# A The probability of finding R out of N bosons in one half of a volume

#### Philip Koeck

Summary
I come up with an expression for the probability of finding R out of a total of N bosons in one half of a volume which depends on the number of available states g in half the volume as long as g is not very large compared to N and 1. I want to know whether this can be understood and explained in a common-sense way.
For the probability of finding R out of N (indistinguishable) bosons in one half of a volume with a total of 2g states (g in each half) I get the following expression:

PR = WR / WT

where WT is the number of ways of distributing N particles in the total volume:

WT = (N+2g-1)! / (N! (2g-1)!)

and WR is the number of ways of distributing R particles in one half of the volume and the remaining N-R in the other half:

WR = ((R+g-1)! (N-R+g-1)!) / (R! (g-1)! (N-R)! (g-1)!)

As long as I don't assume a large number of states g and low occupancy (g >> N), PR depends on the value of g.

My questions:
Does this seem to be correct?
Is there a common-sense way of explaining this dependence on g?

Note that at low occupancy and large g the dependence on g disappears and that for (hypothetical) distinguishable "bosons" PR doesn't depend on g in the first place, whatever the value of g is.

For fermions I can understand that PR depends on g since each fermion blocks a state and alters the situation for all other fermions so that PR must depend on g.
Can one come up with a similar explanation for indistinguishable bosons?

In case my equation is unclear I have a complete derivation here: https://www.researchgate.net/publication/336375268_Probability_of_finding_R_of_N_particles_in_one_half_of_a_volume

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#### PeterDonis

Mentor
Moderator's note: Thread opened after moderator review.

#### vanhees71

Gold Member
I could never come up with an intuitive explanation for bosons. It's of course just the Bose-Einstein-enhancement factor. Fermions are more intuitive since we deal with fermions in our daily experience, i.e., when some piece of matter occupies some space it's usually hard to put more matter into that space, and that's of course due to Pauli blocking.

The BE-enhancement is of course the bosonic analogue of P-blocking for fermions, and it just tells us that bosons like to occupy the same states, i.e., if already bosons are present other bosons of the same kind like to also occupy this space. Formally you can see this from the BE-enhancement factors $(1+f)$ in the collision term of the bosonic Boltzmann-Uehling-Uhlenbeck equation.

Now you can ask further, where this enhancement factor comes from on microscopic grounds, and the only answer is that it's a specific quantum property, discovered by Einstein in 1917, when he rederived the Planck radiation formula from a kinetic point of view: In addition to the usual absorption and induced emission contributions from classical theory he had to assume also spontaneous emission, which is specifically quantum (and in fact the only hint at the necessity of field quantization known before 1925!). In other words the most simple empirical reason for the necessity of quantizing the electromagnetic field is the phenomenon of spontaneous emission. Neither Einstein's photoelectric-effect explanation of 1905 nor Compton scattering of 1923 really needs field quantization but at lowest order perturbation theory (tree-level Feynman diagrams only) but one can derive the corresponding cross sections in the semiclassical way, i.e., quantizing the electrons only and keeping the em. field classical (semi-classical approximation of QED).

#### Philip Koeck

Just a few remarks from me: I realise that my model is very simplified, since I implicitly give every state the same probability to be occupied. I guess this would only be true if there is only a single energy level with 2g states.

About bosons: If there are such things as distinguishable bosons, then they wouldn't actually exhibit the same kind of enhancement as indistinguishable bosons, according to my results. For distinguishable bosons the probability doesn't depend on the number of states even if g is not big, so I would conclude that distinguishable bosons wouldn't influence each other at all.
(On the other hand they would have to be very strange, being distinguishable and able to crowd into the same state without limit.)

Only indistinguishable bosons should have this enhancement factor and therefore the probability depends on g, which becomes most visible for small values of g.
Therefore the dependence of the probability on g ought to have something to do with bosons being indistinguishable as well as being able to crowd into states.

