量子力学
Quantum mechanics
School of Physics and Information Technology
Shaanxi Normal University
Chapter 5
IDENTICAL PARTICLES
5.1 Two-Particle Systems
201
5.2 Atoms
210
5.3 Solids
218
5.4 Quantum Statistical Mechanics
230
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!
Why? Give a example to demonstrate!
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
E = E A + E B + EC =
π 2h 2
2ma 2
(n A2 + nB2 + nC2 ),
where nA,nB, and nC are positive integers. Now suppose, for the sake of
argument, that total energy
E = 363
π 2h 2
2ma
2
,
which is to say,
n A2 + nB2 + nC2 = 363.
nA = ?
nB = ?
nC = ?
Thus (nA,nB,nC ) can be one of the following:
(11, 11, 11)
1
(13, 13, 5) (13, 5, 13) (5, 13, 13)
3
(1, 1, 19) (1, 19, 1) (19, 1, 1)
3
6 (5, 7, 17) (5, 17, 7) (7, 5, 17) (7, 17, 5) (17, 5, 7) (17, 7, 5)
13 total
For example, (nA,nB,nC )=(11,11,11) means nA=11, nB=11, nC=11, and A,B,C in
the single states
ψ A = ψ 11 ( x A ); ψ B = ψ 11 ( xB ); ψ C = ψ 11 ( xC ).
If the particles are distinguishable, the three-particle state is
ψ ( x) = ψ 11 ( x A )ψ 11 ( xB )ψ 11 ( xC )
The total number of probable (nA,nB,nC ) is 13.
The most important quantity is the number of particles in each state, that is, the
occupation number, Nn, for the single state ψ n (x ).
Configuration(排布）: The collection of all occupation numbers for a given
(3-particle) state we will call the configuration.
ψ1, ψ 2 , ψ 3 , ψ 4 , ψ 5 , ψ 6 , ψ 7 , L
(N 1 , N 2 , N 3 , N 4 , N 5 , N 6 , N 7 , L).
1. Distinguishable condition:
If the particles are distinguishable, each of these (nA,nB,nC )
represents a distinct quantum state, and the fundamental
assumption of statistical mechanics says that in thermal
equilibrium they are all equally likely.
If all the three particles are in ψ 11 ( x) , the configuration is
ψ1 ψ 2 ψ 3 ψ 4 L
(0,
ψ11
0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, L ).
N11 = 3
(11, 11, 11) one state
ψ4
ψ3
ψ2
ψ1
If two are in ψ 13 and one is in ψ 5 , the configuration is
ψ1 ψ2
(0,
ψ5
ψ13
ψ13
0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, L).
N 5 = 1, N13 = 2
ψ5
(13, 13, 5) (13, 5, 13) (5, 13, 13)
three different states
If two are in
ψ1
(2,
ψ1
and one is inψ 19 , the configuration is
ψ19
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0 L ).
N1 = 1, N19 = 2
(1, 1, 19) (1, 19, 1) (19, 1, 1)
three different states
If one is in ψ 5 , one in ψ 7 , and one is inψ 17 , the configuration is
ψ7
ψ17
(0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0 L).
ψ5
ψ 7 N 5 = 1, N 7 = 1, N17 = 1 ψ17
(0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0 L).
(5, 7, 17)
C
A
B
A
B
(5, 17, 7)
C
C
B
A
(7, 5, 17)
six different states
C
A
B
(7, 17, 5)
C
B
A
(17, 5, 7)
C
B
A
(17, 7, 5)
ψ5
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
In the third configuration,
two particles are in E1
Probability 3/13
Probability 2/3
3 2 2
× =
P1
13 3 13
E2 :
0
E3 :
0
E4 :
0
E5 : the second configuration
In the second configuration,
one particle is in E5
the fourth configuration
In the fourth configuration,
one particle is in E5
Probability 3/13
Probability 1/3
Probability 6/13
Probability 1/3
3 1 1
× =
13 3 13
6 1 2
× =
13 3 13
1 2
3
+ =
13 13 13
P5
E7 : Only the fourth configuration
In the fourth configuration,
one particle is in E7
E11 : Only the first configuration
In the first configuration,
three particles are in E11
Probability 6/13
Probability 1/3
Probability 1/13
Probability 3/3
6 1 2
P7
× =
13 3 13
1 3 1
× =
P11
13 3 13
Similarly
