- #1
professorscot
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I don't know much about brackets and Hamiltonians, but there is a much simpler aspect of the time / energy uncertainty discussion that confuses me. It has to do with the sign of the uncertainty. [Disclaimer: Purists, please bear with me for a moment as I hash out my point heuristically.]
Here's the typical argument that I hear against the vacuum fluctuation hypothesis of the Big Bang: The energy of a system can be "unknown / inconstant / immeasurable / whatever" within a range of values ΔE, but ONLY FOR LESS TIME than Δt. Larger energy fluctuations are doomed to a shorter lifespan. So a fluctuation the size of the universe would have a Δt of about one googolth of a second. (Perhaps the only opportunity I'll ever have to use the fraction "one googolth"!)
But the damn formula has a GREATER THAN sign in it! On its face mathematically, the inequality shows us that Δt must be LONGER than h/ΔE. Why do discussions of the principle always invoke an upper bound, when the formula gives a lower bound? Is there indeed an upper bound on ΔEΔt? Or is this inequality understood to mean "greater than but approximately equal to?"
Here's the typical argument that I hear against the vacuum fluctuation hypothesis of the Big Bang: The energy of a system can be "unknown / inconstant / immeasurable / whatever" within a range of values ΔE, but ONLY FOR LESS TIME than Δt. Larger energy fluctuations are doomed to a shorter lifespan. So a fluctuation the size of the universe would have a Δt of about one googolth of a second. (Perhaps the only opportunity I'll ever have to use the fraction "one googolth"!)
But the damn formula has a GREATER THAN sign in it! On its face mathematically, the inequality shows us that Δt must be LONGER than h/ΔE. Why do discussions of the principle always invoke an upper bound, when the formula gives a lower bound? Is there indeed an upper bound on ΔEΔt? Or is this inequality understood to mean "greater than but approximately equal to?"