The uncertainty principle and two-particle collisions

It is not possible to simultaneously know the exact position and momentum of a particle as a consequence of non-commutativity of the position and momentum operators. But what if I consider a simple two-particle collision, where I make a simultaneous measurement of the position of one particle and momentum of the other particle?

Suppose I first measure the momenta of both particles and find particle A to be at rest and particle B moving with some specified initial momentum. If I later again measure the momentum of particle A, which used to be at rest, and find it now to move with a certain momentum, I would know that it had scattered with the other, initially moving particle. Because of conservation of momentum, I now know the exact momentum of particle B. If I hence simultaneously measure the position of particle B when I am measuring the momentum of particle A, should this give me a complete description of particle B in violation of the uncertainty principle?

You are thinking way too classically.

Suppose I first measure the momenta of both particles and find particle A to be at rest and particle B moving with some specified initial momentum.

This isn't possible. States with definite momentum are not physical. But let's say your measurements put particle A into a state with really low $$\Delta p_A$$ and particle $$B$$ into a state with really low $$\Delta p_B$$. Then by the HUP $$\Delta x_A$$ and $$\Delta x_B$$ will be fairly large.

If I later again measure the momentum of particle A, which used to be at rest, and find it now to move with a certain momentum, I would know that it had scattered with the other, initially moving particle. Because of conservation of momentum, I now know the exact momentum of particle B.

Not necessarily. Remember, quantum measurements essentially sample a probability distribution. Since each particle has some non-zero $$\Delta p$$, just because we measured some value $$p_0$$ the first time does not mean we will measure the same momentum next time, even if no interaction had occurred.

If I hence simultaneously measure the position of particle B when I am measuring the momentum of particle A, should this give me a complete description of particle B in violation of the uncertainty principle?

Once you have measured the position of particle B you have changed the state of particle B, and this state has high $$\Delta p$$. So you have changed the system. You aren't keeping track of what you think you are keeping track of anymore here.

You can't rig the game here. No states exist where both $$\Delta x=0$$ and $$\Delta p=0$$. It's not like they exist but nature somehow prevents us from accessing them unless we find a way to rig the system. They just don't exist. Period.

• Should it not be possible to setup an experiment with sufficiently large momenta and sufficiently small uncertainties to say that this particle is effectively at rest in comparison to the other, or that it had almost certainly scattered in-between the two measurements (due to a sudden large increase in $p_A$)? Putting aside states with definite momentum or position, should I not be able to at least get a more precise description of one of the particles than permitted by the uncertainty principle? – Daphne Aug 13 at 13:02
• @Daphne No. $\Delta x\Delta p\geq\hbar/2$ is always true. No state exists where this is not true. You seem to be forgetting that measuring one value changes the entire state of the system. It's not like you can nail down a really precise position and then keep that while you measure a really precise momentum. Measurement changes the state. – BioPhysicist Aug 13 at 13:04