Graduate Do bosons contradict basic probability laws?

[Dr.AbeNikIanEdL]

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?).

 

[PeroK]

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.

 

[Philip Koeck]

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).

 

[PeroK]

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.

 

[Dr.AbeNikIanEdL]

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.

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.

 

[Philip Koeck]

What about the following calculation to keep it as simple as possible?
We have two non-interacting bosons in a 1-dimensional, infinitely deep well.
Both are in the same eigen-state.
What is the probability of finding both in the same half of the well.

 

[PeroK]

What about the following calculation to keep it as simple as possible?
We have two non-interacting bosons in a 1-dimensional, infinitely deep well.
Both are in the same eigen-state.
What is the probability of finding both in the same half of the well.

That's $1/2$. I.e. $1/4 + 1/4$.

 

[Philip Koeck]

That's $1/2$. I.e. $1/4 + 1/4$.

I meant the probability for both being in the left half of the well.

 

[Dr.AbeNikIanEdL]

Well, that’s then the 1/4, of which there are two in @PeroK ’s post.

 

[Philip Koeck]

Well, that’s then the 1/4, of which there are two in @PeroK ’s post.

That would be completely classical behaviour as for distinguishable particles. 1/4 for both in the left, 1/4 for both in the right and 1/2 for one in each.

 

[Dr.AbeNikIanEdL]

That would be completely classical behaviour as for distinguishable particles. 1/4 for both in the left, 1/4 for both in the right and 1/2 for one in each.

The difference being that, with indistinguishable particles you can not resolve the last case into 1/4 for “1st left 2nd right” and 1/4 for “1st right 2nd left”. As I said a few times now, nothing particularly QM related is happening in your question.

 

[AndreasC]

OK, first of all, I literally learned what boson/fermion statistics are like 3 days ago after off handedly reading the relevant section in Shankar, so I only have the faintest idea of what I'm talking about, so take it with a massive grain of salt, but I think I understand what confused you. Let me try to figure out what exactly you are having trouble with.

Let's call the state where one boson is in the left department and one in the right as A11. The state where both bosons are in the left department is A20, and the one where both are in the second A02. We can do the same with top and bottom departments and call them B11, B02, and B20. We can assume (and that's probably what you have in mind) that these two are independent. You assumed that the 3 states are of equal probability, having some kind of "uniform" distribution in mind (although it probably does not exactly what you had in mind). You make the assumption that if you now separate it into 4 departments, thus getting 10 states, each of the 10 states should also have equal probability. This is wrong. The right way to do it with your assumption of 1/3 probability for each of the states in experiments A and B would be by multiplying the probabilities for these. So for example, for both particles in the left department you'd have P(A20)*P(B20)+P(A20)*P(B02)+P(A20)*P(B11)=1/3. Same for the right department. And then you'd be left with one more event, namely A11B11, which corresponds to one ball left one ball right, one ball up one ball down, and it also has a probability of 1/3. They all add up to 1 so nothing is wrong here. However while each of the states A20B20, A20B11, etc. corresponds to one state of the new space, so 2020, 2011,etc, the state A11B11 translates to FOUR separate states, namely 1010, 1001, 0101, 0110, so each of these states should have probabilities of 1/12.

This should not come as a surprise, it comes as a direct consequence of the way you chose your distribution to be. It is similar to how 7 is the most common result when you throw two dice instead of all results having equal weight. It's like making two dice so that each one is fair, but then being surprised when you throw them together and they don't give the same weight to each number. It doesn't have much to do with QM or bosons in particular, it has to do with statistics.

 

[PeroK]

That would be completely classical behaviour as for distinguishable particles. 1/4 for both in the left, 1/4 for both in the right and 1/2 for one in each.

It's not completely classical behaviour, as there is no classical analogue to the spatial wave-function and the distribution of position measurements would be non-classical. It's just that we have a symmetry about the centre of the box. Identical bosons in the same eigenstate behave the same as distinguishable bosons for the reason that the simple product wave-function is allowed, as it is symmetric.

If we choose different eigenstates ($a$ and $b$) for the two particles, there will be a difference because we have to symmetrize the wave-function for the indistinguishable case:
$$\psi(x_1, x_2) = \frac 1 {\sqrt 2}(\psi_a(x_1)\psi_b(x_2) + \psi_b(x_1)\psi_a(x_2))$$
The exchange forces are a direct result purely of this symmetrization requirement. Or, in the case of fermions, the anti-symmetrization requirement.

