chap 5. identical particles
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Chap 5. Identical Particles. Two-Particle Systems Atoms Solids Quantum Statistical Mechanics. Two particles having the same physical attributes are equivalent . They behave the same way if subjected to the same treatment. - PowerPoint PPT PresentationTRANSCRIPT
Chap 5. Identical Particles
1. Two-Particle Systems
2. Atoms
3. Solids
4. Quantum Statistical Mechanics
Two particles having the same physical attributes are equivalent.
They behave the same way if subjected to the same treatment.
CM: Equivalent particles are distinguishable since one can
keep track of each particle all the time.
QM: Equivalent particles are indistinguishable since one cannot keep
track of each particle all the time due to the uncertainty principle.
Indistinguishable particle are called identical.
5.1. Two-Particle Systems
2-particle state :
i H
t
1 2, , tr r
2 2
2 21 2 1 2
1 2
, ,2 2
H V tm m
r r
Schrodinger eq.:
2 3 31 2 1 2, , t d dr r r r Probability of finding particle 1 & 2 within
d 3 r1 & d 3 r2 around r1 & r2 , resp.
Normalization : 23 31 2 1 2, , 1 d d tr r r r
1 2,V V r r
1 2 1 2, , H Er r r r
1 2 1 2, , , exp
it E tr r r r
Read Prob 5.1,Do Prob 5.3.
5.1.1. Bosons & Fermions
1 & 2 indistinguishable (identical) :
1 2 1 2, a br r r rDistinguishable particles 1 & 2 in states a & b, resp. :
1 2 1 2 1 2, a b b aAr r r r r rbosonsfermions
Spin statistics theorem :Integer spin bosons
Half-integer spin fermions
No two fermions can occupy the same state.
1 2 1 2 1 2, 0 a a a aAr r r r r r
Pauli exclusion principle
Total :Symmetric
Anti-symm
Exchange Operator
Exchange operator : 1 2 2 1, ,P f fr r r r
21 2 2 1, ,P f P fr r r r 1 2, f r r f 2 1P
Let g be the eigenfunction of P with eigenvalue : P g g
2 P g P g 2 g g
1
For 2 identical particles, 1 2 1 2 2 1 2 1, , , , P H Hr r r r r r r r 1 2 1 2, ,H Pr r r r
P H H P or , 0P H
H & P can, & MUST, share the same eigenstates.
i.e.
1 2 2 1, , r r r r forbosonsfermions
Symmetrization requirement
Example 5.1. Infinite Square Well
Consider 2 noninteracting particles, both of the same mass m, inside an infinite square well of width a.
1-particles states are 2sin
n
nx x
a afor
222
2
nE n
m a2n K
Distinguishable particles :
1 2 1 21 2 1 2, n n n nx x x x
1 2
2 21 2 n nE n n K
E.g., Ground state: 11 1 2 1 2
2, sin sin
x x x xa a a
11 2E K
1st excited state : 12 1 2 1 2
2 2, sin sin
x x x xa a a
21 1 2 1 2
2 2, sin sin
x x x xa a a
12 5E K
21 5E K
Doubly degenerate.
Bosons :
2sin
n
nx x
a a2nE n K
Ground state: 0 1 2 1 2
2, sin sin
x x x xa a a
0 2E K
1st excited state :
1 1 2 1 2 1 2
2 2 2, sin sin sin sin
x x x x x x
a a a a a1 5E K
Fermions :
Ground state:
0 1 2 1 2 1 2
2 2 2, sin sin sin sin
x x x x x x
a a a a a0 5E K
Nondegenerate
5.1.2. Exchange Forces
Particles distinguishable : 1 2 1 2, a bx x x x
Consider 2 particles, one in state a, the other in state b.
