On 5/7/2025 9:03 PM, Alan Grayson wrote:
On Wednesday, May 7, 2025 at 6:58:20 PM UTC-6 Brent Meeker wrote:
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??
*You wrote,* *"So both position and momentum are measured and the
uncertainty in their value arises from instrument limitations, not
from QM" AG*
*You left out the preceding sentence that conditional the "So...".*
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.
*Do you mean, the scatter of each observable is caused by the
unavoidable variations in how the observables are prepared, and then,
by applying QM, one gets the UP? *
*Yes. The measurements the UP applies to are ideal measurements that
leave the system in the measured state, i.e. prepare it.*
*So what has this to do with "instrument limitations"? AG
*
*It doesn't. That's why instrument limitations are noted separately as
swamping the UP in most applications.
Brent
*
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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