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
<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
<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
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