STATISTICS  •  Often the number of particles in a system is prohibitively large

to solve detailed state (from Schrodinger equation) or predict exact motion (using Newton’s laws).

•  Bulk properties (pressure, temperature, average energy, average velocity, etc) can be determined from statistical arguments (combined with physical laws)

•  Goal: Determine the probability a particle/system will have an energy E given a temperature T.

Insigh,ul  Example  •  Consider a system of 6 distinguishable particles with total

energy 8E. Each particle can have an energy

•  A microstate is a particular state specifying the quantum state (and thus energy) of each of the individual particles (such that the net energy is 8E).

•  # of microstates such that there are nj particles in the jth energy state (for all values of j) is given by permutation theory:

•  How many microstates are the for one particle having all of the energy?

Ei = niE, ni = 0, 1, 2, ..., 8

N(n0, n1, n2, ..., n8) =N !

n1!n2!...n8!

•  n0 = 5 •  n8 = 1

•  n0 = 4 •  n1 = 1 •  n7 = 1

N(5, 0, 0, ..., 1) =6!

5!1!= 6

N(4, 1, 0, 0, 0, 0, 0, 1, 0) =6!

4!1!1!= 30

•  n0 = 3 •  n1 =2 •  n6 =1

N =6!

3!2!1!= 60

N =6!

2!2!1!1!= 180

•  Overall, there are 1287 total microstates, and 20 macrostate configurations.

•  Each microstate is just as likely as any other microstate. •  Probability of particular macrostate is given by

•  Avg. # of particles in jth energy state give by: (sum over all 20 macrostates, using the number of particles in the jth state for that macrostate). For example, average number in ground state is:

P = N(n0, ..., n8)/1287

�nj⇥ = n1jP1 + n2

jP2...n20j P20

�n0⇥ = 5�

61287

⇥+ 4

�30

1287

⇥+ 3

�60

1287

⇥+ 2

�1801287

⇥+ ... = 2.3

•  Since there are 6 particles, the probability any particle is in the ground state is P(0) = <n(0)>/N = 2.3/6 = 0.385

•  Do this for the others:

j   <n>   P(j)  

0   2.3   0.385  

1   1.6   0.256  

2   1.0   0.167  

3   0.6   0.098  

-­‐   -­‐   -­‐  

-­‐   -­‐   -­‐  

-­‐   -­‐   -­‐  

8   0.0047   0.00077  

•  Probability that a high-energy state is occupied (particle in high-energy state) is relatively small since there are fewer microstates having a particle in a high-energy state.

•  This trend continues for macroscopic systems (containing many, many particles)

•  Rigorous derivation involves concepts of entropy, heat transfer, and thermodynamic equilibrium

•  Result is the Boltzman distribution function:

Boltzmann  Factor  •  Boltzmann Probability: probability that a particle will be in a

state

•  Since the particle must be in some state, the sum over all states must equal 1, so

P (�j) ⇥ exp(�Ej/kBT )

P (�j) =exp(�Ej/kBT )�j exp(�Ej/kbT )

�j

ParIIon  FuncIon  •  Degeneracy –  Suppose there were gj states having the same energy. –  If you wanted probability that particle has an energy Ej but

don’t care which of the degenerate states it is in,

–  or just sum over unique energies En:

•  Z is the partition function –  It represents the number of “accessible” states.

P (Ej) = gjexp(�Ej/kBT )�j exp(�Ej/kBT )

P (Ej) = gjexp(�Ej/kBT )�

n gn exp(�En/kBT )=

gj exp(�Ej/kBT )Z

•  Example: The hydrogen atoms with T = 20,000 K

•  Let’s calculate P(E3)/P(E2). •  What is g3/g2?

a) 9/4 b) 4/3 c) 3/2 d) 6/4

En = �13.6 eV1n2

P (En)P (Em)

=gn exp(�En/kBT )gm exp(�Em/kBT )

�E = 1.9 eV kBT = 1.72 eVn3

n2= 2.25 exp(�1.9/1.7) = 0.75

P (E3)

P (E2)=

9

4

exp[�(E3 � E2)/kBT ]

•  Overall probability particle is in n=2 state:

•  Since (E2 – E1) = 10.2 eV, and kT =1.7 eV, sum of all terms in square bracket is roughly just g1

•  Distribution Function: The probable number of particles in the quantum state n with energy E is

P (E2) =g2 exp(�E2/kBT )

ZZ = exp(�E1/kBT )[g1 + g2 exp[�(E2 � E1)/kBT ] + g3 exp[�(E3 � E1)/kBT ] + ...]

