量子力学
Quantum mechanics
School of Physics and Information Technology
IDENTICAL PARTICLES
Chapter 5
5.1 Two-Particle Systems 201
5.2 Atoms 210
5.3 Solids 218
5.4 Quantum Statistical Mechanics
If we have a large number of N particles, in thermal equilibrium at
temperature T, what is the probability that a particle would be found to have the specific energy, Ej?
The fundamental assumption of statistical mechanics is that in thermal equilibrium every distinct state with the same total energy, E, is equally probable.
The temperature, T, is a measure of the total energy of a system in thermal equilibrium in classical mechanics. What is the new in quantum mechanics?
How to count the distinct states!
5.4.1 An Example
Suppose we have just have three noninteracting particles, A, B, and C, (all of mass m) in the one-dimensional infinite square well. The total energy is
),
(
2
2 2
2 2
2 2
C B
A C
B
A
n
n
n
ma
E
E
E
E
=
+
+
=
π
+
+
where nA,nB, and nC are positive integers. Now suppose, for the sake of argument, that total energy
argument, that total energy
,
2
363
22 2
ma
E
=
π
which is to say,
.
363
22 2
=
+
+
B CA
n
n
n
?
=
A
Thus (nA,nB,nC ) can be one of the following:
)
11
,
11
,
11
(
) 5 , 13 , 13
( (13, 5, 13) (5, 13, 13)
)
9
1
,
1
,
1
(
(
1
,
19
,
1
)
(
19
,
1
,
1
)
)
7
1
,
7
,
5
(
(
5
,
1
7
,
7
)
(
7
,
5
,
1
7
)
(
7
,
17
,
5
)
(
17
,
5
,
7
)
(
17
,
7
,
5
)
1
3
3
6
13
totalFor example, (nA,nB,nC )=(11,11,11) means nA=11, nB=11, nC=11, and A,B,C in the single states
);
(
11 A
A
ψ
x
ψ
=
ψ
B=
ψ
11(
x
B);
ψ
C=
ψ
11(
x
C).
The total number of probable (nA,nB,nC ) is 13.
If the particles are distinguishable, the three-particle state is
)
(
)
(
)
(
)
(
x
ψ
11x
Aψ
11x
Bψ
11x
C)
(
x
n
ψ
If the particles are distinguishable, each of these (nA,nB,nC )
1. Distinguishable condition:
Configuration(排布）: The collection of all occupation numbers for a given (3-particle) state we will call the configuration.
,
,
,
,
,
,
,
2 3 4 5 6 71
(
N
1,
N
2,
N
3,
N
4,
N
5,
N
6,
N
7,
)
.
2 3 4 The most important quantity is the number of particles in each state, that is, the occupation number, Nn, for the single state .
A B C
represents a distinct quantum state, and the fundamental assumption of statistical mechanics says that in thermal equilibrium they are all equally likely.
(
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
3
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
)
.
If all the three particles are in , the configuration is
ψ
11(
x
)
1 2 3 4 11
3
11=
N
(
11
,
11
,
11
)
one stateIf two are in and one
ψ
13 is in , the configuration isψ
5(
0
,
0
,
0
,
0
,
1
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
2
,
0
,
0
,
0
,
0
,
0
,
)
.
1 2 5 13
2
,
1
13 5=
N
=
N
(13, 13, 5) (13, 5, 13) (5, 13, 13)three different states
5 13
If two are in and one
ψ
1 is in , the configuration isψ
19(
2
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
2
,
0
,
0
,
0
,
0
,
0
,
1
,
0
)
.
1 19
2
,
1
19 1=
N
=
N
(
1
,
1
,
1
9
)
(
1
,
19
,
1
)
(
19
,
1
,
1
)
If one is in , one in , and one
ψ
5ψ
7 is in , the configuration is(
0
,
0
,
0
,
0
,
1
,
0
,
1
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
1
,
0
,
0
,
0
)
.
5 17
1
,
1
,
1
7 175
=
N
=
N
=
N
17
ψ
7
ψ
(
0
,
0
,
0
,
0
,
1
,
0
,
1
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
1
,
0
,
0
,
0
)
.
