Physics
2203,
Fall
2011
Modern
Physics
.
  Monday,
Nov.
5th
,
2012
‐‐‐Sta=s=cal
Physics:
Quantum
sta=s=cs:
Ch.
15
in
our
book.
Notes
from
Ch.
10
in
Serway
‐‐Quantum
Sta=s=cs:
BE
and
FD
‐‐Example
of
BE‐Blackbody
radia=on:
Specific
heat
  Announcements
‐‐‐Term
papers
due
Nov.
19th
our
book.
‐‐‐New
Syllabus
‐‐‐First‐ever
physics‐major
ice
cream
social:
Wed.
Nov.
14th
4:30
pm
library.
Lets
find
the
average
number
of
par=cles
with
some
energy‐‐Ej
n j = n j1 p1 + n j 2 p2 + i i
n j1 is the number of particles found in the j level in 1
p1 is the probablility of observing arrangement 1
Basic
postulate
of
sta=s=cal
mechanics
is
that
there
is
equal
probability
of
find
any
microstate!
Lets
calculated
the
average
number
of
par=cles
with
energy
zero:
Total
number
of
microstates
is
1287
n0 = n01 p1 + n02 p2 + n03 p3 − − − − −
n01 p1 = (5)(6 / 1287)
n02 p2 = (4)(30 / 1287)
find n1
n0 = 2.307
Lets
find
the
average
number
of
par=cles
with
some
energy‐‐Ej
n j = n j1 p1 + n j 2 p2 + i i
n j1 is the number of particles found in the j level in 1
p1 is the probablility of observing arrangement 1
Lets
calculated
the
average
number
of
par=cles
with
energy
zero:
Total
number
of
microstates
is
1287
n0 = n01 p1 + n02 p2 + n03 p3 − − − − −
n0 = (5)(6 / 1287) + (4)(30 / 1287) + (4)(30 / 1287) +
(3)(60 / 1287) + (4)(30 / 1287) + (3)(120 / 1287) +
(2)(60 / 1287) + (4)(15 / 1287) + (3)(120 / 1287) +
(3)(60 / 1287) + (2)(180 / 1287) + (1)(30 / 1287) +
(3)(60 / 1287) + (2)(90 / 1287) + (2)(180 / 1287) +
(1)(120 / 1287) + (0)(60 / 1287) + (2)(15 / 1287) +
(1)(60 / 1287) + (0)(15 / 1287) = 2.307
Probability
of
finding
a
par=cle
with
energy
0,
if
we
reach
randomly
into
any
of
the
boxes
represen=ng
6
par=cles
with
total
energy
8E.
p(0) =
n0 2.307
=
= 0.385
6
6
Prove that p(1E)=0.256
fMB = Ae
−E / kB T
g(E)
is
the
Density
of
States
N
N = ∑ ni → =
V
Determine A.
∞
∫ g(E) f
MB
(E)dE
0
Velocity
Distribu=on
4π N  m 
n(v)dv =


V  2 π k BT 
3/2
2 −mv 2 /2 kB t
ve
dv
Equipar;;on
of
Energy
–a
classical
molecule
in
thermal
equilibrium
at
temperature
T
has
an
average
energy
of
kBT/2
for
each
independent
mode
of
mo;on—degrees
of
freedom
Three dimensional motion
1 2 1 2
1
1
3k T
mv = mvx + + mvy2 + mvz2 = B
2
2
2
2
2m
MB
sta;s;cs
works
when
the
average
distance
d
between
par;cles
is
greater
than
the
quantum
uncertainty.
Δx << d
N
3
<< 1
 
3/2
 V  8(mkBT )

ΔxΔpx ≤
2
Low
density,
high
mass,
high
Temperature
Fermions: ψab
(
 
r1 , r 2 = ψa (r1 )ψb (r2 ) -ψa (r2 )ψb (r1 )
)
Lets try to put two electrons into the same state, a=b
 
ψab r 2 , r1 = ψa (r2 )ψa (r1 ) -ψa (r1 )ψa (r2 )
 
ψab r 2 , r1 = ψa (r2 )ψa (r1 ) -ψa (r2 )ψa (r1 ) ≡ 0
(
(
)
)
Bosons: ψab
(
 
r1 , r 2 = ψa (r1 )ψb (r2 ) +ψa (r2 )ψb (r1 )
)
1945
Nobel
Prize
for
discovery
of
the
Exclusion
Principle.”
Lets try to put two bosons into the same state, a=b
 