That said, I think we have got side-tracked onto "particles in a box" scenarios, that are possibly not relevant to your paper that deals with discrete quantum states. Maybe the particles in a box is the continuum limit of this, with the discrete states becoming a continuum of position eigenstates and supposing a uniform spatial distribution (which can only exist instantaneously in any case). But, to be honest, I'm not sure about that.

 

[AndreasC]

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.

Cool, but in that case the probabilities wouldn't be evenly split at 1/3 for each state. It would be 1/2 for the state S (one particle in side A and one on side B), and 1/4 for each of the two others. I'm going to "pretend" like we're not talking about an actual box with particles inside but for some abstract configuration where we have two discreet possible states. The end result is the same thing if you have a continuous box with a symmetric wavefunction for each of the particles, except the entire methodology I am following here would be completely stupid because in that case, there is not a single state that corresponds to being found in one of the two departments, but rather an infinite collection of states (so you'd also have similar results for fermions, whereas here you'd get that it is impossible for both fermions to be in the same department), and you'd have to do something significantly different and slightly more complex to do what I am trying to do here so aaaaaaaanyways.

To see this, you can form the space of states from the wavefunctions you started with by means of outer product. For each particle, the state will be:
$$ \ket{\psi} = \frac{\ket{A}+\ket{B}}{\sqrt{2}} $$
Where A represents finding the particle in department A and B... Well you get the idea.
Now we can combine the two and get the wavefunction for both particles:

$$ \ket{\psi_{1}\psi_{2}} = \ket{\psi_1} \otimes \ \ket{\psi_2} = (\frac{\ket{A}+\ket{B}}{\sqrt{2}}) \otimes (\frac{\ket{A}+\ket{B}}{\sqrt{2}})$$

If you work that out, it gives us the following:
$$ \ket{\psi_{1}\psi_{2}} = \frac{1}{2}(\ket{A}\otimes\ket{A}+\ket{B}\otimes\ket{B}+\ket{A}\otimes\ket{B}+\ket{B}\otimes\ket{A}) $$

But now we have to bear in mind that since the particles are identical, there are only 3 states, namely $\ket{AA}$, $\ket{BB}$ and $\ket{S}$. The first two are self explanatory, the third is the state where one particle (it doesn't matter which!) is in A, the other is in B. Let's see what exactly these correspond to:
$$ \ket{AA} = \ket{A}\otimes\ket{A} $$
$$ \ket{BB} = \ket{B}\otimes\ket{B} $$
Alright, these were obvious enough, but the particles are identical and they are bosons, so we have to symmetrize the third state:
$$ \ket{S} = \frac{\ket{A}\otimes\ket{B}+\ket{B}\otimes\ket{A}}{\sqrt{2}} $$

Now we can rewrite the state of the system:
$$ \ket{\psi_{1}\psi_{2}} = \frac{1}{2}\ket{AA}+\frac{1}{2}\ket{BB}+\frac{1}{\sqrt{2}}\ket{S}$$

Notice how the coefficients for AA and BB are 1/2, while for S it is 1/√2. These correspond to probabilities of 1/4, 1/4 and 1/2 respectively. So as you can see a "uniform" distribution for each boson doesn't give you the same probabilities for all 3 states. It's almost exactly what you would expect for two balls, except you can't split the state S into two states, AB and BA. That's the sole difference here. If you had fermions instead it would be a bit different because you can't have them both in one state. Of course, as I explained earlier this would not really apply the same way in the example of a continuous box.

Anyways, there is a VERY high chance I am making some mistake here because as I mentioned earlier, I learned this stuff only around 3 days ago so if I am, I'd appreciate if you corrected me. It was nice though that I was given the opportunity to think about these things a bit more because I feel more confident in my understanding now.

