Some questions regarding Statistical and Quantum Mechanics

  • Thread starter Shan K
  • Start date
  • #1
73
0
I was studying statistical mechanics and found some concept that are difficult for me to grasp. Any kind of help will be highly appreciated.
My questions are :
1. In equilibrium which quantity should be maximized? Is it the number of Microstate or the way that a particular Microstate is achieved?
2. What the 'State' means in the 'Density Of States'? Is it the sets of (nx, ny, nz) for a particular 'n' in the energy of the particle in a 3 D box?
3. If Maxwell Boltzmann Statistics is for classical particles then why we are assuming discrete energy levels as well as quantum states in those energy levels to prove the distribution?

Thank you
 

Answers and Replies

  • #2
atyy
Science Advisor
13,985
2,261
(1) In thermodynamics, for an isolated system, the entropy is maximized. But if the system is not isolated, then other quantities such as the various free energies are maximized or minimized. In classical statistical mechanics, the ensemble corresponding to an isolated system is the microcanonical ensemble, in which all states with the energy of the system are equally probable, and can be seen as maximizing the Boltzmann-Gibbs entropy subject to the energy constraint.

(2) "State" in "density of states" usually refers to quantum states with well defined and fixed energies. These are energy levels or energy eigenstates.

(3) The Maxwell-Boltzmann distribution does not hold in quantum statistical mechanics. However, the quantum distributions do become very similar to the Maxwell-Boltzmann distributions at high energy, so one can derive the Maxwell-Boltzmann distribution with discrete energy levels and the assumption that the temperature is high. If you want to see the Maxwell-Boltzmann distribution derived without discrete energy levels, take a look at http://ocw.mit.edu/courses/physics/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2007/lecture-notes/lec13.pdf [Broken] .
 
Last edited by a moderator:
  • #3
73
0
(1) In thermodynamics, for an isolated system, the entropy is maximized. But if the system is not isolated, then other quantities such as the various free energies are maximized or minimized. In classical statistical mechanics, the ensemble corresponding to an isolated system is the microcanonical ensemble, in which all states with the energy of the system are equally probable, and can be seen as maximizing the Boltzmann-Gibbs entropy subject to the energy constraint.
What is Boltzmann-Gibbs entropy? Is it kBln W?

(2) "State" in "density of states" usually refers to quantum states with well defined and fixed energies. These are energy levels or energy eigenstates.
Can more than one state posses same energy?
 
  • #4
jtbell
Mentor
15,613
3,635
Can more than one state posses same energy?
Yes. If you're familiar with the "infinite square well" that everybody studies when learning QM, consider the three-dimensional version, a cubical box. There are three quantum numbers, one for each dimension, and the energies are
$$E_{lmn} = \frac{\pi^2 \hbar^2}{2mL^2} \left( l^2 + m^2 + n^2 \right)$$
The states with (l, m, n) = (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2, 1) all have the same energy. We call these degenerate states.
 
  • #5
atyy
Science Advisor
13,985
2,261
What is Boltzmann-Gibbs entropy? Is it kBln W?

Can more than one state posses same energy?
Since jtbell answered your second question, I'll just answer the first. The Boltzmann-Gibbs entropy is ##-k_B\sum\nolimits p\ln{p}##. In the microcanonical ensemble, all states with the same energy have the same probability, so p is a constant. If there are ##W## states with the same energy, then ##p = 1/W##, and after summing over ##W## number of states, the Boltzmann-Gibbs entropy reduces to ##k_B \ln{W}## for the microcanonical ensemble, which is the formula you wrote.
 
  • #6
73
0
Yes. If you're familiar with the "infinite square well" that everybody studies when learning QM, consider the three-dimensional version, a cubical box. There are three quantum numbers, one for each dimension, and the energies are
$$E_{lmn} = \frac{\pi^2 \hbar^2}{2mL^2} \left( l^2 + m^2 + n^2 \right)$$
The states with (l, m, n) = (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2, 1) all have the same energy. We call these degenerate states.
Ok let me write what I have understood so far:
Each and every set of (l, m, n) gives a single state, and they are called Degenerate if they have same energy otherwise they are said to be non degenerate state.
So the word "state" in the density of states are these Degenerate and Non Degenerate states together, they can have same or different energies.
Is it right?
 
  • #7
73
0
Since jtbell answered your second question, I'll just answer the first. The Boltzmann-Gibbs entropy is ##-k_B\sum\nolimits p\ln{p}##. In the microcanonical ensemble, all states with the same energy have the same probability, so p is a constant. If there are ##W## states with the same energy, then ##p = 1/W##, and after summing over ##W## number of states, the Boltzmann-Gibbs entropy reduces to ##k_B \ln{W}## for the microcanonical ensemble, which is the formula you wrote.
What is the meaning of the ##W## here?
Is it the number of microstate or the number of ways that a microstate can be attended?
 
  • #8
atyy
Science Advisor
13,985
2,261
What is the meaning of the ##W## here?
Is it the number of microstate or the number of ways that a microstate can be attended?
In the microcanonical ensemble, ##W## is the number of microstates with energy E.
 
  • #9
73
0
In the microcanonical ensemble, ##W## is the number of microstates with energy E.
Then why do we maximize the number of ways by which a microstate can be achieved, when we derive the statistical distributions?
 
  • #10
atyy
Science Advisor
13,985
2,261
Then why do we maximize the number of ways by which a microstate can be achieved, when we derive the statistical distributions?
You don't. You maximize the number of ways a particular energy can be achieved.

Let's consider a probability distribution to correspond to an ensemble of systems. If each system has two states A and B with the same energy, one could get a particular energy if every system in the ensemble is in state A. But if 50% of the members of the ensemble are in state A, and 50% are in state B, the entropy of the distribution is maximized.
 
  • Like
Likes 1 person
  • #11
73
0
You don't. You maximize the number of ways a particular energy can be achieved.

Let's consider a probability distribution to correspond to an ensemble of systems. If each system has two states A and B with the same energy, one could get a particular energy if every system in the ensemble is in state A. But if 50% of the members of the ensemble are in state A, and 50% are in state B, the entropy of the distribution is maximized.
Then we can say that at equilibrium with surrounding the system will have only those microstates that are in the favor of the energy that the system has in the equilibrium. We will have other microstates when the system is not in equilibrium but tending to equilibrium. But in equilibrium all the microstates are in favor of the equilibrium energy.
 

Related Threads on Some questions regarding Statistical and Quantum Mechanics

Replies
4
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
1
Views
503
Replies
33
Views
2K
Replies
9
Views
820
Replies
2
Views
1K
Replies
1
Views
1K
Top