ψab r 2 , r1 = ψa (r2 )ψa (r1 ) +ψa (r1 )ψa (r2 )
 
ψab r 2 , r1 = ψa (r2 )ψa (r1 ) +ψa (r2 )ψa (r1 ) ≡ 2ψa (r2 )ψa (r1 )
(
(
)
)
Bose Condensation
Bosons: ψab
(
 
r1 , r 2 = ψa (r1 )ψb (r2 ) +ψa (r2 )ψb (r1 )
)
Bose − Einstein statistics
Integral
Spin
Systems:
Alpha
par=cle
(S=0),
Photons
(S=1),
deuteron
(S=1)
Fermions: ψab
(
 
r1 , r 2 = ψa (r1 )ψb (r2 ) -ψa (r2 )ψb (r1 )
)
Fermi-Dirac Statistics
Half
Integral
Spin
Systems:
Electrons
(S=1/2),
neutron
(S=1/2)
Fermi
Dirac
Important:
par;cles
are
iden;cal
so
each
microstate
has
iden;cal
count.
Now
the
total
number
is
20,
not
1287.
Lets find n 0
1  5 + 4 + 4 + 3 + 4 + 3 + 2 + 4 + 3 + 3 +
n0 = 

2
+1+
3
+
2
+
2
+1+
0
+
2
+1+
0
20 

n 0 = 2.45
n0
p(0) = = 0.408
6
You should be able to determine
p(1E), p(2E), etc.
MB
distribu=on
BE
distribu=on
Important:
par;cles
are
iden;cal
so
each
microstate
has
iden;cal
count.
Fermions:
Only
two
electrons
in
each
state
Lets find n 0
1
n 0 = ( 2 + 2 + 2 ) = 2.00
3
Lets find n1
1
n1 = ( 2 + 2 +1) = 5 / 3
3
n0
n1
p(0) = = 0.333 : p(1E) = = 0.2.78
6
6
What is n 8 ?
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
MB
distribu=on
FD
distribu=on
Ground
rules
are
a
fixed
number
of
par=cles
at
a
fixed
temperature
T.
f(E)
is
the
probability
of
finding
a
par=cle
in
an
energy
state
E
at
temperature
T.
B
and
H
determined
1
1
by
fixing
the
fBE (E) = E / k T
fFD (E) =
B
Be
−1 number
of
par=cles
HeE / kB T +1
n(E)dE = g(E) fBE (E)dE
N
=
 
 V bosons
fBE (E) =
1
e
E / kB T
∞
∫ Be
0
dE
E / kB T
−1
n(E)dE = g(E) fFD (E)dE
N
=
 