( p’cle 1 in a, 2 in b. )
Bosons : 1 2 1 2 1 2
1,
2 a b b ax x x x x x
Fermions : 1 2 1 2 1 2
1,
2 a b b ax x x x x x
We’ll calculate for each case the standard deviation of particle separation
2 2 21 2 1 2 1 22 x x x x x x
( All states are normalized )
Distinguishable Particles
2 * 21 1 2 1 2 1 1 2, , D
x d x d x x x x x x
1 2 1 2, D a bx x x x 2 2 21 2 1 2 1 22 x x x x x x
2a
x * 2 *1 1 1 1 2 2 2 a a b bd x x x x d x x x
(, a , b normalized )
Similarly,2 22
D bx x
*1 2 1 2 1 2 1 2 1 2, , D
x x d x d x x x x x x x
* *1 1 1 1 2 2 2 2 a a b bd x x x x d x x x x
a bx x
2 2 21 2 2
a ba bDx x x x x x
Identical Particles
1 2 1 2 1 2
1,
2 a b b ax x x x x x 2 2 2
1 2 1 2 1 22 x x x x x x
2 2 2 2 2
1 2 1 2 1 2
* * * *1 1 2 2 1 1 2 2
1,
21
2
a b b a
a b b a b a a b
x x x x x x
x x x x x x x x
22 21 1 2 1 2 1,
x d x d x x x x 2 21
2 a b
x x a ,b orthonormal
Similarly,2 2 22
1
2 b a
x x x
2
1 2 1 2 1 2 1 2, x x d x d x x x x x
1 1
2 2 a b b a ab ba ba ab
x x x x x x x x a b ab ba
x x x x
* a babx d x x x x
2 2 21 2 1 2 1 22 x x x x x x2 2 2 2
1 2
1
2 b a
x x x x
1 2
a b ab bax x x x x x
2 2 21 2 2 2
a b ab baa bx x x x x x x x
2
1 2 2 ab baD
x x x x
or 22 22
abDx x x
Bosons are closer & fermions are further apart than the distinguishable case.
2 2
Dx xNote : if particles are far apart so a ,b don’t overlap.
effectiveattractiverepulsive exchange force for
bosonsfermions
Simplified Derivation
A A
2 2 2a ba b
x x x x
2 2 21 2 1 2 1 22 x x x x x x
D a b 1 2 1 2, D a bx x x x ~
2 2 21 2 1 2 1 22
Dx x ab x ab ab x ab ab x x ab
1 2 1 2 1 2
1,
2a b b ax x x x x x 1
2a b b a ~
1
2A ab A ab ba A ba ab A ba ba A ab
2 2 2 2 21 1 1 1 1
1
2x ab x ab ba x ba ab x ba ba x ab
2 2 2 21
2 a b ab bax x x b a x a b
2 21
2 a bx x if 0a b
abA a A b
Similarly,
2 2 22
1
2 a bx x x
1 2
1
2 a b b a ab ba ba abx x x x x x x x x x
a b ab ba
x x x x
2 2 21 2 2 2
a b ab baa bx x x x x x x x
2
1 2 2ab baD
x x x x
2 2 21
1
2 a bx x x
or 22 22
abDx x x
Bosons are closer & fermions are further apart than the distinguishable case.
2 2
Dx xNote : if particles are far apart so a ,b don’t overlap.
effectiveattractiverepulsive exchange force for
bosonsfermions
H2
If e’s were bosons, form bond. e’s are fermions, H2 dissociates
Let the electrons be spinless & in the same state :
In actual ground state of H2 :Spins of the e’s are anti-parallel so the spatial part is symmetric.
Do Prob 5.7
5.2. Atoms
Atom with atomic number Z ( Z protons & Z electrons ) :
2 2 22
1
1
2 2
Z Z
jj j kj j k
Z e eH k k
m r r r
1-e plan of attack :
1.Replace e-e interaction term with single particle potential.
2.Solve the 1-e eigen-problem.
3.Contruct totally anti-symmetric Z-e wave function, including spins.
4.Total energy is just the sum of the 1-e energies.
Nucleonic DoF dropped.See footnote, p.211.
