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Suppose the case that someone in future makes a time machine which can be used to time travel in past. Now s/he can accurately measure momentum of a particle without caring about the particle's position at that time and then s/he can use the time machine to travel back in the past at the same time when s/he measured particle's momentum and now s/he can measure position of the same particle with higher accuracy? So does this mean that time travel to past is impossible or my question is nonsense?

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Contrary to popular belief, the HUP is not a principle about the accuracy of a measurement. The HUP is simply a statement that relates the spread of position measurements to the spread of momentum measurements of similarly prepared systems. It is a statistical principle about multiple measurements and their standard deviations; it is not a principle that applies to single measurements of one system.

Furthermore, this idea that the HUP means that "the particle has an exact position and momentum, but we just don't know what they are" is not correct. The uncertainties discussed in the HUP arise purely from the postulates of QM and have nothing to do with what we know about the particle or how accurately we measure its position or momentum. Many QM interpretations would even say that before measurement the particle doesn't even have a defined position or momentum at all.

Therefore, your premise is flawed purely from a misunderstanding of what the HUP actually says and how it applies to quantum systems.

Of course, it is an interesting thing to think about what would be the outcome of a quantum measurement if you were to travel back in time and repeat the measurement again in precisely the same manner. Unfortunately, I don't think anything like this can be experimentally tested (at least for now ;) ), so anything about this point would be pure speculation.

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  • $\begingroup$ But I want to ask if this statement can be used to prove that it would be impossible to travel in the past? $\endgroup$
    – anantagni
    Nov 13, 2020 at 15:30
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    $\begingroup$ @silver_surfer It isn't a true statement / understanding, so pursuing it any further seems pointless to me. It would be like saying "since gravity doesn't depend on mass, then can we really use gravity to determine the mass of something?" The premise is flawed, so the question doesn't make much sense. $\endgroup$ Nov 13, 2020 at 15:31
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    $\begingroup$ @BioPhysicist I think you are to quick to dismiss OP. The actual question is "is HUP compatible with time-travel?". This is a fair question, even if OP's interpretation of HUP is flawed. You give the correct interpretation, but you didn't really address the core question, i.e., given the correct interpretation, are those two concepts compatible? (Not that I care about that question myself; but it is, to some extent, a reasonable question) $\endgroup$ Nov 14, 2020 at 14:50
  • $\begingroup$ @AccidentalFourierTransform That is why I included my last paragraph. $\endgroup$ Nov 14, 2020 at 15:53
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    $\begingroup$ @BioPhysicist Sure, my point is that I don't think it is just "pure speculation". It is reasonable to ask e.g. how closed timelike curves are handled in the existing models of quantum gravity, in particular in light of the HUP. Again, I don't care about this question myself, I'm just trying to say that it is not 100% speculation: there are frameworks where the question is well defined and interesting. $\endgroup$ Nov 14, 2020 at 16:12
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No, the uncertainty principle is purely a property of the state of a system. Given two observables $A$ and $B$, you have a statement of the form

$$ \Delta A \Delta B\ge k$$

for some constant $k$. This tells you that if you are able to guess the outcome of an $A$ measurement with good accuracy ($\Delta A$ is small), then you wouldn't be able to guess the outcome of a $B$ measurement accurately ($\Delta B$ is large). If you measure $A$ and then you "go back in time" (by which I mean you revert the system to its original state $\psi$), you have no more information of what the outcome of either measurement will be, and the same uncertainty principle will apply. A new measurement of $A$ would yield a new value uncorrelated with the one you previously got, and nothing would change.

In other words, the question is: if I have a system in the state $\psi$ and I measure some observable $A$, then the state of the system becomes an eigenstate of $A$. I now apply some transformation to the system such that its state gets reverted back to $\psi$. Does my ability to guess the outcome of a measurement of an observable $B$ change? Clearly not.

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Please don't accept this answer, I just want to include it here for the next reader/asker of this question.

So @user2723984's answer gets to the heart of the matter mathematically and @biophysicist's answer gives you some intuitive understanding of what that means.

After reading the OP's comments it seem you still wanted to try to recover the thought experiment and make it work using time travel to create some kind of paradox.

The best I could find was we have a system $\psi$ we are observing and at $T$ we highly constraint the standard deviation of the momentum of the wave function $\psi$, then some time passes, and we go back in time and highly constrain the position instead, and declare "oh but at time T both the position and momentum were known!" but that's not very interesting since though the time is shared the timelines themselves are different. It looks to me (though I am not confident if this is true) whether or not you time travel, at each moment of time locally speaking the heisenberg uncertainty principle holds true. I.E. it will take more than just time travel to break it locally.

