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
*