0
1SI unit
4
1 Gaussian
k
Non-Interaction e Model
2 22
1 2
Z
jj j
Z eH k
m rDrop all e-e terms :
1
Z
jj
h r
where 2 2
2
2
Z eh k
m rr ( Hydrogenic hamiltonian )
where nlm and En are obtained from the hydrogen case by setting e2 Ze2 .
nl m n nl mh Er r rThus
In particular : 2
213.6 n
ZE eV
n0
aa
Zand
(Unsymmetrized) solutions to H E are 11
, ,
j j j
Z
Z n l m jj
r r r
with1
j
n
nj
E E
a0 = Bohr radius
5.2.1. Helium
2 2 2 2 22 21 2
1 2 1 2
2 2
2 2
e e e
H k k km r m r r r
1 2 H h h Er rNon-interacting e model :
1 2 1 2, nl m n l mr r r r n nE E E
2
213.6 n
ZE eV
n
Ground state :
0 1 2 100 1 100 2, r r r r 1 230 0
8 2exp
r ra a
3/2 /10 2 r aR a e
00
1
4Y
0a
aZ
20 13.6 2 1 1 E eV 13.6 8 eV 109 eV
Anti-symmetrized total wave function :0 gives only symmetric spatial part spin part must be antisymm (singlet).
Experiment : Total spin is a singlet. E0 79 eV.
Excited States
Long-living excited states :
1 2 1 100 2, nl mr r r r
( both e in excited states quickly turns into an ion + free e. )
Singlet Triplet
Spatial part of parahelium is symmetric
higher e-e interaction
higher energy than orthohelium counterpart
Do Prob 5.10
5.2.2. The Periodic Table
n \ l 0 s 1 p 2 d 3 f 4 g
1
2
3
4
5
6
72(2l+1) 2 6 10 14 18
Filling order of the periodic table.
l – degeneracy lifted by e-e interaction (screening).
Ground State Electron Configuration
Spectral Term :2 1S
JL
Ground state spectral terms are determined by
Hund’s rules (see Prob 5.13 ) :
1.Highest S.
2.Highest L.
3.No more than half filled : J = | L S |.
More than half-filled : J = L+S .
See R.Eisberg,R.Resnick, “Quantum Physics”, 2nd ed., §10-
3.E.g. , C = (2p)2 .
m 1 0 1
sz
30P S = 1, L = 1, J = 0
Do Prob 5.14
5.3. Solids
1. The Free Electron Gas
2. Band Structure
5.3.1. The Free Electron Gas
Solid modelled as a rectangular infinite well with dimensions { li , i = 1, 2, 3 }.
0 0
i ix lV
otherwiser
22
2
E
m
2 2
22
ii i
i
d XE X
m d x
i ii
X xr
ii
E E
Let2
ii
mEk
22
2i
i ii
d Xk X
d x
0 i ix lfor
sin cos i i i i i i iX A k x B k x
2 2
2
1
2 i
i i i
d XE
m X d x
Set
Set
i =1, 2, 3 unless stated otherwise
2 2
2 i
i
kE
m
Boundary Conditions2
ii
mEk
0 0iX 0iB
0iX l
sin cos i i i i i i iX A k x B k x
i i ik l n 1, 2, 3, in
sini i i iX A k x
22 2
0 0
sin i il l
i i i i i id x X A d x k xNormalization :2
2 i
i
lA 2 sin 2
sin2 4
x x
d x x
2i
i
Al
2sin
i
i ii i
nX x
l l
1 2 3
2sin
in n n i
i i i
nx
l lr 1 2 3
1 2 31 2 3 1 2 3
8sin sin sin
n n nx x x
l l l l l l
sin
ii i i
i
nX A x
l
1 2 3
2 2
2 i
n n ni
kE
m
22
2
i
i i
n
m l
22
2
m
k 1 2 3, , k k kk
k-Space Density
1, 2, 3, in
i
ii
nk
l
i i
i
k nl
Volume occupied by one allowed (stationary) state in the 1st quadrant of the k-space is
k ii
k
i il
3
V
= Volume of solid ii
V l
31
8
k
d k
k
33
2
Vd k
k is over allowed k in 1st quadrant of k-space. d 3 k is over all k-space.By symmetry, all 8 quadrants are equivalent.
