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Lecture 20. Continuous Spectrum, the Density of States (Ch. 7), and Equipartition (Ch. 6). - PowerPoint PPT Presentation
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Lecture 20. Continuous Spectrum, the Density of States (Ch. 7), and Equipartition (Ch. 6)
TkZ
dP
B
iii
exp exp
B
gP d d
Z k T
The units of g(): (energy)-1
Typically, it’s easier to work with the integrals rather than the sums. Thus, whenever we consider an energy range which includes many levels (e.g., when kBT >> inter-level spacing), and, especially, when we are dealing with continuous spectra, we’ll replace the sum over a discrete set of energy levels with an integral over a continuum. When doing this, we must replace the level degeneracy with a new variable, the density of states, g(), which is defined as the number of states per unit energy integral:
11,d
ii d,
The Density of States
Let’s consider G() - the number of states per unit volume with energy less than (for a one- and two-dimensional cases, it will be the number of states with energy less than per unit length or area, respectively). Then the density of states is its derivative:
d
dGg
The number of states per unit volume in the energy range – +d : dg
The probability that one of these states is occupied: exp1
Z
The probability that the particle has an energy between and +d : 1
expP d g dZ
0
1 dPSince the partition function is 0
expZ g d
The density of states, g(), is the number of states per unit energy
Page 279, Ch. 7
The Density of States for a Single Free Particle in 1D
The energy spectrum for the translational motion of a molecule in free space is continuous. We will use a trick that is common in quantum mechanics: assume that a particle is in a large box (energy quantization) with zero potential energy (total energy= kin. energy). At the end of the calculation, we’ll allow the size of the box to become infinite, so that the separation of the levels tends to zero. For any macroscopic L, the
The momentum in each of these states:
- the energy (the non-relativistic case)
m
k
mL
nh
m
pE n
n 282
2
2
222
1E2E3E
4E
L x
For a quantum particle confined in a 1D “box”, the stationary solutions of Schrödinger equation are the standing waves with the wavelengths:
,...2,1,22
nkn
L
nn
kL
hnhp
nn
2
the wave numberL
nk
2
k
where
Lk
energy levels are very close to each other (E ~ 1/L2), and the “continuous” description works well.
2
h
In the “k” (i.e., momentum) space, the states are equidistantly positioned (in contrast to the “energy” space).
The Density of States for a Single Free Particle in 1D (cont.)
The total number of states with the wave numbers < k is: kL
L
kkN
/
G(k) - the number of states
mk 21
2L mG
1 2 1
2D dG L s m
gd
additional degeneracy of the states (2s+1) because of spin [e.g., for electrons 2s+1=2]
g()
1D
kLG k
The Density of States for a Particle in 2D and 3D
2D
kx
ky For quantum particles confined in a 2D “box” (e.g., electrons in FET):
22yx
y
yy
x
xx kkk
L
nk
L
nk
22 2
2
1 2
4 4 4 4
x y
k areak k A mN k G k A G
L L
k
# states within ¼ of a circle of radius k
xL
3D
ky
kx
kz
3/ 23 3 3
2 2 2 2
4 / 31 2
8 6 6 6
x y z
k k volume k V V mN k G k G
L L L
- does not depend on
3/ 23 1/ 2
2 2
2 1 2
4D V s m
g
22
2 1
2D A s m
g
g()
2D
Thus, for 3D electrons (2s+1=2): 3/ 2
3 1/ 22 2
2
2D V m
g
g()
3D
Partition Function of a Free Particle in Three Dimensions
The mean energy can be found without evaluating the integral:
The numerator can be integrated by parts:
d
d
ZE
0
2/1
0
2/3
exp
exp
ln
Z1 – a reminder that this is the partition function for a single particle
dd
0
2/1
0
2/3 exp2
3exp (the integrand vanishes at both limits)
TkE B2
3 - in agreement with the equipartition theorem
dmsVms
gdgVZ D
0
2/3
222/1
2/3
223
0
1 exp2
4
122
4
12exp
The Partition Function of a Free Particle (cont.)
dmV
Z
0
2/3
221 exp2
4
2
expexp 2/3
0
2/3
0
TkdxxxTkd BB
Q
B
V
VTmkVZ
2/3
21 2
Tmk
hL
B
Q 2 the quantum length, coincides [within a factor of
()1/2 ] with the de Broglie length of a particle whose kinetic energy is kBT.
