Possible paradox in the uncertainty principle?

[PreposterousUniverse]

In the Heisenberg's uncertainty principle

$\triangle x \triangle p \geqslant \frac{\hbar}{2}$

what happens when the uncertainty in position becomes very small is that the uncertainty in momentum becomes very large. But what happens when the spread of the uncertainty in momentum becomes greater then the speed of light? I suppose then that above formula is no longer valid in the relativistic limit?

 

[PeterDonis]

what happens when the spread of the uncertainty in momentum becomes greater then the speed of light?

This doesn't make sense. Uncertainty in momentum has no upper limit, just as momentum itself has no upper limit, even in relativity. Uncertainty in momentum, like momentum itself, doesn't have units of speed and can't be compared to the speed of light or any other speed.

 

[Dale]

But what happens when the spread of the uncertainty in momentum becomes greater then the speed of light?

This is not a meaningful sentence. Momentum is in different units from speed, so they cannot be compared. This is the “apples and oranges” thing.

 

[PreposterousUniverse]

But the only difference between velocity and momentum is the mass-factor which stays constant regardless. So I assume we are talking about the relativistic momentum which can approach infinity when the particle approaches the speed of light.

 

[Dale]

But the only difference between velocity and momentum is the mass-factor which stays constant regardless.

This is not correct. Speed is bounded, momentum is not. Momentum is conserved, speed is not. Momentum is the spacelike part of a four vector speed is not.

And in any case the comparison still doesn’t make sense even if mass were the only difference. You simply cannot compare quantities with different units, even if the only difference is mass.

 

[PeterDonis]

the only difference between velocity and momentum is the mass-factor which stays constant regardless

Not in relativity. In relativity, momentum is not $m v$, it's $m \gamma v$ (here $m$ is rest mass). So the relationship between velocity and momentum is not linear as it is in non-relativistic physics, and the intuitive reasoning you are trying to use here does not work.

 

[PreposterousUniverse]

Not in relativity. In relativity, momentum is not $m v$, it's $m \gamma v$ (here $m$ is rest mass). So the relationship between velocity and momentum is not linear as it is in non-relativistic physics, and the intuitive reasoning you are trying to use here does not work.

Thanks! I updates my answer above, including the argument about the relativistic momentum. What confused me first was because the textbook I used never mentioned anything about the relativistic effect.

 

[PeterDonis]

Btw, in addition to the issues already raised in this thread, it is worth noting that applying the uncertainty principle in a relativistic setting requires more care, since the proper relativistic quantum theory is quantum field theory, in which particles can be created or destroyed.

 

[PeterDonis]

I updates my answer above, including the argument about the relativistic momentum.

What argument about relativistic momentum? Unless you just mean you now agree that your OP was mistaken because it failed to take into account the proper relativistic formula for momentum?

 

[PeterDonis]

the textbook I used never mentioned anything about the relativistic effect

Lots of textbooks on QM don't go into relativistic effects. If you really want to learn relativistic QM, you need to learn quantum field theory.

 

[PreposterousUniverse]

What argument about relativistic momentum? Unless you just mean you now agree that your OP was mistaken because it failed to take into account the proper relativistic formula for momentum?

" So I assume we are talking about the relativistic momentum which can approach infinity when the particle approaches the speed of light. "

 

[PeterDonis]

the comparison still doesn’t make sense even if mass were the only difference. You simply cannot compare quantities with different units, even if the only difference is mass

I think this, while technically correct, is a red herring. In the non-relativistic approximation, if one is dealing with a quantum system with constant rest mass, one can indeed divide out the mass and construct an uncertainty relation between position and velocity instead of position and momentum. What keeps this from working for arbitrarily high velocities is relativity, because in relativity dividing out the mass, even though it makes the units the same ($\gamma$ is unitless so $\gamma v$ has the same units as $v$), does not make the uncertainty relation a relation between uncertainties in position and velocity; it makes it a relation between uncertainties in position and $\gamma v$. And if you rework that into a relation between the uncertainties in position and velocity (which you could in principle do, though the math would be ugly), the relation is no longer linear in the velocity uncertainty and the OP's argument fails.

 

[Dale]

" So I assume we are talking about the relativistic momentum which can approach infinity when the particle approaches the speed of light. "

Well, you were asking about the relativistic speed limit of c, so only relativistic momentum makes sense in context.

 

[Dale]

one can indeed divide out the mass and construct an uncertainty relation between position and velocity instead of position and momentum

Sure, but you must actually do that to make the comparison. You can not compare quantities of different units. I disagree about it being a red herring.

Checking units is actually the very first physics lesson I ever had, and it has served me well many times. It allows you to catch mistakes early on. If he had done that then he may have realized that the relativistic momentum formula was important in the context.

I agree that it is not the only problem, but I didn’t say it was.

Another similar problem is people comparing the expansion of the universe to c. Again, simply looking at the units would avoid that and have them check their logic.

 

[atyy]

Does the uncertainty principle between position and momentum hold in relativity? That uncertainty principle comes from non-relativistic quantum mechanics.

 

[PeterDonis]

Does the uncertainty principle between position and momentum hold in relativity?

I believe that in QFT, in cases where you can construct the appropriate operators, there are commutation relations between them that more or less correspond to the non-relativistic ones (but with the caveats that have been discussed already in this thread). However, I don't know if the appropriate operators can be constructed in all cases in QFT.

 

[vanhees71]

In the Heisenberg's uncertainty principle

$\triangle x \triangle p \geqslant \frac{\hbar}{2}$

what happens when the uncertainty in position becomes very small is that the uncertainty in momentum becomes very large. But what happens when the spread of the uncertainty in momentum becomes greater then the speed of light? I suppose then that above formula is no longer valid in the relativistic limit?

The formula is valid in the relativistic limit, as far is it is applicable (i.e., for massive particles/quantum fields of any spin and for massless particles of spin 0 and spin 1/2, for which a position operator in the strict sense is definable). The "SRT speed limit" restricts the "localizability" of particles more stringently than in non-relativsitic QM.

The intuitive picture is simple: If you want to localize a particle in a very small region you need correspondingly strong fields confining it in this region. If the corresponding interaction becomes strong enough rather than confining the particle to ever smaller regions you create some new particles (pair creation).

For a more quantitative but still intuitive argument, see the introductory section(s) of Landau&Lifshitz vol. IV and the cited original papers therein.