Lecture 22. Continuous Spectrum, the Density of States, and Equipartition (Ch. 6)

TkZ

dP

B

iii

exp dVTkZ

gdP

B

""exp

“V”, the volume in dimensions: (length)

(1D – length, 2D – area, 3D – volume)

The units of g() in dimensions: (energy)-1(length)-

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 volume 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 number of states between and +d in volume V: dVg

The probability that one of these states is occupied: exp1

ZThe probability that the particle has an energy between and +d : dg

Z

VdP exp

0

1 dPSince the partition function is dgVZ

0

exp

The density of states, g(), is the number of states per unit energy interval per unit

volume

The Density of States for a Single Free Particle in 1DThe 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 nn 282

2222

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.

2h

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 number of states with the wave numbers < k is simply

k

kGkL

L

kkN

/G(k) - the number of states

per unit lengthmk 2

1

Since

m

G2

2

121 ms

d

dGg D

additional degeneracy of the states (2s+1) because of spin [e.g., for electrons 2s+1=2]

g()

1D

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

2

222 2

4

1

444

1

m

Gk

kGareak

LL

kkN

yx

k

# states within ¼ of a circle of radius k

xL

3D

ky

kx

kz

2/3

222

3

2

33 2

6

1

66

3/4

8

1

m

Gk

kGvolumek

LLL

kkN

zyx

- does not depend on

2/12/3

223 2

4

12

ms

g D

2

2

2

12

ms

g D

g()

2D

Thus, for 3D electrons (2s+1=2): 2/12/3

223 2

2

1

m

g D

g()

3D

1D Harmonic Oscillator at High T

Let’s consider the partition function of a harmonic oscillator at high temperatures (kBT >>). The quantized energy levels are equally spaced and nondegenerate:

...1,0,2

1

ihii

The energy can be considered as a continuous variable at a scale >> = h. The density of states (the number of levels per unit energy per unit volume): g()=()-1 =(h)-1.

h

TkhdhdgZ B

1

0

1

0

expexp

Hence, TkhZE B

1lnln

g()=(h)-1=constharmonic oscillator

x x x

U(x)U(x)U(x)

g() increases with g() decreases with

= h

kBT

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

QQ

(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

Partition Functions for Distinguishable Particles

For two non-interacting particles, Etotal =E1+E2 :

212121

21

1 2

expexpexpexp

exp

ZZsEsEsEsE

sEsEZ

s ss

stotal

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. Recallthe multiplicity for an ideal gas:

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:

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

QQ

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:

qq

exp(-x2)

x

E(q)

The Generalized Equipartition TheoremIn general, if the energy is a function of the canonical coordinates qi ,...,...,1 iqqEE

the generalized equipartition theorem looks as follows: Tkq

Eq B

ii

In many cases, a system of coordinates exists in which E can be written as a sum of quadratic functions of the coordinates:

2i

iiqaE so that ii

i

qaq

E2

The average contribution of the term ~ qi2

to the energy is then given by: 22

12 Tk

q

Eqqa B

iiii

The fact that the contribution to the internal energy of the i th degree of freedom is independent of the magnitude of the coefficient ai has a paradoxical (in classical physics) consequence. Suppose that ai tends to 0. So long as it is non-zero, the contribution to U is kBT/2. On the other hand, if ai actually is zero, that degree of freedom does not contribute to U. Quantum theory resolves the paradox: as ai becomes smaller, the corresponding quantum levels become further and further apart. Thus, a higher and higher temperature is needed for the assumption of continuous energy levels to be valid and for equipartition to hold. Consider a molecule with the moment of inertia I:

and the total energy of the system of N particles with f degrees of freedom for each particle: TkfNU B2

1

For a “linear” degree of freedom, the av. energy = kBT (particles in a uniform field, ultra-relativistic particles with the kin. energy E = cp, etc.) – see Pr. 6.31.

IEEjjEIE qcl 2/12/ 200

2

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

TkBTkE 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

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

dvTk

mv

Tk

mNdvvNPvdN

BBxx

2exp

2

22/1

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

0

23

0

8

2exp4

2 m

Tkdv

Tk

mvv

Tk

mdvvPvv B

BB

The most probable speed: m

Tkv

dv

vdP B

vv

20 max

max

The average speed:

P(v)

maxvrmsvv

22.113.113/82max rmsvvv

v

The root-mean-square speed is proportional to the square root of the average energy:

m

Tk

m

EvvmE Brmsrms

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 BB 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