The energy of a particle is a function of its "state"

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Homework Help Overview

The problem involves five identical particles, each capable of having energies of 1, 2, 3, or 4, with the total energy of the system needing to equal 12. The energy of each particle is dependent on its "state," with multiple states corresponding to the same energy level. The task is to determine the number of distinct states of the system based on these constraints.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various combinations of energies that total 12 and explore the implications of identical particles on counting distinct states. Some suggest using combinatorial methods, while others consider generating functions or brute force approaches.

Discussion Status

There is an ongoing exploration of different methods to calculate the number of system states. Some participants have proposed specific energy combinations and are questioning the validity of their counting methods. Others are considering alternative approaches, indicating a productive exchange of ideas without reaching a consensus.

Contextual Notes

Participants note that the problem's combinatorial nature complicates the counting of states due to the indistinguishability of the particles and the multiple states corresponding to the same energy levels. There is also mention of the limitations of traditional combinatorial tools in this context.

quasar987
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This is for a physics homework but the problem is essentially combinatorial in nature.

There are 5 identical particles. Each can have an energy of 1,2,3 or 4.
The energy of a particle is a function of its "state", and it is possible that to two different states correspond the same energy. As a matter of fact, we are told that there is 1 state of energy 1, 3 of energy 2, 4 of energy 3 and 5 of energy 4. If we call state of the system a particular combination of states that the particles are in, how many states of the system are there, provided that the total energy of the system (i.e. the sum of the individual energies of all five particles) is 12?

Note that since the particles are identical, we do not make a distinction btw the subset of system states arising from the case "particle 1 has energy 1 and particle 2 has energy 3" and the subset arising when "particle 1 has energy 3 and partcile 2 has energy 1".

It seems none of the tools of combinatorics are fit for this problem.
 
Last edited:
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For the possible energy combination, I found

(4,4,2,1,1)
(4,3,3,1,1)
(4,3,2,2,1)
(4,2,2,2,2)
(3,3,2,2,2)
(3,3,3,2,1)

Yeah, I'm pretty sure I didn't miss one.
 
Okay, so you have 6 energy combinations, let's call them E1, ..., E6. If f(Ei) is the number of state combinations that give rise to an energy combination Ei, then the number of state combinations that are possible in total is [itex]\sum _{i=1} ^6 f(Ei)[/itex]. If we say E1 = (4,4,2,1,1), then f(E1) is clearly [itex]({{5}\choose{2}} + 5)({{3}\choose{1}})({{1}\choose{2}} + 1)[/itex], right?
 
It feels like there should be a generating function approach to this problem, but I'm not seeing it off hand.
 
I would say f(E_1) is just 5x5x3x1x1, because we don't care which particle has which energy. See the "note" in the OP.
 
I would write out a 4-tuple representing the number of particles at each energy, ie, (a,b,c,d) corresponds to a system with a particles at E=1, b at E=2, and so on. Since there is more than 1 state with a given energy, each 4-tuple actually corresponds to several distinct system states. The number of states corresponding to (a,b,c,d) can be found as follows: first, there is only 1 way to put the a particles in the 1 E=1 state. Next, the number of ways to put b particles into the three E=2 states is equal to the number of ordered triples whose entries are non-negative integers summing to b, which is a relatively simple combinatorics problem. Continuing this way, get a general formula for the number of states represented by (a,b,c,d), then translate what you found in your second post into this language and add it up.
 
Oh right... just f(E_1) = 5x5x3x1x1 counts way too much states.
 
Well, there's always the easy, brute force approach:
Code:
states = [(1, 1), (2, 2), (3, 2), (4, 2), (5, 3), (6, 3), (7, 3), (8, 3), (9, 4), (10, 4), (11, 4), (12, 4), (13, 4)]
allsystem = [(x, y, z, t, u) | (x, a) <- states, (y, b) <- states, (z, c) <- states, (t, d) <- states, (u, e) <- states, a + b + c + d + e == 12 && x >= y && y >= z && z >= t && t >= u]

f = [1..4]
energies = [(a, b, c, d, e) | a<-f, b<-f, c<-f, d<-f, e<-f, a + b + c + d + e == 12, a>=b, b>=c, c>=d, d>=e]
Here I associate each of the thirteen states with its energy level and put that in the variable states, and then find all ordered quintuplets of states such that the corresponding energy levels sum to 12, and each entry in the quintuplet is >= the next element so that I don't get duplicated system states. Finding the energies is done in a similar way.

