# Do bosons contradict basic probability laws?

• A
• Philip Koeck
In summary, the conversation discusses the concept of mutually exclusive events and how they relate to probability theory. It is explained that for distinguishable particles, the probabilities of different distributions are mutually exclusive and add up to the total probability. However, for bosons which are indistinguishable, the probabilities of different distributions do not add up in the same way. This is because the probability distribution depends on the state in which the particles are prepared, and different states can have different probabilities for the same event. The conversation also touches on the idea that what can be measured in an experiment can affect the probabilities observed.
Philip Koeck
TL;DR Summary
The probabilities of mutually exclusive events are additive. For bosons this does not seem to be the case. How can we explain this?
This questions was brought to my attention by Kazu Okayasu.

According to probability theory the probabilities of mutually exclusive events add upp.
As an example we can distribute 2 balls in two boxes with two compartments each.
So there's a box on the left with a lower and an upper compartment and there's an identical box on the right.
The probability that both balls are in the box on the left is 1/4.
If we now also investigate in which compartment each ball is, we find that there are 4 different ways to place the 2 balls in the two compartments of the box on the left, each with a probability of 1/16.
This is in agreement with probability theory: All distributions are mutually exclusive and indeed the 4 probabilities 1/16 add up to 1/4.

For bosons the situation is different since they are indistinguishable.
The probability of finding 2 bosons in the box on the left is 1/3, but there are 3 different ways to place two bosons in the two compartments of the box on the left, each with a probability of 1/10.
These probabilities add up to 3/10 and not 1/3.

How can we understand this?
Does it mean that different distributions of bosons in compartments are not mutually exclusive?
That would mean that there is some sort of overlap between the event of 2 bosons being in the upper compartment and the event of 1 being in the upper and 1 in the lower, somehow both distributions can exist at the same time.

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Delta2
Philip Koeck said:
For bosons the situation is different since they are indistinguishable.
The probability of finding 2 bosons in the box on the left is 1/3, but there are 3 different ways to place two bosons in the two compartments of the box on the left, each with a probability of 1/10.
These probabilities add up to 3/10 and not 1/3.
I am not very strong in QM, but my understanding is if you can measure both the box and the compartment then the probability of both being in the left box is 3/10, not 1/3.

Wouldn't that mean that the probability I get depends on what I chose to measure?
If I just consider which box the particles are in I get a probability of 1/3 for both bosons in the left hand box. If I also consider compartments the probability suddenly reduces to 3/10. The power of the mind? Maybe that's how nature behaves, but it's definitely weird.

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PeroK
Philip Koeck said:
Wouldn't that mean that the probability I get depends on what I chose to measure?
I said it depends on what you can measure, not what you choose to measure.

I believe that if your experiment is set up so that you can measure both box and partition then the probabilities are different than if your experiment is set up so that you can only measure the box.

Philip Koeck said:
The power of the mind?
No. Different experiments have different probabilities. My (possibly flawed) understanding is that in a two-slit experiment if you can measure only location on the screen you have one probability distribution while if you can measure both location and slit then you have a different distribution, even if you choose not to use the slit information. The probabilities depend on your experimental setup and what you can measure with it.

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Philip Koeck
I might be confused by your setup, but here is my take.

The main problem seems to be that you just assume that every option is equally likely, no matter how those options are chosen. This is a bit like saying that I have a 50% chance of winning the lottery, because after all, there are only two options, either my numbers match or they don’t.

Philip Koeck said:
The probability that both balls are in the box on the left is 1/4.
As stated, this is simply wrong, at least if it is supposed to be a conclusion from what you wrote beforehand. Clearly, if I just have the two boxes in front of me, I could just put the balls in the left box and they will be there with probability one. So without stating how you are choosing to distribute the balls, there is no statement about how likely it is that both are on the same side. Here, apparently the choice is to distribute them such that each of the four distinct ways the distinguishable balls can be distributed is equally likely.

Philip Koeck said:
The probability of finding 2 bosons in the box on the left is 1/3
Again, this is supposed to be a choice I assume? There are now three ways how the two balls could be distributed, and you choose to distribute them in a way that each of them is equally likely.

