Timeline for Examples of systems with infinite dimensional Hilbert space, whose energy is bounded from above
Current License: CC BY-SA 4.0
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| Jun 13 at 14:23 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 13 at 14:13 | comment | added | Valter Moretti | Yes, you are right, for real smooth compactly supported wave functions $k$ is the expectation value of the momentum…The cutoff by Wouter was to be understand in this sense so | |
| Jun 13 at 14:00 | comment | added | lcv | Nice answer. Regarding $k$, when $\psi(x)$ is real $k$ is indeed the expectation value of momentum in the state $\psi_k$. | |
| Jun 13 at 11:48 | comment | added | Valter Moretti | @ Michael Seifert Fixed! Many thanks | |
| Jun 13 at 11:47 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 13 at 11:31 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 13 at 11:20 | comment | added | Valter Moretti | I cannot see a relation between $k$ and UV cuttoffs. ($k$ is not the momentum, it has not necessarily physical meaning here.) However, I considered the case of a system described in $\mathbb{R}^{3N}$. Indeed, I admit that I do not know if my argument may be extended to infinite lattices.... | |
| Jun 13 at 8:31 | comment | added | Wouter | Your argument is nice but relies on the possibility of taking $k\rightarrow \infty$. In a discrete space, there would be a UV cutoff I believe though (This would either be in a lattice, or by the existance of a Planck length that bounds the wavelength from below). Infinite dimensional can still be achieved by the spatial domain being infinitely large. Not completely sure what to get from there because still higher bands for the same k may appear perhaps. | |
| Jun 13 at 6:09 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 12 at 22:56 | vote | accept | Sanjana | ||
| Jun 12 at 15:55 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 12 at 13:16 | comment | added | Valter Moretti | Yes, sorry, sometimes I use to think in my mother tongue, where 'every' is correct in place of 'any' in that case. | |
| Jun 12 at 13:15 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 12 at 12:48 | comment | added | Kvothe | Do you mean to say "boundedness from above is not possible for ANY choice of V"? | |
| Jun 12 at 7:18 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 12 at 7:12 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 12 at 7:00 | comment | added | Tobias Fünke | Ah, I see! Thanks | |
| Jun 12 at 7:00 | comment | added | Valter Moretti | Yes but that way you reversed the sign of the kinetic energy! That is not physical. My argument relies upon the fact that the kinetic energy is unbounded form above... | |
| Jun 12 at 6:53 | comment | added | Tobias Fünke | Hi, I have a quick question: Taking $H$ the Hamiltonian of the quantum harmonic oscillator, then $-H$ is bounded above, no? Where does your argument fail there? Or more generally, which assumption is crucial that it works (you also write that "boundedness from above is not possible for every choice of $V$)? Is it related to the domain? | |
| Jun 12 at 6:43 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 12 at 6:10 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 12 at 5:59 | history | edited | Valter Moretti | CC BY-SA 4.0 |
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| Jun 12 at 5:54 | history | answered | Valter Moretti | CC BY-SA 4.0 |