- the quantum volume and density
2/3
22
1
Tmk
Vn B
(assume that s=0)
3QQ LV
VQ is the volume of a box in which the ground state energy would be approximately equal to the thermal energy, and only the lowest few quantum states would be significantly populated. When V VQ, the continuous approximation breaks down and we must take quantization of the states into account. We also need to consider quantum statistics (fermions and bosons)!
2/322/32
2
2
m
h
TmkV
BQ
For unit volume (V=1): QQ
nV
Z 1
1
1 2
1 2 1 2 1 2exp exp exp exptotals s s
Z E s E s E s E s Z Z
Partition Functions for Distinguishable Particles
For two non-interacting particles, Etotal =E1+E2 : s
total sEsEZ 21exp
Ntotal ZZZZ ......21
Ntotal ...21For a microcanonical ensemble, we know the answer for total:How about the canonical ensemble?
TkTkTkZ
BBB
exp1exp
0exp1
The partition function for a single particle:
The states of this two-particle system are: (0,0), (,), (,0), (0,) - distinguishable
21
24
1
exp12
expexp21exp ZTkTkTkTk
ZBBBB
itotal
Example: two non-interacting distinguishable two-state particles, the states 0 and :
all the states for the composite system
non-interacting distinguishable particles
ln,, BkNVUS ZTkNVTF B ln,, Recall the analogy between and Z:
There is nothing in the derivation of the partition function or the Boltzmann factor that restricts it to a microsystem. However, it is often convenient to express the partition function of a combined macrosystem in terms of the single-particle partition function. We restrict ourselves to the case of non-interacting particles (the model of ideal gas).
distinguishable
Partition Functions for Indistinguishable Particles (Low Density Limit)
(For the quantum indistinguishable particles, there would be three microstates if the particles were bosons [(0,0), (0,), (,)] and one microstate [(0,)] if the particles were fermions).
If the particles are indistinguishable, the equationNtotal ZZZZ ......21
should be modified.
If the states (,0) and (0,) of a system of two classical particles are indistinguishable:
21
3
1
2expexp1exp Z
TkTkTkZ
BBB
itotal
=1
2
2
1
Exceptions: the microstates where the particles are in the same state. If we consider particles in the phase space whose volume is the product of accessible volumes in the coordinate space and momentum space (e.g., an ideal gas), the states with the same position and momentum are very rare (unless the density is extremely high)
For a low-density system of classicalindistinguishable particles:
Ntotal Z
NZ 1!
1
)volume"" momentum accessible the(!
13
N
N
N h
V
N
2121 2
1expexp
2
1
1 2
ZZsEsEZs s
total For two indistinguishable particles:
Recall the multiplicity for IG:
Z and Thermodynamic Properties of an Ideal Gas
The partition functionfor an ideal gas:
NZN
Z 1!
1
N
Q
N
BN
V
V
N
TmkV
NZ
!
1
2!
12/3
2
1ln1ln1lnlnlnln
n
nN
NV
VNNVVNZ Q
Note that the derivation of Z for an ideal gas was much easier than that for : we are not constrained by the energy conservation!
1ln1lnln,,
n
nTNk
NV
VTNkZTkNVTF Q
BQ
BB
TNkNV
VNVNZU B
Q
QQ 2
31
2
31lnln
V
TNkV
VTNk
V
FP B
B
NT
ln,
2
5ln1lnln1ln 2/3
,n
nNkT
TT
n
nNk
n
nT
TNk
T
FS Q
BQ
BQ
B
NV
- the “energy” equation of state
- the “pressure” equation of state (“ideal gas” law)
- the Sackur-Tetrode equation
TNkTSFU B2
3
The Equipartition Theorem (continuous spectrum)
Equipartition Theorem: At temperature T, the average energy of any quadratic degree of freedom is ½kBT.
2cqqE We’ll consider a “one-dimensional” system with just one degree of freedom:
qq B
cqTk
qEZ 2expexp
q – a continuous variable, but we “discretize” it by considering small and countable q’s:
dqcqq
qcqq
Zq
22 exp1
exp1 New variable: qcx
cq
dxxdxxcq
Z
1expexp
11 22
The average energy:
22
1
2
11 12/3 TkZ
ZE B
Again, this result is valid only if the degree of freedom is “quadratic” (the limit of high temperatures, when the spacing between the energy levels of an individual system is << kBT).
