Heisenberg Uncertainty Principle to estimate the position uncertainty

Main Question or Discussion Point

A block is at rest. The block is macroscopic so we can see that it is at rest and where it is, i.e. we know its position with certainty. Suppose we use the Heisenberg Uncertainty Principle to estimate the position uncertainty Δ x. Consider the following argument:
Since the engine is at rest, its momentum p is zero. Moreover, since Δ p = 0, the uncertainty in its position is infinite. Therefore, the position of the locomotive is completely unknown.
Resolve the supposed contradiction.
 

Answers and Replies

again the question is the the measurment tool you have used to know the macroscopic block is at rest means P=0 v=0 and you know it under uncertainity principle we can not know at the same moment where is the block and it is momentum and under your argument you have measured the momentum af the particle at that moment by what and if you subtitute the value of the p in the equation you will get the value as unknown and i think the solution is that it is only applicable to the macroscopic world .
 
dst
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I think [tex]\frac{\hbar}{2\pi}[/tex] is quite a tiny quantity. On several orders of magnitude smaller than your locomotive - [tex]1.67840263\times10^{-35} m^2 kgs^{-1}[/tex] according to google. This is hardly worth considering...
 
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You're looking at a classical system; it does not make sense to say the block is at rest with complete certainty.
Basically, because the equipment you used to make the measurement that the block is stationary is either
a) Also classical, and has rather low precision. Think about how small [tex]\hbar[/tex]/2 is. How is your precision related to [tex]\Delta[/tex]p?
b) In the quantum regime, in which case you would certainly see that the block is not stationary; there are electrons and phonons jiggling all over the place! [tex]\Delta[/tex]p is then nonzero.
 
f95toli
Science Advisor
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The block is not at rest, it does not matter if it is macroscopic or not: QM still applies.
Also, note that this is NOT just a thought experiment.
Bar detectors that are used in experiments where they try to detect gravitational waves weigh hundreds of kg or more (they are essentially just big metal bars) and as far as I understand they actually need to be treated as quantum oscillators; i.e. the displacement due to quantum effects is actually larger than the displacement that would be causes by a gravitational wave. Hence QM effects needs to be taken into account.
 
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Since the engine is at rest, its momentum p is zero. Moreover, since Δ p = 0, the uncertainty in its position is infinite.
To echo what cshimmin said, neither the momentum nor the position of any realworld object can ever be known precisely. In other words, things that look at rest will actually have a range or cloud of velocities hovering around zero (about half being very slightly greater than zero and about half being very slightly less than zero).

Also, don't confuse p with Δ p. If p=0, this simply tells us that the mean value of the momentum within the cloud of momentums is zero - this doesn't mean that the range i.e. the Δ of momentums of the cloud is zero.
 
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The reason it's not a paradox is because the uncertainty principle talks about _knowledge_, not actuality (in Copenhagen Interpretation the two are the same). So you cannot _know_ either the position or momentum (or both) with absolute certainty. You can only know it within a range of values, which you do by looking at the macroscopic object with your eyes. But if you started to look very very closely at it you would appear to see wiggling of the molecules.

And another reaosn there is no paradox is that the HUP deals with individual particles, not macroscopic systems. The reaosn macroscopic systems don't implicate the HUP is because on average, all of the quantum jittering cancels out and you're left with what appears to be a stationary object.

The more interesting paradox is that in the rest frame of an object, IT has zero momentum. Does that mean that from the particle's perspective it can be anywhere in the universe?
 
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The more interesting paradox is that in the rest frame of an object, IT has zero momentum. Does that mean that from the particle's perspective it can be anywhere in the universe?
Actually not so much a pardox as a restatement of the same problem.
Still being unable to define that referance frame origin (0,0) wrt any other frame without the same uncertainty about position and momentum as measured from that other frame.
It is still a small uncertainty, nothing to hint that it could be “anywhere” in that other frame or the universe.
 
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Ah, yes of course, so from the particle's persepctive the entire universe is jittering a little bit
 
jtbell
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I would say that the uncertainty principle talks about our ability to predict the results of measuring position and momentum (or other conjugate pairs of variables).

If a particle has a specified Schrödinger wavefunction [itex]\Psi(x,t)[/itex], then we can predict, for a given time [itex]t[/itex], the range [itex]\langle x \rangle \pm \Delta x[/itex] within which we have an approximately 2/3 probability of measuring the value of [itex]x[/itex], and the range [itex]\langle p \rangle \pm \Delta p[/itex] within which we have an approximately 2/3 probability of measuring the value of [itex]p[/itex]. The uncertainty principle tells us that [itex]\Delta x[/itex] and [itex]\Delta p[/itex] are related by

[tex]\Delta x \Delta p \ge \hbar / 2[/tex]

That's all that the uncertainty principle tells us, strictly speaking. Statements about the particle "jittering" within those ranges of position and momentum values are an attempt to interpret the predictions of the uncertainty principle, and should not be confused with the uncertainty principle itself.
 
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[tex]\Delta x \Delta p \ge \hbar / 2[/tex]

That's all that the uncertainty principle tells us, strictly speaking. Statements about the particle "jittering" within those ranges of position and momentum values are an attempt to interpret the predictions of the uncertainty principle, and should not be confused with the uncertainty principle itself.
Good point;
Statements treating "quantum foam" as being a real thing are in the same category.
Popular interpretation of a way understand HUP & microscopic realty, rather than a known fact.
 
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The reason it's not a paradox is because the uncertainty principle talks about _knowledge_, not actuality (in Copenhagen Interpretation the two are the same). So you cannot _know_ either the position or momentum (or both) with absolute certainty. You can only know it within a range of values, which you do by looking at the macroscopic object with your eyes. But if you started to look very very closely at it you would appear to see wiggling of the molecules.

And another reaosn there is no paradox is that the HUP deals with individual particles, not macroscopic systems. The reaosn macroscopic systems don't implicate the HUP is because on average, all of the quantum jittering cancels out and you're left with what appears to be a stationary object.

The more interesting paradox is that in the rest frame of an object, IT has zero momentum. Does that mean that from the particle's perspective it can be anywhere in the universe?
absolutely...if del p is zero then del x would be infinity and that object can exists anywhere in universe..but the thing is there can be no del p = 0 because there is no absolute rest...everything is relative...thats why we couldnt find ether which is said to be at absolute rest and so it should be existing everywhere in the universe but we couldnt find ether...so there is no absolute rest ...
 

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