Philip Koeck said:
3 different ways to place two bosons in the two compartments of the box on the left, each with a probability of 1/10.
And again, this probability of ##1/10## is a choice. With four compartments, there are 10 ways to distribute two indistinguishable balls, but the fact that we prepare a state where each of those 10 ways is equally likely is a choice.

Now out of the 10 options to distribute the ball among the four compartments, four have them on opposite sides, only three have them both in the left or right box, respectively. So the “state” where each of the 10 ways to distribute them among the four compartments is equally likely is a different state then the one where each of the 3 ways to distribute them among the two boxes is equally likely.
In the distinguishable case, it just happened that your second way to prepare a state (“all of the 16 ways to distribute amongst the compartments equally likely”) happened to produce a state that also matched your first description (“equally among the 4 ways to distribute among the boxes”), but you could easily come up with other distributions among the 4 compartments that match the first description of state, but not the second.

So this is completely about what state you prepare things in, it has nothing to do with what you can or choose to measure later.

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PeroK
I'd like to formulate the problem as a thought experiment.

Imagine we have a container with 2 bosons in it.
We have the possibility to insert dividing walls into this container so that we get a left and a right half and alternatively even a top left (TL), bottom left (BL), top right (TR) and bottom right (BR) quarter.
Now we carry out two sets of experiments:

In the first we only insert one wall and find that in 1/3 of the cases both bosons are in the left half.

In the second series of experiments we divide the container into quarters by inserting all the walls and find that in 1/10 of the cases both bosons are in TL, in 1/10 in BL and in 1/10 of the cases one boson is in BL and one in TL.
In sum this means that in 3/10 of the cases both bosons are in the left half.

The only difference between the two experiments is how many dividing walls we put in.

Can we explain this behaviour?

Did you even read my post?

Here are three things, of which at least one is wrong, depending on what additional assumptions you make about how the “bosons” are distributed in the box.

(1)
Philip Koeck said:
In the first we only insert one wall and find that in 1/3 of the cases both bosons are in the left half.
That is wrong as stated. Please derive this 1/3. At least it is not a unique conclusion, you must make further assumptions to conclude this.(2)
Philip Koeck said:
In the second series of experiments we divide the container into quarters by inserting all the walls and find that in 1/10 of the cases both bosons are in TL, in 1/10 in BL and in 1/10 of the cases one boson is in BL and one in TL.
That is wrong as stated. Please derive this 1/10. At least it is not a unique conclusion, you must make further assumptions to conclude this.(3)
Philip Koeck said:
The only difference between the two experiments is how many dividing walls we put in.
The assumptions that lead to (1) are not the same assumptions that lead to (2), so if you choose to make either of those assumptions this is wrong.Note, afaict this has little to nothing to do with QM, billiard balls with equal colors will work just fine.

PeroK
Philip Koeck said:
Can we explain this behaviour?

This is hypothetical behaviour that you have invented. There's no obligation on QM, for example, to be able to explain such behaviour. QM cannot be tested against just any hypothetical particle behaviour.

If, for example, we have two non-interacting identical bosons in an infinite square well, then we can compute various probabilities using the ground-state wave-function for the system. That would be a test of QM.

Dr.AbeNikIanEdL said:
Did you even read my post?

Here are three things, of which at least one is wrong, depending on what additional assumptions you make about how the “bosons” are distributed in the box.

(1)

That is wrong as stated. Please derive this 1/3. At least it is not a unique conclusion, you must make further assumptions to conclude this.

The probability 1/3 in the first case is quite accepted I think and it follows from some general expressions I've derived. You can read about it in my text on Research Gate: https://www.researchgate.net/publication/336375268_Probability_of_finding_R_of_N_particles_in_one_half_of_a_volume

Look at the end of the section on indistinguishable bosons.

The 1/10 for case 2 is harder to show, I think. Maybe someone can point to a derivation.

I'm not sure if I'm making any hidden assumptions. Maybe.

Dale said:
I said it depends on what you can measure, not what you choose to measure.

I believe that if your experiment is set up so that you can measure both box and partition then the probabilities are different than if your experiment is set up so that you can only measure the box.