P (E2) ⇥g2

g1exp[�(E2 � E1)/kBT ] = 0.011

fB( n) = NP ( n) = A exp(�E/kT )

GRE  QuesIon  

Density  of  States  •  Sometimes the allowed quantum-mechanical systems are

continuous (or modeled to be). Summations turn into integrals

•  number of states within range dE is given by

•  What are the units of D(E)? •  Partition function becomes

•  Average energy

g(Ei)� D(E)

D(E) dE

Z =�

D(E) exp(�E/kBT ) dE

⇥E⇤ =⇥

EP (E) dE =�

ED(E) exp(�E/kBT ) dE

Z

•  Example: Non-interacting particles in a cube of length L.

•  What are the energy states?

•  What is the density of states, D(E)? –  Notice that we can think of the energy depending on n,

where

•  How  many  quantum  states  are  within  a  shell  of  radius  “n”  and  thickness  dn?  

⇥nx,ny,nz = C sin(nx�x/L) sin(ny�y/L) sin(nz�z/L)

Enx,ny,nz =�2�2

2mL2[n2

x + n2y + n2

z]

� ~22m

r2 (x, y, z) = E

n =qn2x

+ n2y

+ n2z

E =~2⇡2

2mL2n2 dE , dn

•  Total number of states in range dn is volume of shell since

there is one quantum state per unit volume Δn3. •  Thus the total number states in range dE is

•  This expression reduces to

1

84⇡n2 dn = D(E) dE

D(E) =1

84⇡n2 dn

dE

E =~2⇡2

2mL2n2

D(E) =1p2⇡2

m3/2

~3 V E1/2 (Not  including  spin  degeneracy)  

Maxwell-­‐Boltzmann  DistribuIon    

P (E) dE =D(E) exp(�E/kBT ) dE

Z=⇥

E exp(�E/kBT ) dE� ⇥

E exp(�E/kBT ) dE

P (E) dE =2⇥�

(kBT )�3/2⇥

E exp(�E/kBT ) dE

1 2 3 4 5

0.1

0.2

0.3

0.4

0.5

1 2 3 4 5

0.1

0.2

0.3

0.4

0.5

E/kT  

�E⇥ =�

P (E) dE =32kBT

Z 1

0

pE exp(�E/kBT ) dE =

p⇡

2

(kTB)3/2

Average  Energy:  

Maxwell-­‐Boltzmann  DistribuIon:  

Maxwell  Speed  DistribuIon  

•  What does P(v) look like?

1. 

2.  3. 

4. 

E =12mv2

dE = mv dv

P (E) dE =2⇥�

(kBT )�3/2⇥

E exp(�E/kBT ) dE

P (v) = P (E)dE

dv

P (v) / v exp(�mv/kBT )

P (v) / v2 exp(�mv/kBT )

P (v) / v2 exp(�mv2/2kBT )

P (v) / v3/2 exp(�mv2/2kBT )

Maxwell  Velocity  DistribuIon  

–  which atoms for a given temperature will have higher average speed?

1.  High-mass atoms 2.  Low-mass atoms 3.  Average speed is independent of mass.

P (v) =

p2⇡m

kBTv2 exp

✓� mv2

2kBT

◆

•  Atmosphere loss: Where did all the hydrogen go on Earth?

•  Doppler shift causes emission lines to spread out. Width of line depends on dispersion of velocities.

�v⇥ =�

8kBT

�m

vrms =�

3kBT

m

Specific  Heat  

Cv / dU

dtkBT = 0.025

T

293 KeV

•  When should you use quantum statistics? –  When do wave functions overlap (assuming energies

appropriate to Boltzmann statistics)?

–  At room temperature for electrons:

–  If average spacing of particles is greater than the de Broglie wavelength, than Boltzmann statistics works fine.

