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 E i = n i E, n i =0, 1, 2,..., 8 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 n j particles in the j th energy state (for all values of j) is given by permutation theory: N(n 0,n 1,n 2,..., n 8 )= N! n 1!n 2!...n 8! How many microstates are the for one particle having all of the energy?
n 0 = 5 n 8 = 1 N(5, 0, 0,..., 1) = 6! 5!1! =6 n 0 = 4 n 1 = 1 n 7 = 1 N(4, 1, 0, 0, 0, 0, 0, 1, 0) = 6! 4!1!1! = 30
n 0 = 3 n 1 =2 n 6 =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 P = N(n 0,..., n 8 )/1287 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: n j = n 1 jp 1 + n 2 jp 2...n 20 j P 20 n 0 =5 6 1287 +4 30 1287 +3 60 1287 +2 180 1287 +... =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 j P ( j ) exp( E j /k B T ) Since the particle must be in some state, the sum over all states must equal 1, so P ( j )= exp( E j/k B T ) j exp( E j/k b T )
ParIIon FuncIon Degeneracy Suppose there were g j states having the same energy. If you wanted probability that particle has an energy E j but don t care which of the degenerate states it is in, exp( E j /k B T ) P (E j )=g j j exp( E j/k B T ) or just sum over unique energies E n : exp( E j /k B T ) P (E j )=g j n g n exp( E n /k B T ) = g j exp( E j /k B T ) Z Z is the partition function It represents the number of accessible states.
Example: The hydrogen atoms with T = 20,000 K E n = 13.6 ev 1 n 2 P (E n ) P (E m ) = g n exp( E n /k B T ) g m exp( E m /k B T ) Let s calculate P(E 3 )/P(E 2 ). What is g 3 /g 2? a) 9/4 b) 4/3 c) 3/2 d) 6/4 P (E 3 ) P (E 2 ) = 9 4 exp[ (E 3 E 2 )/k B T ] E =1.9 ev k B T =1.72 ev n 3 =2.25 exp( n 2 1.9/1.7) = 0.75
Overall probability particle is in n=2 state: P (E 2 )= g 2 exp( E 2 /k B T ) Z Z = exp( E 1 /k B T )[g 1 + g 2 exp[ (E 2 E 1 )/k B T ]+g 3 exp[ (E 3 E 1 )/k B T ]+...] Since (E 2 E 1 ) = 10.2 ev, and kt =1.7 ev, sum of all terms in square bracket is roughly just g 1 P (E 2 ) g 2 g 1 exp[ (E 2 E 1 )/k B T ]=0.011 Distribution Function: The probable number of particles in the quantum state n with energy E is f B ( 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 g(e i ) D(E) number of states within range de is given by D(E) de What are the units of D(E)? Partition function becomes Z = D(E) exp( E/k B T ) de Average energy E = EP(E) de = ED(E) exp( E/k BT ) de Z
Example: Non-interacting particles in a cube of length L. ~ 2 2m r2 (x, y, z) =E nx,n y,n z = C sin(n x x/l) sin(n y y/l) sin(n z z/l) What are the energy states? E nx,n y,n z = 2 2 2mL 2 [n2 x + n 2 y + n 2 z] What is the density of states, D(E)? Notice that we can qthink of the energy depending on n, where n = n 2 x + n 2 y + n 2 z E = ~2 2 2mL 2 n2 de, dn How many quantum states are within a shell of radius n and thickness dn?
E = ~2 2 2mL 2 n2 Total number of states in range dn is volume of shell since there is one quantum state per unit volume Δn 3. Thus the total number states in range de is 1 8 4 n2 dn = D(E) de D(E) = 1 8 4 n2 dn de This expression reduces to D(E) = 1 p 2 2 m 3/2 ~ 3 VE1/2 (Not including spin degeneracy)
Maxwell- Boltzmann DistribuIon P (E) de = D(E) exp( E/k BT ) de Z = E exp( E exp( E/kB T ) de E/kB T ) de Z 1 0 0.5 p E exp( E/kB T ) de = p 2 (kt B) 3/2 Maxwell- Boltzmann DistribuIon: P (E) de = 2 (k B T ) 3/2 E exp( E/k B T ) de 0.4 0.3 0.2 Average Energy: 0.1 E = P (E) de = 3 2 k BT 1 2 3 4 5 E/kT
Maxwell Speed DistribuIon P (E) de = 2 (k B T ) 3/2 E exp( E = 1 2 mv2 What does P(v) look like? 1. de = mv dv P (v) / v exp( mv/k B T ) E/k B T ) de P (v) =P (E) de dv 2. 3. P (v) / v 2 exp( mv/k B T ) P (v) / v 2 exp( mv 2 /2k B T ) 4. P (v) / v 3/2 exp( mv 2 /2k B T )
Maxwell Velocity DistribuIon P (v) = p 2 m mv 2 k B T v2 exp 2k B T 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.
