Transcript To find the average number of particles in each state

We’ve spent quite a bit of time learning about how the
individual fundamental particles that compose the
universe behave.
Naturally, I am sure you are wondering:
Can we start with that “microscopic”
knowledge and learn something about bulk
properties of matter such as temperature?
Will the peculiar rules governing the
behavior of an individual particle or small
group of particles
(such as the Pauli principle) influence
the properties of the matter they
compose?
Particles have identical physical
properties…but can be
distinguished by following their
(well defined) classical paths.
ASSUMPTIONS
In equilibrium, the energy distribution
of the particles will converge
to the most probable allowed.
In principle, there is no
limit on the number of particles
occupying each state.
Imagine 6 particles with 9 indivisible quanta of energy divided among them.
•If the particles are indistinguishable, we only care about how many particles
are in each state, and there are 26 unique ways to distribute the energy
among them—26 unique combinations.
•If the particles are distinguishable (we make a distinction as to which particle
is in which energy state), there are 2002 unique permutations.
Remember that for now we have assumed that these particles are distinguishable.
If you are making choices from n objects, then on your first pick you have n
choices. On your second pick, you have n-1 choices, n-2 for your third choice
and so forth. As illustrated before for 5 objects, the number of ways to pick
from 5 objects is 5! .
Suppose you are going to pick a subset r out of the total number of objects n,
like drawing 5 cards from a deck of 52. For the first pick, you have n choices,
then n-1 and so on down to n-r+1 for the last pick. The number of ways you
can do it is:
n!
n(n  1)( n  2)...( n  r  1) 
 n Pr
(n  r )!
If we care about which
particle is in which
‘state’, there are six
different states where
one particle one particle
has all of the energy.
Here we have to choose
more than one particle
for each state, and we
can distinguish between
different combinations
of particles in each
state, so the
multiplicity gets bigger.
If each of these “microstates” is equally likely (and we assumed that we
will converge on the most probable), it seems that nature doesn’t favor the
situation where one particle has all of the energy.
We assume that each “microstate” (unique permutation) is equally probable.
In other words, if there are 180 permutations that will produce a particular
energy distribution, then that distribution is more probable than a distribution
that can only be produced by six permutations.
Energy
level
Average
number
0
2.143
1
1.484
2
0.989
3
0.629
4
0.378
5
0.210
6
0.105
7
0.045
8
0.015
9
0.003
To find the average number of particles in each state
average number of
particles in the
jth energy level
n j  n j1 p1  n j 2 p2  
count the
number of
particles in
each state for
this
distribution
multiply by the number
of permutations that
can produce this
distribution divided by
the total number of
permutation for all
distributions
There are only 6 (out of 2002) permutations that can
produce a situation where one particle has all 9E.
Apparently it’s not very probable.
y

v  constant
y
x
z
z

dv
x
rms 
3k BT
m
The Maxwell-Boltzmann distribution
can be shown graphically as the plot of
the number of molecules traveling at a
given speed versus the speed. As the
temperature increases, this curve
broadens and extends to higher
speeds.
Using:
rms 
3k BT
m
The equipartition theorem
follows
1
3
m 2  K  k BT
2
2
A classical molecule in thermal
equilibrium has an average energy
of kT/2 per degree of freedom.
But generally, there are more than three degrees of freedom (more than just the
translational motion in each of x, y, and z):
Erot 
1 2
I
2
1
1 2
2
E  m x  kx
2
2
For molecules that can rotate, you can have a rotational
degree of freedom.
A one dimensional harmonic oscillator has two
degrees of freedom, one corresponding to its
potential energy, the other to its kinetic energy.
There is an energy kT/2 associated with each of these degrees of freedom.
The first two pictures give the same outcome. Even though a and b are
identical, you can tell them apart by following them along their unique paths.
a
b
q
q
a
b
a
b
a
??
pq
b
a
Quantum mechanically, each particle has some probability of being
somewhere at a particular time, which overlaps greatly at the collision point.
Which particle emerges where? In wave terms, they interfered.
b
Since we know that particles are really “wavicles” and Maxwell Boltzmann
statistics is only good for distinguishable particles, what good is it?
The Maxwell-Boltzmann distribution assumes that the particles they describe
are distinguishable. Two particles can be considered distinguishable if the
distance separating the particles is large compared to their DeBroglie
wavelength.
Put in mathematical terms, if their average separation is larger than the
the uncertainty in their momentum. d  x
px 
Using the uncertainty principle:
At thermal equilibrium:
the particles are moving randomly and the
directions will cancel each other
px  0
p x2  0
px2 k BT
KE 