Then in the haskell interpreter I find that energies = [(3,3,2,2,2),(3,3,3,2,1),(4,2,2,2,2),(4,3,2,2,1),(4,3,3,1,1),(4,4,2,1,1)] (as you had done by hand) and that the length of allsystem is 450. Specifically the set of all system states is:
Code:
[(5,5,2,2,2),(5,5,3,2,2),(5,5,3,3,2),(5,5,3,3,3),(5,5,4,2,2),(5,5,4,3,2),(5,5,4,
3,3),(5,5,4,4,2),(5,5,4,4,3),(5,5,4,4,4),(5,5,5,2,1),(5,5,5,3,1),(5,5,5,4,1),(6,
5,2,2,2),(6,5,3,2,2),(6,5,3,3,2),(6,5,3,3,3),(6,5,4,2,2),(6,5,4,3,2),(6,5,4,3,3)
,(6,5,4,4,2),(6,5,4,4,3),(6,5,4,4,4),(6,5,5,2,1),(6,5,5,3,1),(6,5,5,4,1),(6,6,2,
2,2),(6,6,3,2,2),(6,6,3,3,2),(6,6,3,3,3),(6,6,4,2,2),(6,6,4,3,2),(6,6,4,3,3),(6,
6,4,4,2),(6,6,4,4,3),(6,6,4,4,4),(6,6,5,2,1),(6,6,5,3,1),(6,6,5,4,1),(6,6,6,2,1)
,(6,6,6,3,1),(6,6,6,4,1),(7,5,2,2,2),(7,5,3,2,2),(7,5,3,3,2),(7,5,3,3,3),(7,5,4,
2,2),(7,5,4,3,2),(7,5,4,3,3),(7,5,4,4,2),(7,5,4,4,3),(7,5,4,4,4),(7,5,5,2,1),(7,
5,5,3,1),(7,5,5,4,1),(7,6,2,2,2),(7,6,3,2,2),(7,6,3,3,2),(7,6,3,3,3),(7,6,4,2,2)
,(7,6,4,3,2),(7,6,4,3,3),(7,6,4,4,2),(7,6,4,4,3),(7,6,4,4,4),(7,6,5,2,1),(7,6,5,
3,1),(7,6,5,4,1),(7,6,6,2,1),(7,6,6,3,1),(7,6,6,4,1),(7,7,2,2,2),(7,7,3,2,2),(7,
7,3,3,2),(7,7,3,3,3),(7,7,4,2,2),(7,7,4,3,2),(7,7,4,3,3),(7,7,4,4,2),(7,7,4,4,3)
,(7,7,4,4,4),(7,7,5,2,1),(7,7,5,3,1),(7,7,5,4,1),(7,7,6,2,1),(7,7,6,3,1),(7,7,6,
4,1),(7,7,7,2,1),(7,7,7,3,1),(7,7,7,4,1),(8,5,2,2,2),(8,5,3,2,2),(8,5,3,3,2),(8,
5,3,3,3),(8,5,4,2,2),(8,5,4,3,2),(8,5,4,3,3),(8,5,4,4,2),(8,5,4,4,3),(8,5,4,4,4)
,(8,5,5,2,1),(8,5,5,3,1),(8,5,5,4,1),(8,6,2,2,2),(8,6,3,2,2),(8,6,3,3,2),(8,6,3,
3,3),(8,6,4,2,2),(8,6,4,3,2),(8,6,4,3,3),(8,6,4,4,2),(8,6,4,4,3),(8,6,4,4,4),(8,
6,5,2,1),(8,6,5,3,1),(8,6,5,4,1),(8,6,6,2,1),(8,6,6,3,1),(8,6,6,4,1),(8,7,2,2,2)
,(8,7,3,2,2),(8,7,3,3,2),(8,7,3,3,3),(8,7,4,2,2),(8,7,4,3,2),(8,7,4,3,3),(8,7,4,
4,2),(8,7,4,4,3),(8,7,4,4,4),(8,7,5,2,1),(8,7,5,3,1),(8,7,5,4,1),(8,7,6,2,1),(8,
7,6,3,1),(8,7,6,4,1),(8,7,7,2,1),(8,7,7,3,1),(8,7,7,4,1),(8,8,2,2,2),(8,8,3,2,2)
,(8,8,3,3,2),(8,8,3,3,3),(8,8,4,2,2),(8,8,4,3,2),(8,8,4,3,3),(8,8,4,4,2),(8,8,4,
