Most Probable State Determination (II)
Assuming Harmonic Oscillator energy levels (i.e. equi-spaced), with separation A, so that the energies are 0, 2A, 4A, etc., and assuming we have 6 oscillators to distribute amongst these states, obtain the populations of each of the states in the most probable `complection/configuration', i.e., the one where S = k ln(W) is defined. Assume that the total energy is 12A.
(See page 10, Dykstra's 'Physical Chemistry, A Modern Introduction')
Number in zero state (E=0)=
0
1
2
3
4
5
6
7
8
9
Number in first excited state (E=2A) =
0
1
2
3
4
5
6
7
8
9
Number in second excited state (E=4A) =
0
1
2
3
4
5
6
7
8
9
Number in third excited state (E=6A) =
0
1
2
3
4
5
6
7
8
9
Number in fourth excited state (E=8A) =
0
1
2
3
4
5
6
7
8
9
Number in fifth excited state (E=10A) =
0
1
2
3
4
5
6
7
8
9
Number in sixth excited state (E=12A) =
0
1
2
3
4
5
6
7
8
9
Number in seventh excited state (E=14A) =
0
1
2
3
4
5
6
7
8
9
Query, are the above entries correct?