# Uncertainty principle and relativity

1. Feb 3, 2015

### barefeet

Consider a particle and you measure the position with a relatively small uncertainty in a box Δx2 . And then you measure the momentum very precisely, lets say with infinite precision.

Is the position then completely undetermined or is it still in a large ball of 4/3π(cΔt)3?
Does the wave function extend to + and -∞ or should it die out at + and -cΔt?

2. Feb 3, 2015

### BvU

No can do. Even in a simple square well the momentum has uncertainty. See here.

Don't quite understand where your c comes from. But outside the box the wave function has to be zero (follows from the Schroedinger equation).

3. Feb 3, 2015

### barefeet

No, I was talking about after measuring the momentum.

4. Feb 3, 2015

### jfizzix

After you measure the momentum, the wavefunction is in a superposition of energy states associated to the value of the momentum you measured. In doing so, the future measurement of the position would be unpredictable.

In short the uncertainty principle tells us:
By measuring the momentum, you make future measurements of the position unpredictable.
By measuring the position, you make future measurements of the momentum unpredictable.

Hypothetically, you could make the box very small and in that way localize the position of a particle with arbitrary precision, and then measure the momentum to an arbitrary precision, breaking the uncertainty principle, quantum mechanics, and possibly destroying the universe.

The reason that this does not happen (thankfully) is that the exact particle in a box (i.e., the infinite square well) Hamiltonian is unphysical. This doesn't happen if you consider instead the finite square well, where there is always a small likelihood to find the particle outside the box (even with arbitrarily high, but still finite sides).

5. Feb 3, 2015

### barefeet

I think my question is misunderstood. After measuring the momentum it doesn't matter if there is a box or anything. My question is, if you have measured the position at time t1 and concluded that it is inside a volume of Δx2, at time t2 you measure the momentum very precisely( I don't care if there is a box now or not), is at time t3 the position completely undetermined and the wavefunction extends to + and - ∞ or does it die out at + and -cΔt?
With c the speed of light and Δt = t3-t1

6. Feb 3, 2015

### jfizzix

Interesting question!

So in non-relativistic quantum mechanics,
we could, say, measure the position to be within some two limits,
then measure the momentum with arbitrary precision,
and then thereafter have a state that is arbitrarily wide in position space

Being arbitrarily wide in position space, there would be a nonzero likelihood of measuring a particle in a place you wouldn't have otherwise due to events outside the light cone of the momentum measurement event (or even also the initial position measurement event).

In relativistic quantum mechanics, I think there are extra conditions that enforce locality built in as axioms into the formalism. In particular, the positions at times $t_{1}$ and $t_{3}$, and the momentum at time $t_{2}$, are observables associated to different events in spacetime.

Because of that, I don't think it follows that an arbitrarily precise measurement of momentum at time $t_{2}$ implies an arbitrarily wide position distribution at any other point in spacetime.

I would like to hear what a proper field theory expert has to say about this, though.

7. Feb 3, 2015

### barefeet

Yes ,that is what I meant. I don't know much about relativistic quantum mechanics though. I'd like to hear about these conditions as well.

8. Feb 4, 2015

### Staff: Mentor

An arbitrary precision will take a very long time - more time the more precise the measurement should be. Your position uncertainty in the third measurement will come from that timescale.

9. Feb 4, 2015

### BvU

Re post #5: Sorry, I also misinterpreted your question.
Perhaps you want to read up on the double slit experiment (e.g. in the Feynman lectures).
Position measurement = knowing through what slit it goes
Width of diffraction pattern (now from a single slit) = momentum uncertainty

Put me right if I'm still interpreting it wrongly :)

10. Feb 4, 2015

### barefeet

Interesting, practically or fundamentally? If fundamentally, how is the measurement time and the measurement precision related? And where does it originate from?

11. Feb 4, 2015

### Staff: Mentor

Relatively fundamentally. How do you measure momentum? You can track the position over time - but then you get the problem with the uncertainty relation again. You can measure energy - that will take time (there is an effective uncertainty relation between energy and time as well). In general, to not disturb the momentum, the interaction you use should be very low-energetic, but that means long wavelengths and timescales for the measurement.

12. Feb 5, 2015

### barefeet

Maybe I am seeing it wrong but it seems to me you are referring to specific measurment techniques. What if you propose a new measurement technique with an arbitrary measurement precision, is there theoretically something that dictates the measurement duration?
By the way, if your measurement of an observable takes a long time, what have you exactly measured, the observable at the end of your experiment or the observable at the beginning of your experiment or something else?

I never quite understood the uncertainty in time well enough. Could you give an example just like the example of bhobba dividing a number of equally prepared systems? What are you are you then measuring in one of the lots. Because time is not a property of the system, right?

Why is it important not to disturb it? It doesn't really matter if you don't know the momentum now, all that matters is that you know the momentum at a certain time (potentially in the past) then my question still stands.

Could you also give a similar example for the energy-time relationship?

