Position,velocity cannot be found simultaneously with appreciable

[prudhvi mohan]

position,velocity cannot be found simultaneously with appreciable accuracy,why is it so?

 

[adaptation]

A particle travels in a wave. Its momentum is related to its wavelength. To determine its wavelength we need to measure the distance between crests (or troughs) in the wave.

The particle can be at any nonzero point along the wave.

The more waves we measure, the better we know the particle's wavelength, and thus its momentum. Measuring more waves means there are more places for the particle to be in the wave.

The fewer waves we have, the better we can determine the particle's position because there are fewer places for the particle to be. But if we have fewer waves, it's harder for us to accurately measure wavelength, and thus momentum.

That is simply my understanding. If I've made errors, I trust they'll be identified soon enough.

 

[vanhees71]

Well, the problem is to define, what you mean by "measuring the difference between the crests of the wave". This you can do for classical waves, where the wave field is an observable quantity like the position of the surface of a lake where you can measure the wavelength of the water waves directly or in the case of (Laser) light, where you can measure the wavelength by doing a refraction experiment.

All this is very different for the wave function in quantum theory. According to the standard interpretation, (nearly) all phycist agree upon (the socalled minimal statistical interpretation), is due to Born and says that the modules squared, $$|\psi(t,\vec{x})|^2$$, is the probability density for finding at time $$t$$ a particle at position $$\vec{x}$$. With one particle you never are able to measure this complete probability density, since you always find one particle at a position (within the spatial resolution of your particle detector which is always finite).

The only way, according to the minimal interpration of QT, is to measure the position of many particles that are equally and independently from each other prepared in the state described by the wave function $$\psi$$.

 

[adaptation]

I thought it was obvious since we're talking about uncertainty in the QP forum that I did not mean classical waves. Perhaps I should have said wave packet?

The only way, according to the minimal interpration of QT, is to measure the position of many particles that are equally and independently from each other prepared in the state described by the wave function $$\psi$$.

What do you mean by equally and independently from each other prepared? I'm sorry I didn't get that. Maybe you could provide a link.

 

[billschnieder]

position,velocity cannot be found simultaneously with appreciable accuracy,why is it so?

because velocity is defined as $${\Delta x}\over{\Delta t}$$. It is obvious that $$\Delta x$$ is undefined at a precise instant/point, since you need more than one point to determine its value. Therefore the velocity is also undefined at a precise instant/point.

Usually that is overcome by inferring backwards. That is you use a series of points in time, obtain your average velocity and then you go back and conclude that the velocity at one of those points was the value you got.

 

[vanhees71]

I thought it was obvious since we're talking about uncertainty in the QP forum that I did not mean classical waves. Perhaps I should have said wave packet?


What do you mean by equally and independently from each other prepared? I'm sorry I didn't get that. Maybe you could provide a link.

The point is that you said you want to measure properties of the wave function like its wave length. This suggested that you have a (too) classical picture about those waves in mind and that's why I reminded you about the Minimal Interpretation, which in my opinion is the only interpretation which is free of intrinsic contradictions and free from esoteric "mumbo jumbo".

According to this interpretation the wave function (or better quantum state) is interpreted probabilistically according to the Born rule, and in this sense describes only ensembles of quantum systems, but not the behavior of individual particles.

On the other hand, of course, a single-particle state refers to one particle, i.e., it is to be associated with single particles. The resolution of the apparent contradiction is again to think about the concrete meaning of the association of the state with the real-world particle. This association is the preparation of a particle in this state, i.e., we must be able to perpare many single particles in this state to check the (probabilistic!) predictions of quantum theory. As in any statistical experiment, we have to make sure that we always prepare the particle in this state and that there are no hidden correlations in the preparation process between the individual experiments forming the ensemble. I hope, now it's clear what I meant in my posting before.

The minimal interpretation is due to L. Ballentine and can be found in

Ballentine, Leslie E.: The Statistical Interpretation of Quantum Mechanics, Rev. Mod. Phys. 42, volume 42, APS, 358–381, 1970

He has also written a very nice textbook about quantum theory:

Ballentine, Leslie E.: Quantum Mechanics, World Scientific, 1998

 

[adaptation]

Thanks for the reply vanhees71. It's a lot more clear. After reading a bit about the Ensemble interpretation, I like the fact that it assumes less, however I'm not sure I'm ready to convert just yet.

On the page "en.wikipedia.org/wiki/Uncertainty_Principal"[/URL] it clearly states that the momentum of a particle is proportional to the wavelength of the wave. I see now according to the Ensemble interpretation that this is not entirely accurate since in EI the wave function does not apply to a single particle.

As I understand it, the most widely accepted interpretation is still the Copenhagen interpretation. I believe my summary is acceptable according to that interpretation. (If it is not, please explain to me why.)

Has the Ensemble interpretation replaced the Copenhagen interpretation?