5
ψ
7 17A
B
C
A
C
B
)
7
1
,
7
,
5
(
)
7
,
7
1
,
5
(
A
C
B
A
C
B
A
C
B
A
C
B
A
C
B
)
7
,
7
1
,
5
(
)
7
1
,
5
,
7
(
)
5
,
17
,
7
(
)
7
,
5
,
17
(
)
5
,
7
,
17
(
Of course, the last is the most probable configuration, because it can be
achieved in six different ways, whereas the middle two occur three ways, and the first only one.
Under the above condition, if we select one of these three particles at random, what is the probability (Pn)of getting a specific (allowed) energy En ？
E1 : Only the third configuration Probability 3/13
In the third configuration,
two particles are in E1 Probability 2/3
13
2
3
2
13
3
=
×
E2 :
0
E3 :
0
1
P
E5 : the second configuration Probability 3/13
In the second configuration,
one particle is in E5 Probability 1/3
13
1
3
1
13
3
=
×
the fourth configuration Probability 6/13
In the fourth configuration,
one particle is in E5 Probability 1/3
13
2
3
1
13
6
=
×
13
3
13
2
13
1
=
+
E4 :
0
5
E7 : Only the fourth configuration Probability 6/13
In the fourth configuration,
one particle is in E7 Probability 1/3
13
2
3
1
13
6
=
×
P
7E11 : Only the first configuration Probability 1/13
In the first configuration,
three particles are in E11 Probability 3/3
13
1
3
3
13
1
=
×
P
11=
13
P
Similarly
13
2
3
2
13
3
=
×
P
17=
13
2
3
1
13
6
=
×
P
19=
13
1
3
1
13
3
=
×
We can check this by total probability
=
+
+
+
+
+
+
=
P
1P
5P
7P
11P
13P
17P
19P
113 1 13
2 13
2 13
1 13
2 13
3 13
2
= +
+ +
+ +
+
Above analysis is based on the assumption that the three particles are
2. Identical fermions:
For fermions, no two particles are in the same state. This antisymmetrization
requirement exclude the configurations where two particles are in the same state. Only the fourth configuration is available now!
E1 :
0
(
0
,
0
,
0
,
0
,
1
,
0
,
1
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
0
,
1
,
0
,
0
,
0
)
.
5
ψ
7 17E5 : Only one configuration Probability 1
In the fourth configuration,
one particle is in E5 Probability 1/3 3
1 3
1
1× =
P
5E7 : In this configuration, one particle is in E7 Probability 1/3
P
73. Identical bosons:
For bosons, each configuration enables one state, so
E1 : The third configurations Probability 1/4
In this configuration, two
particles are in E1 Probability 2/3 12
1 3 1 4 1 =
×
P
1E5 : the second configuration Probability 1/4
In the second configuration,
one particle is in E Probability 1/3
12
1
3
1
4
1
=
×
1
1
1
one particle is in E5 Probability 1/3
4
3
12
the fourth configuration Probability 1/4
In the fourth configuration,
one particle is in E5 Probability 1/3
12
1
3
1
4
1
=
×
6
1
12
1
12
1
=
+
5P
=
13P
Similarly6
1
3
2
4
1
=
×
=
17P
P
19=
=
7P
12
1
3
1
4
1
=
×
P
11=
Conclusion:
(1) This example shows that the nature of the particles determines the counting properties, or the statistical properties! The number of internal distinct states is different and the probability of getting specific energy is different too.
(2) This example gives a system of three particles. If the number of
5.4.2 The General Case
Now consider an arbitrary potential, for which one particle energies are
(
E
1,
E
2,
E
3,
,
E
n,
)
,
with degeneracies
(
d
1,
d
2,
d
3,
,
d
n,
)
.
Suppose we put N particles (all with the same mass) into this potential; we are interested in the configuration
(
N
1,
N
2,
N
3,
,
N
n,
)
,
for which there are N1 particles with energy E1, N2 particles with energy E2, and so on.
The answer: The number of the distinct states Q(N1,N2,N3,……) depends on whether the particles are distinguishable, identical fermions, or
identical bosons.
