The Wavefunction when Mass is Much Greater than Planck's Constant

In summary, when it comes to measuring the position of a macroscopic object, the uncertainty is low because the object's center of mass is known with great accuracy. However, when it comes to measuring the positions of individual atoms, the uncertainty is high because there is a lot of room for variation.
  • #1
JohnH
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TL;DR Summary
What does the wavefunction look like if the position is known precisely, but there are so many particles involved such that the certainty in position is high since the mass is much greater than Planck's constant?
I understand that the uncertainty is low when you're dealing with a "macro" scale area that is much bigger than Planck's constant. But what's confusing to me is when you know with extreme precision the location, but there's so many particles involved that there is little uncertainty since the mass involved is so much greater than Planck's constant. For example, let's say there were a bowling ball that had its position measured with extreme accuracy. We wouldn't get hardly any uncertainty in either its position or momentum because its mass is so much greater than Planck's constant, but what does that mean in terms of the wavefunction? And you might say that the bowling ball takes up a lot of space such that my premise is flawed, but what if we throw the bowling ball into a black hole where it's crushed down to size?
 
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  • #2
JohnH said:
let's say there were a bowling ball that had its position measured with extreme accuracy.
Measuring the position of the bowling ball, a macroscopic object, is not the same as measuring the position of every single atom in it. The position of the bowling ball is basically the position of its center of mass. Even if that position is measured with extreme accuracy, it still leaves a lot of room for uncertainty in the positions of the individual atoms.

JohnH said:
what if we throw the bowling ball into a black hole where it's crushed down to size?
We can no longer observe the ball once it falls into the hole.
 
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  • #3
Okay, I was confused about how multi-particle systems worked, but with some research, I understand now. You can essentially treat each particle as if it were in its own dimensions of space.
 
  • #4
The difference is that with bigger objects you know where the object will be, or can calculate with excellent certainty. Not quite so with quantum scale objects.
As you implied, it's about size.
 
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  • #5
JohnH said:
Okay, I was confused about how multi-particle systems worked, but with some research, I understand now. You can essentially treat each particle as if it were in its own dimensions of space.
How would you measure the position of a bowling ball to within the radius of an atom?

Even if you could, if you do the maths, then the uncertainty in momentum is such that the uncertainty in velocity is small.

Try doing the maths.
 
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  • #6
Last but not least already the subject line of this thread doesn't make sense. Mass has the dimension mass (in SI units kg), while Planck's constant has the dimension of an action, i.e., length times momentum or energy times time (in SI units Js). So you cannot say "the mass is much larger than Planck's constant".
 
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  • #7
JohnH said:
You can essentially treat each particle as if it were in its own dimensions of space.
Of Hilbert space. Not ordinary space.
 
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  • #8
CoolMint said:
The difference is that with bigger objects you know where the object will be, or can calculate with excellent certainty. Not quite so with quantum scale objects.
As you implied, it's about size.
I was thinking about this the other day, and got confused again when I tried to think about this in terms of mass (on another forum I had heard mass mentioned in this context which is what got me confused in the first place), but I was thinking about this and came to the same conclusion. It's just about the shape of the wavefunction and the degree to which it approaches a Dirac delta, orthoganality taking care of the rest.

Thank you all for the replies.
 
  • #9
Before we go too far afield, everybody knows that Palnck's Constamnt is not a measure of mass, so saying that the mass is more than Planck's constant does not convey what you think. Is blue greater than a meter?
 
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  • #10
Vanadium 50 said:
Before we go too far afield, everybody knows that Palnck's Constamnt is not a measure of mass, so saying that the mass is more than Planck's constant does not convey what you think. Is blue greater than a meter
Yes, but mc^2=E and mc=p, etc. That's what I meant by mass.
 
  • #11
Well as long as you've explained it to your satisfaction...all is copacetic.
 
  • #12
JohnH said:
Yes, but mc^2=E and mc=p, etc. That's what I meant by mass.
Neither energy nor momentum can be compared with Planck's constant either, so this doesn't really help.
 
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  • #13
JohnH said:
It's just about the shape of the wavefunction and the degree to which it approaches a Dirac delta,
For macroscopic objects the overall wave function looks nothing at all like a Dirac delta. The wave function has a very, very small uncertainty in one degree of freedom, the center of mass position, but that doesn't make it a Dirac delta.

JohnH said:
orthoganality taking care of the rest.
Orthogonality of what?
 
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  • #14
Okay, so apparently I still don't quite get it. So first of all, some more context: I read a without-math description online of someone talking about the uncertainty principle and at first he gave the common example of rope which makes it clear enough how the uncertainty in position relates to the Hup. But then he gave the example of a bowling ball and if memory serves me correctly, he said that the bowling ball had momentum mc.
PeterDonis said:
Neither energy nor momentum can be compared with Planck's constant either, so this doesn't really help.
And so if we put that in for the uncertainty relation via delta mc multiplied by delta x is greater than or equal to Planck's constant, then seemingly because the bowling ball is much more massive, it has relatively little uncertainty and behaves in a macro way.