No. Different experiments have different probabilities. My (possibly flawed) understanding is that in a two-slit experiment if you can measure only location on the screen you have one probability distribution while if you can measure both location and slit then you have a different distribution, even if you choose not to use the slit information. The probabilities depend on your experimental setup and what you can measure with it.
I like the comparison with the double slit: As long as we only know that the particle went through either of the slits we get an interference pattern. As soon as we know/measure which slit the particle went through we get the diffraction from a single slit. So we change the probability distribution simply by changing how well we localize the particle (or its path).

Dale
Dr.AbeNikIanEdL said:
The main problem seems to be that you just assume that every option is equally likely, no matter how those options are chosen. This is a bit like saying that I have a 50% chance of winning the lottery, because after all, there are only two options, either my numbers match or they don’t.
.
.
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So this is completely about what state you prepare things in, it has nothing to do with what you can or choose to measure later.
I'm not actually thinking about preparing a state.
A very simple example: If I have a single particle (distinguishable or indistinguishable) in a container and then I place a partition wall in the middle at random time points, then the particle should be found in the left half in about half the measurements. Other than putting in the wall i leave the particle alone.

Philip Koeck said:
I'm not sure if I'm making any hidden assumptions. Maybe.

You have to, otherwise you can not possibly arrive at any conclusion. I can clearly just put both balls in the left box always and your statements will be wrong. You can not just replace “unknown distribution” with “uniform distribution in whatever I please to choose”.

Philip Koeck said:
If I have a single particle (distinguishable or indistinguishable) in a container and then I place a partition wall in the middle at random time points, then the particle should be found in the left half in about half the measurements. Other than putting in the wall i leave the particle alone.

Get a box and a billiard ball (or whatever object), place the ball in the bottom left corner, and insert a wall in the middle. Repeat this 100 times (or dont, if you see where this is going). According to what you say, you should find the ball in the right half of the box in about 50 cases.

Philip Koeck said:
The only difference between the two experiments is how many dividing walls we put in.

Can we explain this behaviour?
Why are you surprised that changing the experimental setup would change the probability? The potential function for a particle in a box is different from the potential function for a particle in two boxes or four boxes.

Dr.AbeNikIanEdL said:
You have to, otherwise you can not possibly arrive at any conclusion. I can clearly just put both balls in the left box always and your statements will be wrong. You can not just replace “unknown distribution” with “uniform distribution in whatever I please to choose”.
Get a box and a billiard ball (or whatever object), place the ball in the bottom left corner, and insert a wall in the middle. Repeat this 100 times (or dont, if you see where this is going). According to what you say, you should find the ball in the right half of the box in about 50 cases.
Well, I was thinking more in terms of photons in a cavity or maybe Helium atoms, not macroscopic objects that stay where they are.

Philip Koeck said:
I'm not actually thinking about preparing a state.
A very simple example: If I have a single particle (distinguishable or indistinguishable) in a container and then I place a partition wall in the middle at random time points, then the particle should be found in the left half in about half the measurements. Other than putting in the wall i leave the particle alone.
If you are talking about QM and the particle is in the ground state, it will be found according to the wave-function:
$$\psi(x) = \sqrt{\frac 2 L} \sin(\frac \pi L x)$$
Where ##L## is the width of the box. In other words, the position measurements will not be uniformly distributed, as I believe you are assuming. That's why your analysis is not quantum mechanical in the first place.
The ground state for two bosons (where the Pauli exclusion does not apply) would be the simple product:
$$\psi(x_1, x_2) = \psi(x_1)\psi(x_2)$$
Which wouldn't obey your hypothetical model.

Dale
Philip Koeck said:
Well, I was thinking more in terms of photons in a cavity or maybe Helium atoms, not macroscopic objects that stay where they are.

Thats besides my point. As I said, I don't see anything QM related happening here. Make the things moving around if you want, you can not tell how likely it is that they are in the left side of the box if you don't know how they are moving. I repeat

Dr.AbeNikIanEdL said:
You can not just replace “unknown distribution” with “uniform distribution in whatever I please to choose”.

PeroK
PeroK said:
That's why your analysis is not quantum mechanical in the first place.

Well, we don't have to talk about the ground state. You can certainly (at least in principle) prepare QM states that match descriptions like “the distribution in space is uniform” or “all three option (both left), (both right), (left right) are equally likely” or “all ten options you get with 4 walls are equally likely”, they just are not (necessarily) the same state, so you should not expect the same probabilities from them.