The partition function for this system:
exp(-x2)
x
E(q)
Equipartition Theorem and a Quantum Oscillator
Classical oscillator: TkTkTk
Ekx
m
pE B
BB 2222
22
Quantum oscillator:m
knn
2
1
00 exp12
expexp
2expexp
nZ nthe equipartition function:
...2expexp1
exp1ln2
ln Z
the average energy of the oscillator:
1exp
1
2
1
exp1
exp
2ln
ZE
the limit of high T:
1exp1
TkB
TkE B
11
2
1
the limit of low T: 1
TkB 2
E (the ground state is by far
the most probable state)
the assumption of a continuous energy spectrum was violated
The Maxwell Speed Distribution (ideal gas)
Now let’s look at the speed distribution for these particles:
The probability that a molecule has an energy between and +d is:
dTk
gn
dTk
gZ
dPBQB
exp
1exp
1
1
QD n
mg 2/32/12/1
2/3
223 22
4
1
TkP
B
exp2 2/12/3
0
TkB
exp
0
g() Cg
=
0
P()
2
2
1mv
The probability to find a particle with the speed between v and v+dv, irrespective of the direction of its velocity, is the same as that finding it between and +d where d = mvdv: mvdvPdPdvvP
Tk
mvv
Tk
mmvvmvP
BB 2exp4
22
1exp
2 22
2/3
222/32/1
Maxwelldistribution
Note that Planck’s constant has vanished from the equation – it is a classical result.
The Maxwell Speed Distribution (cont.)
vx
vy
vz
dvv
dvvv24
volume""
v
10
dvvP2/3
2
Tk
mC
B
dvvTk
mv
Tk
mdvvP
BB
222/3
42
exp2
The structure of this equation is transparent: the Boltzmann factor is multiplied by the number of states between v and v+dv. The constant can be found from the normalization:
dvTk
mvv
Tk
mNdvvNPvdN
BB
2exp4
2
22
2/3
dTk
NdNPdNB
exp
1/ 2 2
exp2 2
xx x x
B B
mvmdN v NP v dv N dv
k T k T
energy distribution, N – the total # of particles
speed distribution
distribution for the projection of velocity, vx
P(vx)
vx
v
P(v)
Tk
mvv
Tk
mvP
BB 2exp4
2
22
2/3
Because Maxwell distribution is skewed (not symmetric in v), the root mean square speed is not equal to the most probable speed:
The Characteristic Values of Speed
3
223
0 0
84 exp
2 2B
B B
k Tm mvv v P v dv v dv
k T k T m
The most probable speed: m
Tkv
dv
vdP B
vv
20 max
max
The average speed:
P(v)
maxvrmsvv
22.1:13.1:13:/8:2::max rmsvvv
v
The root-mean-square speed is proportional to the square root of the average energy:
m
Tk
m
EvvmE B
rmsrms
32
2
1 2
Problem (Maxwell distr.)
Consider a mixture of Hydrogen and Helium at T=300 K. Find the speed at which the Maxwell distributions for these gases have the same value.
Tk
vmv
Tk
m
Tk
vmv
Tk
m
BBBB 2exp4
22exp4
2
222
2/3
22
12
2/3
1
Tk
vmm
Tk
vmm
BB 2ln
2
3
2ln
2
3 22
2
21
1
km/s 6.1107.12
2ln3001038.13ln3
2ln
2
327
23
21
2
1
21
2
2
1
mm
mm
Tk
vmmTk
v
m
m B
B
Tk
mvv
Tk
mmTvP
BB 2exp4
2,,
22
2/3
Problem (final 2005, MB speed distribution)Consider an ideal gas of atoms with mass m at temperature T. (a) Using the Maxwell-Boltzmann distribution for the speed v, find the
corresponding distribution for the kinetic energy (don’t forget to transform dv into d).
(b) Find the most probable value of the kinetic energy.(c) Does this value of energy correspond to the most probable value of speed?
Explain.
dvTk
mvv
Tk
mNdvvNPvdN
BB
2exp4
2
22
2/3
dmTkmTk
mNd
mdv
mvd
d
dvNPdN
BB
2
2
1exp
8
2
2
2
122/3
TkmTk
mP
BB
exp24
2 2/3
2/12/3
0
1expexp
2
10 2/12/1
TkTkTk
P
BBB
TkTk B
B 2
1
2
1max
max
the most probable value of speed
m
Tkv
dv
vdP B
vv
20 max
max
the kin. energy that corresponds to the most probable value of speed TkB
(a)
(b)
(c)doesn’t
correspond
