On 5/7/2025 5:46 PM, Alan Grayson wrote:
On Tuesday, May 6, 2025 at 9:53:17 PM UTC-6 Brent Meeker wrote:
On 5/6/2025 7:47 PM, Alan Grayson wrote:
Maybe someone can explain this; if, say, the momentum operator
always returns an eigenvalue of the momentum of the system being
measured, then when used in the UP, how can there be an
uncertainty in momentum to give a statistical variance? TY, AG
You misunderstand what the HUP refers to. There is often
confusion between/preparing/ a system in state, which is limited
by the uncertainty principle, and making a destructive measurement
on the system which can be more precise (but not more accurate)
that the uncertainty principle. In the literature the former, a
preparation, is referred to as an ideal measurement…but then the
“ideal” gets dropped and people assume that it applies to any
measurement. Then there’s a confusion between precision of single
measurements and the scatter of measurements of the same system
state.
Heisenberg’s uncertainty principle is commonly misinterpreted as
saying you cannot make a precise (i.e. to arbitrarily many decimal
places) measurement of both the x-axis momentum and the position
along the x-axis at the same time or on the same particle. This
is untrue. It comes from confusing the concept of preparing
particles in a state and measuring the state of a particle.
Heisenberg actually contributed to this confusion with his
microscope thought experiment
The theory only says you cannot prepare a particle so that it has
precise values of both momentum and position. The distinction is
that you can measure both x and p and get precise values, but when
you repeat the process with exactly the same preparation of the
particle the measurement will yield different values. So even
though you measure precise values there is no sense in which you
can say the particle /*had*/ those values independent of the
measurement. Your measurement has been precise, but not
accurate. And if you repeat the experiment many times, the
scatter in the measured values will satisfy the uncertainty
principle. See Ballantine, /“Quantum Mechanics, A Modern
Development”/ pp 225–227 for more complete exposition.
To illustrate, you can prepare particles so that their position
has only a small uncertainty and when you measure their position
and momentum you will get a scatter plot like the blue points below.
Each point is a precise measurement of both momentum, /*p*/, and
position, /*q*/. But because the scatter in position is small the
scatter in momentum will be big - and the uncertainty principle
will the satisfied. The green points illustrate the complementary
case in which the particles have a small scatter in momentum, but
a big scatter in position. The red points illustrate an
intermediate case, a preparation in which the momentum and
position scatters are similar.
It might also mention that in practice, i.e. in the laboratory and
in colliders like the LHC, the error scatter due to instrument
uncertainties is usually much bigger than that due to the
theoretical limit of the Heisenberg uncertainty, So both position
and momentum are measured and the uncertainty in their value
arises from instrument limitations, not from QM
So the scatter does not arise from QM,
Where did I say that?? It arises because in QM you cannot prepare a
particle to have zero scatter in both position and the conjugate
momentum. But that doesn't mean you can't measure them both precisely.
The scatter is an ensemble property.
Brent
and yet, as I recall, that when QM is developed axiomatically, the UP
follows. I'm not sure I get it. AG
Brent
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