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

[JohnH]

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?

 

[PeterDonis]

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.

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.

 

[JohnH]

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.

 

[CoolMint]

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.

 

[PeroK]

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.

 

[vanhees71]

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".

 

[PeterDonis]

You can essentially treat each particle as if it were in its own dimensions of space.

Of Hilbert space. Not ordinary space.

 

[JohnH]

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.

 

[Vanadium 50]

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?

 

[JohnH]

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.

 

[hutchphd]

Well as long as you've explained it to your satisfaction...all is copacetic.

 

[PeterDonis]

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.

 

[PeterDonis]

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.

orthoganality taking care of the rest.

Orthogonality of what?

 

[JohnH]

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.

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

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.

Orthogonality of what?

And that's what I meant by orthogonal.

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

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.

 

[PeterDonis]

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.

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.

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.

I thought that meant that each particle in a Hilbert space is definitionally orthogonal to every other

This doesn't even make sense.

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.

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.

 

[PeroK]

@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.

 

[JohnH]

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.

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.

 

[JohnH]

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?

 

[gentzen]

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 ...

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?

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.

 

[PeterDonis]

mean the wavefunction for a multi-particle system has different arguments for each particle

Yes.

all of which are orthogonal to each other.

This still doesn't make sense.

 

[vanhees71]

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.

 

[JohnH]

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."