Transcript Chapter 13 Classical & Quantum Statistics

Chapter 13 Classical & Quantum
Statistics
13.1 Boltzmann Statistics
• It deals distinguishable, non-interacting particles.
• There are two constrains
NJ = N
NJ · ε J = U
where NJ is the number of particles with
single-particle energy εJ.
• The goal: find the occupation number of each
energy level when the thermodynamic
probability is a maximum (i.e. the configuration
of an equilibrium state)!
• The number of ways of selecting NJ particles
from a total of N to be placed in the j level is
• For level 1:
Wj 
N!
N j !( N  N j )!
W1 
N!
N1!( N  N1 )!
• For level 2, there are only (N-N1) particles left,
thus
( N  N )!
W2 
1
N 2 !( N  N1  N 2 )!
• We consider that each energy level may contain more
then one quantum state (degeneracy, g ≥1 )!
• Using gj to represent the number of quantum state on
energy level j.
• Therefore, there are g1 quantum state on level 1, where
each of the N1 particles would have g1 choices.
Therefore, the total possibility would be g1N1.
After considering the arrangement of these N1 particles
W1 becomes
N!( g1 ) N
W1 
1
N1!( N  N1 )!
For the second energy level,
W2 becomes
( N  N1 )!( g 2 ) N 2
W2 
N 2!( N  N1  N 2 )!
The thermodynamic probability for a system
with n energy levels is
N !( g1 ) N1
( N  N1 )!( g 2 ) N 2
W


N1!( N  N1 )! N 2 !( N  N1  N 2 )!
n
 N ! (
j
gj
Nj
N j!
)
The above equation is subjected to two
constraints listed at the beginning of this
section
13.2 Lagrange multiplier
• Suppose that there are only two energy levels in a
system, the thermodynamic probability, W, can be
expressed based on the number of particles on each
level
W = W (N1, N2)
• The arrangement of N1 and N2, which gives the largest
value of W can be found via differentiating the above
equation against N1 and N2, respectively.
W
W
dW  (
) N 2 dN1  (
) N1 dN 2
N1
N 2
at the maximum
dw = 0
If N1 and N2 are independent variables, one
has ( W )  0
and ( W )  0
N1
N 2
N2
N1
However, N1 and N2 are connected through
N = N1 + N 2
For N = N(N1, N2), one has
dN  (
N
N
) N 2 dN1  (
) N1 dN 2
N1
N 2
Thus, one gets
=
Let the ratio be a constant
=α
Therefore,
–α
=0
• For a system with n energy levels, there will be n
differential equations.
• Note that when taking the derivative against N1 ,
N2 , N3 ,… Nn are kept constant!
When there are two constrains, two parameters
are needed!
For example: N = NJ
U = NJ · EJ
therefore
–α
–β
=0
• α and β are the Lagrange multipliers.
13.3
Boltzmann Distribution
Since ln W is a monotonically increasing function of W, maximizing ln W is
equivalent to maximizing W. Obviously, the logarithm is much easy to
work with!
W = N!
)
ln W = lnN! + ln(
) --- using ln (x/y) = lnx + lny
ln W = lnN! + ln · gJNJ – ln NJ! --- using
ln (x/y) = lnx – lny
ln W =lnN! + ln · gJNJ – ln(NJ!)--- using, again,
ln (x/y) = lnx - lny
Using stirlings application to the last term…
ln W =lnN! +
(NJlngJ) – ( NJlnNJ –
NJ)
Using the method of Lagrange multiplier
+α
+β·
=0
Therefore, lngJ – lnNJ + α + βEJ = 0 (see
chalkboard)
ln
= 0 + α + βEJ
= e α + βE
J
The ratio of NJ/gJ is called the Boltzmann
distribution, which indicates the number of
particles per quantum state.
Now we relate α and β to some physical
properties!
Since
lngJ – lnNJ + α + βEJ = 0,
NJ·lngJ – NJ·lnNJ + αNJ + βNJεJ = 0
Sum the above eqn over all energy levels
NJ·lngJ – NJ·lnNJ + αNJ + βNJεJ = 0
NJ·lngJ – NJ·lnNJ + α·N + β·U = 0
The first two term can be replaced by
lnW – lnN! –N + α·N + β·U = 0 (see inclass derivation)
lnW = lnN! + N - α·N - β·U
Since
S = k·lnW
S = k(lnN + N - αN – βU)
s = klnN - αkN -βkU
But
klnN + αkN = so
Therefore,
s = so - βkU
Because T·ds = dU + P·dV =
ds =
+
·dV =
·dU +…
therefore,
=
Since s = so + k βU,
= -kβ
thus,
= -k·β
 -β =
= e-α·e
e-α =
·e NJ = gJ e-α · e
NJ = gJ·e-α · e
e-α =
=
·e
is called partition function!