1. Distinguishable particles:
(1) Choose N1 from N for energy bin: the binomial coefficient
)!
(
!
!
1 1
1
N
N
N
N
N
N
−
≡
(
N
1,
N
2,
N
3,
,
N
n,
)
1
N
2
N
2
d
)!
(
!
)
1
(
)
2
)(
1
(
1 1
N
N
N
N
N
N
N
N
−
=
+
−
−
−
(2) Arrangement of the N1 particles within the bin on the degenerate d1 states:
!
1
N
!
(
)!
!
1
1
N
N
N
N
−
( )
1 1N
d
1
(3) Thus the number of ways to put N1 particles, selected from a total population of N, into a bin containing d1 distinct option, is
( )
)!
(
!
!
1 1 1 1N
N
N
d
N
N−
(4) The same goes for energy bin E2, of course, except that there are now only
N-N1 particles left to work with:
(
1) ( )
!
2 2
d
N
N
−
N(
) ( )
;
)!
(
!
!
2 1 2 2 1N
N
N
N
d
N
N
−
−
−
(5) Finally, it follows that
=
)
,
,
,
(
N
1N
2N
3Q
( )
)!
(
!
!
1 1 1 1N
N
N
d
N
N−
(
) ( )
)!
(
!
!
2 1 2 2 1 2N
N
N
N
d
N
N
N−
−
−
( ) ( )
( )
∏
∞ ==
=
1 2 1 2 1!
!
!
!
!
1 22. Identical fermions:
)!
(
!
!
1 11 1
N
d
N
d
N
d
C
dN−
=
=
1
N
(1) The particles are identical.
1
N
1
d
1
1
N
d
≥
(2) The antisymmetrization requires that only one particle can occupy any given state.
Here we pick N1 draws from d1 draws to locate particles.
1 2
(
N
1,
N
2,
N
3,
,
N
n,
)
)!
(
!
1 11 1
1
N
d
N
N
C
d−
=
=
=
2 2
2 2
N
d
C
dN=
n n N
d
N
d
C
nn
=
)
,
,
,
(
N
1N
2N
3Q
∏
∞
=1
!
(
−
)!
!
n n n n
n
N
d
N
3. Identical bosons: 1
N
1d
1 2
(1) The particles are identical.
N
1(2) Although the wave function of the N-particle state is symmetry, more than one particles can occupy the draws in certain bin.
)!
1
(
11
+
d
−
N
!
N
(
d
−
1
)!
−
+
=
−
−
+
1 1 1 1 1 1 11
)!
1
(
!
)!
1
(
N
d
N
d
N
d
N
(
N
1,
N
2,
N
3,
,
N
n,
)
)
1
(
11
+
d
−
N
!
1
N
(
d
1−
1
)!
1 1 1=
)
,
,
,
(
N
1N
2N
3Q
−
+
=
−
−
+
n n n n n n nN
d
N
d
N
d
N
1
)!
1
(
!
)!
1
(
∏
∞ =−
−
+
1
!
(
1
)!
)!
1
(
n n n
5.4.3 The Most Probable Configuration
In thermal equilibrium, every state with a given total energy E and a given particle number N is equally likely. So the most probable configuration (N1,
N2, N3, ……) is the one that can be achieved in the largest number of different ways—— it is that particular configuration for which
Q(N1,N2,N3,……..) is a maximum, subject to the constraints
N
N
n=
∞
=
.
E
N
E
n n=
∞
=
n=1 n=1
The problem of maximizing a function F(x1, x2, x3,……) of several variables, subject to the constraints f (x1, x2, x3,……)=0, f (x1, x2, x3,……)=0, etc., is most conveniently handled by the method of Lagrange multipliers. We introduce the new function
,
)
,
,
,
and set all its derivatives equal to zero:
;
0
=
∂
∂
n
x
G
.
