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A first look at Quantum Distributions
Some definitions
εi
ni
gi
ni / gi
energy of the i’th energy level
number of particles in the i’th energy level
degeneracy of the i’th energy level (number of different states with the same energy)
occupation index
A distribution is a set of numbers of particles {ni} in each energy level
An arrangement is a particular way of putting the particles into each state of a particular
distribution
If particles are indistinguishable, two arrangements that differ only in the exchange of two
particles would be counted as one arrangement.
Types of particles
Distinguishable: classical particles - Boltzmann statistics
Indistinguishable
One particle per state: Fermi-Dirac statistics
(Pauli exclusion: electrons, protons, other half integer spin particles)
Many particles per state: Bose-Einstein statistics
(photons, helium-4 nuclei, other integer spin particles)
We wish to find the average distribution for a group of particles:
⎧⎪
⎪⎩
⎫⎪
⎪⎭
{n1,n 2 ,…} = ⎨∑ [n1,j* (Probability of distribution) j ],∑ [n 2 ,j*(Probability of distribution) j ],…⎬
j
j
assuming that each arrangement is equally probable.
(Probability of distribution) j =
number of arrangements for distribution “j”
total number of arrangements of all distributions
A simple example:
Equally spaced energy levels: ε1 = 0 ; ε2 = E ; ε 3 = 2E ; ε4 = 3E ; ε5 = 4E ; ε 6 = 5E
Various degeneracies: g1 = 1 ; g2 = 2 ; g3 = 2 ; g4 = 3 ; g5 = 3 ; g6 = 4
Four particles with: E total = 5E
Two arrangements of the same distribution
ε1
ε2
ε3
ε4
ε5
ε6
All possible distributions of four particles
Distributions
Arrangements
n1 n2 n3 n4 n5 n6
a
b
c
d
e
f
3
2
2
1
0
1
0
1
0
2
3
1
0
0
1
0
1
2
0
0
1
1
0
0
0
1
0
0
0
0
1
0
0
0
0
0
Boltz
FD
BE
16
72
72
144
64
96
464
0
0
0
3
0
2
5
4
6
6
9
8
6
39
Probabilities
Boltz
FD
16/464 0
72/464 0
72/464 0
144/464 3/5
64/464 0
96/464 2/5
BE
4/39
6/39
6/39
9/36
8/39
6/39
Explanation of row a
Check total energy: E total = 3ε1 + 0ε2 + 0ε 3 + 0ε4 + 0ε5 + 1ε 6 = 5E √
Bose-Einstein particles (4 arrangements): states are distinguishable, the particles are not.
Fermi-Dirac particles (0 arrangements)
No arrangements possible (only one particle per state allowed)
Boltzmann particles (16 arrangements)
Here are 4 arrangements with particle #4 in the 6th energy level
123
4
123
123
4
123
There would also be 4 arrangements with particle #3 in the 6th energy level.
The same for particle #1 and #2.
Explanation of row d
Check total energy: E total = 1ε1 + 2ε2 + 0ε 3 + 1ε4 + 0ε5 + 0ε 6 = 5E √
Bose-Einstein particles (9 arrangements)
4
4
Fermi-Dirac particles (3 arrangements)
Boltzmann particles
Here are 3 arrangements with particle #4 in the 4th energy level
1
2 3
4
1
2 3
1
2 3
# arrangements
4
4
Switch 2&3 and cycle 4 in the 4th energy level
1
3 2
4
Back to original, switch 1&2 and cycle 4 in the 4th energy level
2
1 3
4
Switch 1&3 and cycle 4 in the 4th energy level
2
3 1
4
Back to original, switch 1&3 and cycle 4 in the 4th energy level
3
2 1
4
Switch 1&2 and cycle 4 in the 4th energy level
3
1 2
4
Now start with
1
23
4
and cycle 4 in the 4th energy level
Switch 1&2 and cycle 4 in the 4th energy level
2
13
4
Switch 1&3 and cycle 4 in the 4th energy level
3
21
4
Do the same in the other half of the second energy level
1
23
4
2
13
4
3
21
4
3
3
3
3
3
3
3
3
3
3
3
3
So far, we have 36 arrangements. We could do the above with particle #3 in the 4th energy level,
followed by particle #2, followed by particle #1.
There are, thus, 4*36=144 arrangements
Done with row d!!! You should try to justify some of the other rows.
From the table, what are the average distributions for the three types of particles?
For the 1st energy level and Boltzmann particles:
⎛ 16 ⎞ ⎛ 72 ⎞ ⎛ 72 ⎞ ⎛ 144 ⎞ ⎛ 64 ⎞ ⎛ 96 ⎞ 576
n1 = 3⎜
⎟ + 2⎜
⎟ + 2⎜
⎟ + 1⎜
⎟ + 0⎜
⎟ + 1⎜
⎟=
⎝ 464 ⎠ ⎝ 464 ⎠ ⎝ 464 ⎠ ⎝ 464 ⎠ ⎝ 464 ⎠ ⎝ 464 ⎠ 464
For the 3rd energy level and Fermi-Dirac particles:
⎛ 0⎞ ⎛ 0⎞ ⎛ 0⎞ ⎛ 3⎞ ⎛ 0⎞ ⎛ 2⎞ 4
n3 = 0⎜ ⎟ + 0⎜ ⎟ + 1⎜ ⎟ + 0⎜ ⎟ + 1⎜ ⎟ + 2⎜ ⎟ =
⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠ 5
That is, the average distributions are
Boltzmann
Bose-Einstein
Fermi-Dirac
n1
576/464
51/39
1
n2
648/464
54/39
8/5
n3
n4
328/464 216/464
26/39
15/39
4/5
3/5
n5
72/464
6/39
0
n6
16/464
4/39
0
Notice that n2 is larger than 1 for all three types of particles.
The average occupation index is
Boltzmann
Bose-Einstein
Fermi-Dirac
n1 / g1
576/464
51/39
1
n2 / g2
324/464
27/39
4/5
n3 / g3
164/464
13/39
2/5
n4 / g4
72/464
5/39
1/5
n5 / g5
24/464
2/39
0
n6 / g6
4/464
1/39
0
n2 / g2
0.70
0.69
0.80
n3 / g3
0.35
0.33
0.40
n4 / g4
0.16
0.13
0.20
n5 / g5
0.05
0.05
0.00
n6 / g6
0.01
0.03
0.00
As rounded real numbers
Boltzmann
Bose-Einstein
Fermi-Dirac
n1 / g1
1.24
1.31
1.00
Notice that the average occupation index for Boltzmann particles is always between that for BoseEinstein particles and Fermi-Dirac particles.
Simple probability
How many ways can you select 1 marble from 4 different marbles?
How many ways can you select 2 marbles from 4 different marbles?
A,B A,C A,D ; B,C B,D ; C,D
N!
4!
=
=4
n!(N − n)! 1!3!
4!
4 * 3* 2*1
=
=6
2!2! (2 *1) *(2 *1)
How many ways can you select 3 marbles from 4 different marbles?
A,B,C A,B,D A,C,D ; B,C,D
How many ways can you select 4 marbles from 4 different marbles?
How many ways can you permute (place in different order) 3 different marbles?
A,B,C ; A,C,B ; B,C,A ; B,A,C ; C,A,B ; C,B,A
4!
=4
3!1!
4!
=1
4!0!
N! = 3! = 6