= k-space density = density of allowed states 32V
with ni 1.
Fermi Energy
Consider a solid of N atoms, each contributing q free electrons.In the ground state, these qN electrons will occupy the lowest qN states.
Since2
2
2
E
mn kthese states occupy a sphere centered at the origin of the k-space.
Let the radius of this sphere be kF .
3
3
42
32
F
Vk qN ( Factor 2 comes from spin degeneracy. )
1/323
F
qNk
V 1/323
qN
V= free electron density
22
2
F FE k
m
22/323
2
m
Fermi energy
Total Energy
Total energy of the ground state :
totsmallest
E Enn
2 23
3222
Fk k
V kd k
m
2 22
30
2 422
Fk
V kk d k
m
25
210
F
Vk
m
1/323 Fk 5/32
22
310
tot
V qNE
m V
25/32 2/3
23
10
qN V
m
Compressing the solid increases its engergy.
Work need be done to compress it.
Electrons exerts outward pressure on the solid boundary.
totd EP
dV
2
3 tot
tot
Ed E dV
V
2
3 totE
V
25
2
2
3 10
Fk
m
2/32 2
5/33
5
mdegeneracy pressure
( Caused by Pauli exclusion )
Miscellaneous
25
210tot F
VE k
m
3 2 2
25 2F FV k k
m
1/323 e
F
Nk
V
2 2
2F
F
kE
m
3
5tot
Fe
EE
N Energy per e.
2
3 totPV EIdeal gas :
BN k T
2
3 totPV E2
5 e FN E
2
5F
eB
ET
k
5
16
2
8.617 10 /
6.582 10
0.511 /
B
e
k eV K
eV s
m MeV c
e gas :
Do Prob 5.16and for Al
5.3.2. Band Structure
1-D crystal with periodic potential V x a V x a = lattice constant
2 2
22
d
H V x Em d x
Simplest model : 1-D Dirac comb
For a finite solid with N atoms, strict mathematical periodicity can be achieved by imposing the periodic boundary condition V x Na V x
Bloch’s Theorem
V x a V xBloch’s theorem : i K ax a e x K = const
Proof : Let D be the displacement operator : D f x f x a
d x d x a
Dd x d x a
d x a
d x
dD x
d x
D V x x V x a x a V x D x
2 2
2 2
d d
D Dd x d x
D H H D i.e.
DH HD , 0D Hor
, 0D H D & H can share the same eigenstates.
Let D x x x a x
Since 0, setting i K ae completes the proof.
Periodic Boundary Condition
V x a V x i K ax a e x
For a 1-D solid with N atoms, imposing the periodic BC x Na x
gives 1i K N ae
2
n
KN a
0, 1, 2, n
K is real 2 2 x a x As expected
Dirac Comb
1-D Dirac comb :
1
0
N
j
V x x ja
For x ja :2 2
22
d
Em d x
22
2
d
kd x
2
mE
k
sin cos x A k x B k x
Periodic BC
sin cos x A k x B k x
Impose periodic BC : x Na x
Bloch’s theorem i K ax a e x2
n
KN a
0, 1, 2, n
Let
i K ax e x a
sin cos x A k x B k x
x ja
continuous at x = 0 sin cos i K aB e A ka B ka
0 < x < a
a < x < 0 sin cos i K ae A k x a B k x a
2
20
d d m
d x d x 2
2cos sin
i K a mA k e k A ka k B ka B
sin cos i K aB e A ka B ka 2
2cos sin
i K a mA k e k A ka k B ka B
sin cos 1 0 i K a i K ae ka A e ka B
2
21 cos sin 0
i K a i K a mk e ka A ke ka B
Condition for consistency is
2
sin cos 102
1 cos sin
i K a i K a
i K a i K a
e ka e ka
mk e ka ke ka
2
2sin sin 1 cos cos 1 0
i K a i K a i K a i K amke ka e ka k e ka e ka
2 2 2 22
2sin sin 1 2 cos cos 0
i K a i K a i K a i K amke ka e ka k e ka e ka
22
2sin 1 2 cos 0
i K a i K a i K ame ka k e ka e
22
2sin 1 2 cos 0
i K a i K a i K ame ka k e ka e
2
2sin 2cos 0
i K a i K amka k e ka e
2
2sin 2 cos cos 0
m
ka k Ka ka
2cos cos sin
m
Ka ka kak
Let z ka 2
m a
sincos cos
zKa z f z
z
Since cos 1Ka there’s no solution for | f (z) | > 1 ( band gaps ).