If someone thinks they can violate the heisenberg uncertainty principle assuming access to time travel i'd be curious to see it in the comments.

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    $\begingroup$ There is a flaw here. Your "constraining" in each situation makes new quantum states, so they aren't comparable with the HUP $\endgroup$ Nov 14, 2020 at 1:50
  • $\begingroup$ in my (downvoted) answer I sketched how you could use time travel to create an ensemble which violates H.U. $\endgroup$
    – lalala
    Jan 2, 2021 at 21:23
  • $\begingroup$ @lalala I’ve made a comment on your answer, it is addressed by the body of this answer I made as well. $\endgroup$ Jan 3, 2021 at 6:24
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The key point about the Hiesenberg uncertainty principle is that the measurement, itself, changes the quantum state of the system. So, if you went and collapsed the system onto a momentum state at time $t_0$, and then went back to some time $t_{-1} < t_0$ and tried to collapse the same system onto a position state, you would decohere the momentum of the particle, and the second experiment to measure the momentum of the particle would give a different answer.

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    $\begingroup$ "The key point about the Hiesenberg uncertainty principle is that the measurement, itself, changes the quantum state of the system." I disagree, the uncertainty principle is not about consecutive measurements, it is about how sure you can be of the outcome before the measurement, the post measurement state never enters the picture. $\endgroup$ Nov 13, 2020 at 15:37
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    $\begingroup$ @user2723984: the measurement itself is a projection operator from one basis set to another, though, right? You're projecting onto a different basis of the Hilbert space, where a different set of operators is well-defined. There's no way around that, the final state is important. If this wasn't true, the quantum zeno effect wouldn't work. $\endgroup$ Nov 13, 2020 at 15:40
  • $\begingroup$ Yes, but what does that have to do with the uncertainty principle? "In an eigenstate of $A$ I cannot be sure of a measurement outcome of $B$" is just a consequence of the uncertainty principle, which more generally says "If I can guess the outcome of a measurement of $A$ well, I can't do the same for a measurement of $B$". The connection with a post-measurement state can be done if you consider that after measuring $A$ the system is in an $A$ eigenstate, hence you can perfectly guess the outcome of an $A$ measurement, but the principle applies even to other states, even if you never measured. $\endgroup$ Nov 13, 2020 at 15:44
  • $\begingroup$ what I mean is that the fact that the measurement changes the state is not important for the HUP, because you can even verify it with many copies of a single system, without ever measuring any of them twice and throwing them away immediately after you measure them. $\endgroup$ Nov 13, 2020 at 15:50
  • $\begingroup$ @user2723984 but the uncertainty principle is just a statement that projectors onto different eigenstate bases don't commute -- if i measure with $X$, and then with $P$, it's different than measuring with $P$ and then with $X$. The OP is proposing to do exactly this. $\endgroup$ Nov 13, 2020 at 15:53
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That is a beautiful idea. And I believe it is true. If time travel were possible, you could measure (for each) particle observable A and remember which ones lead to what result. Then travel back and do the same for observable B. Then travel back again and you remove the particles from the ensemble based on your knowledge. This will allow you to create an ensemble which violates H.U.

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  • $\begingroup$ why the downvote? $\endgroup$
    – lalala
    Nov 16, 2020 at 8:38
  • $\begingroup$ Your example here isn’t really violating heisenberg’s uncertainty principle since when you travel back in time to get momentum, by definition, you are on A DIFFERENT timeline, one that didn’t measure position. At no point in time (which is now a loop) did both truly simultaneously happen. As a result even though one could argue “I went back in time and measured both at the same time coordinate” locally speaking, meaning looking at your measurements at the instant they were taken: heisenbergs uncertainty principle was respected. And it was respected at all stages of your time travel. $\endgroup$ Jan 3, 2021 at 6:22
  • $\begingroup$ for the record, i didn't downvote this, but i believe that might explain the source of some of the downvotes $\endgroup$ Jan 3, 2021 at 6:29
  • $\begingroup$ @frogeyedpeas thanks for the feedback. So it goes to what do we mean by timetravel, doesnt it.To make it clearer: Lets say I have a machine which counts radioactive decay (or anything QM), you are saying when I go back in time and then watch (dont touch) do I get the same result when waiting until present, or do I get different (because different timeline)? $\endgroup$
    – lalala
    Jan 3, 2021 at 13:16

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