Number of individual states with the same occupation numbers

Click For Summary
SUMMARY

The discussion focuses on calculating the number of individual microscopic states with the same occupation numbers, denoted as ##n_k##, for a system of noninteracting particles. The formula used is $$\frac{N!}{n_1!n_2! \dots n_k! \dots}$$, which accounts for the indistinguishability of particles within each group. A user attempted to apply this formula to a system of 13 particles with specific occupation numbers but received an unexpected result. Clarifications were sought regarding the correct interpretation of the problem statement and the distinction between bosonic and fermionic particles.

PREREQUISITES
  • Understanding of combinatorial mathematics and factorial notation
  • Familiarity with the concept of occupation numbers in quantum mechanics
  • Knowledge of the differences between bosonic and fermionic particles
  • Basic grasp of statistical mechanics principles
NEXT STEPS
  • Study the principles of statistical mechanics, focusing on particle statistics
  • Learn about the implications of bosonic and fermionic statistics on wave functions
  • Explore combinatorial methods in physics, particularly in quantum systems
  • Investigate the application of the multinomial coefficient in particle distribution problems
USEFUL FOR

Students and researchers in physics, particularly those studying statistical mechanics, quantum mechanics, and particle physics, will benefit from this discussion.

WeiShan Ng
Messages
36
Reaction score
2

Homework Statement


A state of a system of many noninteracting particles can be specified by listing which particle is in which of the accessible single particle states. In each microscopic state we can identify the number of particles in a given single particle state ##k##. This number is called the occupation number, of state k and is denoted by ##n_k##.
How many individual microscopic states have the same set of occupation numbers ##n_k##?

Homework Equations


Solution given:
The number of states with the same set of ##n_k## is the number of ways the N particles can be distributed in groups of ##n_k## each. It is the combinatorial factor that expresses the fact that all N particles can be interchanged, but interchanges of particles within each group do not produce new states. Hence the number is
$$\frac{N!}{n_1!n_2! \dots n_k! \dots}$$

The Attempt at a Solution


So I tried using this equation to see if I can find the correct answer. I took a system of ##N=13## particles
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
k & 1 & 2 & 3 & 4 & 5 \\
n_k & 3 & 2 & 1 & 4 & 3 \\
\hline
\end{array}
$$
where ##k## are the individual microscopic states and ##n_k## is their occupation numbers respectively
And when I applied the equation, I get
$$\frac{13!}{3!2!1!4!}=21621600$$ which is so far off from the number expected (2)
I must have understand the equation wrong. Can someone please explain to me how exactly does this equation work?
 
Physics news on Phys.org
So a couple of things:

1) I'm not sure about your problem statement. You say that k is the number of particles in a given state. This should be n_k. In occupation number representation the number operator n_k returns the number of particles in the k^{th} momentum state.

2) Are you using bosonic particles or fermionic particles? Because the statistics change depending on which; for example you can have two fermions in a state (spin up and down), and exchanging them is non trivial and creates a different wave function. For bosonic particles you can have infinitely many in a state (infinite dimensional Hilbert space) and swapping them leaves the wavefunction unchanged.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K