P13 =
3 2 2
× =
13 3 13
P17 =
6 1 2
× =
13 3 13
P19 =
3 1 1
× =
13 3 13
We can check this by total probability
2 3 2 1 2 2 1
P = P1 + P5 + P7 + P11 + P13 + P17 + P19 = + + + + + + = 1
13 13 13 13 13 13 13
Above analysis is based on the assumption that the three particles are
distinguishable!
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!
ψ7
ψ17
(0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0 L).
ψ5
E1 :
0
E5 : Only one configuration
In the fourth configuration,
one particle is in E5
Probability 1
1 1
1× =
3 3
Probability 1/3
E7 : In this configuration, one particle is in E7
E17 : In this configuration, one particle is in E17
Probability 1/3
P5
P7
Probability 1/3 P17
3. Identical bosons:
For bosons, each configuration enables one state, so
E1 : The third configurations
In this configuration, two
particles are in E1
E5 : the second configuration
In the second configuration,
one particle is in E5
the fourth configuration
In the fourth configuration,
one particle is in E5
Probability 1/4
Probability 2/3
Probability 1/4
Probability 1/3
Probability 1/4
Probability 1/3
1 1 1
× =
4 3 12
1 1 1
× =
4 3 12
1 1 1
× =
4 3 12
Similarly
1 1 1
P7 = × =
4 3 12
1 1 1
P17 = × =
4 3 12
1 3 1
1 2 1
P11 = × =
P13 = × =
4 3 4
4 3 6
1 1 1
P19 = × =
4 3 12
P1
1 1 1
+ =
12 12 6
P5
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
particles in huge, we can conclude: The distribution of individual particle
energies, at equilibrium, is simply their distribution in the most probable
configuration.
5.4.2 The General Case
Now consider an arbitrary potential, for which one particle energies are
(E1 ,
E 2 , E 3 , L L , E n , L L ),
with degeneracies
(d1 ,
d 2 , d 3 , L L , d n , L L ).
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 , L L , N n , L L ),
for which there are N1 particles with energy E1, N2 particles with energy E2,
and so on.
Now we consider general question: how many distinct states correspond
to this particular configuration?
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: ( N 1 , N 2 , N 3 , L L , N n , L L )
(1) Choose N1 from N for energy bin: the binomial coefficient
d2 N2
N
N!
  ≡
 N1  N1!( N − N1 )!
N!
N ( N − 1)( N − 2) L( N − N1 + 1) =
( N − N1 )!
N1!
N1
1 2
N!
N1!( N − N1 )!
d1
(2) Arrangement of the N1 particles within the bin on the degenerate d1 states:
(d1 )N
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
N ! (d1 ) 1
N1!( N − N1 )!
N
(4) The same goes for energy bin E2, of course, except that there are now only
N-N1 particles left to work with:
(N − N1 )! (d 2 )N
2
N 2 !( N − N1 − N 2 )!
;
LLL
(5) Finally, it follows that
(
N ! (d1 ) 1
N − N1 )! (d 2 ) 2
Q ( N1 , N 2 , N 3 , L) =
LLL
N1!( N − N1 )! N 2 !( N − N1 − N 2 )!
N
N
∞
(
dn )
N ! (d1 ) 1 (d 2 ) 2 L
=
= N !∏
N1! N 2 !L
N n!
n =1
N
N
Nn
2. Identical fermions: ( N 1 , N 2 , N 3 , L L , N n , L L )
(1) The particles are identical. N1
(2) The antisymmetrization requires that only one particle can occupy any
given state.
d1 ≥ N1
Here we pick N1 draws from d1 draws to locate particles.
C
C
N2
d2
N1
d1
 d1 
d1!
=   =
 N1  N1!(d1 − N1 )!
 d2 
=  
 N2 
LLL C
∞
Nn
dn
 dn 
=  
 Nn 
d n!
Q( N1 , N 2 , N 3 , L) = ∏
n =1 N n !( d n − N n )!
1 2
d1
N1
3. Identical bosons:
(N 1 ,
N 2 , N 3 , LL , N n , LL)
(1) The particles are identical. N1
1 2
d1
(2) Although the wave function of the N-particle
state is symmetry, more than one particles can
occupy the draws in certain bin.
N1
N1 + (d1 − 1)!
( N1 + d1 − 1)!  N1 + d1 − 1