Dr.AbeNikIanEdL said:
You can not just replace “unknown distribution” with “uniform distribution in whatever I please to choose”.
Actually, you can, particularly with Bayesian probability. Even if someone is non-randomly placing classical balls in classical containers, if I have no knowledge of their non-random choice then the way I express that ignorance is precisely by using a uniform distribution. It is perfectly valid in many cases (though I am sure not all) to replace an unknown distribution with a uniform distribution.

In any case, it is possible to construct a box/partitions/particles scenario where the probabilities are uniform.

Dr.AbeNikIanEdL said:
Well, we don't have to talk about the ground state. You can certainly (at least in principle) prepare QM states that match descriptions like “the distribution in space is uniform” or “all three option (both left), (both right), (left right) are equally likely” or “all ten options you get with 4 walls are equally likely”, they just are not (necessarily) the same state, so you should not expect the same probabilities from them.
They're not eigenstates, so you can't have a stable uniform distribution like you can classically.

PeroK said:
They're not eigenstates, so you can't have a stable uniform distribution like you can classically.
Does it need to be stable?

Dale said:
Does it need to be stable?
I was thinking about the "random time" element. I guess, in principle, you could prepare a uniform distribution. In any case, I don't believe the two-bosons-in-a-box would behave as required for this analysis.

Dale said:
Why are you surprised that changing the experimental setup would change the probability? The potential function for a particle in a box is different from the potential function for a particle in two boxes or four boxes.
What about if we don't use partition walls at all. We have two heavy bosons in the container and let them do whatever they are doing and once in a while we record where they are optically with minimal disturbance.
We have two ways of doing this: Either we just look how many particles are in the left half or we look how many are in the BL and how many in the TL.
I would say the potential is always that of a single container (apart from minimal disturbance while we do the measurement).
Still we would get 1/3 in the first case and 3/10 in the second, just as with the walls.

PeroK
Philip Koeck said:
What about if we don't use partition walls at all. We have two heavy bosons in the container and let them do whatever they are doing and once in a while we record where they are optically with minimal disturbance.
We have two ways of doing this: Either we just look how many particles are in the left half or we look how many are in the BL and how many in the TL.
I would say the potential is always that of a single container (apart from minimal disturbance while we do the measurement).
Still we would get 1/3 in the first case and 3/10 in the second, just as with the walls.
You probably need to learn some QM!

Dale said:
In any case, it is possible to construct a box/partitions/particles scenario where the probabilities are uniform.

Uniform in what? Uniform in (both left), (both right), (left right) is not the same as uniform in “the ten options with four walls”.

PeroK said:
If you are talking about QM and the particle is in the ground state, it will be found according to the wave-function:
$$\psi(x) = \sqrt{\frac 2 L} \sin(\frac \pi L x)$$
Where ##L## is the width of the box. In other words, the position measurements will not be uniformly distributed, as I believe you are assuming. That's why your analysis is not quantum mechanical in the first place.
The ground state for two bosons (where the Pauli exclusion does not apply) would be the simple product:
$$\psi(x_1, x_2) = \psi(x_1)\psi(x_2)$$
Which wouldn't obey your hypothetical model.
All wave functions in an infinitely deep well have a symmetric absolute square, so the probability of finding 1 particle in one half of the well should be 1/2. I don't see any contradiction with my assumptions.

Dr.AbeNikIanEdL said:
Uniform in what? Uniform in (both left), (both right), (left right) is not the same as uniform in “the ten options with four walls”.
I think the experiment as stated assumes:

We put two identical bosons in a box with a mixture of energy states that gives them (each?) a uniform distribution (or would do if there was only one of them). Then we measure their position and find they are in opposite halves ##1/3## of the time. I.e. we would have a repulsive exchange force of sorts.

I don't believe this would be the case.

PeroK said:
They're not eigenstates, so you can't have a stable uniform distribution like you can classically.

Fair point, if you want to consider time evolution QM just makes this unnecessarily complicated, and as I said I think it is best to leave it completely out of this.

PeroK said:
I think the experiment as stated assumes:

We put two identical bosons in a box with a mixture of energy states that gives them (each?) a uniform distribution (or would do if there was only one of them). Then we measure their position and find they are in opposite halves ##2/3## of the time. I.e. we would have a repulsive exchange force of sorts.