2m
2

 x 
2 mkBT

2
but their magnitudes aren’t zero
for each degree of freedom
V 
using d   
N
1
3
 p x 
p x2   p x   mkBT

N
 1
 
3
 V  8 mkBT  2
2
Maxwell Boltzmann
statistics are valid for
low density and high
temperature and
particle mass
Before, we had distinguishable particles….
   a (r1 ) b (r2 )
   a (r2 ) b (r1 )
the probabililty of both particles
being in the same state is then…
assuming they are in two different states, you get two
distinct wavefunctions, where r1 and r2 are the positions
of the toe particles…
 *   a* (r1 ) a* (r2 ) a (r1 ) a (r2 )   a (r1 )  a (r2 )
2
2
…but indistinguishable particles require a more complicated wavefunction.

1
[ a (r1 ) b (r2 )  a (r2 ) b (r1 )]
2
or

1
[ a (r1 ) b (r2 )  a (r2 ) b (r1 )]
2
If both particles are in the same state these become…
1

[ a (r1 ) a (r2 )  a (r2 ) a (r1 )]  2  a (r1 ) a (r2 )
2
“bosons” as they are called, are
more likely to be found in the
same energy state than apart…
  *  2 a (r1 )  a (r2 )  2 *
2
1

[ a (r1 ) a (r2 )  a (r2 ) a (r1 )]  0
2
2
“fermions”, however, can never be found
in the same state! The uncertainty
principle…again.
We are now considering indistinguishable particles we are now only counting
how many particles are in each energy state, rather than which particles are
in each energy state.
However, our assumption that there is no theoretical limit on the number of
particles occupying each state still holds.
Now each of the 26 energy combinations show occur with equal probability.
Average
number
MaxwellBoltzmann
Average
number
BoseEinstein
0
2.143
2.269
1
1.484
1.538
2
0.989
0.885
3
0.629
0.538
4
0.378
0.269
5
0.210
0.192
6
0.105
0.115
7
0.045
0.077
8
0.015
0.038
9
0.003
0.038
Energy
level
n j  n j1 p1  n j 2 p2  
This probability is now
the same for each
energy distribution, just
1 divided by the total
number of distributions.
Not surprisingly, the distribution for
bosons show a higher probability at
the extremes, where there were fewer
permutations producing the observed
energy spectrum.
The specific heat is the amount of heat per unit mass
required to raise the temperature by one degree
Celsius.
dU
C
dT
Thermal energy per atom:
number of degrees of freedom per
dimension (kinetic and potential)
kBT / 2
here, U is the thermal energy
in real life, there are
three dimensions
 2  3  3kBT
energy per degree of freedom
(from the equipartition theorem)
To get the energy per mole, multiply by
Avagadro’s number:
U  3N AkBT  3RT
dU
d
3RT   3R  5.97 cal mol  K

dT dT
Note that the specific heat predicted by
the classical equipartition theorem is
constant with temperature!
This prediction
disagrees with the
data at low
temperature.
Einstein’s solution:
The energy is then:
The specific heat is then:

E

e
k BT
1
Treat each atom as
an independent
quantum simple
harmonic oscillator,
and quantize its
energy.
this approaches the classical
result, kT, at high temperatures
as we know it should (the
correspondence principle)

2
  
dU
e k BT

C
 3R
2
 / k B T
dT
k
T
e

1
 B 


Einstein’s approach worked (mostly). Some fine tuning was required, however.
Debye pointed out that the atoms in a solid do not move independently but interacts
with its neighbors. The result - continuous vibrational waves-sound.
•Once again we assume that the particles are distinguishable.
•Thanks to the Pauli exclusion principle, we must, however, remove
assumption that there is no theoretical limit on the number of particles in
each state!
As in the case for bosons, each
allowed energy distribution has an
equal probability.
Unlike the boson case, only five
energy distributions are allowed!
All others have more than the
permitted two particles in each
state in violation of the exclusion
principle.
E
Average
number
MaxwellBoltzmann
Average
number
BoseEinstein
Average
number
FermiDirac
0
2.143
2.269
1.8
1
1.484
1.538
1.6
2
0.989
0.885
1.2
3
0.629
0.538
0.8
4
0.378
0.269
0.4
5
0.210
0.192
0.2
6
0.105
0.115
0
7
0.045
0.077
0
8
0.015
0.038
0
9
0.003
0.038
0
n j  n j1 p1  n j 2 p2  
Now there are few allowed
states. Since the particles are
indistinguishable, each state is
again equally likely.
Note that the energy distribution
will flatten out at higher energies.
If one particle has all of the
energy state, that would require
all of the other particle to be in the
same energy state – zero - in
violation of the Pauli principle.