= 
N1
N1!(d1 − 1)! 

N1! (d1 − 1)!
( N n + d n − 1)!  N n + d n − 1

= 
Nn
N n !( d n − 1)! 

∞
( N n + d n − 1)!
Q( N1 , N 2 , N 3 ,L) = ∏
n =1 N n !( d n − 1)!
N1 + (d1 − 1)
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 =1
∞
n
=N
∑E N
n
n
= E.
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
G ( x1 , x2 , x3 , L) ≡ F + λ1 f1 + λ2 f 2 + L ,
and set all its derivatives equal to zero:
∂G
= 0;
∂xn
∂G
= 0.
∂λn
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
∞
∞




G ≡ ln(Q ) + α  N − ∑ N n  + β  E − ∑ N n En ,
n =1
n =1




where α and β are the Lagrange multipliers.
∂G
=0⇒
∂α
∂G
=0⇒
∂β
∂G
= 0;
∂N n
1. Distinguishable particles:
∞
Q ( N1 , N 2 , N 3 ,L) = N !∏
(d n )Nn
n =1
 ∞ (d n ) N n
G = ln( N !) + ln ∏
 n =1 N n !
N n!
∞
∞





 + α  N − ∑ N n  + β  E − ∑ N n En ,
n =1
n =1





∞
G = ln( N !) + ∑ [N n ln(d n ) − ln( N n !)]
n =1
∞
∞




+ α  N − ∑ N n  + β  E − ∑ N n En  ,
n =1
n =1




Assuming the relevant occupation numbers (Nn) are large, we can invoke
stirling’s approximation:
ln (z!) ≈ z ln z − z
for z >> 1
∞
G ≈ ∑ [N n ln(d n ) − N n ln( N n ) + N n − αN n − β En N n ]+ ln( N !) + αN + β E
n =1
It follows that
∂G
= ln(d n ) − ln( N n ) − α − β En = 0
∂N n
The most probable occupation numbers, for distinguishable particles, are
ln( N n ) = ln(d n ) − α − β En
N n = d n e − (α + βEn ).
2. Identical fermions:
∞
dn!
Q ( N1 , N 2 , N 3 ,L) = ∏
n =1 N n !( d n − N n )!
∞
G = ∑ [ln(d n !) − ln( N n !) − ln[(d n − N n )!]]
n =1
∞
∞




+ α  N − ∑ N n  + β  E − ∑ N n En  ,
n =1
n =1




Assume Nn>>1 and dn>>Nn, so the stirling’s approximation applies
∞
G ≈ ∑[ln(dn!) − Nn ln(Nn ) + Nn − (dn − Nn ) ln(dn − Nn ) + (dn − Nn ) − αNn − βEn Nn ]
n=1
+ αN + βE
∂G
= − ln( N n ) + ln(d n − N n ) − α − β En = 0
∂N n
The most probable occupation numbers, for identical fermions, are
ln( N n ) = ln(d n − N n ) − α − β En
N n = ( d n − N n ) e − (α + β E n )
Nn =
dn
e
(α + βEn )
+1
.
3. Identical bosons:
∞
( N n + d n − 1)!
Q( N1 , N 2 , N 3 ,L) = ∏
n =1 N n !( d n − 1)!
∞
G = ∑ [ln[( N n + d n − 1)!] − ln( N n !) − ln[(d n − 1)!]]
n =1
∞
∞