 

[Philip Koeck]

.
.
Now we can rewrite the state of the system:
$$ \ket{\psi_{1}\psi_{2}} = \frac{1}{2}\ket{AA}+\frac{1}{2}\ket{BB}+\frac{1}{\sqrt{2}}\ket{S}$$

Notice how the coefficients for AA and BB are 1/2, while for S it is 1/√2. These correspond to probabilities of 1/4, 1/4 and 1/2 respectively. So as you can see a "uniform" distribution for each boson doesn't give you the same probabilities for all 3 states. It's almost exactly what you would expect for two balls, except you can't split the state S into two states, AB and BA. That's the sole difference here. If you had fermions instead it would be a bit different because you can't have them both in one state. Of course, as I explained earlier this would not really apply the same way in the example of a continuous box.

If I understand your post correctly you are saying that for a 2-state system the 3 possible distributions of 2 bosons don't all have the same probability of 1/3 and for a 4 state system the 10 possible distributions don't all have probability 10, all based on QM-calculations. Clearly if you are right my original problem disappears.
In the microcanonical derivation of BE-statistics, however, there is an implicit assumption, I think, that all distributions of n bosons among g states are equally likely (see https://en.wikipedia.org/wiki/Bose–Einstein_statistics or my own texts on ResearchGate).
What's going on here? Is there a real discrepancy between QM and statistical physics?

 

[Kazu]

Hi, every one.
Do you know Probability Axioms (by Kolmogorov) ?

Event : E⊆Ω　　　Probability : Pr(E)
Axiom 1 0≦Pr(E)≦1
Axiom 2 Pr(Ω)=1
Axiom 3 Ei∩Ej=∅ (i≠j)
⇒ Pr(E1∪E2∪･･･∪En)=Pr(E1)+Pr(E2)+ ･･･+Pr(En)

Two particles in the container.
[left, right] (left up, left low, right up, right low)

Ω={[2,0],[1,1],[0,2]}
Ω={(2,0,0,0),(1,1,0,0),(0,2,0,0),
(1,0,1,0),(1,0,0,1),(0,1,1,0),(0,1,0,1),
(0,0,2,0),(0,0,1,1),(0,0,0,2)}

Method A:Two particles are distinguishable.
→Pr({[2,0]})=Pr({[0,2]}), Pr({1,1]})=2Pr({[2,0]})
Pr({(2,0,0,0)})=Pr({(0,2,0,0)})=Pr({(0,0,2,0)})=Pr({(0,0,0,2)}),
Pr({(1,1,0,0)})=Pr({(1,0,1,0)})=Pr({(1,0,0,1)})=Pr({(0,1,1,0)})
=Pr({(0,1,0,1)})=Pr({(0,0,1,1)})=2Pr({(2,0,0,0)})
Method B:Two particles are indistinguishable.
→Pr({[2,0]})=Pr({[0,2]}), Pr({1,1]})=Pr({[2,0]})
Pr({(2,0,0,0)})=Pr({(0,2,0,0)})=Pr({(0,0,2,0)})=Pr({(0,0,0,2)})
=Pr({(1,1,0,0)})=Pr({(1,0,1,0)})=Pr({(1,0,0,1)})=Pr({(0,1,1,0)})
=Pr({(0,1,0,1)})=Pr({(0,0,1,1)})

By using Method A,
1=Pr(Ω)=4Pr({[2,0]}) ∴Pr({[2,0]})=1/4
1=Pr(Ω)=16Pr({(2,0,0,0)}) ∴Pr({(2,0,0,0)})=1/16
Accordingly
Pr({(2,0,0,0)})+Pr({(1,1,0,0)})+Pr({(0,2,0,0)})=1/4
By the way,
{[2,0]}={(2,0,0,0)}+{(1,1,0,0)}+{(0,2,0,0)}
And
Pr({[2,0]})= Pr({(2,0,0,0)})+Pr({(1,1,0,0)})+Pr({(0,2,0,0)})
It fits Axom 3.

By using Method B,
1=Pr(Ω)=3Pr({[2,0]}) ∴Pr({[2,0]})=1/3
1=Pr(Ω)=10Pr({(2,0,0,0)}) ∴Pr({(2,0,0,0)})=1/10
Accordingly
Pr({(2,0,0,0)})+Pr({(1,1,0,0)})+Pr({(0,2,0,0)})=3/10
By the way,
{[2,0]}={(2,0,0,0)}+{(1,1,0,0)}+{(0,2,0,0)}
But
Pr({[2,0]})≠ Pr({(2,0,0,0)})+Pr({(1,1,0,0)})+Pr({(0,2,0,0)})
It doesn't fit Axom 3.