About fermions: I just realized that I get the same expression for P for both distinguishable and indistinguishable fermions, even without approximating for large g.
It seems to me that it is fundamentally impossible to decide whether a system consists of distinguishable or indistinguishable fermions. Could that be true, or are there experiments that would answer that question?

Thanks again.

Philip

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#### Kazu

For example, 2 dice.
・We throw 2 dice.
・We calculate the probabilities seeing the next numerical combinations.
Combination Ⅰ: odd number and odd number
Combination Ⅱ: odd number and even number
Combination Ⅲ: even number and even number

Distinguishable 2 dice (A and B)

calculation 1
Ⅰ odd+odd
(A,B)=
( odd, odd)
number of cases = 1
probability = 1/4
Ⅱ odd+even
(A,B)=
( odd,even)
(even, odd)
number of cases = 2
probability = 1/2
Ⅲ even+even
(A,B)=
(even,even)
number of cases = 1
probability = 1/4

calculation 2
Ⅰ odd+odd
(A,B)=
(1,1) (1,3) (1,5)
(3,1) (3,3) (3,5)
(5,1) (5,3) (5,5)
number of cases = 9
probability = 1/4
Ⅱ odd+even
(A,B)=
(1,2) (1,4) (1,6)
(3,2) (3,4) (3,6)
(5,2) (5,4) (5,6)
(2,1) (2,3) (2,5)
(4,1) (4,3) (4,6)
(6,1) (6,3) (6,5)
number of cases = 18
probability = 1/2
Ⅲ even+even
(A,B)=
(2,2) (2,4) (2,6)
(4,2) (4,4) (4,6)
(6,2) (6,4) (6,6)
number of cases = 9
probability = 1/4

Calculation 2 agree with calculation 1. It is reasonable.

Indistinguishable 2 dice

calculation 3
Ⅰ odd+odd
( odd, odd)
number of cases = 1
probability = 1/3
Ⅱ odd+even
( odd,even)
number of cases = 1
probability = 1/3
Ⅲ even+even
(even,even)
number of cases = 1
probability = 1/3

calculation 4
Ⅰ odd+odd
(1,1) (1,3) (1,5)
(3,3) (3,5) (5,5)
number of cases = 6
probability = 2/7
Ⅱ odd+even
(1,2) (1,4) (1,6)
(3,2) (3,4) (3,6)
(5,2) (5,4) (5,6)
number of cases = 9
probability = 3/7
Ⅲ even+even
(2,2) (2,4) (2,6)
(4,4) (4,6) (6,6)
number of cases = 6
probability = 2/7

Calculation 4 give us the different probabilities from calculation 3.

• Stephen Tashi

#### Philip Koeck

Your example illustrates that for indistinguishable bosons the probabilities depend on the number of states associated with, for example, being even. If being even is just one state you get one set of probabilities, whereas if you associate 3 states with being even, you get different probabilities.
It's certainly interesting that indistinguishable bosons behave like that, but I don't see anything contradictory or illogical about it.

#### Kazu

We do not have to use the character of particle to calculate the probability about the dice.
Let's suppose two indistinguishable dice whose faces with odd number are blue and with even number are red.
We roll the 2 dice many times and count the combination Ⅰ, Ⅱ, Ⅲ.
We get the next probabilities if we see the color and count them.
Ⅰ:1/3, Ⅱ:1/3, Ⅲ:1/3
But if we see the number and count them, we get next probabilities.
Ⅰ:2/7, Ⅱ:3/7, Ⅲ:2/7

#### Swamp Thing

This discussion reminded me of an article I read long ago (but still can't understand fully).
It may be of interest to you.

