Problem with the idea of identical particles in QM

[QuasarBoy543298]

assume i have a gass made from N identical particles in a box and i want to calculate the probability for k out of N particles to be in the left side of the box.
the problem is ,that if we treat the N particles as identical , each state in which exacly k of the N particles are in the left side of the box is considered to be the same state (thats the meaning of identical particles from qm as i understood - the configuration in which particle A is on the left side and particle B on the right side and the configuration in which particle A is on the right side and particle B is on the left side are considered as one configuration ).
that does not seem very reasonable considering the fact that we now have not so neglegable probability for all the particles to go to the left side of the box

its kinda silly but i can't figure out where the mistake is

help me I am confused :(((

 

[PeterDonis]

that does not seem very reasonable considering the fact that we now have not so neglegable probability for all the particles to go to the left side of the box

Why do you think the probability would not be negligible? Have you tried to calculate the fraction of states that have exactly $k$ particles on the left side of the box?

 

[tnich]

I am not entirely clear about what your point of confusion is, but I suspect you are thinking something like this: "How can I count the number of combinations in which k particles are on the left side of the box if I cannot distinguish anyone of those combinations from another?"
If that is the case, imagine this process: starting with a bag of n identical particles, you draw out the particles one by one and with equal probability put each one in either the left or the right side of the (initially empty) box. Can you now calculate the probability of k particles in the left side of the box?

 

[QuasarBoy543298]

Why do you think the probability would not be negligible? Have you tried to calculate the fraction of states that have exactly $k$ particles on the left side of the box?

if what i said is true. the probability of having k particles on the left side of the box is simply 1/(N+1) (independent of k).

 

[QuasarBoy543298]

I am not entirely clear about what your point of confusion is, but I suspect you are thinking something like this: "How can I count the number of combinations in which k particles are on the left side of the box if I cannot distinguish anyone of those combinations from another?"
If that is the case, imagine this process: starting with a bag of n identical particles, you draw out the particles one by one and with equal probability put each one in either the left or the right side of the (initially empty) box. Can you now calculate the probability of k particles in the left side of the box?

correct , i am unsure why the idea of identicle partices does not states that we basically need to count all the possible configurations where there are k particles in one side as one configuration
(for example ,why the configurations left- A,B right - C and left -A C , right-B can be treated as diffrenet configurations and be counted as two diffrent possibilities although A,B,C are undistinguishable)

and yes, with this process there are 1/2^N times N choose k probability , but I am still not sure why can you think of the problem like that, or why what i said is wronge

and thanks for the quick response :):):)

 

[vanhees71]

For a heuristic treatment of quantum statistics of indistinguishable particles in connection to kinetic theory, see Sect. 1.8

https://itp.uni-frankfurt.de/~hees/publ/kolkata.pdf

 

[PeterDonis]

if what i said is true. the probability of having k particles on the left side of the box is simply 1/(N+1) (independent of k).

No, it isn't. For a simple counterexample, suppose N = 2 (2 particles) with the particles indistinguishable. Then the probability of having 1 particle in the left half of the box (and the other in the right half) is 1/2, not 1/3; but the probability of having 2 particles (i.e., both of them) in the left half of the box is 1/4, not 1/3. So (a) the probability is not independent of k, and (b) the probability is never 1/(N+1) as you claim.

How much do you know about the basics of probability theory?

 

[PeterDonis]

why the configurations left- A,B right - C and left -A C , right-B can be treated as diffrenet configurations and be counted as two diffrent possibilities

If the particles are indistinguishable, these are the same configuration, not different ones.

What references (textbooks or peer-reviewed papers) are you looking at to learn about how QM treats indistinguishable particles?

 

[QuasarBoy543298]

unfortunately I am currently learning statistical mechanics without a proper background in qm (which i will get in the next semester),so we had to take the concept of identical particles as given.
i still can't understand why on the one hand you said that the two configurations are the same but on the other hand you said that for 2 particles there is 1/4 probability for 2 particles to be on the same side (that means that we have 1/2 probabilty for the two particles to be on different sides, hence 2 configurations if we give each configuration the same probability)

 

[vanhees71]

A very good standard text is

A. L. Fetter, J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill
Book Company, New York (1971).