The elementary particles are mysterious.
The observation result may change for one's heart.
But, this is a discussion of the mathematics purely.

 

[PeroK]

If I understand your post correctly you are saying that for a 2-state system the 3 possible distributions of 2 bosons don't all have the same probability of 1/3 and for a 4 state system the 10 possible distributions don't all have probability 10, all based on QM-calculations. Clearly if you are right my original problem disappears.
In the microcanonical derivation of BE-statistics, however, there is an implicit assumption, I think, that all distributions of n bosons among g states are equally likely (see https://en.wikipedia.org/wiki/Bose–Einstein_statistics or my own texts on ResearchGate).
What's going on here? Is there a real discrepancy between QM and statistical physics?

That Wikipedia page is talking about the distribution of energy states in thermodynamic equilibrium. That's a different scenario from the one you are looking at.

 

[Philip Koeck]

That Wikipedia page is talking about the distribution of energy states in thermodynamic equilibrium. That's a different scenario from the one you are looking at.

As I read it, it's talking about distributions of particles among states with the same energy. This is then extended to several energy levels. The most likely distribution of particles among energy levels is simply found by looking for the maximum number of distributions given the constraints of total energy and particle number.
Interestingly there is no discussion of different probabilities for different distributions.

 

[PeroK]

As I read it, it's talking about distributions of particles among states with the same energy. This is then extended to several energy levels. The most likely distribution of particles among energy levels is simply found by looking for the maximum number of distributions given the constraints of total energy and particle number.
Interestingly there is no discussion of different probabilities for different distributions.

You must be able to see your problem. If you simply look for particles in different positions you are dealing with a common wave-function or common distribution of states. That doesn't change so that all options become equally likely, depending on what you want to look for. I.e. if you decide notionally to subdivide the box into thirds, that doesn't create three "states". Particle measurements are indepenent of any notional subdivisions. Your false assumption is that by notionally deciding the box is divided into thirds, you have created three possible states.

If, however, you physically subdivide the system into thirds, then you have a different system from the original. And, if you allow the system to reach equilibrium, you may use the Bose-Einstein statistics.

In other words, the change from a possible two to a possible three states must be a real physical change. Not just an arbitrary decision to categorise the results of where you find particles when you look for them.

 

[AndreasC]

I believe you are still making the mistake of conflating the "states" (again, can't really split AB and BA) AA, BB, AB and BA being of equal probability with the states AA, BB and S being of equal probability.

EDIT: Nah, I believe PeroK's reply is probably the correct take.

 

[PeroK]

In other words, the change from a possible two to a possible three states must be a real physical change. Not just an arbitrary decision to categorise the results of where you find particles when you look for them.

PS to see the problem more clearly, imagine you gather the data first. Then you analyse it. You have only one set of data and it cannot meet all your equally likely criteria.

If, however, you physically subdivide the box, then you can only obtain data for that one configuration. And, there is no need for that configuration to be compatible in with a different physical configuration.

 

[Philip Koeck]

Let's take a close look at the derivation on Wikipedia. They just give an expression for the number of actually different distributions of n particles among g states on the same energy level. Clearly particle 1 in state 1 and 2 in 2 is the same as 1 in 2 and 2 in 1, since they are bosons. They do this for each energy level.
Then they just look for the distribution of particles among energy levels with the largest total number of distributions among states with fixed total energy and particle number. This is reported as the most likely distribution, without considering that different distributions of particles among states might have different probabilities. Isn't that right?

 

[AndreasC]

To understand a bit better what I believe Pero is saying, think of fermions. Let's say we have two identical electrons in a box. Let's call the state of them being in the box A. So they're both in state A. Except that's impossible. Now let's say we call the left part of the box A and the right B. Well, suddenly they can both be in the box by one being in state A and the other in B. What's happening? Well, neither A or B are real states. Instead, an infinite number of states can correspond to the particle being found in the box or in the left department or in the right department. There has to be some extra physical change to really separate the system into 2 distinguishable states, and the whole discussion is a bit confusing because we're jumping back and forth from a continuous particle in a box example to distinguishable states to thermodynamic equilibrium etc.