4,3),(8,8,4,4,4),(8,8,5,2,1),(8,8,5,3,1),(8,8,5,4,1),(8,8,6,2,1),(8,8,6,3,1),(8,
8,6,4,1),(8,8,7,2,1),(8,8,7,3,1),(8,8,7,4,1),(8,8,8,2,1),(8,8,8,3,1),(8,8,8,4,1)
,(9,2,2,2,2),(9,3,2,2,2),(9,3,3,2,2),(9,3,3,3,2),(9,3,3,3,3),(9,4,2,2,2),(9,4,3,
2,2),(9,4,3,3,2),(9,4,3,3,3),(9,4,4,2,2),(9,4,4,3,2),(9,4,4,3,3),(9,4,4,4,2),(9,
4,4,4,3),(9,4,4,4,4),(9,5,2,2,1),(9,5,3,2,1),(9,5,3,3,1),(9,5,4,2,1),(9,5,4,3,1)
,(9,5,4,4,1),(9,5,5,1,1),(9,6,2,2,1),(9,6,3,2,1),(9,6,3,3,1),(9,6,4,2,1),(9,6,4,
3,1),(9,6,4,4,1),(9,6,5,1,1),(9,6,6,1,1),(9,7,2,2,1),(9,7,3,2,1),(9,7,3,3,1),(9,
7,4,2,1),(9,7,4,3,1),(9,7,4,4,1),(9,7,5,1,1),(9,7,6,1,1),(9,7,7,1,1),(9,8,2,2,1)
,(9,8,3,2,1),(9,8,3,3,1),(9,8,4,2,1),(9,8,4,3,1),(9,8,4,4,1),(9,8,5,1,1),(9,8,6,
1,1),(9,8,7,1,1),(9,8,8,1,1),(9,9,2,1,1),(9,9,3,1,1),(9,9,4,1,1),(10,2,2,2,2),(1
0,3,2,2,2),(10,3,3,2,2),(10,3,3,3,2),(10,3,3,3,3),(10,4,2,2,2),(10,4,3,2,2),(10,
4,3,3,2),(10,4,3,3,3),(10,4,4,2,2),(10,4,4,3,2),(10,4,4,3,3),(10,4,4,4,2),(10,4,
4,4,3),(10,4,4,4,4),(10,5,2,2,1),(10,5,3,2,1),(10,5,3,3,1),(10,5,4,2,1),(10,5,4,
3,1),(10,5,4,4,1),(10,5,5,1,1),(10,6,2,2,1),(10,6,3,2,1),(10,6,3,3,1),(10,6,4,2,
1),(10,6,4,3,1),(10,6,4,4,1),(10,6,5,1,1),(10,6,6,1,1),(10,7,2,2,1),(10,7,3,2,1)
,(10,7,3,3,1),(10,7,4,2,1),(10,7,4,3,1),(10,7,4,4,1),(10,7,5,1,1),(10,7,6,1,1),(
10,7,7,1,1),(10,8,2,2,1),(10,8,3,2,1),(10,8,3,3,1),(10,8,4,2,1),(10,8,4,3,1),(10
,8,4,4,1),(10,8,5,1,1),(10,8,6,1,1),(10,8,7,1,1),(10,8,8,1,1),(10,9,2,1,1),(10,9
,3,1,1),(10,9,4,1,1),(10,10,2,1,1),(10,10,3,1,1),(10,10,4,1,1),(11,2,2,2,2),(11,
3,2,2,2),(11,3,3,2,2),(11,3,3,3,2),(11,3,3,3,3),(11,4,2,2,2),(11,4,3,2,2),(11,4,
3,3,2),(11,4,3,3,3),(11,4,4,2,2),(11,4,4,3,2),(11,4,4,3,3),(11,4,4,4,2),(11,4,4,
4,3),(11,4,4,4,4),(11,5,2,2,1),(11,5,3,2,1),(11,5,3,3,1),(11,5,4,2,1),(11,5,4,3,
1),(11,5,4,4,1),(11,5,5,1,1),(11,6,2,2,1),(11,6,3,2,1),(11,6,3,3,1),(11,6,4,2,1)
,(11,6,4,3,1),(11,6,4,4,1),(11,6,5,1,1),(11,6,6,1,1),(11,7,2,2,1),(11,7,3,2,1),(
11,7,3,3,1),(11,7,4,2,1),(11,7,4,3,1),(11,7,4,4,1),(11,7,5,1,1),(11,7,6,1,1),(11