�d =h

p� h⇥

2mkBT

�d � 8 nm

FERMI-­‐DIRAC  STATISTICS    •  Fermions –  half-integer spin –  Pauli exclusion principle allows only two particles per

energy state (spin up and spin down) –  particles are indistinguishable

IndisInguishable  Fermions  •  Consider again the 6 particles with total energy 8E. –  Only two fermions in each spatial state. –  Now there are only three microstates!

�n0⇥ = 2(1/3) + 2(1/3) + 2(1/3) = 2

P (m = 0) = 2/6 = 0.33

m   <n>   P(m)  

0   2   0.33  

1   5/3   0.28  

2   1   0.16  

3   1   0.16  

4   1/3   0.05  

Doesn’t drop off as quickly initially compared to Boltzmann probability distribution.

Comparison  •  Maxwell-Boltzmann on left, Fermi-Dirac on right

FERMI-­‐DIRAC  DISTRIBUTION  •  A formal derivation of the distribution function yields

–  can be thought as a normalization constant. It is called the Fermi energy.

–  Distribution function gives the expected number of particles occupying a particular quantum state j. Consistent with the Pauli Exclusion principle, the maximum value of of this function is 1.

–  Distribution function is proportional to the probability distribution.

�f

fFD(Ej) =1

exp[(Ej � �f )/kBT ] + 1

Fermi-­‐Dirac  DistribuIon  FuncIon  

μ  is  Fermi  energy  

fFD(Ej) =1

exp[(Ej � �f )/kBT ] + 1

Return  to  “Insigh,ul  Example”  

•  Very often –  This means essentially all energy states below Fermi

energy are occupied and all above are empty. •  We assumed this distribution in discussing ground-

state configurations of multi-electron atoms.

�f/(kT )� 1

GRE  QuesIon  

Clicker  QuesIon  •  How does the average kinetic energy for a gas of

indistinguishable fermions compare to that predicted by Maxwell-Boltzmann statistics (assuming εf >> kT)? 1.  It is higher 2.  It is lower 3.  It is the same

Consider a collection of 4 identical particles obeying quantum mechanics. The particles can occupy a set of equally spaced energy levels: At a temperature of T = 0 K, what would be the average energy of these 4 particles if they behaved like spin-½ particles? A) 2 eV B) 3 eV C) 4 eV D) 5 eV E) 6 eV

33  

8  eV  6  eV  4  eV  2  eV  

JITT  Comment  •  Fermions have spin =1/2 and must obey the Pauli exclusion

principle. A maximum of two per energy level, with 4 particles 2 can occupy the 2eV level and the next two in the 4eV. The average is 3eV, at the ground state configuration.

Fermi  Energy  •  Consider again N non-interacting particles in a cube of length

L, but now let them be spin 1/2 fermions. •  Let’s determine the Fermi energy and average energy. –  Use “normalization” to determine Fermi energy:

–  As we did earlier, convert to an integral using density of states (this time account for degeneracy due to spin)

N =X

j

g(Ej)fFD(Ej)

N =X

j

g(Ej)fFD(Ej) !Z

D(E)fFD(E) dE

•  Density of states is given by (2s+1) times what we derived last time.

•  Integral would have to be done numerically if we want an

exact answer, but often it is a very good approximation to assume that the Fermi energy is much greater than kT. Thus

N =�

D(E)fFD(E) dE

D(E) dE = (2s + 1)V

�3

�m3

2�4

�E dE

fFD(E) = 1

fFD(E) = 0

E ✏F

E > ✏F

fFD(Ej) =1

exp[(Ej � �f )/kBT ] + 1

•  Thus,

•  How do you expect the Fermi energy to depend on the number density, N/V? 1.  Proportional to density 2.  Proportional to density to the 2/3 power 3.  Proportional to density to -3/2 power

–  This is independent of geometry.

N =� �f

0

(2s + 1)m3/2V

�2�3�

2

�E dE

�f =h2

8m

�3⇥

N

V

⇥2/3

•  Consider a conducting metal, with 1 valence electron per nucleus.

•  At room temperature Fermi energy much higher than kT!

