B Molecular uncertainty

  • Thread starter name123
  • Start date
444
3
Apologies in advance if my understandings are simply incorrect.

As I understand it, there is a limit to what can be known about both a particle's position and momentum, and in some interpretations this is because there is no position or momentum until measurement only a probability. What I am not clear on is how big the particle can be with such an interpretation, can it include a molecule. For example as I understand it, even at room temperature water molecules can evaporate if they are near the surface if they gain enough energy, and that they can gain this energy through collisions. But then for the evaporated molecule to be measured, would it not be thought to have of collided at a certain position at a certain momentum with another water molecule prior to the measurement? Thus be thought to have had a certain position and momentum (at the point of collision) prior to measurement. Or is it just assigned a probability of evaporating, or perhaps there being no "it" to assign it to, but there just being a probability of measuring such an effect given the probabilities within the system? Sorry if I am not phrasing this correctly. If not I would appreciate any help on how to best phrase it (assuming the respondent can understand the questions).
 
24,312
5,990
What I am not clear on is how big the particle can be with such an interpretation, can it include a molecule
Strictly speaking, the uncertainty principle should apply to everything, regardless of its size. However, since the uncertainty is in position and momentum, the mass of the object makes a difference; for an object with very large mass, the uncertainty in position and velocity can be very small and still meet the uncertainty principle requirement, because momentum is mass times velocity, so the product of uncertainties ##\Delta x \Delta p## becomes ##\Delta x m \Delta v##, and a big ##m## allows ##\Delta x## and ##\Delta v## to both be very small.

A water molecule's mass is not very large in these terms, so there might still be a significant uncertainty in position and velocity. But it's good to keep the above in mind.

for the evaporated molecule to be measured, would it not be thought to have of collided at a certain position at a certain momentum with another water molecule prior to the measurement?
No. Why would this be necessary? Knowing that the molecule evaporated doesn't require knowing exactly where it evaporated or how fast it was moving when it evaporated.
 
444
3
for the evaporated molecule to be measured, would it not be thought to have of collided at a certain position at a certain momentum with another water molecule prior to the measurement?
No. Why would this be necessary? Knowing that the molecule evaporated doesn't require knowing exactly where it evaporated or how fast it was moving when it evaporated.
I wasn't wondering whether knowing that the molecule evaporated required knowing exactly its position or its momentum. What I was wondering was in the interpretation where it doesn't have an exact position or momentum until measured (the wave packet collapsing), is it thought not to have collided in order to get the energy to evaporate, or is it perhaps thought to have collided but not at a specific position with a specific momentum? I wasn't thinking that it was supposed to be the case that the values were actual but just hidden from our knowledge.
 
Last edited:
24,312
5,990
in the interpretation where it doesn't have an exact position or momentum until measured (the wave packet collapsing), is it thought not to have collided in order to get the energy to evaporate, or is it perhaps thought to have collided but not at a specific position with a specific momentum?
Since we aren't measuring the individual water molecules, we can't tell how they are exchanging energy. Saying they "collided" doesn't help because they aren't little billiard balls. They interact, and their interactions exchange energy between them, but we aren't measuring the individual interactions so we can't say anthing about exactly what happens during them. All we can say is that the molecules interact and exchange energy, and as a result some molecules end up escaping from the liquid to the gas (and also some molecules end up being captured from the gas into the liquid).
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
Thus be thought to have had a certain position and momentum (at the point of collision) prior to measurement
Approximate position and momentum is enough, and both are guaranteed with a high probability.
 
444
3
Since we aren't measuring the individual water molecules, we can't tell how they are exchanging energy. Saying they "collided" doesn't help because they aren't little billiard balls. They interact, and their interactions exchange energy between them, but we aren't measuring the individual interactions so we can't say anthing about exactly what happens during them. All we can say is that the molecules interact and exchange energy, and as a result some molecules end up escaping from the liquid to the gas (and also some molecules end up being captured from the gas into the liquid).
Ok so if I substitute the word "collided" for the word "interacted" would you be saying that when prior to the evaporated molecule's evaporation, at any point in time during the evaporated molecule's prior interaction with other molecules, it would have had a specific position and momentum, even if we don't know what those values would have been. Or would you be saying that at no point in time during that molecule's interactions with other molecules did it have a specific position or momentum. (I am wondering what is being said with the wave collapse on measurement type interpretations).
 
444
3
Thus be thought to have had a certain position and momentum (at the point of collision) prior to measurement.
Approximate position and momentum is enough, and both are guaranteed with a high probability.
Are you saying it would be thought to have had an approximate position and momentum prior to measurement? If so, I am not certain I understand what that means. I assume it means that during the interactions with other molecules that gave it the energy required to evaporate it was not thought to have ever existed at a specific position with a specific momentum.
 
24,312
5,990
if I substitute the word "collided" for the word "interacted" would you be saying that when prior to the evaporated molecule's evaporation, at any point in time during the evaporated molecule's prior interaction with other molecules, it would have had a specific position and momentum, even if we don't know what those values would have been
No, because molecules are not little billiard balls. They're quantum objects. If you're not measuring the position or momentum of a quantum object, you can't say it has a definite position or momentum.