,7,7,1,1),(11,8,2,2,1),(11,8,3,2,1),(11,8,3,3,1),(11,8,4,2,1),(11,8,4,3,1),(11,8
,4,4,1),(11,8,5,1,1),(11,8,6,1,1),(11,8,7,1,1),(11,8,8,1,1),(11,9,2,1,1),(11,9,3
,1,1),(11,9,4,1,1),(11,10,2,1,1),(11,10,3,1,1),(11,10,4,1,1),(11,11,2,1,1),(11,1
1,3,1,1),(11,11,4,1,1),(12,2,2,2,2),(12,3,2,2,2),(12,3,3,2,2),(12,3,3,3,2),(12,3
,3,3,3),(12,4,2,2,2),(12,4,3,2,2),(12,4,3,3,2),(12,4,3,3,3),(12,4,4,2,2),(12,4,4
,3,2),(12,4,4,3,3),(12,4,4,4,2),(12,4,4,4,3),(12,4,4,4,4),(12,5,2,2,1),(12,5,3,2
,1),(12,5,3,3,1),(12,5,4,2,1),(12,5,4,3,1),(12,5,4,4,1),(12,5,5,1,1),(12,6,2,2,1
),(12,6,3,2,1),(12,6,3,3,1),(12,6,4,2,1),(12,6,4,3,1),(12,6,4,4,1),(12,6,5,1,1),
(12,6,6,1,1),(12,7,2,2,1),(12,7,3,2,1),(12,7,3,3,1),(12,7,4,2,1),(12,7,4,3,1),(1
2,7,4,4,1),(12,7,5,1,1),(12,7,6,1,1),(12,7,7,1,1),(12,8,2,2,1),(12,8,3,2,1),(12,
8,3,3,1),(12,8,4,2,1),(12,8,4,3,1),(12,8,4,4,1),(12,8,5,1,1),(12,8,6,1,1),(12,8,
7,1,1),(12,8,8,1,1),(12,9,2,1,1),(12,9,3,1,1),(12,9,4,1,1),(12,10,2,1,1),(12,10,
3,1,1),(12,10,4,1,1),(12,11,2,1,1),(12,11,3,1,1),(12,11,4,1,1),(12,12,2,1,1),(12
,12,3,1,1),(12,12,4,1,1),(13,2,2,2,2),(13,3,2,2,2),(13,3,3,2,2),(13,3,3,3,2),(13
,3,3,3,3),(13,4,2,2,2),(13,4,3,2,2),(13,4,3,3,2),(13,4,3,3,3),(13,4,4,2,2),(13,4
,4,3,2),(13,4,4,3,3),(13,4,4,4,2),(13,4,4,4,3),(13,4,4,4,4),(13,5,2,2,1),(13,5,3
,2,1),(13,5,3,3,1),(13,5,4,2,1),(13,5,4,3,1),(13,5,4,4,1),(13,5,5,1,1),(13,6,2,2
,1),(13,6,3,2,1),(13,6,3,3,1),(13,6,4,2,1),(13,6,4,3,1),(13,6,4,4,1),(13,6,5,1,1
),(13,6,6,1,1),(13,7,2,2,1),(13,7,3,2,1),(13,7,3,3,1),(13,7,4,2,1),(13,7,4,3,1),
(13,7,4,4,1),(13,7,5,1,1),(13,7,6,1,1),(13,7,7,1,1),(13,8,2,2,1),(13,8,3,2,1),(1
3,8,3,3,1),(13,8,4,2,1),(13,8,4,3,1),(13,8,4,4,1),(13,8,5,1,1),(13,8,6,1,1),(13,
8,7,1,1),(13,8,8,1,1),(13,9,2,1,1),(13,9,3,1,1),(13,9,4,1,1),(13,10,2,1,1),(13,1
0,3,1,1),(13,10,4,1,1),(13,11,2,1,1),(13,11,3,1,1),(13,11,4,1,1),(13,12,2,1,1),(
13,12,3,1,1),(13,12,4,1,1),(13,13,2,1,1),(13,13,3,1,1),(13,13,4,1,1)]
 
Last edited:
Amazing. And 450 is right.
 

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