2 2 2 2
22 2
k zE
m ma
10
Do Prob 5.19with 20
5.4. Quantum Statistical Mechanics
1. An Example
2. The General Case
3. The Most Probable Configuration
4. Physical Significance of and
5. The Blackbody Spectrum
Fundamental assumption of statistical mechanics :In thermal equilibrium, each distinct system configuration of the same E is equally likely to occur.
( Can also be stated in terms of ensemble. )
5.4.1. An Example
Consider 3 non-interacting particles ( all of mass m ) in an 1-D infinite square well.
A B CE E E E22
2 2 2( )2
A B Cn n n
m a
Let 2 2 2 363 A B Cn n n
There’re 13 combinations of (nA , nB , nC ) that can satisfy the condition :
(11, 11, 11),
(13, 13, 5), (13, 5, 13), (5, 13, 13),
(1, 1, 19), (1, 19, 1), (19, 1, 1),
(5, 7, 17), (5, 17, 7), (7, 5, 17), (7, 17, 5), (17, 5, 7), (17, 7, 5).
Occupation Number
Specification of system configuration:
(nA , nB , nC ) occupation number
(11, 11, 11) ( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... ) N11 = 3
(13, 13, 5), (13, 5, 13), (5, 13, 13) ( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... ) N5 = 1N13 = 2
(1, 1, 19), (1, 19, 1), (19, 1, 1), ( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... ) N1 = 2N19 = 1
(5, 7, 17), (5, 17, 7), (7, 5, 17), (7, 17, 5), (17, 5, 7), (17, 7, 5).
( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... ) N5 = 1N7 = 1N17 = 1
Pn : Particles Distinguishable
Pn = Probability of finding a particle with energy En .
P1 (3/13) (2/3) = 2/13
P5 (9/13) (1/3) = 3/13
P7 (6/13) (1/3) = 2/13
P11 (1/13) (3/3) = 1/13
P13 (3/13) (2/3) = 2/13
P17 (6/13) (1/3) = 2/13
P19 (3/13) (1/3) = 1/13
Sum = 1
Particles are distinguishable each configurations is distinct.
Each is equally likely when system is in equilibrium.
(nA , nB , nC )
(11, 11, 11)
(13, 13, 5), (13, 5, 13), (5, 13, 13)
(1, 1, 19), (1, 19, 1), (19, 1, 1),
(5, 7, 17), (5, 17, 7), (7, 5, 17), (7, 17, 5), (17, 5, 7), (17, 7, 5).
Total = 13 distinct configurations
Pn : Fermions
Pn = Probability of finding a particle with energy En .
P1 0
P5 (1/1) (1/3) = 1/3
P7 (1/1) (1/3) = 1/3
P11 0
P13 0
P17 (1/1) (1/3) = 1/3
P19 0
Sum = 1
No state can be doubly occupied.Allowed distinct configurations are :
( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... ) N5 = 1, N7 = 1, N17 = 1
occupation number
( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... ) N11 = 3
( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... ) N5 = 1N13 = 2
( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... ) N1 = 2N19 = 1
( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... ) N5 = 1N7 = 1N17 = 1
Total = 1 distinct configurations
Pn : Bosons
Pn = Probability of finding a particle with energy En .