But for a long time I've been puzzling over what a more mathematical description of that looks like and when I read about the uncertainty principle it never mentioned how a mult-particle system might collectively have an uncertainty in position that is very low such that it essentially behaves in a way that adheres to classical mechanics. Before posting, I thought it had to do with the number of particles or quanta involved such that they collectively defied the uncertainty principle, but then I read
PeterDonis said:
Measuring the position of the bowling ball, a macroscopic object, is not the same as measuring the position of every single atom in it.
And so I thought that meant that each particle in a Hilbert space is definitionally orthogonal to every other such that the Hup essentially applied to each particle individually such that "it's about size" and not about the number of particles and not about the overall mass of all the particles in a multi-particle system.
PeterDonis said:
Orthogonality of what?
And that's what I meant by orthogonal.

PeterDonis said:
For macroscopic objects the overall wave function looks nothing at all like a Dirac delta. The wave function has a very, very small uncertainty in one degree of freedom, the center of mass position, but that doesn't make it a Dirac delta.
But now you seem to be saying, and correct me if I'm wrong (I am, of course, not "copacetic" about being wrong), that you can think about this in one degree of freedom such that you are thinking of all the particles collectively contributing to a very low uncertainty, something similar to what I thought in the first place. But I'm still confused as to what that looks like, and again, I didn't say it was a Dirac delta. I said
JohnH said:
the degree to which it approaches a Dirac delta

So I hope this gives you a better perspective on where I'm at, so that you can help eliminate my confusion.
 
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  • #15
JohnH said:
I read a without-math description online
Where? Please give a specific reference. We can't comment on something we can't read for ourselves.

JohnH said:
the uncertainty relation via delta mc multiplied by delta x is greater than or equal to Planck's constant
For a single quantum particle, yes, that is the uncertainty relation (although momentum would usually be written ##m v##, not ##m c##, since we're not talking about light). But we're not talking about a single quantum particle. We're talking about a macroscopic object with perhaps ##10^{25}## atoms. So you can't just wave your hands and say "position" and "momentum". The object doesn't have a single "position" or "momentum"; it has ##10^{25}## particles, each of which has their own position and momentum.

You can talk about the center of mass position and momentum, but that just means you have rearranged your description of the object slightly so you have 1 degree of freedom that you are explicitly dealing with (the center of mass), and ##10^{25} - 1## other degrees of freedom (the position/momentum of all but one of the object's particles, since the last one is fully determined by the center of mass plus all the others) that you ignore. You can write down an uncertainty relation for the center of mass, but that only applies to the center of mass degree of freedom; it doesn't mean you've pinned down the exact position and momentum of every particle in the ball to that level of accuracy.

JohnH said:
for a long time I've been puzzling over what a more mathematical description of that looks like
It looks like the classical equations for the center of mass position and momentum, derived under the assumption that whatever quantum uncertainty is involved with the ##10^{25} - 1## other degrees of freedom doesn't affect the classical behavior of the center of mass equations. The center of mass equations themselves are classical because, if you just look at the center of mass degree of freedom by itself, you can (at least heuristically) apply the uncertainty principle using the overall mass of the macroscopic object, so you can indeed show that the uncertainty in the center of mass position and momentum is so small that the center of mass behaves classically to a good enough approximation for all practical purposes.

Depending on what source you are reading, this approach might be simply asserted without argument, given a hand-wavy heuristic justification, or justified by something like the Ehrenfest theorem based on some reasonable attempt at writing down actual mathematical expressions for the center of mass position and momentum observables.

JohnH said:
I thought that meant that each particle in a Hilbert space is definitionally orthogonal to every other
This doesn't even make sense.

JohnH said:
such that the Hup essentially applied to each particle individually
It does. But you don't need to deal with this if all you're interested in is the center of mass behavior.

JohnH said:
you can think about this in one degree of freedom such that you are thinking of all the particles collectively contributing to a very low uncertainty
Sort of. See above for a better description.
 
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  • #16
@JohnH Even if we put QM to one side, there is a fundamental difference between the nature and behaviour of a bowling ball and the nature and behaviour of its constituent molecules. Take, for example, the kinetic theory of heat. The bowling ball may be static, yet all its molecules are moving in random directions. This randomness averages out. Even classically, therefore, there is classical uncertainty about what any molecule is doing. But that does not imply uncertainty about the motion of the ball.
 
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  • #17
PeterDonis said:
This doesn't even make sense.
Imprecise language. I mean the wavefunction for a multi-particle system has different arguments for each particle all of which are orthogonal to each other.

PeterDonis said:
It does. But you don't need to deal with this if all you're interested in is the center of mass behavior.
I found this paper: https://iopscience.iop.org/article/10.1088/1367-2630/aa719a. It seems to be saying more or less the same thing you guys are, PeterDonis and PeroK.
 
  • #18
PeterDonis said:
justified by something like the Ehrenfest theorem based on some reasonable attempt at writing down actual mathematical expressions for the center of mass position and momentum observables.
How successful is this approach?
 
