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KFC
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Hi all,
I am reading an introduction on classical and quantum-mechanical statistics. The material considers a 4-particle system with discrete energy level 0E, 1E, 2E, 3E, 4E, 5E and 6E. It is said that the classical particle is indistinguishable but you can identify the different particle by their position or trajectory. And for quantum particle (the material takes bosons as example), they are totally indistinguishable.
Assuming the total energy of the system is 6E. For classical system, to find the possible configuration of those particles giving total 6E energy, we have
1) 2 particle2 on 2E level, 2 particles on 1E level. 6 possible ways to get this configuration.
2) 3 particles on 2E level, 1 particle on 0E level and 2 particles on 0E level. Total 4 ways.
3) 1 particle on 3E level, 3 particles on 1E level. Total 4 ways.
4) 1 particle on 3E level, 1 particle on 2E level, 1 particle on 1E level, 1 particle on 0E level. Total 24 ways.
5) 2 particles on 3E level, 2 particles on 0E level. Total 6 ways.
6) 1 particle on 4E level, 2 particles on 1E level, 1 particle on 0E level. Total 12 ways.
7) 1 particle on 4E level, 1 particle on 2E level, 2 particles on 0E level. Total 12 ways.
8) 1 particle on 5E level, 1 particle on 1E level, 2 particles on 0E level. Total 12 ways.
9) 1 particle on 6E level, 3 particles on 0E level. Total 4 ways.
For each configuration, giving 6E energy, it is called macroscopic states, each way to get that configuration is called microscopic state (?). We got total 9 different configuration (or 9 macroscopic states). There are total 84 microscopic states.
We want to find what's probability to find the particle in 0E level while total 6E energy is given. Here is my math. There are 84 microscopic states, the average number of particles found on 0E level is
##N = (0)\frac{6}{84} +
(1)\frac{4}{84} +
(0)\frac{4}{84} +
(1)\frac{24}{84} +
(2)\frac{6}{84} +
(1)\frac{12}{84} +
(2)\frac{12}{84} +
(2)\frac{12}{84} +
(3)\frac{4}{84} = 1.619##
The number if parenthesis denotes the number of particles in 0E level. There are total 4 particles, so the probability to find that particles in 0E level is ##1.619/6 \simeq 26.98\%##
Now let's consider the quantum case (bosons), the material said all particles are indistinguishable, so the author only counts the macroscopic state. The average number of particles found to be in 0E level is
##(0)\frac{1}{9} +
(1)\frac{1}{9} +
(0)\frac{1}{9} +
(1)\frac{1}{9} +
(2)\frac{1}{9} +
(1)\frac{1}{9} +
(2)\frac{1}{9} +
(2)\frac{1}{9} +
(3)\frac{1}{9} =1.333##
so the probability to find the particle in 0E level is ##1.333/6 \simeq 22.22\% ##.
Here is my question. I don't quite understand why we only counts the number of macroscopic state for quantum case. And in classical system, what makes (for example in the first configuration) it to count all 6 possible microstates? That is to ask where is that "6" come from in the classical case when 2 particles in 2E level and 2 in 1E level? To answer my question, I am trying to give the following reasoning. In classical case, we label each particle as p1, p2, p3 and p4. We could pick p1-p2, or p1-p3 or p1-p4 or p2-p3 or p2-p4 or p3-p4 to fill 2E level and leave the rest to the 1E level, this will give 6 ways to pick them up. But in this sense, we agree that each particles is unique to others. If that's the case, when we pick 2 out of 4 to fill 2E, says p1 and p2, since p1 is different from p2, why don't we worry about p1-p2 and p2-p1 is not the same configuration while filling 2E level?
I will appreciate if you could confirm my reasoning is correct (at least on finding the number of microstates).
And for the quantum case, using the same example as above, since we cannot tell which one is p1, p2, p3 and p4. So all those 6 microstates are the same, there is only 1 state actually. So does it mean in quantum system, there is no difference between macrostate and microstate?
To extend the bosonic system to Fermi system, I think we have the similar procedure except that we need to consider no more than 2 particles can fill on the same level, right?
