4
$\begingroup$

In the case of a free particle, if you have perfect knowledge of its momentum, the uncertainty in its position becomes infinitely large.

Does this imply that there is a finite probability of finding the free particle anywhere in the entire universe?

$\endgroup$
2
  • 3
    $\begingroup$ Interestingly, I find the question itself problematic. It is true that in non-relativistic QM a wavefunction (in position representation) with a perfectly well-defined momentum is a planewave with "infinite wavepacket size," but when it comes to your question, what do you mean by "perfect knowledge?" Where does this perfect knowledge come from? Presumably, the knowledge would come from a prior measurement or a preparation of the quantum state of some sort. I'm not criticizing the question, it's just that I find this point interesting. $\endgroup$ Jun 23, 2023 at 17:00
  • $\begingroup$ Another point along the lines of the above comment: there is no state in your Hilbert space that has definite momentum. Momentum eigenstates are not normalizable. Put differently, suppose it had a finite probability of being in any open set in your position space. If that position space is not compact, you'll find that it has probability > 1 of being in some compact subset of position space, which seems problematic. $\endgroup$ Jun 25, 2023 at 3:03

3 Answers 3

11
$\begingroup$

It is impossible to perfectly know the momentum of a particle in quantum mechanics. This isn't just a practical limitation. Even theoretically, states with infinitely precise position or momentum cannot be included in the theory without causing problems. For example, in the situation you're asking about, perfect knowledge of momentum would require that the particle is equally likely to be found anywhere in the universe. However, assuming the universe is infinite, there is no way to assign a nonzero probably to every point in space while also requiring that all the probabilities add up to 100%. In the language of quantum mechanics, this state is not normalizable.

Having said that, in theoretical calculations, we often pretend that perfect momentum or position states are allowed. Usually, this is fine because we can get arbitrarily close to a perfect momentumn or position state (as long as there is still a little uncertainty). So for your example, it is true that, if you have an extremely precise measurement of momentum, then it might be true that there is a nonzero probably for the particle to be located across a vast portion of the universe. Good luck making such a measurement in real life though! Usually, if you're doing an experiment measuring a particle, you already know that the particle is located in your lab, so there is a limit on how precisely your equipment can measure momentum.

$\endgroup$
6
  • $\begingroup$ Note, "finite probability of being anywhere" != "finite probability anywhere of being" $\endgroup$ Jun 24, 2023 at 8:56
  • $\begingroup$ @OverLordGoldDragon how does the video showcase that statement? Is it about the sentence he made in the video about "something can be possible but has probability of 0"? $\endgroup$
    – justhalf
    Jun 25, 2023 at 5:33
  • $\begingroup$ @justhalf Yes, the odds of landing at any specific point on the dartboard are zero, but the dart lands somewhere. In simple (and incomplete) terms, it's since "probability" is defined as an average. $\endgroup$ Jun 25, 2023 at 5:47
  • $\begingroup$ @OverLordGoldDragon Yes. However, in this case, the probably density would have to be nonzero everywhere. It is impossible to have a constant, nonzero probability density everywhere while still having the total probably integrate to 1 $\endgroup$ Jun 25, 2023 at 11:44
  • $\begingroup$ Not on an infinite radius dartboard, which is incidentally your answer's premise. $\endgroup$ Jun 25, 2023 at 12:34
6
$\begingroup$

Having a precisely defined momentum means the wavefunction $\psi(\vec{r})$ is an eigenfunction of the momentum operator $-i\hbar\vec{\nabla}$. That means it is $$\psi(\vec{r})=Ae^{i\vec{k}\vec{r}}$$ and the eigenvalue $\vec{p}=\hbar\vec{k}$ is the measured momentum without any uncertainty.

From the wavefunction above you get the probability that the particle is within a volume element $d^3r$ at position $\vec{r}$ is $$|\psi(\vec{r})|^2 d^3r=|A|^2d^3r .$$ So the probability is indeed the same at every position $\vec{r}$, i.e. everywhere in the universe.

$\endgroup$
1
  • $\begingroup$ That doesn't answer the question whether it is finite. What is $A$? $\endgroup$
    – tobi_s
    Jun 25, 2023 at 11:24
5
$\begingroup$

Yes. Yes it does.

But quantum mechanics is inherently non-relativistic, so it's not really applicable for this question. (The math is fairly simple: Fourier transform of $\phi(p) \rightarrow \psi(x)$).

Now relativistic quantum field theory is another story. It has ways to keep everything in the forward light cone, but that's a different question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.