0
=
∂
∂
n
G
λ
In our case it’s a little easier to work with the logarithm of Q, instead of Q
itself——this turns the products into sums. Since the logarithm is a
monotonic function of its argument, the maxima of Q and ln(Q) occur at the same point. So we let
same point. So we let
,
)
ln(
1 1
−
+
−
+
≡
∞
= ∞
= n
n n n
n
E
N
E
N
N
Q
G
α
β
where and are the Lagrange multipliers.
;
0
=
∂
∂
n
N
G
=
∂
∂
0
α
G
=
∂
∂
0
β
1. Distinguishable particles:
=
)
,
,
,
(
N
1N
2N
3Q
∏
( )
∞
=1
!
!
n n Nn nN
d
N
+
=
∏
∞=1
!
)
(
ln
)
!
ln(
n n N nN
d
N
G
n,
1 1−
+
−
+
∞ = ∞ = n n n nn
E
N
E
N
N
β
α
[
]
∞−
+
=
ln(
N
!
)
N
ln(
d
)
ln(
N
!
)
G
[
]
=
−
+
=
1)
!
ln(
)
ln(
)
!
ln(
n n nn
d
N
N
N
G
,
1 1−
+
−
+
∞ = ∞ = n n n nn
E
N
E
N
N
β
α
Assuming the relevant occupation numbers (Nn) are large, we can invoke
stirling’s approximation:
[
]
∞
=
−
−
+
−
≈
1
)
ln(
)
ln(
n
n n n
n n
n n
n
d
N
N
N
N
E
N
N
G
α
β
+
ln(
N
!
)
+
α
N
+
β
E
It follows that
0
)
ln(
)
ln(
−
−
−
=
=
∂
∂
n n
n n
E
N
d
N
G
β
α
The most probable occupation numbers, for distinguishable particles, are
n n
n
d
E
N
)
=
ln(
)
−
α
−
β
ln(
(
)
.
n
E n
n
d
e
2. Identical fermions:
=
)
,
,
,
(
N
1N
2N
3Q
∏
∞
=1
!
(
−
)!
!
n n n n
n
N
d
N
d
[
]
∞ =−
−
−
=
1]
)!
ln[(
)
!
ln(
)
!
ln(
n n n nn
N
d
N
d
G
,
−
+
−
+
∞ ∞ n nn
E
N
E
N
N
β
α
1 1 = = n nAssume Nn>>1 and dn>>Nn, so the stirling’s approximation applies
[
]
∞ =−
−
−
+
−
−
−
+
−
≈
1)
(
)
ln(
)
(
)
ln(
)
!
ln(
n n n n n n n n n n n n nn
N
N
N
d
N
d
N
d
N
N
E
N
d
G
α
β
E
N
β
0
)
ln(
)
ln(
+
−
−
−
=
−
=
∂
∂
n n
n n
n
E
N
d
N
N
G
β
α
The most probable occupation numbers, for identical fermions, are
n n
n
n
d
N
E
N
)
=
ln(
−
)
−
α
−
β
ln(
(
)
.
1
+
=
+n
E n n
e
d
N
α β) (
)
(
Enn n
n
d
N
e
3. Identical bosons:
=
)
,
,
,
(
N
1N
2N
3Q
∏
∞
=
−
−
+
1
!
(
1
)!
)!
1
(
n n n
n n
d
N
d
N
[
]
∞ =−
−
−
−
+
=
1]
)!
1
ln[(
)
!
ln(
]
)!
1
ln[(
n n n nn
d
N
d
N
G
,
−
+
−
+
∞ ∞E
N
E
N
N
β
α
,
1 1−
+
−
+
= = n n n nn
E
N
E
N
N
β
α
Assuming Nn>>1 and using stirling’s approximation
[
]
∞ = − − − − + − − + − − + − + ≈ 1 ] )! 1 ln[( ) ln( ) 1 ( ) 1 ln( ) 1 ( n n n n n n n n n n n n nn d N d N d N N N d N E N
N
G α β
E
N β
0
)
ln(
)
1
ln(
+
−
−
−
−
=
=
∂
∂
n n
n n
n
E
N
d
N
N
G
β
α
)
(
)
1
ln(
)
ln(
N
n=
N
n+
d
n−
−
α
+
β
E
n) (
)
1
(
Enn n
n
N
d
e
N
=
+
−
− α+β(
)
1
1
−
−
=
+n
E n n
e
d
N
α β(
)
1
−
=
+n
E n n
e
d
N
α β5.4.4 Physical significance of and
The parameters and came into the story as Lagrange multipliers, associated with the total number of particles and the total energy.
