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
*