I am reading an introduction on classical and quantum-mechanical statistics. The material considers a 4-particle system with discrete energy level 0E, 1E, 2E, 3E, 4E, 5E and 6E. It is said that the classical particle is indistinguishable but you can identify the different particle by their position or trajectory. And for quantum particle (the material takes bosons as example), they are totally indistinguishable.
Assuming the total energy of the system is 6E. For classical system, to find the possible configuration of those particles giving total 6E energy, we have
1) 2 particle2 on 2E level, 2 particles on 1E level. 6 possible ways to get this configuration.
2) 3 particles on 2E level, 1 particle on 0E level and 2 particles on 0E level. Total 4 ways.
3) 1 particle on 3E level, 3 particles on 1E level. Total 4 ways.
4) 1 particle on 3E level, 1 particle on 2E level, 1 particle on 1E level, 1 particle on 0E level. Total 24 ways.
5) 2 particles on 3E level, 2 particles on 0E level. Total 6 ways.
6) 1 particle on 4E level, 2 particles on 1E level, 1 particle on 0E level. Total 12 ways.
7) 1 particle on 4E level, 1 particle on 2E level, 2 particles on 0E level. Total 12 ways.
8) 1 particle on 5E level, 1 particle on 1E level, 2 particles on 0E level. Total 12 ways.
9) 1 particle on 6E level, 3 particles on 0E level. Total 4 ways.
For each configuration, giving 6E energy, it is called macroscopic states, each way to get that configuration is called microscopic state (?). We got total 9 different configuration (or 9 macroscopic states). There are total 84 microscopic states.
We want to find what's probability to find the particle in 0E level while total 6E energy is given. Here is my math. There are 84 microscopic states, the average number of particles found on 0E level is
##N = (0)\frac{6}{84} +
(1)\frac{4}{84} +
(0)\frac{4}{84} +
(1)\frac{24}{84} +
(2)\frac{6}{84} +
(1)\frac{12}{84} +
(2)\frac{12}{84} +
(2)\frac{12}{84} +
(3)\frac{4}{84} = 1.619##
The number if parenthesis denotes the number of particles in 0E level. There are total 4 particles, so the probability to find that particles in 0E level is ##1.619/6 \simeq 26.98\%##
Now let's consider the quantum case (bosons), the material said all particles are indistinguishable, so the author only counts the macroscopic state. The average number of particles found to be in 0E level is
##(0)\frac{1}{9} +
(1)\frac{1}{9} +
(0)\frac{1}{9} +
(1)\frac{1}{9} +
(2)\frac{1}{9} +
(1)\frac{1}{9} +
(2)\frac{1}{9} +
(2)\frac{1}{9} +
(3)\frac{1}{9} =1.333##
so the probability to find the particle in 0E level is ##1.333/6 \simeq 22.22\% ##.
Here is my question. I don't quite understand why we only counts the number of macroscopic state for quantum case. And in classical system, what makes (for example in the first configuration) it to count all 6 possible microstates? That is to ask where is that "6" come from in the classical case when 2 particles in 2E level and 2 in 1E level? To answer my question, I am trying to give the following reasoning. In classical case, we label each particle as p1, p2, p3 and p4. We could pick p1-p2, or p1-p3 or p1-p4 or p2-p3 or p2-p4 or p3-p4 to fill 2E level and leave the rest to the 1E level, this will give 6 ways to pick them up. But in this sense, we agree that each particles is unique to others. If that's the case, when we pick 2 out of 4 to fill 2E, says p1 and p2, since p1 is different from p2, why don't we worry about p1-p2 and p2-p1 is not the same configuration while filling 2E level?
I will appreciate if you could confirm my reasoning is correct (at least on finding the number of microstates).
And for the quantum case, using the same example as above, since we cannot tell which one is p1, p2, p3 and p4. So all those 6 microstates are the same, there is only 1 state actually. So does it mean in quantum system, there is no difference between macrostate and microstate?
To extend the bosonic system to Fermi system, I think we have the similar procedure except that we need to consider no more than 2 particles can fill on the same level, right?