Transcript Quantum Statistics Applications

Quantum Statistics:Applications
• Determine
P(E) = D(E) x n(E)
probability(E) = density of states x prob. per state
• electron in Hydrogen atom. What is the relative
probaility to be in the n=1 vs n=2 level?
D=2 for n=1
D=8 for n=2
as density of electrons is low can use Boltzman:
nFD 
•
1
e
( E  E F ) / kT
1
e
 E / kT
E  EF
can determine relative probability
P(n  2) 8 e  E2 / kT
 ( E2  E1 ) / kT


4
e
P(n  1) 2 e  E1 / kT
 4e 10.2 / .026  10171 (!)
for E  10.2eV T  3000 K
• If want the ratio of number in 2S+2P to 1S to be .1
you need T = 32,000 degrees. (measuring the
relative intensity of absorption lines in a star’s
atmosphere or a interstellar gas cloud gives T)
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1D Harmonic Oscillator
• Equally spaced energy levels. Number of states at
each is 2s+1. Assume s=0 and so 1 state/energy
level
• Density of states
D( E)  NE  1
• N = total number of “objects” (particles) gives
normalization factor for n(E)

N   D( E )n( E )dE
0
N 

A  E / kT
0 

e
dE 
AkT

N
A
kT
• note dependence on N and T
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1D H.O. : BE and FD
• Do same for Bose-Einstein and Fermi-Dirac
1
1
nBE 
n

FD
BeE / kT  1
BeE / kT  1
N 

1
0 
n( E )dE 
1
BBE 
BFD 
1
(  e  E F / kT )
1  e  N / kT
e N / kT  1
• “normalization” varies with T. Fermi-Dirac easier
to generalize
• T=0 all lower states fill up to Fermi Energy
1
1
n( E  E F )  1 ( E  EF ) / kT

(T  0)
e
1 0 1
1
1
n( E  E F )  0

(T  0)
( E  E F ) / kT
e
1  1
N 
EF
0
1

 1 dE 
EF

or E F  N
• In materials, EF tends to vary slowly with energy
(see BFD for terms). Determining at T=0 often
“easy” and is often used. Always where n(E)=1/2
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Density of States “Gases”
• # of available states (“nodes”) for any wavelength
• wavelength --> momentum --> energy
• “standing wave” counting often holds:often called
“gas” but can be solid/liquid. Solve Scrd. Eq. In 1D
d2
dx 2
  a  0
n 
L
 ( x  0)   ( x  L)  0 0
n 2
L
2 L
 kn 

 n  k 
L n

2
2L
n
• go to 3D. ni>0 and look at 1/8 of sphere
kx 
n x
L
ky 
n y
L
k z  Lnz
1 4n3
# nodes
 Degeneracy
8 3
1 4
( 2 L) 3

 Deg  3
take derivative
8 3

4V
 N ( )d  4  Degd

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Density of States II
• The degenracy is usually 2s+1 where s=spin. But
photons have only 2 polarization states (as m=0)
• convert to momentum
p
h


hn
2L
# nodes
1 4
8 3
(2s  1)(
2 Lp 3
h
)
D( p)dp  2 (2s  1)( 2hL )3 p 2 dp
• convert to energy depends on kinematics
E  pc dE  cdp
relativistic
D( E )dE  2 (2s  1)( 2hcL )3 E 2 dE
8V 2
 3 3 E dE
ch
for 
• non-realtivistic
2
p
E
2m
p
 dE  dp
m

2L 3
m
D ( E )dE  (2 s  1)( ) 2m E
dE
2
h
2m E

2L 3
 (2 s  1)( ) 2m 3 / 2 E 1/ 2 dE
2
h
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Plank Blackbody Radiation
• Photon gas - spin 1 Bosons - dervied from just stat.
Mech. (and not for a particular case) by S.N. Bose
in 1924
• Probability(E)=no. photons(E) = P(E) = D(E)*n(E)
• density of state = D(E) = # quantum states per
energy interval =
2
8 V E dE
 3
3
c
h
• n(E) = probability per quantum state.
Normalization: number of photons isn’t fixed and
so a single higher E can convert to many lower E
n
1

e e
E / kT
1

1
e
E / kT
1
  if E  0
• energy per volume per energy interval =
8 E 3
T ( E )  E D( E )n( E )  3 3 E / kT
c h (e
 1)
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Phonon Gas and Heat Capacity
• Heat capacity of a solid depends on vibrational
modes of atoms
• electron’s energy levels forced high by Pauli Ex.
And so do not contribute
• most naturally explained using phonons
- spin 1 psuedoparticles
- correspond to each vibrational node
- velocity depends on material
• acoustical wave <---> EM wave
phonon
<---> photon
• almost identical statistical treatment as photons in
Plank distribution. Use Bose statistics
• done in E&R Sect 11-5, determines
dE
cV 
dT
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