+ α  N − ∑ N n  + β  E − ∑ N n En  ,
n =1
n =1




Assuming Nn>>1 and using stirling’s approximation
∞
G ≈ ∑[( Nn + dn −1) ln(Nn + dn −1) − ( Nn + dn −1) − Nn ln(Nn ) + Nn − ln[(dn −1)!] − αNn − βEn Nn ]
n=1
+ αN + βE
∂G
= ln( N n + d n − 1) − ln( N n ) − α − β En = 0
∂N n
ln( N n ) = ln( N n + d n − 1) − (α + β En )
N n = ( N n + d n − 1)e − (α + βEn )
dn −1
N n = (α + βEn )
−1
e
dn>>1
Nn =
dn
e
(α + βEn )
−1
5.4.4 Physical significance of α and β
The parameters α and β 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 = d n e − (α + βEn )
Nn =
Nn =
dn
e
(α + βEn )
e
∑N
n
=N
n =1
+1
dn
(α + βEn )
∞
α and β
∞
∑E N
n
n
=E
n =1
−1
To carry out the summation, En and dn should be know for particular potential.
By 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
h 2
Ek =
k where
2m
r  πn πn y πn
k = x ,
, z
 l
 x l y lz
l

.


z
ly
lx
A shell of thickness dk contains a volume
dτ
1
dτ = 4π k 2 dk ,
8
(
)
so the “degeneracy” is (the number of
electron states in the shell )
dτ
V 2
dk = 3
=
k dk .
2
π / V 2π
(
)
∫
0
For distinguishable particles, the first constraint becomes
N n = d ne
∑N
n
x e
− ax
∞
∞
− (α + β E n )
∞
N = ∑ dke
=N
n
n =1
n!
dx = n +1
a
− (α + βE k )
n =1
In the k-space, the sum will be converted into an integral, treating k as a
continuous variable, then
V 2
dk =
k dk ,
2
2π
2
h k
Ek =
,
2m
V −α
N = ∫ dN =
e ∫ e
2
0
2π
∞
∫
∞
0
2 n − ax 2
x e
2
h 2k 2
−β
2m
V −α
dN =
e e
2
2π
h2k 2
−β
2m
 m 

k dk = Ve 
2 
 2πβh 
−α
2
1 ⋅ 3 ⋅ 5 ⋅ L ⋅ (2n − 1) π
dx =
2 n +1 a n
a
∫
∞
0
2
e − ax x 2 dx =
k 2 dk .
3/ 2
⋅
π 1
 
16  a 
3/ 2
so
e
−α
N  2πβh
= 
V  m
2



3/ 2
⋅
∞
The second constraint
∑E N
n
n
= E becomes
n =1
2
V −α h ∞
E = ∫ dE =
e
e
2
∫
2π
2m 0
Or, putting in
e −α :
−β
h2k 2
2m
3N
E=
2β
3V −α  m 

k dk =
e 
2 
2β
 2πβh 
4
3/ 2
⋅
E
3
=
N 2β
The average kinetic energy of an atom at temperature T, in classical mechanic,
is
E 3
= k BT
N 2
Boltzmann constant
This suggests that β is related to the temperature:
1
β=
k BT
Different substances in thermal equilibrium with one another have the same
value of β , and which can be adopted as s definition of T.
Then
N  2πh 

