# Uncertainty equal to zero

1. Nov 7, 2012

### IWantToLearn

could we able to measure the velocity for a free particle with uncertainty equal to zero, could we able to have a certain values for its velocity, and let the uncertainty of the position to be infinity

is there any theoretical restrictions on having a certain value for the velocity?
is there any practical restrictions?

Last edited: Nov 7, 2012
2. Nov 7, 2012

### Vorde

If you have no idea where the particle is, how are you going to measure it?

3. Nov 7, 2012

### IWantToLearn

that is means that, there have to be some uncertainty (not equal to zero and not equal to infinity) in the position to measure some uncertainty (not equal to zero and not equal to infinity) in velocity

and the bottom line is that we can't be have a measurement for velocity with zero uncertainty

am i right?

4. Nov 7, 2012

### Vorde

Yes.

I think you have the right idea in the sentence before, but it's worded a little weirdly. Basically to be able to take any measurement of an object, you have to have some idea of where the object is, and that means you have a finite uncertainty in its position.

5. Nov 7, 2012

### IWantToLearn

yes the wording wasn't clear
it should be like that
"The bottom line is, we can't do a measurement for velocity with zero uncertainty"

6. Nov 8, 2012

### DrChinese

There is no theoretical limit to measuring momentum or position with complete certainty.

7. Nov 8, 2012

### IWantToLearn

But this will mean that the particle has a single value of velocity, there is no distribution whatsoever of this velocity, this would be a particle that has a single value of momentum or in the wave aspect a plane wave that fills the whole universe, a particle that exist by the same probability in everywhere in the universe.
When we solve Schrodinger equation in the time independent case for a free particle we would get a solution that is a plane wave with a definite momentum..A e^_{ixp/h} but this solution is not normalized, or in other words, the sum of probabilities of finding the particle somewhere is not one but infinite
the free particle with one single value of velocity or momentum is a myth that does not exist.

8. Nov 8, 2012

### DrChinese

Not necessarily that it is equally likely to be anywhere, but certainly a wide range of source positions. That is the wave picture.

9. Nov 9, 2012

### king vitamin

This is at least an overstatement, right? I make a local measurement of a particle's momentum, then the position is certainly bound by a light cone for a given time afterwards. Doesn't this imply a lower bound on my initial momentum measurement?

10. Nov 9, 2012

### DrChinese

Not sure I see that. When you specify how are making the measurement, I think it will be clear that the limits are experimental/practical - not theoretical.

IMPORTANT NOTE: Of course when I say there are no theoretical limits to measuring momentum or position with complete certainty, I do NOT mean they can be determined at the same time. It is often restated as follows:

There are no theoretical limits to preparing a particle in a certain eigenstate of momentum or position with complete certainty, however it cannot be in a certain eigenstate of both momentum or position at the same time.

11. Nov 9, 2012

### Thoros

Yes, in principle we can measure the velocity (actually we prefer to use momentum, ie mass times velocity) with complete certainty.

In theory, no. The only downside is that by measuring the momentum to a perfect accuracy, we lose complete knowledge of it's position. Ofcourse that means, that the particle, whose momentum you measured, is effectively at "nowhere".

Yes and quite severe ones. The only waveform with a single frequency (and wavelength) is an infinite plane-wave. Such a thing cannot exist in reality since it would span from infinity to infinity. Now because of the particle-wave duality, your particle, whose velocity you measure with complete certainty, would collapse into such a state, which is impossible. From this we can deduce that in practice, you cannot achieve such accuracy no matter how well you design your apparatus.

12. Nov 9, 2012

### IWantToLearn

it is like you are saying that mathematically it is possible to measure the momentum with zero uncertainty , but in physics this is impossible

saying that the free particle is never exist is much like saying that nature refuse a complete certainty for a measurement of the momentum or the position

so i can say that (a measurement with a complete certainty is impossible and "this is an inherited feature in nature")

Last edited: Nov 9, 2012
13. Nov 9, 2012

### Staff: Mentor

I think you are confused about matters of principle and what can be done in practice. In practice it is impossible to know any real number with 100% accuracy because that would involve an infinite decimal expansion but in principle QM allows position and momentum to be measured to any degree of accuracy - just not simultaneously.

Thanks
Bill

14. Nov 9, 2012

### IWantToLearn

i understand that the uncertainty principle prevent us from doing a measurement with complete certainty for pairs of physical quantities (position and momentum) (energy and time) and that there is some kind of trade off in the uncertainty of the measurement for each quantity of them

Ok, now this is not my point

my point is regardless of any technological advances or accurate measurements, i claim that to have a complete certainty, this means that we have a free particle, exist in all of the universe with the same probability everywhere and not affected by any field
in my view this is a myth
since the universe is a grand field full of potentialities, and in fact there is no meaning for the term nothing, because always there is something, so our particle will never be free
so no complete certainty even theoretically

this is my claim, correct me of i am mistaken

15. Nov 9, 2012

### Staff: Mentor

A free particle, like knowing position exactly, is a conceptualization, an idealization pertaining to a particle in an inertial frame. Even in deepest space an exact inertial frame does not exist but as a conceptualization is very valuable. It is similar to a point being defined as having position and no size - such does not exist either but as a way of stripping away the inessential and getting to the heart of what a point is it is vital.

Thanks
Bill

16. Nov 9, 2012

"nature abhors infinitey" - in this case infinite precision - might be the general principle...

17. Nov 10, 2012

### Thoros

If i understand you correctly, you are right. The free particle with definite momentum will be spread across the universe with the same probabilty amplitude. This is the infinite plane-wave that exists everywhere simultaneously.

Now, you are right, it is a myth, but the simplest answer would be this: Any real physical solution to the Schrödinger equation must be normalizable. An infinite plane-wave is NOT normalizable. You can check it yourself by trying to normalize a sin function that spans infinity.

So to conclude: a free particle with definite energy (energy and momentum for a free particle are the same thing) cannot exist.

PS: Just to be clear why the position and momentum behave as such. They are infact the same thing in principle. They are related by a fourier transform. Therefore knowing the momentum means knowing the position. But in Quantum Mechanics, all you can know, is the probability distribution. And this is what we have been talking about.

In this example, your distribution of momentum would be a delta function. The fourier transform of a delta function, that is the distribution of position, would in turn be a constant across space. And vice versa.

And just to make sure i don't leave room for misunderstandings: As i've said, we only know probabilities that an experiment will give some real result. To know the momentum to an absolute accuracy, you would have to conduct the experiment on an identical particle in identical situations a lot of times (essentially infinite experiments to know it at 100% accuracy). But no matter how hard you try, your results will be spread out a little. For reasons i've already tried to explain.

18. Nov 10, 2012

### Staff: Mentor

That is true - it is a myth - but if it is normalizable or not depends on the formalism you are working with - in the Rigged Hilbert Space formulation it can be accommodated.

Thanks
Bill

19. Nov 10, 2012