v = 8k BT m 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 rms = 3k BT m
Specific Heat C v / du dt k B T =0.025 T 293 K ev
When should you use quantum statistics? When do wave functions overlap (assuming energies appropriate to Boltzmann statistics)? d = h p h 2mkB T At room temperature for electrons: d 8 nm If average spacing of particles is greater than the de Broglie wavelength, than Boltzmann statistics works fine.
Fermions half-integer spin FERMI- DIRAC STATISTICS 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! n 0 = 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 f FD (E j )= 1 exp[(e j f )/k B T ]+1 f 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.
Fermi- Dirac DistribuIon FuncIon f FD (E j )= 1 exp[(e j f )/k B T ]+1 μ is Fermi energy
Return to Insigh,ul Example
Very often This means essentially all energy states below Fermi energy are occupied and all above are empty. f /(kt) 1 We assumed this distribution in discussing groundstate configurations of multi-electron atoms.
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 ε f >> kt)? 1. It is higher 2. It is lower 3. It is the same
33 Consider a collection of 4 identical particles obeying quantum mechanics. The particles can occupy a set of equally spaced energy levels: 8 ev 6 ev 4 ev 2 ev 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
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: N = X j g(e j )f FD (E j ) As we did earlier, convert to an integral using density of states (this time account for degeneracy due to spin) N = X j g(e j )f FD (E j )! Z D(E)f FD (E) de
N = D(E)f FD (E) de f FD (E j )= 1 exp[(e j f )/k B T ]+1 Density of states is given by (2s+1) times what we derived last time. D(E) de = (2s + 1) V 3 m 3 2 4 EdE 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 f FD (E) =1 f FD (E) =0 E apple F E> F
Thus, N = 0 f (2s + 1)m 3/2 V 2 3 2 EdE 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 f = h2 8m 3 N V 2/3 This is independent of geometry.
Consider a conducting metal, with 1 valence electron per nucleus. N/V = / m nuc 5 103 kg/m 3 10 29 m 3 20 m p f 10 18 J 10 ev T f = f /k B 70, 000 K At room temperature Fermi energy much higher than kt!
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: E = E = EP(E) de = f E 3/2 de 0 f E 0 1/2 de = 3 5 f ED(E) de N sill assuming f = 1 Degeneracy Pressure: U P = V U = N E V 2/3 P deg = U V = 2 3 E V = 2 5 N V f
P deg = U V = 2 3 E V = 2 5 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.
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 2 k 2 (1/ d ) 2 squishing (decreasing volume) causes spacing to decrease, which means average kinetic energy of electrons increases. Work is required to squish, so there is pressure.
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
n 0 = 5(1/20) + 4(1/20) + 3(1/20) +... =2.45 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 f BE (E m )= (not 2.3) 1 exp(e m µ)/k B T 1
Comparison of DistribuIon FuncIons f MB (E) = 1 Ae E/k BT 0 E f MB (E) f BE (E) = 1 Ae E/k BT 1 1.0 f BE (E) 0 E f FD (E) = 1 Ae E/k BT +1 1.0 f FD (E) 0 E
f BE (E m )= Blackbody RadiaIon 1 exp(e m µ)/k B T 1 Chemical potential can also be thought as a normalization constant. A =exp( µ/k B T ) µ µ 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 D(E) = 1 8 4 n2 dn de This doesn t include the factor of two for spin The energy of a photon is given by = pc = ~kc = ~n /a ) n = ae ~ Including spin, we find that D(E) = 8 V h 3 c 3 E2
f BE (E m )= D(E) = 8 V h 3 c 3 E2 Blackbody RadiaIon 1 exp(e m µ)/k B T 1 The energy density of photons with the interval de is u(e) de = Ef BE (E)D(E)dE/V u( ) = 8 3 (hc) 3 e /kt 1
f BE (E m )= 1 exp(e m µ)/k B T 1 f BE (E 0 )= 1 exp[(e 0 µ)/k B T ] 1 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 E 0 3. µ kt 4. µ > E 0 5. None of these!