= 
V  k BTm 
2
e
−α
3/ 2
It is customary to replace α by the so-called chemical potential,
µ (T ) ≡ −α k BT
α =−
µ (T )
k BT
.
By using the chemical potential, we can rewrite the most probable number of
particles in a particular (one-particle) state with energy ε :
N n = d ne
− (α + β E n )
Nn
n(ε ) =
dn
= e − (ε − µ )/ k BT
Maxwell-Boltzmann distribution
Nn =
Nn =
dn
e
(α + βEn )
dn
e
(α + βEn )
+1
Nn
n(ε ) =
dn
=
1
e
(ε − µ ) / k BT
+1
Fermi-Dirac distribution
−1
Nn
n(ε ) =
dn
=
1
e
(ε − µ ) / k BT
−1
Bose-Einstein distribution
The Maxwell-Boltzmann distribution is the classical result, for distinguishable
particles; the Fermi-Dirac distribution applies to identical fermions, and the
Bose-Einstein distribution is for identical bosons.
Maxwell-Boltzmann distribution
n(ε ) = e
− (ε − µ ) / k BT
T →0
(ε −µ) / kBT
e
0, if ε < µ (0),
→ 
∞, if ε > µ (0).
−(ε −µ) / kBT
e
0, if ε > µ (0),
→
∞, if ε < µ (0).
1 2
ε = mv
2
n(v ) = e (
)
− mv 2 − µ / 2 k BT
The Fermi-Dirac distribution has a particularly simple behavior as T
n(ε ) =
1
e
(ε − µ ) / k BT
+1
.
kB
As
(ε −µ) / kBT
e
→0 :
0, if ε < µ (0),
→
∞, if ε > µ (0).
1, if ε < µ (0),
n(ε ) → 
0, if ε > µ (0).
µ ( 0)
kB
All states are filled, up to an energy µ(0), and none are occupied for energies
above µ(0). Evidently the chemical potential at absolute zero is precisely the
Fermi energy:
µ ( 0) = E F
Bose-Einstein distribution
n(ε ) =
n(ε ) > 0
1
e (ε − µ )/ k BT − 1
e(ε −µ )/ kBT −1 > 0
e (ε − µ )/ k BT > 1
ε > µ (T )
µ (T ) < ε
hk 2
ε=
2m
V →0
hk 2
ε=
→0
2m
ε <0
Returning to the special case of an ideal gas, for distinguishable particles we
found that the total energy at temperature T is
3
E = Nk BT
2
and the chemical potential is
N  2πh 

= 
V  k BTm 
2
e
−α
µ (T ) ≡ −α k BT
3/ 2
  N  3  2πh 2
µ (T ) = k BT ln   + ln
  V  2  mk BT



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:
E = hν = hω
(2) Wave number: k = 2π/λ = ω / c
(1) Energy:
(3) Spin: two spin states occur, m=1 or –1.
(4) The number of the photons are not conserved:α
Nn =
dk
e
(α + βEk )
−1
Nn =
=0
dk
e hω / k B T − 1
For free photons in a box of volume V, dk is given by
V 2
dk =
k dk ,
2
2π
multiplied by 2 for two spin states, and expressed in terms of
ω = kc
V
d k = 2 3 ω 2 dω.
π c
So the energy density,
V
2
ω
dω
2 3
π c
E N n hω
=
= hω / k B T
V
V
e
That is
3
h
ω
hω
dω
= 2 3 hω / k B T
− 1)
− 1 V π c (e
hω 3
E (ω ) = 2 3 hω / k BT
dω
π c (e
− 1)
We introduce energy density per unit frequency:
hω 3
ρ (ω ) = 2 3 hω / k BT
π c (e
− 1)
This is Plank’s famous formula for the blackbody spectrum, giving the energy
per unit volume, per unit frequency, for an electromagnetic field in equilibrium at
temperature T.
ω = 2πν = kc =
2πc
dω = −
λ
2πc
2
λ
dλ
hω 3
E (ω ) = ρ (ω ) dω = 2 3 hω / k BT
dω
− 1)
π c (e
E (λ ) =
8πh
λ3 (e 2πhc / k Tλ
B
16π 2 hc
dλ = 5 2πhc / k BTλ
dλ = ρ ( λ ) dλ
2
− 1)
− 1) λ
λ (e
2πc
16π 2 hc
ρ (λ ) = 5 2πhc / k BTλ
λ (e
− 1)