Boundary terms in hilbert space goes vanish

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SUMMARY

The discussion centers on the behavior of square integrable functions in Hilbert space, specifically addressing the misconception that these functions must approach zero as the variable approaches infinity. The participants clarify that while the claim is incorrect, it holds true under certain conditions, such as when the position operator Q applied to the function is square integrable. Additionally, there is speculation that the same may apply to the momentum operator P. The need for precise references, particularly from Google Books, is also highlighted.

PREREQUISITES
  • Understanding of Hilbert space concepts
  • Familiarity with square integrable functions
  • Knowledge of quantum mechanics operators, specifically position operator (Q) and momentum operator (P)
  • Basic grasp of mathematical limits and convergence
NEXT STEPS
  • Research the properties of square integrable functions in Hilbert space
  • Study the implications of the position operator Q on function behavior
  • Explore the momentum operator P and its effects on square integrable functions
  • Investigate counterexamples to the claim regarding function limits at infinity
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Mathematicians, physicists, and students of quantum mechanics who are exploring the properties of functions in Hilbert space and the implications of operator theory.

notojosh
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Thant helped. thank you!
 
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Can you give a more exact reference? A link to the exact page at google books would be the best way to reference it.

I assume it has something to do with the common claim that square integrable functions must go to zero as the variable goes to infinity ([itex]\psi(x)\rightarrow 0[/itex] when [itex]x\rightarrow\infty[/itex]), which is actually wrong. (There are counterexamples. See this thread). However, I think [itex]\psi(x)[/itex] must go to zero as x goes to infinity if [itex]Q\psi[/itex] (where Q is the position operator) is square integrable. Maybe it also has to go to zero if [itex]P\psi[/itex] (where P is the momentum operator) is square integrable? (I don't have time to think that through right now).
 
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