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5.62 Physical Chemistry II
Spring 2008
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Transformed Q from sum over states of an entire N-molecule assembly to sum over states of an individual molecule
Q
= ∑ e j
−
E j kT =
(
∑ e i
−ε j
)
= q
N sum over states sum over states of assembly of a molecule
∑ Ω
( { } ) e
−
( { } ) kT = ∑
∏ n j
N! j
!
e
−∑ i n i
ε i kT sum over all sets of occupation numbers
FOR INDEPENDENT, DISTINGUISHABLE PARTICLES
BIG PROBLEM: Identical molecules are INDISTINGUISHABLE!!
We have implicitly been assuming that we can distinguish particle 1 from particle 2, but quantum mechanics tells us that two identical particles are not distinguishable. [Even if we could label individual atoms by their position at t
1
, then follow each of the atoms until t
2
, these positional labels will be corrupted each time there is a collision.] Exceptions? Molecules in a crystal or in spatially confined traps on the surface of a solid.
Thus we will need to modify the above result for molecules, since molecules of the same type are indistinguishable. We had shown that
( { n i
} ) =
N!
t
n i
=
1 i
!
is the number of ways of putting N distinguishable particles into t states with

5.62 Spring 2008 Lecture 6, Page 2 occupation numbers {n i
}. But if all the particles are identical, the number of ways is just 1. How do we know this to be true? Do example of 3 molecules in 2 states.
That is, for INDISTINGUISHABLE PARTICLES, Ω ({n i
}) = 1
However
Ω
= N!/
Π n i
! is a multinomial coefficient that was needed to produce this “tidy” result relating Q to q:
Q
= ∑ j e
−
E j kT = ∑
{ }
( { n j
} ) e
− E
( { n j
} ) kT =
(
∑ e
−ε i kT i
)
N = q
N
No way to exploit this simplification if
Ω
{n i
} = 1. That is the reason why we did the distinguishable particle case first. Now we have to take this
"tidy" result and fix it for indistinguishability. Problem is that each term in this sum is too large by a factor N!/
Π n i
! Will solve this problem by dividing
Q by N!/
Π n i
! for a special limiting case (which turns out to be almost universally applicable).
CORRECTED BOLTZMANN STATISTICS – AN APPROXIMATE
CORRECTION FOR INDISTINGUISHABILITY
Define n i
≡
average occupation number of i th molecular state in ensemble or ensemble average # of molecules in i th state probability of finding n molecules in state i of j i th assembly n i
=
( ) n j
P j n i i
( j) i number of molecules in th state of j th assembly sum over assemblies
If the number of molecular states i is much greater than N, then n i
<<
1 revised 1/9/08 9:36 AM

5.62 Spring 2008 Lecture 6, Page 3
The average occupation number will be much less than 1. So, most of the occupation numbers will be n i
= 0 or n i
= 1 with n i
> 1 occurring very rarely.
Under these conditions because 1! = 0! = 1
Ω =
N! n i
=
N!
! 1
=
N! when n i i
<<
1
For n i
 1 , Ω is too large by a factor of N! in each term of
Q
=
Ω e
−
E /kT owing to our neglect of indistinguishability.
Remember, we want
Ω
= 1. So, take the distinguishable molecule result
Q = q N and divide by N!. This division will make
Ω
= 1 and is a valid correction as long as all n– i
 1. Therefore by dividing by N!, we have
“corrected” for indistinguishability
Q
(
N,V, T
) = q N N!
corrected for indistinguishability
Boltzmann statistics
In other words, we used
Ω =
N!
n i i
!
earlier to get a “tidy” result and now we are dividing it out. But the value we're dividing by is N! Not
N!
∏ n i
!
because
– i
 1. So, we have fixed the “canonical” partition function to account for indistinguishability. How generally can we depend on this to be true? revised 1/9/08 9:36 AM

5.62 Spring 2008 Lecture 6, Page 4
i

i
. Express the ensemble average in terms of sets of state occupation numbers rather than single state occupation numbers. n i
=
P j j
( ) n i
( j)
= number of molecules in i th state in j th assembly
P
{ }
( { n j
} ) n i
( { n j
} ) averaged over all sets of occupation numbers in ensemble number of molecules in i th molecular state for the specified set of occupation numbers {n j
}
Need P({n j
})
P(E)
=
(E)e
Q
−
E /kT
P
( { } ) =
( { } ) e
− E
( { } ) kT
Q
P
( { n i
} ) =
N!
n j j
!
exp
{
n i
i
/ kT
}
Q i
( j
}
) into the above equation for n i
;
β =
1 / kT calculating average value of n i n i
=
{ }
⎜
⎛
⎝ n
1
!n
2
!
N! ⎞
… n i
!
… ⎠
⎟ e
−β n
1
ε
1 e
−β n
2
ε
2
… n i e
−β n i
ε i
…
/ Q partition out factor N from N! and factor e
–
βε i from product
=
{ }
⎛
⎜⎝ n
1
!n
2
N(N
!
… (
− n i
1)!
−
1
)
!
… ⎠
⎞
⎟ e
−β n
1
ε
1 e
−β n
2
ε
2
…
⎝
⎛
⎜ e
−β ( n i
− 1
) ε i
−βε i  removed term
⎠
⎟
⎞ …
/ Q n i
=
Ne
−ε i
Q
/kT
{n j
}
⎝⎜
⎛ n
1
!n
2 cancelled factor of n i
in numerator
(
N
−
1
)
!
…
(n i
!
−
1)!
… ⎠⎟
⎞ e
− n
1
ε
1 kT e
− n
2
ε
2 kT … e
− (n i
− 1) ε i kT …
The sum over {n j
} looks like Q(N–1, V, T), but we will need to check this. revised 1/9/08 9:36 AM

5.62 Spring 2008 Lecture 6, Page 5 n i
=
Q
Ne
(
−ε i
/kT
N,V, T
)
(
1,V, T
) check to be sure n– i
is correctly normalized
∑ i n i
=
N
∑ i n i
=
NQ(N
−
1,V, T)
Q(N,V, T)
∑ i e
−ε i kT =
N
( q
N − 1
(
N
−
1
)
!
( q
N
N!
)
) q
= q
Nq
N
N
−
1
N!
(N
−
1)!
q
=
N
2 too large by a factor of N.
So original equation for n i
must be divided by N. So we get n i
= e
−ε i kT
⎣⎢
⎡ Q(N
−
1,V, T) ⎤
Q(N,V, T) ⎦⎥ n i
=
Ne
– ε i
/kT q
N–1 q
N
– ε i
/kT
=
Ne q
−ε j
/kT <
(because
ε j
≥ 0) n i
 1 if q  N
N
 1 q condition for validity of corrected Boltzmann statistics
Thus whenever the value of the single particle partition function is larger than the number of molecules in the system, it is OK to correct for particle indistinguishability merely by dividing q N by N! But is this ever true?
Typically N ~ 10 23 . How can q be so much larger than this? revised 1/9/08 9:36 AM