# Can heisenberg uncertainty principle be beaten by fast measurements?

1. Jan 11, 2014

### htyj6g9jv1ev6

As I understand it the HUP prevents good accuracy of a small object's momentum and position at the same time.

Assume trying to measure the momentum and position of a single atom.

In theory, can't we get a very good idea of both of these values by first measuring the position and then, (almost) 1 Planck time after, measuring the momentum?

If not, how about the following:

In theory, measure the position of the atom, then go back in time and measure the momentum instead? (I read that backwards time travel is not rendered impossible by the current mathematical models of reality).

Wouldn't we then combine the values to determine both its momentum and position?

If not, how about the following:

In theory, measure the momentum of the atom. Then set up another identical experiment and instead measure the position. Wouldn't this give the momentum and position?

Sorry if it is a dumb question.

Last edited: Jan 11, 2014
2. Jan 11, 2014

### ZapperZ

Staff Emeritus
In other words, you are asking that if we violate other parts of physics, can we violate the HUP?

Does this appear rational to you?

Zz.

3. Jan 11, 2014

### htyj6g9jv1ev6

Are you suggesting that taking a measurement in a few Planck time units is impossible?
Your post probably needs more explanation.

Last edited: Jan 11, 2014
4. Jan 11, 2014

### ZapperZ

Staff Emeritus
That, and going back in time to do a measurement. I want to see those done! (It is the curse of the experimentalist in me)

And before you tell me that this is allowed in theory, I'd like you to cite actual, verified formalism that UNIQUELY describe these phenomena.

Zz.

5. Jan 11, 2014

### Staff: Mentor

For a momentum measurement with a reasonable precision (note: you need that), yes it is impossible.

The uncertainty principle is not a measurement issue - it is an intrinsic property of the particles. No measurements can ever change this.

6. Jan 11, 2014

### atyy

Even if these were possible, the theory says that a measurement of momentum immediately disturbs position, and a measurement of position immediately disturbs momentum. The theory does allow exceptions for certain states of a particle, but not for an arbitrary state of a particle.

This is not possible by current models, contrary to what you read.

Yes, this is the most common way of phrasing Heisenberg's uncertainty principle, since in this way the position or momentum of a particle is not disturbed by any previous measurement. The theory says that even in this case of perfect measurements, the distribution in the values of positions measured in one set of experiments and the distribution of momenta measured in the other set of experiments will obey the uncertainty principle. The uncertainty is not due to the measurements, but is intrinsic to any state.

Last edited: Jan 11, 2014
7. Jan 11, 2014

### edguy99

Thats kind of what Einstein said to Bohr back in 1933:

Suppose two particles are set in motion towards each other with the same, very large, momentum, and that they interact with each other for a very short time when they pass at known positions.
... Consider now an observer who gets hold of one of the particles, far away from the region of interaction, and measures its momentum; then, from the conditions of the experiment, he will obviously be able to deduce the momentum of the other particle. If, however, he chooses to measure the position of the first particle, he will be able to tell where the other particle is.
... How can the final state of the second particle be influenced by a measurement performed on the first, after all physical interaction has ceased between them?"

BTW, These guys are trying to do what I think is the essence of your question, doing a number of measurements that don't disrupt the system too much. In their words: "exploiting a more general class of quantum measurements than the class of projective measurements."

http://www.nature.com/srep/2013/130717/srep02221/full/srep02221.html

8. Jan 11, 2014

### atyy

Just a note that what the authors of http://www.nature.com/srep/2013/130717/srep02221/full/srep02221.html mean by violation of the "uncertainty relations" refers to the measurement-disturbance relations. Depending on how one defines "uncertainty", there are several possible measurement disturbance relations possible within quantum mechanics.
http://arxiv.org/abs/1304.2071
http://arxiv.org/abs/1312.4392

However, within quantum mechanics, the intrinsic uncertainty cannot be violated, regardless of any measurements. Because the various uncertainty relations are all often loosely called "Heisenberg's uncertainty principle", the intrinsic uncertainty principle is sometimes referred to as "Kennard's uncertainty relation" or "Robertson's uncertainty relation" to reduce confusion.

Last edited: Jan 11, 2014
9. Jan 11, 2014

### Staff: Mentor

To the OP - it's in the formalism of QM - no escaping.

If QM is correct you can't do it - if you can you have disproved QM.

If you read the lay literature it's often expressed in the form of thought experiments like Bohr and Einstein tussled with. However that is not its actual basis - its actual basis is that QM implies it by the mathematical nature of non-commuting observables.

Thanks
Bill