If I have a two-body system (particles A and B) which I measure the total momentum of, can I measure the position of particle A very accurately and the momentum of particle B very accureately, and then use the momentum of particle B and the total (conserved) momentum to give me the momentum of particle A with uncertainty satisfying $\Delta x \Delta p < \hbar/2$, violating the uncertainty principle?

A similar trick can be played with any conserved quantity and it's conjugate variable. What step or assumption is incorrect? I think that maybe the difference between measuring the whole system and measuring constituent subsystems is maybe not as trivial as I have assumed.

Following on from @AaronStevens comment, I suppose the question may boil down to 'Does the wavefunction of a particle change if we obtain new information about it without measuring it directly? If not, why not?'.

I think a similar question has been asked here (Violation of the uncertainty principle) but it wasn't quite formulated properly. This is perhaps what that author meant by their question?

  • $\begingroup$ The $\Delta$ values are not the precision or accuracy in your measurements here. $\endgroup$ – Aaron Stevens Aug 1 '19 at 13:31
  • $\begingroup$ @AaronStevens Is the implication that the momentum wavefunction of particle A is unchanged by a measurement on particle B? How can that be the case if we have new information about A? $\endgroup$ – WuRuyi Aug 1 '19 at 13:35
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    $\begingroup$ You haven't obtained new information. Momentum is conserved on average. Measuring the momentum of one particle doesn't measure the momentum of the other particle. You're putting some classical thinking into your quantum thinking. $\endgroup$ – Aaron Stevens Aug 1 '19 at 13:51
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    $\begingroup$ Possible duplicate of Violation of the uncertainty principle The answer there is what you want. $\endgroup$ – Aaron Stevens Aug 1 '19 at 13:52
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    $\begingroup$ Yes, but not for one event, for an accumulation of the same boundary condition events.that is what probability distributions( coming from wavefunctions) mean. $\endgroup$ – anna v Aug 1 '19 at 16:06

Take the simplest possible example, where each particle has only two possible positions and two possible momenta. Write $M_1$ and $M_2$ for the eigenstates of the momentum operator. Write the state of the system as $$\sum_{i,j}\alpha_{ij}M_i\otimes M_j$$

Then if you make a momentum measurement on particle $A$, you'll get $M_1$ with probability $$p=\sum_j\alpha_{1j}^2/\sum\alpha_{ij}^2$$

On the other hand, you could first make a momentum measurement on particle $B$ and only then make a momentum measurement on particle A. The first measurement puts the system into one of the states $\sum\alpha_{i1}M_i\otimes M_1$ or $\sum_i\alpha_{i2}M_i\otimes M_2$ with probabilities $q_1,q_2$ proportional to $\sum_i\alpha_{i1}^2$ and $\sum_i\alpha_{i2}^2$. In each of these cases, you can compute the probability $r_i$ that your second measurement yields $M_1$. So the probability that your second measurement yields $M_1$ is $q_1r_1+q_2r_2$, which turns out (do the arithmetic!) to be exactly the same as $p$.

Bottom line: Measuring the second particle can't give you any information about the probability distribution of outcomes for a measurement on the second.

Your mistake: Assuming each particle had both a well-defined momentum and a well-defined position in the first place.


can I measure the position of particle A very accurately and the momentum of particle B very accureately, and then use the momentum of particle B and the total (conserved) momentum to give me the momentum of particle A

I might be reading this incorrectly but, it seems to me, the problem with this question is the assumption underlying the bolded section.

It's a little unclear precisely what you have in mind but, regardless of how you imagine how these measurements are made, there seems to be an assumption that the state of A just after these measurements is a state of (essentially) definite momentum and position.

But there is no such state that I am aware of. In particular, if A is in a state of essentially definite position, A simply doesn't have a well defined momentum to "give you".

If you had something else in mind, please clarify your question and I will amend my answer.

  • $\begingroup$ Sigh, I didn't see WillO's answer until after I posted. Since I took the time to write, I'll just leave it here. $\endgroup$ – Alfred Centauri Aug 1 '19 at 14:57
  • $\begingroup$ I think you are right that’s probably where my argument breaks down (it surely does break down somewhere) but I’m not really sure why. Surely given an exact measurement of the total momentum and of the momentum of B (which are both valid operations given we don’t care about position certainty), repeated measurements on A would consistently give the classically expected value? Is this last step not true? If it is untrue, why does it not violate conservation of momentum? $\endgroup$ – WuRuyi Aug 1 '19 at 14:58
  • $\begingroup$ Also it seems feasible that once we measure the total momentum and the momentum of B we should be able to establish the momentum of A; we have two particles and two measurements. I know there is a difference between classical values and quantum expectation values but the point still stands. $\endgroup$ – WuRuyi Aug 1 '19 at 14:59
  • $\begingroup$ Also, more answers are always better :) $\endgroup$ – WuRuyi Aug 1 '19 at 14:59
  • $\begingroup$ @WuRuyi, if A is in a state of definition position, and B is in a state of definite momentum, is it possible for the system state to be one of definite total momentum? $\endgroup$ – Alfred Centauri Aug 1 '19 at 15:17

There are two cases. 1. particles are entangled 2.particles are not entangled.

Let's consider the first case. For example assume that an electron and positron collide such that the total momentum before collision is zero, so we expect that after annihilation two created photon will move such that total momentum become zero again, because total momentum is conserved. That would mean if we measure one photon's momentum, we would automatically acquire the other's momentum too! However, because they are entangled, if we measure one, the wavefunction of the system will collapse and so the position of both of them will become unknown completely. Well, after all in these kind of systems you can't have two seprate wavefunctions there is only and only for wavefunction for entire system. You can say the same thing for pair creation

As for second case, there is no entanglement. Particles have independent wavefunctions. In this case, you can't define total momentum for the system, unless you measure each particle's momentum seprately. And you know, If you do that, you can't measure their position afterward, because it becomes uknown according to Heisenberg principle. In this case, their momentum are not related and there is no way to find one, from the other.

If you want, I can go into math details. But @WillO 's answer cover the second case completely.

  • $\begingroup$ So in the second case, are you saying there is no way of measuring the total momentum of the system to a high precision without measuring each individually ? $\endgroup$ – WuRuyi Aug 1 '19 at 16:02
  • $\begingroup$ @WuRuyi If you think there is a way, I would like to hear that! We can check to see if it's possible or not. Of course you can assign a probabilistic total momentum. But if you do that, how are you going to use conservation of momentum? $\endgroup$ – Paradoxy Aug 1 '19 at 16:05
  • $\begingroup$ There's no need to divide this into two cases. Case 2 is a special case of Case 1. $\endgroup$ – WillO Aug 1 '19 at 16:26
  • $\begingroup$ @WillO ,well yes. In second case you are able to divide wavefunction of entire system to two wavefunctions for particles, while in the first one it's not possible. I just wanted to point out that there is indeed a way to know A's momentum from B's momentum (in entangled particles) but it does not violate Heisenberg pricinple. $\endgroup$ – Paradoxy Aug 1 '19 at 16:31

This was a comment, but I'm promoting it to a second answer.

There is absolutely nothing in this question that requires two particles. You could ask exactly the same question about a single particle in an eigenstate of momentum. Measure that particle's position, which could be anything. Now that it's in an eigenstate of position, measure its momentum --- which could be anything.

This example has all the relevant features of your original question, with the distractions stripped away.

Now your question is: "If the momentum has been measured beforehand, how could it have changed?". The answer is: It could have changed if quantum mechanics is true.


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