N/V = �/⇤mnuc⌅ ⇥5� 103 kg/m3

20 mp⇥ 1029 m�3

�f � 10�18 J � 10 eV

Tf = �f/kB � 70, 000 K

•  Average Energy –  If D(E) is uniform, you would expect average to be

one-half of the Fermi energy. –  But D(E) is not uniform:

•  Degeneracy Pressure:

�E⇥ =⇥

E P (E) dE =�

E D(E) dE

N

sIll  assuming    f  =  1  �E⇥ =

� �f

0 E3/2 dE� �f

0 E1/2 dE=

35�f

P = ��U

�V

U = N⇥E⇤ � V �2/3Pdeg = �⇥U

⇥V=

23⇥E⇤V

=25

N

V�f

•  For metals and systems in which density is very high, this degeneracy pressure is much higher than thermal pressure, and can dominate. –  Young’s Bulk modulus of metals –  SupportsWhite Dwarfs & Neutron Stars from

collapsing due to gravity.

Pdeg = �⇥U

⇥V=

23⇥E⇤V

=25

N

V�f

•  Conceptually, what is the cause of degeneracy pressure? –  Exclusion principle forbids two electrons with identical

quantum numbers to be on top of each other. –  physical spacing must be larger than de Broglie

wavelength –  kinetic energy is proportional to –  squishing (decreasing volume) causes spacing to

decrease, which means average kinetic energy of electrons increases.

–  Work is required to squish, so there is pressure.

�2k2 � (1/�d)2

BOSE-­‐EINSTEIN  STATISTICS  •  Bosons –  Particles with integer spin

•  No limit to # of particles in a quantum state –  Indistinguishable

•  Re-visit 6 particles with energy 8E –  Same 20 possible configurations as with Boltzmann

statistics, but now only 1 microstate for each configuration since the particles are indistinguishable

�n0⇥ = 5(1/20) + 4(1/20) + 3(1/20) + ... = 2.45(not  2.3)  

m   P(m)   P(m)  Boltzmann  

0   0.41   0.38  

1   0.26   0.26  

2   0.15   0.17  

3   0.08   0.10  

4   0.05  

8   0.008   0.0008  

•  More  parIcles  expected  in  lower  energy  states  as  compared  to  the  Boltzmann  distribuIon.  

•  Proper  derivaIon  yields  fBE(Em) =

1exp(Em � µ)/kBT � 1

Comparison  of  DistribuIon  FuncIons  

   0                                    E  

fMB(E)  

   0                                    E  

1.0  fBE(E)  

   0                                    E  

1.0  

fFD(E)  

fMB(E) =1

AeE/kBT

fBE(E) =1

AeE/kBT � 1

fFD(E) =1

AeE/kBT + 1

Blackbody  RadiaIon  

•  Chemical potential can also be thought as a normalization constant. –  For photons, is zero (number of photons is not

conserved). –  Consider standing waves in a cubic box of width a. As we

derived before, the density of states is given by

–  The energy of a photon is given by

–  Including spin, we find that

fBE(Em) =1

exp(Em � µ)/kBT � 1

µ

µA = exp(�µ/kBT )

D(E) =1

84⇡n2 dn

dE This  doesn’t  include  the  factor  of  two  for  spin  

✏ = pc = ~kc = ~n⇡/a

D(E) =8⇡V

h3c3E2

) n =aE

~⇡

Blackbody  RadiaIon  

•  The energy density of photons with the interval dE is

u(✏) =8⇡

(hc)3✏3

e✏/kT � 1

D(E) =8⇡V

h3c3E2

fBE(Em) =1

exp(Em � µ)/kBT � 1

u(E) dE = EfBE(E)D(E)dE/V

•  Consider a system of N particles, where N>>1. What is µ for the case when T is nearly zero? 1.  µ ≈ 0 2.  µ ≈ E0 3.  µ ≈ kT 4.  µ > E0 5.  None of these!

fBE(Em) =1

exp(Em � µ)/kBT � 1

fBE(E0) =1

exp[(E0 � µ)/kBT ]� 1

•  Distribution is very sharply peaked at ground state for kT<< (E1-E0). (In fact, as T goes to zero, ALL particles go to ground state.) –  Macroscopic system with essentially all particles in the

same quantum state. •  Bose-Einstein condensation

fBE(Em) =1

exp(Em � µ)/kBT � 1

fBE(E0) =1

exp[(E0 � µ)/kBT ]� 1

Consider a collection of 4 identical particles obeying quantum mechanics. The particles can occupy a set of equally spaced energy levels: At a temperature of T = 0 K, what would be the average energy of these 4 particles if they behaved like spin-1 particles? A) 2 eV B) 3 eV C) 4 eV D) 5 eV E) 6 eV

49  

8  eV  6  eV  4  eV  2  eV  

JITT  Comment  •  The limit of N(E) goes to infinity as T goes to zero. I think that

this means that the number of particles in the lowest energy state is at its highest. Therefore 2 eV.