At least, that's how the minimal interpretation of QM works; i.e., that's how QM is used in practice. Other particular interpretations might make more claims about quantum objects having definite values. But any such claims will be interpretation dependent.

I am wondering what is being said with the wave collapse on measurement type interpretations
What does measurement have to do with it? When was the last time you measured a single water molecule evaporating?

If you want to set up a particular scenario in which a physicist actually does make such a measurement, then you need to specify the scenario--how the measurement is made. Outside of a specific measurement scenario, your questions are not well-defined and can't be answered.
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
Are you saying it would be thought to have had an approximate position and momentum prior to measurement? If so, I am not certain I understand what that means. I assume it means that during the interactions with other molecules that gave it the energy required to evaporate it was not thought to have ever existed at a specific position with a specific momentum.
You may think of the particles as being little moving clouds, with rough position and momentum but not specific position and specific momentum. This is also how particles are observed - never with exact postion or momentum!
 
444
3
No, because molecules are not little billiard balls. They're quantum objects. If you're not measuring the position or momentum of a quantum object, you can't say it has a definite position or momentum.
So during the interactions with the other molecules that gave it the energy required to evaporate it was not thought to have ever existed at a specific position with a specific momentum because it was not a little billiard ball but was a quantum object?
 
Last edited:
444
3
You may think of the particles as being little moving clouds, with rough position and momentum but not specific position and specific momentum. This is also how particles are observed - never with exact postion or momentum!
In the double slit experiment, aren't the dots on the screen supposed to indicate the position (presumably at the centre of the dot)?
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
In the double slit experiment, aren't the dots on the screen supposed to indicate the position (presumably at the centre of the dot)?
These dots are big and provide a fuzzy location only. Their center has no special meaning.
 

PeroK

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
9,072
3,278
So during the interactions with the other molecules that gave it the energy required to evaporate it was not thought to have ever existed at a specific position with a specific momentum because it was not a little billiard ball but was a quantum object?
It's not easy in general to apply QM to systems like the evaporation of water, for example. The water is a system with an very large number of molecules and each molecule is a system of a number of atoms and their constituent particles.

In general, you need to look at how a small number of quantum objects behave when they interact and then apply statistical techniques to simulate the behaviour of a large number of particles.

In this case, therefore, the real issue is to understand the nature of a quantum collision, as opposed to a classical collision. What you are struggling to understand is how two particles can "collide" if they don't have well-defined classical trajectories. That's really the heart of the matter here.

The short answer is that when you model particles according to the laws of QM then you get interactions similar in nature to classical collisions. The simplest example is, perhaps, a particle interacting with a barrier.

A classical example could be a ball rolling towards a hill of a certain height. The particle has definite position and velocity at all times and if it has sufficient energy it passes over the hill every time and if it has insufficient energy it rolls part way up the hill, stops and rolls back down and back the way it came.

The QM model is different. The hill could be a potential barrier and the particle can have a fairly precise initial energy. If you look for the particle at any time there is now only a probability of where you will find it and a probability of the velocity your will find.

However, those probabilities change with time. To begin with the particle would almost certainly be found moving towards the barrier; and after a certain time, the particle would almost certainly be found moving away from the barrier. (QM being QM, there is also a probability that the particle will be found on the far side of the barrier, having "tunnelled" through it. Although that's not the issue here.)

The point is that something similar to a classical collision with the barrier has happened: the particle usually bounces off the barrier, in some sense. But, it does this without a well-defined classical trajectory. The particle never was anywhere definitely until you eventually looked for it.

If you ask whether I (or anyone) can genuinely intuitively imagine what's going on, then I don't know. With QM you have to develop a different understanding based on the underpinning mathematics, which predicts this behaviour.

Your example of water evaporation is going to be significantly more complicated, although ultimately it is a statistical combination of a large number of quantum collisions that do not require well-defined trajectories.
 
444
3
It's not easy in general to apply QM to systems like the evaporation of water, for example. The water is a system with an very large number of molecules and each molecule is a system of a number of atoms and their constituent particles.

In general, you need to look at how a small number of quantum objects behave when they interact and then apply statistical techniques to simulate the behaviour of a large number of particles.

In this case, therefore, the real issue is to understand the nature of a quantum collision, as opposed to a classical collision. What you are struggling to understand is how two particles can "collide" if they don't have well-defined classical trajectories. That's really the heart of the matter here.

The short answer is that when you model particles according to the laws of QM then you get interactions similar in nature to classical collisions. The simplest example is, perhaps, a particle interacting with a barrier.

A classical example could be a ball rolling towards a hill of a certain height. The particle has definite position and velocity at all times and if it has sufficient energy it passes over the hill every time and if it has insufficient energy it rolls part way up the hill, stops and rolls back down and back the way it came.