 

[Dr.AbeNikIanEdL]

That

the whole discussion is a bit confusing because we're jumping back and forth from a continuous particle in a box example to distinguishable states to thermodynamic equilibrium etc.

@Philip Koeck , you seem to just take isolated blocks from one of those scenarios and assume that it is true in a completely different one, just because one can broadly headline both with "has something to do with QM". What is the scenario we are actually supposed to discuss?

Particles in a box, prepared in a state that yet has to be specified?
If yes, please specify that state.

Particles in a system with n discrete states, distributed according to a distribution that yet has to be specified?
If yes, please specify that distribution.

Particles in a system with n discrete states, distributed according to thermal equilibrium?
If yes, please state that.

 

[Philip Koeck]

That

@Philip Koeck , you seem to just take isolated blocks from one of those scenarios and assume that it is true in a completely different one, just because one can broadly headline both with "has something to do with QM". What is the scenario we are actually supposed to discuss?

Particles in a box, prepared in a state that yet has to be specified?
If yes, please specify that state.

Particles in a system with n discrete states, distributed according to a distribution that yet has to be specified?
If yes, please specify that distribution.

Particles in a system with n discrete states, distributed according to thermal equilibrium?
If yes, please state that.

I'll go for the last one. I think that fits the discussion of BE-statistics best.

 

[Dr.AbeNikIanEdL]

In that case the last few answers apply. What you actually want to have is something like a system with 2 discrete states, you then make an actual physical change to that system so that it has 4 discrete states afterwards. You then wait until thermal equilibrium is reached again. After this, probabilities have changed, which should be expected since you changed the system, and the 2 states you started with are not occupied in the same way as before.

 

[Philip Koeck]

PS to see the problem more clearly, imagine you gather the data first. Then you analyse it. You have only one set of data and it cannot meet all your equally likely criteria.

If, however, you physically subdivide the box, then you can only obtain data for that one configuration. And, there is no need for that configuration to be compatible in with a different physical configuration.

Let's turn the problem around:
I have 2 bosons and 4 states with the same energy. According to the assumptions implicit in the derivation of BE-statistics the 10 possible distributions all have the same probability of 1/10.
If I carry out a series of measurements I'll find that both bosons are in the states 1 or 2 with a probability of 3/10 (or 3/10ths of the time).
Now I do a second set of measurements with a less accurate instrument. I cannot say whether a particle is in state 1 or 2, and likewise I cannot say whether a particle is in state 3 or 4. Effectively the instrument only distinguishes between 2 states.
Should I now get the probability 1/3 for finding both bosons in states 1 or 2 (which I can't tell apart) or should I get 3/10.
Notice that I haven't changed the physics of the system. I've only changed the accuracy of my measurement.

 

[Dr.AbeNikIanEdL]

If you don't change the system, the probability for a given state to find both bosons in state 1 or 2 does not change. Whether you actually resolve between state 1 and 2 does not matter for this.

 

[PeroK]

Let's turn the problem around:
I have 2 bosons and 4 states with the same energy. According to the assumptions implicit in the derivation of BE-statistics the 10 possible distributions all have the same probability of 1/10.
If I carry out a series of measurements I'll find that both bosons are in the states 1 or 2 with a probability of 3/10 (or 3/10ths of the time).
Now I do a second set of measurements with a less accurate instrument. I cannot say whether a particle is in state 1 or 2, and likewise I cannot say whether a particle is in state 3 or 4. Effectively the instrument only distinguishes between 2 states.
Should I now get the probability 1/3 for finding both bosons in states 1 or 2 (which I can't tell apart) or should I get 3/10.
Notice that I haven't changed the physics of the system. I've only changed the accuracy of my measurement.

This is going off at a tangent now. To take an example. Suppose we have 9 states that all have similar energies and one outlier. Let's say with equal probability. An accurate measurement might identify all ten states and the outlier would be found 1/10.

If we now have less sensitive equipment that cannot distinguish the nine states, then we still get the outlier only 1/10 and the amalgamation of the other states 9/10.

 

[Philip Koeck]

I think I'm satisfied with the take home message of this discussion: States are real things and not a matter of how one groups them together or how accurately one can measure. What led to the original problem in my first post is that Kazu and I grouped states together and treated these groups as if they were physical states.
Thank you all very much for input.