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
*
