On Tuesday, April 28, 2020 at 7:38:12 PM UTC-6, Brent wrote: > > > > On 4/28/2020 5:59 PM, Alan Grayson wrote: > > > > On Monday, April 27, 2020 at 6:47:39 PM UTC-6, Alan Grayson wrote: >> >> >> >> On Monday, April 27, 2020 at 4:45:02 PM UTC-6, Brent wrote: >>> >>> >>> >>> On 4/26/2020 6:37 PM, Alan Grayson wrote: >>> >>> >>> >>> On Sunday, April 26, 2020 at 6:39:15 PM UTC-6, Brent wrote: >>>> >>>> >>>> >>>> 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. AG >>>>> >>>>> >>>>> The UP doesn't apply to virtual particles because it refers to the >>>>> result of conjugate measurement (projection) operators. You can't >>>>> measure >>>>> virtual particles. >>>>> >>>>> Brent >>>>> >>>> >>>> In 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. >>>> >>>> Brent >>>> >>> >>> Can you give an example of the ensembles used in applying the >>> time-energy form of the UP? TIA, AG >>> >>> >>> https://arxiv.org/pdf/quant-ph/0511245.pdf >>> >> >> This article seems to establish a lower bound on time, but nothing >> related to ensembles. I have no idea about the meaning of the terms in the >> time-energy form of the UP. AG >> >>> >>> >>> There's also an interesting discussion of how to measure time in QM. >>> Since time is not an operator you have to construct a clock which defines >>> the physical meaning of time. >>> http://www.god-does-not-play-dice.net/clock_peres.pdf >>> >>> Brent >>> >> > Since the "uncertainty" in the UP is a statistical entity with a > well-defined definition, aka "the standard deviation", how large must the > sample size be, to calculate it? TIA, AG > > > You mean to experimentally estimate it from the scatter of results? That > depends on how accurately you want to estimate. The error scales as > 1/sqrt(N). In most experiments with photons or electrons, it's easy to > make N big. But it's also hard to eliminate other sources of scatter that > have nothing to do with the UP. So only experiments deliberately designed > for maximum precision are going to push the UP bounds for simultaneous > measurements. > > Brent >
If the experiment is designed for max precision, how large does N have to be to satisfy the UP? TIA, AG > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected] <javascript:>. > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/2e37b187-cb85-48ab-843e-1e8939a3ec63%40googlegroups.com > > <https://groups.google.com/d/msgid/everything-list/2e37b187-cb85-48ab-843e-1e8939a3ec63%40googlegroups.com?utm_medium=email&utm_source=footer> > . > > > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/da73c6d1-6720-4a6d-a6a4-87116b058dba%40googlegroups.com.
