3 ms·

Any physicist willing to comment? Sure, the spin matrices were built to deal with three spatial axes. Is there more to the paper than that?

by nyeah 1y ago

n4r9 1y ago
> the spin matrices were built to deal with three spatial axes If I understand correctly, it kinda happened the other way around. First the Pauli matrices were introduced to explain unexpected degrees of freedom in experimental observations; then the term "spin" was proposed because the operators related to each other in the same way as classical angular momentum operators. See e.g. https://www.sciencedirect.com/science/article/abs/pii/S1355219808000269 https://www.sciencedirect.com/science/article/abs/pii/S13552...
naasking 1y ago
Hossenfelder actually did a video on this just yesterday: https://www.youtube.com/watch?v=7See8OhtN-k https://www.youtube.com/watch?v=7See8OhtN-k
- deleted 1y ago
  [deleted]
dawnofdusk 1y ago
Not in this field directly, but first of all, talking about the "geometry of space" is more than just saying there are three spatial dimensions: geometry involves the local curvature of the object. Historically the Pauli matrices are discovered by assuming certain symmetries of spacetime. This paper shows the other direction also makes sense: if we assume certain structure on quantum observables, measurable only by temporal measurements and independent of the content of the quantum state (i.e., a measurement of any system will do), we can get the spatial symmetries we want. I suppose the ideal outcome is that there is some sort of exotic algebra of observables which is well motivated somehow by purely quantum considerations and by serendipity induces all the usual spacetime symmetries + extra stuff we didn't know about. This paper itself is cute but not sure if it's very impactful, I would defer to domain experts.
- nyeah 1y ago
  Only I feel like it says Euclidean 3-space in the abstract.
  - dawnofdusk 1y ago
    Yes but Euclidean 3-space means you need to test 1) that it's 3-space, by counting dimensions and 2) that it's Euclidean, which is a question about local curvature

Any physicist willing to comment? Sure, the spin matrices were built to deal with three spatial axes. Is there more to the paper than that?

by nyeah 1y ago

Any physicist willing to comment? Sure, the spin matrices were built to deal with three spatial axes. Is there more to the paper than that?

n4r9 1y ago

> the spin matrices were built to deal with three spatial axes If I understand correctly, it kinda happened the other way around. First the Pauli matrices were introduced to explain unexpected degrees of freedom in experimental observations; then the term "spin" was proposed because the operators related to each other in the same way as classical angular momentum operators. See e.g. https://www.sciencedirect.com/science/article/abs/pii/S1355219808000269 https://www.sciencedirect.com/science/article/abs/pii/S13552...

naasking 1y ago

Hossenfelder actually did a video on this just yesterday: https://www.youtube.com/watch?v=7See8OhtN-k https://www.youtube.com/watch?v=7See8OhtN-k

deleted 1y ago
[deleted]

dawnofdusk 1y ago

Not in this field directly, but first of all, talking about the "geometry of space" is more than just saying there are three spatial dimensions: geometry involves the local curvature of the object. Historically the Pauli matrices are discovered by assuming certain symmetries of spacetime. This paper shows the other direction also makes sense: if we assume certain structure on quantum observables, measurable only by temporal measurements and independent of the content of the quantum state (i.e., a measurement of any system will do), we can get the spatial symmetries we want. I suppose the ideal outcome is that there is some sort of exotic algebra of observables which is well motivated somehow by purely quantum considerations and by serendipity induces all the usual spacetime symmetries + extra stuff we didn't know about. This paper itself is cute but not sure if it's very impactful, I would defer to domain experts.

nyeah 1y ago
Only I feel like it says Euclidean 3-space in the abstract.
- dawnofdusk 1y ago
  Yes but Euclidean 3-space means you need to test 1) that it's 3-space, by counting dimensions and 2) that it's Euclidean, which is a question about local curvature