Timeline for What's the point of Pauli's Exclusion Principle if time and space are continuous?
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13 events
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| Nov 2, 2019 at 15:44 | comment | added | cmaster - reinstate monica | @R.M. Oh, two electrons can occupy the same space all right. It's just that one of them needs to be in a higher energy state to do so. So, when you sit on your chair, your electrons are pushed into the chairs' electron clouds, but they lack the energy so they push back on you. If you were supplying sufficient energy by moving towards your chair at relativistic speeds, for example, your electron clouds would easily pass through the chair's electron clouds. They'd also interact a bit, providing your atoms with some random movement that you won't like, but hey, you pass through the chair! | |
| Oct 30, 2016 at 14:53 | comment | added | Kevin Fegan | @R.M. - "... hence the electron clouds of your feet don't fall through those of the floor." There's an implicit assumption that location equals quantum state there". Location is irrelevant. You can move your feet and the floor anywhere you like and you will still see the same behavior. | |
| Oct 26, 2016 at 0:48 | comment | added | ACuriousMind♦ | @YogiDMT Yes, the $\lvert k \rangle$ aren't a basis, either. The continuity of time has nothing to do with anything, the Schrödinger equation at its core already assumes time is continuous and wavefunctions are differentiable w.r.t. it. | |
| Oct 25, 2016 at 19:01 | comment | added | user107153 | @JgL I didn't say infinite-dimensional, I said uncountable, which is a very different thing: infinite but countable bases are one thing, infinite but uncountable ones are very different. Sliding between these two things without being clear about it the 'clever physicist's trick' I referred to, and is a mathematical horror. | |
| Oct 25, 2016 at 18:24 | comment | added | Ruslan | I don't quite get your "$|x\rangle$ don't form a basis": does $|k\rangle$ then also not form a basis? If yes, then how does Fourier transform work? | |
| Oct 25, 2016 at 17:53 | comment | added | Yogi DMT | What about time though? If time is continuous wouldn't simultaneity be impossible? | |
| Oct 25, 2016 at 16:36 | comment | added | ACuriousMind♦ | @JgL Note the "uncountable" in tfb's comment. The $\lvert x\rangle$ really don't form a basis of the Hilbert space in the standard mathematical sense - any Hilbert/Schauder basis should be countable, and Hamel bases are rather useless. | |
| Oct 25, 2016 at 16:29 | comment | added | Jeroen | @tfb, there is nothing mathematically ill-defined about using an infinite dimensional basis. This is actually an important part of the mathematics underlying various aspects of e.g. differential operators. You can start reading here: math.lsa.umich.edu/~kesmith/infinite.pdf | |
| Oct 25, 2016 at 16:03 | comment | added | R.M. | One lingering point of uncertainty here is that the Pauli exclusion principle is invoked for more than just the levels of hydrogen. For example, for why you don't fall through the floor (example site only) - there's a hand-wavy "electrons can't occupy the same quantum state, so they can't occupy the same 'spot', so therefore they can't occupy the same space, hence the electron clouds of your feet don't fall through those of the floor." There's an implicit assumption that location equals quantum state there. | |
| Oct 25, 2016 at 14:07 | vote | accept | Yogi DMT | ||
| Oct 25, 2016 at 13:49 | history | edited | ACuriousMind♦ | CC BY-SA 3.0 |
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| Oct 25, 2016 at 13:48 | comment | added | user107153 | This is a really important point: physicists do this clever trick where you quietly assume that you can use an uncountable basis, and you can't actually do that because the underlying maths falls apart horribly. | |
| Oct 25, 2016 at 13:45 | history | answered | ACuriousMind♦ | CC BY-SA 3.0 |