f BE (E m )= f BE (E 0 )= 1 exp(e m µ)/k B T 1 1 exp[(e 0 µ)/k B T ] 1 Distribution is very sharply peaked at ground state for kt<< (E 1 -E 0 ). (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
49 Consider a collection of 4 identical particles obeying quantum mechanics. The particles can occupy a set of equally spaced energy levels: 8 ev 6 ev 4 ev 2 ev 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
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 T C = N 2.61V 2/3 h 2 2 mk Condensation temperature: T C =1.6 10 4 n 10 21 m 3 mp m K
First BEC Achievement Wienman and Cornell (CU Boulder) 1995 2,000 rubidium-87 atoms Volume is 2 micrometers cubed T c = 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 E 0 and E 1 ) 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 R up. 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. R up = B 01 N 0 u(e) Similarly, the rate of transitions via stimulated emission is given by R stim = B 10 N 1 u(e) The rate of spontaneous decays is independent of the radiation field, so R spon = AN 1
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. R up = R stim + R spon B 01 N 0 u(e) =(B 10 u(e)+a)n 1 The relative population of particles in each state is given by the Boltzmann factor: N 1 N 0 =exp[ (E 1 E 0 )/k B T ] Thus exp[ (E 1 E 0 )/k B T ]= B 01u(E) B 10 u(e)+a Note: E = E 1 E 0
Einstein s A and B Coefficients exp[ (E 1 E 0 )/k B T ]= B 01u(E) B 10 u(e)+a We can use this equation to solve for u: u(e) = A/B 01 exp[e/k B T ] B 10 /B 01 But in thermal equilibrium, the energy density of the radiation is a blackbody, and is described by the Planck formula u(e) = 8 E2 h 3 c 3 E exp[e/k B T ] 1 Thus, B 10 = B 01 A = 8 E3 (hc) 3 B
Clicker QuesIon Einstein s coefficients were derived assuming the system is in thermodynamic equilibrium. What if the system isn t in equilibrium? Does B 01 and B 02 change? 1. No, they don t change. 2. Yes, they change: B 01 becomes greater than B 10 3. Yes, they change: B 01 becomes less than B 10
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 u(e) = A/B 01 exp[e/k B T ] B 10 /B 01 From the previous equations, we can show that A Bu(E) = ee/k BT 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: 1
Clicker QuesIon Is it ever possible that the rate of stimulated emission is greater than the rate of absorption? 1. Yes 2. No
In thermal equilibrium, N 1 PopulaIon Inversion N 0 =exp[ (E 1 E 0 )/k B T ] 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 N 1 >N 2. 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 (solidstate lasers) and electric discharge pumping (gas lasers).
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 (N 3 N 1 ). 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 N 2 > N 1. 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. n = 2L n Resonant cavity reduces the bandwidth. The resonant cavity also significantly reduces the opening angle of the beam. R = 50% 95%
Bandwidth Frequency of resonant modes are n = c = nc n 2L These modes are separated by = c 2L = 150 MHz 1m L Consider a He-Ne laser at 632.8 nm (4.7 x 10 8 MHz) Bandwidth is typically 1.5 10 9 Hz For a 1 m laser, then there are roughly resonant modes. 10 Relative width of each mode: 0 30 MHz 4.7 10 8 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 R For a 5 mm aperture and a wavelength of 500 nm 0.1 mrad At 1 meter away, beam size is A=10-8 m 2 (.1mm radius) For a 5 mw pointer laser, the radiant intensity is I 5 10 5 W/sr https://www.youtube.com/watch? v=d1pxhhtuaso&feature=iv&src_vid=woitedskprk&annotation_id=annotati on_822873 https://www.youtube.com/watch?v=jixugt-qiei 10 8 sr
hjp://science.energy.gov/~/media/bes/csgb/pdf/docs/reports%20and %20AcIviIes/Sauul_report_final.pdf