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
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