I guess my question is what happens if you would want to measure first the momentum and then the position of the same lot? I understand that the standard deviation of the position of this lot is not the same as of the other lot, because obviously the measurement of momentum has effected the systems, but can you say anything about the standard deviation of the position of this lot in relation with the standard deviation of the momenta of this lot?

On another note, what if the number of identically prepared systems are prior to any measurement known to be in a finite volume V and I measure the momentum of one lot precisely. What I understand from your explanation is that the other lot should now be spread over the entire space. But if my measurement of the momenta has been done in a finite time this would violate the principles of relativity.

If you theoretically can measure the momentum in a finite time, I still see a problem regarding the principles of relativity.

13. Feb 5, 2015

### Staff: Mentor

The time needed for measurement techniques always depends on the measurement techniques.
If you do not want to disturb the momentum, yes, but the details how that comes into play will depend on the measurement.
The usual idea of measurements is to not disturb the value you want to measure, otherwise you run into this question (but if you know the time-evolution in between, it might be possible to factor that out).

And how do you measure how exactly you disturbed it?
I don't think there is one. The position/momentum uncertainty is a fundamental property of the particles independent of measurements, the energy/time one is an effective one that comes from the way we measure things.
You can try to measure one set of particles "faster" and you'll see a larger spread of the energies there. As an example, take the masses of very short-living particles: you can reconstruct them via the energies of the decay products. You'll note that those measurements do not always give the same answer, even with a perfect detector. The shorter the lifetime of the particle, the larger the width of the mass spectrum you get.

If you do that with a large set, you'll get the uncertainty relation again.
You cannot exactly measure momentum in finite time.

14. Feb 8, 2015

### barefeet

This is the answer to (a part of) my question. Nevertheless there are still a few cases where my question can be relevant.

If there is no upper limit for the relation between measurement time and technique, I still see a problem. Even though you can't measure infinitely precise within a finite time, if my time increases my precision exponentially, principles of relativity would still be violated. So it must be bounded (probably your precision can at most increase linearly with your measurement time) regardless which measurement technique you use.

If the time evolution is known exactly and it doesn't decrease your accuracy, I still see a problem. If for example you know at t1 the particle must be in a volume V, at t2 you start your measurement and ended it after a long time at t3. Then you know relatively precise what the momentum is at t3 and the time evolution gives you the momentum at t2 with the same precision as t3, then how long the measurement takes, does't really matter.

Lastly, if you change my original question a bit and measure the position precisely, would the momentum be completely undetermined or would it be bounded by c and E? As I have read just now, you can't measure the position with an uncertainty less then the Compton wavelength. So the statement that you can measure arbitrarily precise isn't entirely correct and I was wondering if the same could be said about measuring the momentum.

As a practical example, if you measure the momentum by applying a magnetic field and measuring the radius, where does the uncertainty come from and how would you improve that by measuring for a longer time?

[Mentor's note: edited to remove quotation of deleted post]

Last edited by a moderator: Feb 8, 2015
15. Feb 8, 2015

### Staff: Mentor

Yes, non-relativistic quantum mechanics violates principles of relativity. That's what "non-relativistic" means, and that doesn't make it any more wrong than non-relativistic classical mechanics is wrong. If you want something that works when relativistic effects are significant, you have to use QFT.

It comes from position uncertainty. The magnetic field sets particles of slightly different momentum on slightly diverging paths. The longer we allow them to diverge before we measure the positions, the smaller the momentum difference corresponding to a given amount of position difference.

16. Feb 9, 2015

### Demystifier

If you measured the position first, then you cannot achieve a very precise measurement of momentum after an arbitrarily short time. The reason is, as you probably suspect, the theory of relativity.

17. Feb 9, 2015

### Staff: Mentor

It would be true without relativity as well.

18. Feb 11, 2015

### barefeet

Yes, I could have guessed that, but I wasn't trying to prove QM wrong. That QM is wrong in relativistic situations, doesn't mean that all the principles of QM are incompatible with SR/GR. It is interesting to see what exactly is incompatible with SR/GR and in this case I was asking about the uncertainty principle. Many here seem to disagree that this is incompatible with SR, but as it turns out, it isn't the uncertainty principle rather the uncertainty principle combined with the statement that you can measure arbitrarily precise the position or momentum as long as the uncertainty principle is upheld, that is incompatible with relativity. For example, you can not measure the position with an uncertainty less than the Compton wavelength. Now I am just wondering if something could be said about the momentum uncertainty as well.

Ok, but after a few revolutions, measuring for a longer time would be useless.

Isn't there a difference between precision as in how accurate your measurement is and precision as in how much uncertainty your measurements have? I thought that you could measure as accurately as you want, but it would not be reproducible, because it would be rather random.

19. Feb 11, 2015

### Staff: Mentor

Depends on your setup. If you want very precise momentum measurements, you have your particles exit the magnetic field at different angles instead of just making them go in circles.

20. Feb 11, 2015

### Demystifier

Yes, but not always during an arbitrarily short time.

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