Mathematically, they are determined by substituting the most probable occupation numbers Nn back into the constraints.
N
N
=
∞
(
En)
n
n
d
e
N
=
− α+βN
N
n
n
=
=1
E
N
E
n
n
n
=
∞
=1
(
)
1
−
=
+n
E n n
e
d
N
α β(
)
1
+
=
+n
E n n
e
d
N
α β andBy using an example to do this: ideal gas
Idea gas: a large number of noninteracting particles, all with the same mass, in the three dimensional infinite square well—— a box!
We know that the allowed energies of the particle are
2 2
2
m
k
E
k=
where=
,
,
.
z z
y y
x x
l
n
l
n
l
n
k
π
π
π
lx ly
l
z
ly
τ
d
A shell of thickness dk contains a volume
(
4
)
,
8
1
2dk
k
d
τ
=
π
so the “degeneracy” is (the number of electron states in the shell )
(
/
)
2
.
2 2
3
k
dk
V
V
d
d
kπ
π
τ
For distinguishable particles, the first constraint becomes
In the k-space, the sum will be converted into an integral, treating k as a continuous variable, then
,
2 2
k
E
=
(
En)
n
n
d
e
N
=
− α+βN
N
n n
=
∞ =1(
)
∞ = + −=
1 n E k ke
d
N
α β,
2
dk
k
V
d
=
2 2.
2 2
dk
k
e
e
V
dN
m k β α − −=
∞ + −=
0 1!
n ax na
n
dx
e
x
,
2
m
E
k=
,
2
2k
dk
d
kπ
=
.
2
2 22
e
e
k
dk
V
dN
α mπ
−=
⋅
=
=
=
− ∞ − − 2 / 3 2 0 2 2 22
2
2 2πβ
π
α β αm
Ve
dk
k
e
e
V
dN
N
m k 2 / 3 0 21
16
2=
∞ −a
dx
x
e
axπ
∞ + −
⋅
⋅
⋅
⋅
−
=
0 1 22
)
1
2
(
5
3
1
2a
a
n
dx
e
The second constraint becomes so
⋅
=
−
2 / 3 2
2
m
V
N
e
απβ
E N
E
n
n
n =
∞
=1
⋅
=
=
=
− ∞ − −2 / 3
2 0
4 2
2
2
2
2
3
2
2
2 2
πβ
β
π
α β
α
m
e
V
dk
k
e
m
e
V
dE
E
mk
Or, putting in :
e
−αβ
2
3
N
E
=
The average kinetic energy of an atom at temperature T, in classical mechanic, is
T
k
N
E
B
2
3
=
β
2
3
=
N
E
This suggests that is related to the temperature:
T
k
B1
=
β
Different substances in thermal equilibrium with one another have the same value of , and which can be adopted as s definition of T.
Then
2 / 3 2
2
=
−
Tm
k
V
N
e
B
π
α
Then
It is customary to replace by the so-called chemical potential,
T
k
T
)
B(
α
µ
≡
−
(
)
.
T
k
T
B
µ
By using the chemical potential, we can rewrite the most probable number of particles in a particular (one-particle) state with energy :
(
En)
n
n
d
e
N
=
− α+β(
)
1
+
=
+n
E n n
e
d
N
α β(
)
kBTe
− ε−µ /=
(
)
1
1
/
+
=
− k TB
e
ε µn n
d
N
n
(
ε
)
=
n n
d
N
n
(
ε
)
=
Maxwell-Boltzmann distribution
(
)
1
−
=
+n
E n n
e
d
N
α β1
+
e
ne
+
1
n n
d
N
n
(
ε
)
=
(
)
1
1
/
−
=
− k TB
e
ε µFermi-Dirac distribution
Bose-Einstein distribution
(
)
kBTe
n
(
ε
)
=
− ε−µ />
∞
<
).