  • #19
JohnH said:
But for a long time I've been puzzling over what a more mathematical description of that looks like and when I read about the uncertainty principle it never mentioned how a mult-particle system might collectively have an uncertainty in position that is very low such that it essentially behaves in a way that adheres to classical mechanics. Before posting, I thought it had to do with the number of particles or quanta involved such that they collectively defied the uncertainty principle, but then I read
... at least the center of mass got mentioned multiple times ... as did the uncertainty relation between position and momentum ...

PeterDonis said:
justified by something like the Ehrenfest theorem based on some reasonable attempt at writing down actual mathematical expressions for the center of mass position and momentum observables.
JohnH said:
How successful is this approach?
For "fully describable" model systems in non-relativistic QM, it just works. All you need to do is change the description from the one where the absolute global coordinate of every particle is given to the one where the absolute global coordinate of the center of mass is given, and the coordinates of every particle relative to that center of mass. One can check that this change of description (if done correctly) is a "canonical transformation". The notion of "canonical transformation" comes from Hamilton mechanics, and it is special in that it leaves the basic form of the description of the dynamics basically unchanged. One can prove that this is not just true in Hamilton mechanics, but also in quantum mechanics.

And this allows one to derive (in that specific context) the uncertainty relation between position and momentum for the case where the position is the center of mass, and the momentum is the total momentum.
 
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  • #20
JohnH said:
mean the wavefunction for a multi-particle system has different arguments for each particle
Yes.

JohnH said:
all of which are orthogonal to each other.
This still doesn't make sense.
 
  • #21
An ##N##-particle wave function has ##N## arguments ##\xi_j##, which mean the eigenvalues of a complete set of compatible single-particle observables, e.g., ##\xi_j=(\vec{x}_j,\sigma_j)##, where ##\vec{x}_j \in \mathbb{R}^3## are the position-vector components and ##\sigma_j \in \{-s_j,\ldots,s_j\}##, where ##s_j \in \{0,1/2,\ldots \}## are the spin quantum numbers of these particles.
 
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  • #22
vanhees71 said:
An ##N##-particle wave function has ##N## arguments ##\xi_j##, which mean the eigenvalues of a complete set of compatible single-particle observables, e.g., ##\xi_j=(\vec{x}_j,\sigma_j)##, where ##\vec{x}_j \in \mathbb{R}^3## are the position-vector components and ##\sigma_j \in \{-s_j,\ldots,s_j\}##, where ##s_j \in \{0,1/2,\ldots \}## are the spin quantum numbers of these particles.
Yeah, that's what I was trying to say in a hamfisted, less precise way.

The following excerpt from the linked paper, seems to sum things up nicely:
"The use of the center of mass for establishing the classicality of a quantum state has some promising advantages. The first one is related to the description of the initial conditions. Fixing the initial position and velocity of a classical particle seems unproblematic, while it is forbidden for a quantum particle due to the uncertainty principle [1, 14]. The use of the center of mass relaxes this contradiction: it is reasonable to expect that two experiments with the same preparation for the wave function will give quite similar values for the initial position and velocity of the center of mass when a large number of particles is considered, although the microscopic distribution of the positions and velocities for all (Bohmian) particles will be quite different in each experiment.

The second advantage is that it provides a natural coarse-grained definition of a classical trajectory that coexists with the underlying microscopic quantum reality. One can reasonably expect that the Bohmian trajectory of the center of mass of a large number of particles can follow a classical trajectory, without implying that each individual particle becomes classical. Therefore, the use of the center of mass allows a definition of the quantum-to-classical transition, while keeping a pure quantum behavior for each individual particle."
 
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FAQ: The Wavefunction when Mass is Much Greater than Planck's Constant

1. What is the significance of the wavefunction when mass is much greater than Planck's constant?

The wavefunction is a mathematical representation of the quantum state of a particle. When the mass of a particle is much greater than Planck's constant, it means that the particle is in a macroscopic state and its behavior is best described by classical mechanics rather than quantum mechanics.

2. How does the wavefunction change when mass is much greater than Planck's constant?

When the mass is much greater than Planck's constant, the wavefunction becomes highly localized, meaning that the particle's position is well-defined and its momentum is uncertain. This is in contrast to the wave-like behavior of particles with smaller masses, where the wavefunction is more spread out and the position and momentum are both uncertain.

3. What is the relationship between the wavefunction and Planck's constant?

Planck's constant is a fundamental constant in quantum mechanics that relates the energy of a particle to its frequency. The wavefunction is a mathematical function that describes the probability of finding a particle in a certain state. When the mass is much greater than Planck's constant, the wavefunction becomes negligible and classical mechanics is a more accurate description of the particle's behavior.

4. Can the wavefunction still be used to describe particles with large masses?

Yes, the wavefunction can still be used to describe particles with large masses, but its effects become negligible and the particle's behavior is better described by classical mechanics. This is because the wavefunction becomes highly localized and the particle's position and momentum can be accurately determined using classical equations.

5. How does the wavefunction behave in the limit of infinite mass?

In the limit of infinite mass, the wavefunction becomes completely localized, meaning that the particle's position is known with absolute certainty. This is because the particle's momentum becomes negligible and it behaves like a classical object. In this limit, the wavefunction is no longer a useful tool for describing the particle's behavior and classical mechanics is the best approach.

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