 V Fermions
H = e−EF / kB T
−1 EF → Fermi Energy
∞
∫ He
0
dE
E / kB T
+1
fFD (E) =
1
e
( E −EF ) / kB T
+1
This
defines
the
Fermi
energy!
It
is
the
energy
between
filled
and
empty
states—keeps
the
system
neutral.
Fermi‐Dirac
Distribu=on
at
T=0
n(E)dE = g(E) fFD (E)dE
Fermi Energy can be a function of T
Fermi‐Dirac
Distribu=on
at
T≠0
1
High
T
fFD ( E ) = ( E − E ) / kT
e F
+1
1
fFD ( E << EF ) = −∞
=1
e +1
1
fFD ( E >> EF ) = ∞
=0
e +1
The
Fermi‐Dirac
func=on
is
symmetric.
What
you
lose
below
EF
you
gain
above
EF.
kT > EF
fFD ( E ) =
1
eE / kT + 1
= e− E / kT
BE at high temperature
fBE (E) =
1
e E / kB T
e−E / kB T
−E / kB T
=
≈
e
−1 1 − e−E / kB T
FD at high temperature
E>>E F : k B T>>E F
fFD (E) =
1
( E −EF ) / kB T
=
e
− ( E −EF ) / kB T
e
+1 1+ e
fFD (E >> EF ) ≈ e−E / kB T
− ( E −EF ) / kB T
Fermions ψ (1, 2 ) = −ψ ( 2,1)
Bosons ψ (1, 2 ) = +ψ ( 2,1)
In
the
Quantum
region
we
use
the
Pauli
principle
for
and
put
everything
in
the
lowest
state
for
Bosons.
At
an
elevated
temperature
we
should
go
over
to
a
classical
Maxwell‐Boltzmann
distribui=on
Today
we
will
talk
about
heat
capacity
in
solids.
The
electrons
are
and
the
laece
vibra=ons
are
Bosons.
Photons
in
a
box
at
Temperature
T.
BE
sta=s=cs
n(E)dE = g(E) fBE (E)eE
Energy density u(E)dE is
g(E)EdE
u(E)dE=En(E)dE= e/ k T −1
e B
g(E) =
Need to find g(E)
We have done this
Standing waves in box.
8π E 2
(hc)
3
u(E)dE=
8π h
f3
u( f ,T ) = 3 hf / k T
c e B −1
E = hf
8π
(hc)
3
E 3dE
ee/ kB T −1
Just
BE
(a)
Find
an
expression
for
the
number
of
photons/unit
volume
with
energies
between
E
and
E+dE
in
a
cavity
of
temperature
T
n(E)dE=
8π
(hc)
3
E 2 dE
ee/ kB T −1
(b)
Find
an
expression
for
the
total
number
of
photons
per
unit
volume
(all
energies).
N
=
V
∞
∞
∫ n(E)dE = ∫
0
N 8 π (kBT )3
=
3
V
(hc)
0
∞
∫
0
8π
(hc)
3
E 2 dE
ee/ kB T −1
( E / k BT )
2
dE / kBT
N
(kBT )3
= 8π
3
V
hc
( )
∞
∫
0
z 2 dz
ez −1
ee/ kB T −1
(c)
Find
the
number
of
photons/cm3
inside
a
cavity
whose
walls
are
T=3000
K
∞
∫
0
z 2 dz
≈ 2.40
ez−1
3
 (8l62x10 −5 eV / K )(3000K ) 
N
11
3
= 8 π (2.40)
=
5.47x10
photons
/
cm

V
1.24x10 −4 eV icm


The
molar
specific
heat
C,
is
determined
by
the
change
in
thermal
energy
U
dU
added
to
a
mole
of
substance,
as
a
func=on
of
T.
C=
dT
Model
our
solid
as
a
collec=on
of
independent
harmonic
oscillators.
A
one
dimensional
harmonic
oscillator
has
two
degrees
of
freedom:
poten=al
energy
and
kine=c
energy
Equipartition Theorem:
k BT
2
U = 3N A kBT = 3RT
R universal gas constant
R=N A kB = 8.31J / moliK
1-D HO is 2
k BT
for each degree of freedom
2
Low
temperature
wrong:
Einstein
solved
this
in
1907
d(3RT )
C=
= 3R = 5.97cal / moliK
dT
Three dimensional
3kBT
Einstein
assumed
as
we
did
that
a
solid
can
be
represented
by
independent
harmonic
oscillators.
He
showed
that
in
1‐D
system
the
average
energy
is:
E=
ω
Atoms independent so U=3N A E =
eω / kB T −1
2
 ω 
dU
e ω / k B T
C=
= 3R 
 ω / k B T
dT
−1)2
 kBT  (e
when ω <<k BT
ω
E = ω / k T
≈ k BT
B
e
−1
One paramater ω
ω = k BTE : Einstein Temp.
3N A ω
eω / kB T −1
d(3RT )
C=
= 3R Dulong-Petit Law
dT
2
 ω 
dU
e ω / k B T
C=
= 3R 
 ω / k B T
2
dT
k
T
(e
−1)
 B 
Equipartition Theorem:
k BT
for each degree of freedom
2
3kBT
for three translational degrees of freedom.
2
E=
3kBT 1 2 1 2 1
1
+ Iω x + Iω y + µ (dr / dt)2 + k(r − r0 )2
2
2
2
2
2
(a)
Calculate
the
vibra=on
frequency
of
the
carbon
atoms
in
diamond
if
the
Einstein
temperature
is
1300K.
Find
the
energy‐level
spacing.
k T
ω = k BTE : ω = B E

ω = 0.112eV
8.62x10
(
=
−5
eV / K ) (1300K )
6.58x10 −16 eV is
= 1.70x1014 Hz
(b)
Calculate
the
average
oscillator
energy
at
room
temperature
and
at
1500K,
compare
to
level
spacing.
E(300) =
ω
e
ω / kB T
−1
E(300)
= 0.01
ω
= 0.00149eV
E(1500) =
ω
e
ω / kB T
−1
E(1500)
=1
ω
= 0.0813eV