0
(
if
,
),
0
(
if
,
0
µ
ε
µ
ε
( )
→
− kBT
e
ε µ /0
→
T
Maxwell-Boltzmann distribution
( )
→
−
− kBT
e
ε µ /<
∞
>
).
0
(
if
,
),
0
(
if
,
0
µ
ε
µ
ε
2
2
1
mv
=
ε
(
mv)
kBTe
v
The Fermi-Dirac distribution has a particularly simple behavior as :
T
→
0
( )
→
− kBT
e
ε µ /(
)
.
1
1
)
(
/+
=
− k TB
e
n
ε
ε µ>
∞
<
).
0
(
if
,
),
0
(
if
,
0
µ
ε
µ
ε
As kB
→
)
(
ε
n
>
<
).
0
(
if
,
0
),
0
(
if
,
1
µ
ε
µ
ε
All states are filled, up to an energy (0), and none are occupied for energies above (0). Evidently the chemical potential at absolute zero is precisely the Fermi energy:
F
E
=
)
0
(
µ
B
k
) 0 (
Bose-Einstein distribution
( )
(
)
1
1
/
−
=
− k TB
e
n
ε
ε µ( )
ε
>
0
n
( )/1
0
>
−
− kBT
e
ε µ(
)
1
/
>
− kBT
e
ε µ>
1
ε
>
µ
(
T
)
e
ε
>
µ
(
)
ε
µ
(
T
)
<
m
k
2
2=
ε
0
→
V
0
2
2
→
=
m
k
ε
0
<
Returning to the special case of an ideal gas, for distinguishable particles we found that the total energy at temperature T is
T
Nk
E
B2
3
=
and the chemical potential is
2 / 3 2
2
=
−
Tm
k
V
N
e
απ
2
=
Tm
k
V
e
B
T
k
T
)
B(
α
µ
≡
−
+
=
T
mk
V
N
T
k
T
B B
2
2
ln
2
3
ln
)
(
π
5.4.5 The Blackbody Spectrum
Photons (quantum of the electromagnetic field) are identical bosons with spin 1, but they are very special, because they are massless particles, and hence intrinsically relativistic. There are four properties belong to
nonrelativistic quantum mechanics:
(1) Energy:
(2) Wave number:
ω
ν
=
=
h
E
c
/
k
=
2
=
ω
/
(3) Spin: two spin states occur, m=1 or –1.
(4) The number of the photons are not conserved:
1
/
−
=
k kTn B
e
d
N
ω0
=
α
(
)
1
−
=
+k
E k n
e
d
For free photons in a box of volume V, dk is given by
,
2
2 2
k
dk
V
d
kπ
=
multiplied by 2 for two spin states, and expressed in terms of
ω
=
kc
.
2 3
2
ω
ω
π
c
d
V
d
k=
So the energy density, So the energy density,
V
e
V
N
V
E
T k n
B
ω
ω
ω
1
/
−
=
=
ω
ω
π
c
d
V
23 2
ω
π
ω
ω
d
e
c
3(
/kBT1
)
23
−
=
That is
ω
π
ω
ω
ωd
e
c
E
k TB
1
)
(
)
(
2 3 /3
)
1
(
)
(
2 3 /3
−
=
k TB
e
c
ωπ
ω
ω
ρ
We introduce energy density per unit frequency:
λ
π
πν
ω
=
2
=
kc
=
2
c
λ
λ
π
ω
c
d
d
=
−
2
2( )
ω
π
ω
ω
ω
ρ
ω
ωd
e
c
d
E
k TB
1
)
(
)
(
2 3 /3
−
=
=
( )
λ
λ
π
λ
π
λ
π λc
d
e
E
c k TB 2
/ 2 3
2
)
1
(
8
−
=
λ
ρ
λ
λ
λ
π
λ
π
d
d
e
c
T k
c B
(
)
)
1
(
16
/ 2 5
2
=
−
=
)
1
(
16
)
(
5 2 /2
−
=
π λλ
π
λ
ρ
c k TB
e