I don't believe this would be the case.
The bosons are also non-interacting. I don't want any forces between them.
They're not in opposite halves 2/3 of the time. They are both in the left half 1/3 of the time.
2/3 of the time they are not both in the left half.

Dr.AbeNikIanEdL said:
Uniform in what? Uniform in (both left), (both right), (left right) is not the same as uniform in “the ten options with four walls”.
Agreed. He has to be clear about uniform in what.

Philip Koeck said:
The bosons are also non-interacting. I don't want any forces between them.
They're not in opposite halves 2/3 of the time. They are both in the left half 1/3 of the time.
2/3 of the time they are not both in the left half.

I've assumed the bosons are non-interacting. Exchange forces come with the territory - it's a purely QM phenomenon.

I guess what you're saying is that if you can create a state where the distribution of a boson is uniform and you put two such bosons in a box, then the attractive exchange force means you find them a) generally closer together than classical particles and b) both in the left half 1/3 of the time?

Is that what you have proved in your paper?

Dale
PeroK said:
We put two identical bosons in a box with a mixture of energy states that gives them (each?) a uniform distribution (or would do if there was only one of them).

So uniform in position, right (@Philip Koeck this is an assumption that would allow to do an actual calculation, is if what you want to discuss)? But if for each particle on its own the position is uniformly distributed, the probability that the two are in opposite sides of the box is not 1/3 (or did I go crazy?).

Dr.AbeNikIanEdL said:
So uniform in position, right (@Philip Koeck this is an assumption that would allow to do an actual calculation, is if what you want to discuss)? But if for each particle on its own the position is uniformly distributed, the probability that the two are in opposite sides of the box is not 1/3 (or did I go crazy?).
Here's a qualitative analysis. 1) If we have two bosons with different energy levels, then the exchange force means we tend to find them closer together - hence both in the left half more than ##1/4## of the time (?). 2) If we have each boson in a mixture of energy states to approximate a uniform distribution, then we find them in the left half ##1/3## of the time(?) It would be an interesting calculation.

Dale
PeroK said:
I've assumed the bosons are non-interacting. Exchange forces come with the territory - it's a purely QM phenomenon.

I guess what you're saying is that if you can create a state where the distribution of a boson is uniform and you put two such bosons in a box, then the attractive exchange force means you find them a) generally closer together than classical particles and b) both in the left half 1/3 of the time?

Is that what you have proved in your paper?
In a way I guess. I don't use the term exchange force, but I show that indistinguishable and distinguishable particles have different probabilities of being in one half of a container together (1/3 and 1/4 for example).

Philip Koeck said:
In a way I guess. I don't use the term exchange force, but I show that indistinguishable and distinguishable particles have different probabilities of being in one half of a container together (1/3 and 1/4 for example).
Looking at what you've done, you got the same answer for distinguishable and indistinguishable fermions. That can't be correct, because identical fermions (or to be precise, fermions with a symmetric spin state, hence anti-symmetric spatial wave-function) have a repulsive exchange force, so would tend to be found further apart.

The fact is that without properly applying quantum theory, I can't see how your results can be valid in general.

PeroK said:
Here's a qualitative analysis. 1) If we have two bosons with different energy levels, then the exchange force means we tend to find them closer together - hence both in the left half more than ##1/4## of the time (?). 2) If we have each boson in a mixture of energy states to approximate a uniform distribution, then we find them in the left half ##1/3## of the time(?) It would be an interesting calculation.

I am not sure I follow. When do you do the measurement?

So we set the particles up so that the position for each is uniformly distributed. This will probably not be a stationary state. Now do we

(1) immediately measure. In that case, exchange forces are irrelevant, and we do not get the 1/3 probability for the 3 options.

(2) have some time evolution, and then measure (and average over time?). It might or might not turn out that the particles are more or less likely to be in the same side, I don't know, but they would not remain uniform in space. This seems to me to deviate quite far from the original post.

Philip Koeck said:
I don't use the term exchange force, but I show that indistinguishable and distinguishable particles have different probabilities of being in one half of a container together (1/3 and 1/4 for example).

No, you show that there are 3 and 4 distinct ways how two indistinguishable and distinguishable particles can be distributed in the two halves. You then assert that each of them is equally likely to get to the 1/3 and 1/4 probability.

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