On 4/26/2020 3:22 PM, Alan Grayson wrote:
On Sunday, April 26, 2020 at 1:46:59 PM UTC-6, Brent wrote: On 4/26/2020 9:24 AM, Alan Grayson wrote:On Sunday, April 26, 2020 at 9:48:45 AM UTC-6, John Clark wrote: On Sat, Apr 25, 2020 at 12:49 PM Alan Grayson <[email protected]> wrote: /> How does QM tell us that conservation of energy can be violated for brief durations? If you apply the time-energy form of the UP for your proof, please state the context of your proof, that is, exactly what do E and t stand for./ The shorter the time (t) a system is under observation the larger the amount of energy (E) could pop into existence from nothing without direct detection, enough energy to create virtual particles. And you can calculate how large the indirect effects these virtual particles would have on the system. As I understand the UP, it's a statistical statement about an ensemble of observations, say for position and momentum of identical particles. It says nothing about the result of events, say for the position and momentum of a single particle or event. Doing some arithmetic to get the time-energy form of the UP does not change this reality. As a result, your description of what happens to a single particle, virtual or not, is not intelligible. Please try again. AGThe UP doesn't apply to virtual particles because it refers to the result of conjugate measurement (projection) operators. You can't measure virtual particles. BrentIn its usual form, does the UP allow us to measure position and momentum *simultaneously*, or must we measure each variable independently (for an ensemble of identical particles, of course)? What is proper interpretation of the time/energy form of the principle in statistical terms? TIA, AG
You can measure them simultaneously; but when you repeat the pair of measurements on many identically prepared particles you find that there is a scatter in the position and a scatter in the momentum such that the HUP is satisfied.
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