P1 (1/4) (2/3) = 1/6
P5 (1/4) (1/3) + (1/4) (1/3) = 1/6
P7 (1/4) (1/3) = 1/12
P11 (1/4) (3/3) = 1/4
P13 (1/4) (2/3) = 1/6
P17 (1/4) (1/3) = 1/12
P19 (1/4) (1/3) = 1/12
Sum = 1
Allowed distinct configurations are :
( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... ) N11 = 3( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... ) N5 = 1, N13 = 2( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... ) N1 = 2, N19 = 1( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... ) N5 = 1, N7 = 1, N17 = 1
occupation number
( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... ) N11 = 3
( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... ) N5 = 1N13 = 2
( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... ) N1 = 2N19 = 1
( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... ) N5 = 1N7 = 1N17 = 1
Total = 4 distinct configurations
5.4.2. The General Case
Consider a system whose 1-particle energies are Ei with degeneracy di , i = 1,2,3,...
Now, N particles are put into the system such that Ni particles have energy Ei .
Question :
For a given configuration (N1 , N2 , N3 ,... ), what is the number, Q(N1 , N2 , N3 ,... ),
of distinct states allowed?
Distinguishable Particles
Each energy level Ei corresponds to a bin with di compartments.
The problem is equivalent to finding the number of distinct ways to put N particles,
with Ni of them going into the di compartments of the ith bin.
Ans.: 1. There’re 1
1 1
!
! !N
N
NC
N N N
ways, without regard of picking order, to choose N1 particles from the whole N particles.
2. When putting a particle into the 1st bin, there’re d1 choices of compartments.
the number of ways to put N1 particles into the 1st bin is 1
1
1 1
!
! !
NN d
N N N
321
1 2 1 2 311 2 3
1 1 2 1 2 3 1 2 3
! !!, , ,
! ! ! ! ! !
NNN N N d N N N dN dQ N N N
N N N N N N N N N N N N
3. For the 2nd bin, one starts with N N1 particles so one gets
2
1 2
2 1 2
!
! !
NN N d
N N N N
31 21 2 3
1 2 3
!
! ! !
NN NN d d d
N N N 1
!!
iN
i
i i
dN
N
Fermions
1. Fermions are indistinguishable, so it doesn’t matter which ones are going to which
bin.
2. Each bin compartment can accept atmost one fermion.
So the number of way to place Ni fermions into the ith bin is
!
! !i
i i i
d
N d Nif i id N
1 2 31
!, , ,
! !
i
i i i i
dQ N N N
N d N
Bosons
1. Bosons are indistinguishable, so it doesn’t matter which ones are going to which bin.
2. Each bin compartment can accept any number of bosons.
Consider placing Ni bosons into the di compartments of the ith bin.
Let the bosons be represented by Ni dots on a line.
By inserting di 1 partitions, the dots are “placed” into di compartments.
If the dots & partitions are all distinct, the number of all possible arrangement is 1 ! i iN d
However, the dots & partitions are indistinguishable among themselves. Hence
1 2 3
1
1 !, , ,
1 ! !
i i
i i i
N dQ N N N
d N
5.4.3. The Most Probable Configuration
An isolated system in thermal equilibrium has a fixed total energy E,
and fixed number of particle N , i.e.,
1
ii
N N1
i ii
N E E
Since each possible way to share E among N particles has the same probability to exist,
the most probable configuration (N1 , N2 , N3 ,... )
is the one with the maximum Q(N1 , N2 , N3 ,... ).
Since Q(N1 , N2 , N3 ,... ) involves a lot of factorials, it’s easier to work with lnQ.
Lagrange Multipliers
Problem is to maximize ln Q, subject to constraints1
i ii
N E E1
ii
N Nand
Using Lagrange’s multiplier method, we maximize, without constraint,
1 1
ln
i i ii i
G Q N N E N E
i.e., we set 0
i
G
N
Note : 0
G 0
Gare just the original constraints.
Distinguishable Particles
1
!!
iN
i
i i
dQ N
N
1
ln ln ! ln ln !
i i ii
Q N N d N
Stirling’s approximation : ln ! ln z z z z for z >>1
1
ln ln ln ln
i i i i ii
Q N N N N d N N N
1 1
ln
i i ii i
G Q N N E N E
1 1 1
ln ln ln
i i i i i i i ii i i
G N N N N d N N N N N E N E
ln ln
i i ii
Gd N E
N0
exp i i iN d E
Fermions 1 1
ln
i i ii i
G Q N N E N E
1
!