JITT  Comments  •  I didn't seem to follow the reasoning behind the beta and mu

values or what they stood for. Or why the factor 2s+1 for photons is only 2 not three.

Bose-­‐Einstein  CondensaIon  

•  Condensation temperature:

TC =�

N

2.61V

⇥2/3 h2

2�mk

TC = 1.6⇥ 10�4⇣ n

1021 m�3

⌘ mp

mK

•  First BEC Achievement –  Wienman and Cornell (CU Boulder) 1995 –  2,000 rubidium-87 atoms –  Volume is 2 micrometers cubed –  Tc = 5 x 10-7 K –  They achieved 20 nK

Velocity  DistribuIon  

•  BECs have now been show for fermions! Fermions can pair up and act like bosons. Understanding this is probably key to understanding superconductivity.

The  Laser  •  Laser stands for light amplification by stimulated emission •  Properties of laser light: –  It is coherent.

•  Monochromatic •  Unidirectional •  In phase

Einstein’s  A  and  B  Coefficients  •  Recall that there are three types of transitions

between atomic states involving radiation. –  Spontaneous emission –  Stimulated emission –  Absorption

•  Consider a collection of two-state systems (with energies E0 and E1) that is in thermal equilibrium. –  The rate of transitions from state 0 to 1

must equal the rate of transitions from 1 to 0. •  Thus the rate of absorption must equal

the rate of stimulated emission plus spontaneous decay.

Einstein’s  A  and  B  Coefficients  •  Let’s denote the rate of absorption as Rup. We

expect it to depend on the number of atoms in the lower state and the amount of radiation having the right frequency/energy to be absorbed.

•  Similarly, the rate of transitions via stimulated emission is given by

•  The rate of spontaneous decays is independent of the radiation field, so

Rup = B01N0u(E)

Rstim = B10N1u(E)

Rspon

= AN1

Einstein’s  A  and  B  Coefficients  

•  The rate of transitions from state 0 to 1 must equal the rate of transitions from 1 to 0.

•  The relative population of particles in each

state is given by the Boltzmann factor:

•  Thus •  Note: E = E1 – E0

Rup

= Rstim

+Rspon

B01N0u(E) = (B10u(E) +A)N1

exp[�(E1 � E0)/kBT ] =B01u(E)

B10u(E) +A

N1

N0= exp[�(E1 � E0)/kBT ]

Einstein’s  A  and  B  Coefficients  

•  We can use this equation to solve for u:

•  But in thermal equilibrium, the energy density of the radiation is a blackbody, and is described by the Planck formula

•  Thus,

exp[�(E1 � E0)/kBT ] =B01u(E)

B10u(E) +A

u(E) =

A/B01

exp[E/kBT ]�B10/B01

u(E) =

8⇡E2

h3c3E

exp[E/kBT ]� 1

B10 = B01 A =8⇡E3

(hc)3B

Clicker  QuesIon  •  Einstein’s coefficients were derived assuming the system is in

thermodynamic equilibrium. What if the system isn’t in equilibrium? Does B01 and B02 change? 1.  No, they don’t change. 2.  Yes, they change: B01 becomes greater than B10

3.  Yes, they change: B01 becomes less than B10

Clicker  QuesIon  •  In thermodynamic equilibrium, is it ever possible that the rate

of stimulated emission is greater than the rate of absorption? 1.  Yes 2.  No

Spontaneous  vs  SImulated  Emission  

•  From the previous equations, we can show that

•  Thus spontaneous emission is far more probable than

stimulated emission for E>>kT •  Stimulated emission dominates for E<<kT •  Now compare rate of emission to rate of absorption:

A

Bu(E)= eE/kBT � 1

u(E) =

A/B01

exp[E/kBT ]�B10/B01

Clicker  QuesIon  •  Is it ever possible that the rate of stimulated emission is

greater than the rate of absorption? 1.  Yes 2.  No

PopulaIon  Inversion  •  In thermal equilibrium,

•  The only way to get the rate of stimulated emission higher

than the rate of absorption (such that the density of lasing photons can increase with time) is to have N1>N2. –  Thus the system needs to be out of thermal equilibrium.