• Stephen Tashi and Philip Koeck

#### Philip Koeck

Thanks for sharing this interesting paper. It fits very well into what I've been thinking about and what we've discussed so far.
I've started reading it and I get stuck at the beginning of section 3 where it states:
"For distinguishable children obeying classical statistics, statistical independence implies that the outcome of the remaining n−1 children are not aﬀected by the outcome of the ﬁrst child."
I can't make sense of that.
If there are 100 children and the most likely distribution is 50/50 to start with, then, after removing one child, the most likely distribution must be 49/50, I would think.
Even for classical children the distribution after removing one child is affected by whether you remove a boy or a girl.
Do you agree?

#### Stephen Tashi

If there are 100 children and the most likely distribution is 50/50 to start with, then, after removing one child, the most likely distribution must be 49/50, I would think.
I think that would be true if you knew for certain that the actual distribution was going to be close to the most likely distribution. However, we don't know that. (The "most likely" outcome may not be very likely at all.)

If we assign each possible distribution its correct probability under the assumption of independent trials, the most probable distributions for the 99 remaining trials should be 49-to-50 and 50-to-49, each having the same probability.

#### Philip Koeck

I'm trying to make sense of this so let's take a different example.
In a large dance hall for singles there are 1000 people and the organizers know that on average there are 500 women and 500 men.
Towards the end of the evening people start leaving randomly and somebody keeps track of who is leaving. To his surprise the first 100 people to leave are all men, but there is no obvious reason for this. It seems to be just by chance.
What is the most likely distribution of man and women left in the dance hall now?

#### Stephen Tashi

I'm trying to make sense of this so let's take a different example.
In a large dance hall for singles there are 1000 people and the organizers know that on average there are 500 women and 500 men.
Stating the average of a random variable doesn't define its distribution. So we don't have a specific mathematical problem. For example, one possibility is that each ratio: 0 to 1000, 1 to 999, 2 to 998, ... ,1000 to 0 has the same probabiltity. Another possibility is that the ratios 101 to 899, 500 to 500, and 899 to 101 each have probability 1/3 and the other ratios have probability zero.

#### Philip Koeck

Stating the average of a random variable doesn't define its distribution. So we don't have a specific mathematical problem.
We can be very specific: The organizers of the dance know the exact distribution from the past. It's a Gaussian with a maximum at 500 men and 500 women. The standard deviation is 50.
Now 100 men leave early one evening. What is the most likely number of men afterwards?

#### Stephen Tashi

We can be very specific: The organizers of the dance know the exact distribution from the past. It's a Gaussian with a maximum at 500 men and 500 women. The standard deviation is 50.
Now 100 men leave early one evening. What is the most likely number of men afterwards?
I don't know what to do with a Gaussian. It's easier for me to think of a discrete distribution. Let $f(k)=$ probability that there are $k$ total men out of 1000

Let $g(k,m) =$ probability that there are $k$ total men out of 1000 given that $m$ randomly selected people are men.

Let $h(m,k) =$ probability that $m$ randomly selected people are men given there are $k$ total men out of 1000.

$g(k,100) = 0$ if $k < 100$
if $k \ge 100$, $g(k,100) = h(100,k) f(k)/ ( \sum_{j=100}^{1000} h(100,j) f(j))$.

We want to find the value of $x$ that maximizes $g(x,100)$. We can ignore the constant factor $\sum_{j=100}^{1000} h(100,j) f(j))$ and consider only the function $h(100,x) f(x)$ for $x = 100,101,...1000$

$h(100,x) = (x/1000)((x-1)/1000)((x-2)/1000)....((x-99)/1000)$
$= (x!)/ ( ( x-100)\ !1000) )$.

[Edit: No, it should be $h(100,x) = (x/1000)((x-1)/(1000-1))((x-2)/(1000-2)....((x-99)/(1000-99))$ ]

The maxium of $h(100,x) f(x)$ depends on how $h(100, x)$ alters the shape of $f(x)$.

I suppose the approach of statistical physics would be to consider numbers much larger than 100 and 1000 and introduce Sterlings approximation.

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• Philip Koeck