On 5/18/2025 10:02 AM, Alan Grayson wrote:
On Tuesday, May 13, 2025 at 4:54:55 AM UTC-6 Alan Grayson wrote:
On Monday, May 12, 2025 at 4:15:52 PM UTC-6 Brent Meeker wrote:
On 5/12/2025 1:58 PM, Alan Grayson wrote:
On Friday, May 9, 2025 at 10:40:42 PM UTC-6 Brent Meeker wrote:
On 5/9/2025 7:08 PM, Alan Grayson wrote:
*I can see that the measurement spreads due to
instrument limitations are usually immensely larger than
the much smaller spreads accounted for by the UP, but
what causes these much smaller spreads? Is this a
quantum effect? AG*
Yes. Quantum evolution is unitary, i.e. the state vector
just rotates in a complex Hilbert space so that
probability is preserved. Consequently the infinitesimal
time translation operator is U=1+e6/6t or in common
notation 1-i(e/h)H where H=ih6/6t and h is just
conversion factor because we measure energy in different
units than inverse time. It's not mathematics, but an
empirical fact that h is a universal constant.
Brent
*If one wants to prepare a system in some momentum state to
be measured, doesn't this imply a pre-measurement measurement, *
Right, given that it's an ideal measurement. Most
measurements don't leave the system in the eigenstate that is
the measurement result. An ideal measurement is one that
leaves the system in the state that the measurement yielded.
*and the observable to be measured remains in that state on
subsequent measurements? *
Only if they're ideal measurements of that same variable or of
other variables that commute with it.
*If so, how can the unitary operator, which just changes the
state of the system's wf, create the quantum spread? *
You don't need a change in the wf to "create the quantum
spread". Having prepared in an eigenstate of A just measure
some other variable B that doesn't commute with A. In general
A will be a superposition of other variables, say A=xC+yD;
that's just a change of coordinates. But the system is not in
an eigenstate of C or D.
Brent
*Sorry, I really don't get it. Not at all! If we want to prepare a
particle with some momentum p, why would we measure it with some
non-commuting operator, and why would this, if done repeatedly,
result in a spread of momentum? And what has this to do with a
unitary operator which advances time? TY, AG *
*
*
*Is the spread in momentum caused by an imprecision in preparing a
particle in some particular momentum? Generally speaking, how is that
done? TY, AG
*
*The HUP doesn't limit how precisely you can prepare a particle's
momentum. The HUP just says that the more precisely the momentum is
determined the less precisely defined will be the conjugate position.
Think of a searchlight. How sharply you can focus the beam isn't
limited in principle, it just depends on how big you make the
reflector. So the lateral spread of the beam is "caused" by the finite
size, if you want to call that an "imprecision".
I don't think there is any "generally speaking way it's done". Generally
experiments do not attempt to reach the HUP limit. For example in the
LHC they're aiming for the highest possible energy. The accelerating
frequency causes the groups of protons to have a certain length and
hence a certain scatter in energy, which is small but not necessarily at
the HUP limit.
Brent*
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