points eventually. Thank you for the walk through! | |
| Jul 8, 2015 at 16:26 | history | edited | CR Drost | CC BY-SA 3.0 |
added 290 characters in body
|
| Jul 8, 2015 at 16:21 | history | edited | CR Drost | CC BY-SA 3.0 |
added 290 characters in body
|
| Jul 8, 2015 at 16:13 | history | answered | CR Drost | CC BY-SA 3.0 |