Timeline for Stern-Gerlach-Experiment with j=1 atoms?
Current License: CC BY-SA 3.0
10 events
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| Jul 13, 2011 at 8:55 | comment | added | shark.dp | Okay, I got it. Thanks for pointing out the difference. But for the "classical" outcome of the experiment, it would mean that the atoms where uniformly distributed on the screen, because the deflection is proportional to $J_z$. | |
| Jul 12, 2011 at 15:19 | comment | added | David Bar Moshe | No, he is correct. $\theta$ isn't uniformly distributed with respect to the surface elements of the sphere, but $J_z$ which is proportional to $cos(\theta)$ is uniformly distributed. You can verify that according to the random variable distribution transformation law: $p_{J_z}= \frac{p_{\theta}}{|\frac{d(J_z)}{d\theta}|}$, where $p$ denotes the distribution density | |
| Jul 12, 2011 at 15:07 | comment | added | shark.dp | One more remark: it is interesting to see that Otto Stern probably did exactly the same mistake as me: See springerlink.com/content/wm65pq5t22706x62/fulltext.pdf, page 252. He writes: "Nun ist die Anzahl der Atome, für die $\theta$ einen bestimmten Wert hat, proportional $\sin(\theta)$". Translated: "The number of atoms for which $\theta$ has a certain value is proportional to $\sin(\theta)$". | |
| Jul 12, 2011 at 12:33 | comment | added | shark.dp | Trying to compute this on my own, I found a much simpler derivation for this surface: With $z = R \cos(\theta)$ one can do the integration in spherical coordinates: $A = R \int_0^{2\pi} \text{d} \phi \int_{\arccos(a/R)}^{\arccos(b/R)} \text{d} \theta \sin(\theta)$ $=$ $2 \pi R \int_a^b \text{d} \left(\cos(\theta)\right) = 2 \pi R (b-a)$. Thanks again! | |
| Jul 12, 2011 at 12:18 | vote | accept | shark.dp | ||
| Jul 12, 2011 at 12:18 | comment | added | shark.dp | Okay, this came really surprising for me. I didn't expect this. Thanks for the hint and the URL. I would upvote you're answer, if I were allowed to ;-) | |
| Jul 12, 2011 at 12:10 | comment | added | David Bar Moshe | Please see the derivation in: mathworld.wolfram.com/Zone.html | |
| Jul 12, 2011 at 12:04 | comment | added | shark.dp | Thanks for the detailed answer. I fully agree on the quantum mechanical part. But I don't understand the reasoning for the classical counterpart: shouldn't the R in the formular for S be the "in-plane" radius (distance from the z-axis, not from the origin)? $S = 2\pi \times \sin(\theta) h$? | |
| Jul 12, 2011 at 11:37 | history | edited | David Bar Moshe | CC BY-SA 3.0 |
added 35 characters in body
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| Jul 12, 2011 at 11:30 | history | answered | David Bar Moshe | CC BY-SA 3.0 |