! !
i
i i i i
dQ
N d N 1
ln ln ! ln ! ln !
i i i ii
Q d N d N
di , Ni >>1 : 1
ln ln ln ln
i i i i i i i ii
Q d d N N d N d N
1 1 1
ln ln ln
i i i i i i i i i i ii i i
G d d N N d N d N N N E N E
ln ln
i i i ii
GN d N E
N0
1 exp ii
i
dE
N
exp 1
i
ii
dN
E
Bosons 1 1
ln
i i ii i
G Q N N E N E
1
ln ln 1 ! ln 1 ! ln !
i i i ii
Q N d d N
di , Ni >>1 :
1 1 1
1 ln 1 1 ln 1 ln
i i i i i i i i i i ii i i
G N d N d d d N N N N E N E
ln 1 ln
i i i ii
GN d N E
N0
11 exp
ii
i
dE
N
exp 1
i
ii
dN
E
1
1 !
1 ! !
i i
i i i
N dQ
d N
1
ln 1 ln 1 1 ln 1 ln
i i i i i i i ii
Q N d N d d d N N
di 1 di
Do Prob 5.26, 5.27
5.4.4. Physical Significance of and
Ideal gas in 3-D infinite square well.
2 2
2
k
kE
m, ,
yx z
x y z
nn n
l l lk
3
32
V
d kk
Taking the “bin” as the spherical shell between k and k+dk, the degeneracy dk isjust the number of states in the shell :
2
3 42
k
Vd k d k 2
22
Vk d k
Distinguishable Particles
exp i i iN d E
222
k
Vd k dk
1
ii
N N 22
0
exp2
k
VN dk k E
22 2
20
exp2 2
V
e dk k km
3/2
22
mV e
2
10
1 1 1
2 2
n x
n
ndx x e
1
2
3/2
2 2
2
2 4
V me
3/222
Ne
V m
1
i ii
N E E2 2
4 22
0
exp2 2 2
V
E e dk k km m
5/22
2 2
3 2
4 4
V me
m3/2
2
3
2 2
V mE e
3
2
N
cf. kinetic theory result : 3
2 BE N k T
1 Bk T universal
Distributions
Setexp
exp 1
exp 1
ii i
B
ii
i
B
ii
i
B
EN d
k T
dN
Ek T
dN
Ek T
Bk T
D
F
B
exp
1
exp 1
1
exp 1
B
B
B
nk T
n
k T
n
k T
ii
i
Nn
d
Maxwell – Boltzmann
Fermi – Dirac
Bose - Einstein
Chemical potential
n() Most probable number of particles in a state with energy . occupation number
Fermi-Dirac as T 0 1
exp 1
B
n
k TT 0 :
0 0exp
0
Bk T
1 0
0 0
n (0) EF
T 0
T > 0
Ideal Gas
Distinguishable particles 3
2 BE N k T
3/222
Ne
V m
3/222ln
B
B
Nk T
V mk T
23 2ln ln
2
B
B
Nk T
V mk T
Indistinguishable particles :
3
32
k
VN d k n
2
2 2 202 1
exp 12
B
V kdk
kk T m
22
02
k
Vdk k n F
B
3
32
k k
VE d k n
2 4
2 2 202 2 1
exp 12
B
V kdk
m kk T m
FB
fixes
gives
EC
TRead Prob. 5.29 Do Prob 5.28
5.4.5. The Blackbody Spectrum
Photon quantum of EM field
spin 1, massless boson with v c.
Only m 1 occurs.
Number not conserved 0.
E h
2
kc
exp 1
B
dN
k T/
k k cd d 2
2 32
2
Vd
c
N
dV
Energy density in range d
3
2 3
1
exp 1
B
c
k T
Black body spectrum
Blackbody Spectrum
6000 K
4000 K
2000 K
Visible region
Do Prob 5.31Read Prob 5.30