The QM model is different. The hill could be a potential barrier and the particle can have a fairly precise initial energy. If you look for the particle at any time there is now only a probability of where you will find it and a probability of the velocity your will find.

However, those probabilities change with time. To begin with the particle would almost certainly be found moving towards the barrier; and after a certain time, the particle would almost certainly be found moving away from the barrier. (QM being QM, there is also a probability that the particle will be found on the far side of the barrier, having "tunnelled" through it. Although that's not the issue here.)

The point is that something similar to a classical collision with the barrier has happened: the particle usually bounces off the barrier, in some sense. But, it does this without a well-defined classical trajectory. The particle never was anywhere definitely until you eventually looked for it.

If you ask whether I (or anyone) can genuinely intuitively imagine what's going on, then I don't know. With QM you have to develop a different understanding based on the underpinning mathematics, which predicts this behaviour.

Your example of water evaporation is going to be significantly more complicated, although ultimately it is a statistical combination of a large number of quantum collisions that do not require well-defined trajectories.
So with the QM barrier type example when the particle is interacts with the barrier and bounces off, it isn't thought to have ever existed at a specific position with a specific momentum during the interaction because it was not like a classical particle but was a quantum object.

With the water example, is there a probability (however ridiculously small) that all the water molecules could evaporate at the same time, or is there an energy constraint on such an event happening?
 

PeroK

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
9,072
3,278
So with the QM barrier type example when the particle is interacts with the barrier and bounces off, it isn't thought to have ever existed at a specific position with a specific momentum during the interaction because it was not like a classical particle but was a quantum object.
Yes, exactly. The way the interaction is mathematically described is in terms of an evolving wave function. The wave function does not say where a particle is or what its velocity is at any time. It only says what the probability is of finding a particle in any position or with any velocity at a given time.

With the water example, is there a probability (however ridiculously small) that all the water molecules could evaporate at the same time, or is there an energy constraint on such an event happening?
That question is not unique to QM. The classical model of evaporation is statistical as well. In both models there is no fixed time for a certain amount of water to evaporate. But, the statistics of large numbers being what it is, you'll find that there is almost no variation. If the average is 5 minutes, say, for a certain amount of evaporation, then even having this happen (under the same conditions) in 4 minutes is next to impossible.

It's interesting you mention the energy constraint, because I think that adds an additional element of impossibility to instantaneous evaporation. In any case, it's a question of no importance IMO.
 
444
3
With the water example, is there a probability (however ridiculously small) that all the water molecules could evaporate at the same time, or is there an energy constraint on such an event happening?

That question is not unique to QM. The classical model of evaporation is statistical as well. In both models there is no fixed time for a certain amount of water to evaporate. But, the statistics of large numbers being what it is, you'll find that there is almost no variation. If the average is 5 minutes, say, for a certain amount of evaporation, then even having this happen (under the same conditions) in 4 minutes is next to impossible.

It's interesting you mention the energy constraint, because I think that adds an additional element of impossibility to instantaneous evaporation. In any case, it's a question of no importance IMO.
As I understand it in the classical model but the molecules are thought to have a specific momentum and position, and energy, and to exchange energy upon collision. The statistical model being simply used for pragmatic modelling. There though presumably the water molecules could not all evaporate as while they may exchange energy during collisions that amounts to a redistribution of energy, but for them all to evaporate would require an increase in energy.

You mentioned that you don't find the question interesting with the quantum model, and you did not answer it. I personally would be interested in reading the answer because if the answer is "yes" then it would seem that the energy of the system of the water in the bowl would not be conserved in such a situation. I am not interested in how rare such a situation would be.
 

PeroK

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
9,072
3,278
You mentioned that you don't find the question interesting with the quantum model, and you did not answer it. I personally would be interested in reading the answer because if the answer is "yes" then it would seem that the energy of the system of the water in the bowl would not be conserved in such a situation. I am not interested in how rare such a situation would be.
For example, statistically everyone in the UK could die tomorrow. Is that something to debate or think about? Is it even relevant to anything?

Also, these questions are always (IMO) the wrong way round. You say IF water could evaporate instantaneously, THEN energy is not conserved in QM. But that's not right. Energy is conserved in QM. Whether a rare event can happen has nothing to do with that. It just means that the system must have - by a statistical anomaly - got a lot of energy from somewhere else. It is the same in the classical case. The water is not a closed system and exchanges energy with its environment.
 
444
3
Also, these questions are always (IMO) the wrong way round. You say IF water could evaporate instantaneously, THEN energy is not conserved in QM. But that's not right. Energy is conserved in QM. Whether a rare event can happen has nothing to do with that. It just means that the system must have - by a statistical anomaly - got a lot of energy from somewhere else. It is the same in the classical case. The water is not a closed system and exchanges energy with its environment.
Ok, thank you. So if there was a closed system of water in a room, then if the water all evaporated, then that would provide you with the information that the rest of the room had lost that energy. There is 0 chance that all the molecules in the room could gain energy for example.
 

Want to reply to this thread?

"Molecular uncertainty" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top