•  Need a method to maintain the population inversion of states. –  Most common methods are called optical pumping (solid-

state lasers) and electric discharge pumping (gas lasers).

N1

N0= exp[�(E1 � E0)/kBT ]

Three-­‐Level  Laser  •  All lasers involve a metastable state –  Typical lifespan of state is 10-3 s (rather than 10-9 s).

–  Intense pumping is applied. The n=3 becomes highly

populated (N3 ≈ N1). –  After pumping, electrons in the n=3 quickly decay mostly

to the metastable state, producing a population inversion between n=2 and n=1.

–  Light can be amplified via stimulated emission until the inversion goes away. The output is pulsed.

•  http://www.slideshare.net/HebaRageh/laser-physics-lect1-1

hjp://www.nature.com/arIcles/srep11342  

Four-­‐Level  Laser  •  The efficiency of three-level lasers is very poor since a lot of

power is expended to keep N2 > N1. Every decay leads to increasing the population of the lower state.

•  The lasing transition is n=2 to n=1. The n=1 state is always nearly empty due to the quick timescale for electrons to decay to the ground state. Thus it is much easier to achieve a population inversion.

Pumping  Schemes  •  Optical Pumping –  Atoms are excited via absorption of a powerful source of

light, such as an incoherent lamp. –  Better suited for solid-state lasers

•  Electrical Pumping –  Free electrons are accelerated by electric field –  Collisions with atoms cause excitations.

Examples  •  Helium-Neon Laser

–  Gas laser –  Uses Resonant Energy Transfer: •  Helium is excited via collisions with free electrons to

the singlet 2s metastable energy state. •  The excited state of helium is very close to excited

energy levels of neon. Collisions between neon and helium cause a transfer of energy, and electrons in the ground state of neon are excited to these levels. •  This resonant transfer causes a population inversion between the 3s and 2p states of neon.

The  Resonant  Cavity  •  In order to get a high-power output beam, the light needs to

pass through the gain medium multiple times. •  Cavity length is chosen such that standing waves of output

radiation can form in resonant cavity. •  Resonant cavity reduces the bandwidth. •  The resonant cavity also significantly reduces the opening

angle of the beam.

�n =2L

n

R = 50% ⇠ 95%

Bandwidth  •  Frequency of resonant modes are

–  These modes are separated by

–  Consider a He-Ne laser at 632.8 nm (4.7 x 108 MHz)

–  Bandwidth is typically

–  For a 1 m laser, then there are roughly resonant modes.

–  Relative width of each mode:

⌫n =c

�n=

nc

2L

�⌫ =c

2L= 150 MHz

✓1 m

L

◆

�⌫ ⇠ 1.5⇥ 109 Hz

�⌫

�⌫⇠ 10

�⌫

⌫0⇠ 30 MHz

4.7⇥ 108 MHz⇠ 6.4⇥ 10�8

Clicker  QuesIon  •  If you want to design a single-mode operation of a laser, do

you make the resonant cavity long or short? 1.  Long 2.  Short 3.  Shouldn’t matter!

Angular  Spread  and  Intensity  •  The limitation of the angular spread of a beam is typically the

diffraction limit

–  For a 5 mm aperture and a wavelength of 500 nm

–  At 1 meter away, beam size is A=10-8 m2 (.1mm radius) –  For a 5 mW pointer laser, the radiant intensity is

–  https://www.youtube.com/watch?v=d1pXhhtUaso&feature=iv&src_vid=woiTedSKPrk&annotation_id=annotation_822873

–  https://www.youtube.com/watch?v=jIxugT-QiEI

�✓ ⇠ �

R

�✓ ⇠ 0.1 mrad �⌦ ⇠ 10�8sr

I ⇠ 5⇥ 105 W/sr

hjp://science.energy.gov/~/media/bes/csgb/pdf/docs/Reports%20and%20AcIviIes/Sauul_report_final.pdf