Chap 5. Identical Particles 1.Two-Particle Systems 2.Atoms 3.Solids 4.Quantum Statistical Mechanics

Two particles having the same physical attributes are equivalent. They behave the same way if subjected to the same treatment. CM: Equivalent particles are distinguishable since one can keep track of each particle all the time. QM: Equivalent particles are indistinguishable since one cannot keep track of each particle all the time due to the uncertainty principle. Indistinguishable particle are called identical.

5.1. Two-Particle Systems 2-particle state : Schrodinger eq.:  Probability of finding particle 1 & 2 within d 3 r 1 & d 3 r 2 around r 1 & r 2, resp. Normalization :  Read Prob 5.1, Do Prob 5.3.

Bosons & Fermions 1 & 2 indistinguishable (identical) : Distinguishable particles 1 & 2 in states a & b, resp. : bosons fermions Spin statistics theorem : Integer spin  bosons Half-integer spin  fermions  No two fermions can occupy the same state. Pauli exclusion principle Total  : Symmetric Anti-symm

Exchange Operator Exchange operator :  f  f   Let g be the eigenfunction of P with eigenvalue :  For 2 identical particles,     or  H & P can, & MUST, share the same eigenstates. i.e.for bosons fermions Symmetrization requirement

Example 5.1.Infinite Square Well Consider 2 noninteracting particles, both of the same mass m, inside an infinite square well of width a. 1-particles states arefor Distinguishable particles : E.g., Ground state: 1 st excited state : Doubly degenerate.

Bosons : Ground state: 1 st excited state : Fermions : Ground state: Nondegenerate

Exchange Forces Particles distinguishable : Consider 2 particles, one in state a, the other in state b. ( p’cle 1 in a, 2 in b. ) Bosons : Fermions : We’ll calculate for each case the standard deviation of particle separation ( All states are normalized )

Distinguishable Particles ( ,  a,  b normalized ) Similarly, 

Identical Particles  a,  b orthonormal Similarly,

 or  Bosons are closer & fermions are further apart than the distinguishable case. Note : if particles are far apart so  a,  b don’t overlap.  effective attractive repulsive exchange force for bosons fermions

Simplified Derivation ~ ~ if

Similarly, or  Bosons are closer & fermions are further apart than the distinguishable case. Note : if particles are far apart so  a,  b don’t overlap.  effective attractive repulsive exchange force for bosons fermions

H2H2 If e’s were bosons, form bond. e’s are fermions, H 2 dissociates Let the electrons be spinless & in the same state : In actual ground state of H 2 : Spins of the e’s are anti-parallel so the spatial part is symmetric. Do Prob 5.7

5.2. Atoms Atom with atomic number Z ( Z protons & Z electrons ) : 1-e plan of attack : 1.Replace e-e interaction term with single particle potential. 2.Solve the 1-e eigen-problem. 3.Contruct totally anti-symmetric Z-e wave function, including spins. 4.Total energy is just the sum of the 1-e energies. Nucleonic DoF dropped. See footnote, p.211.

Non-Interaction e Model Drop all e-e terms : where( Hydrogenic hamiltonian ) where  nlm and E n are obtained from the hydrogen case by setting e 2  Ze 2. Thus In particular :and (Unsymmetrized) solutions to are with a 0 = Bohr radius

Helium Non-interacting e model :  Ground state : Anti-symmetrized total wave function :  0 gives only symmetric spatial part  spin part must be antisymm (singlet). Experiment : Total spin is a singlet. E 0   79 eV.

Excited States Long-living excited states : ( both e in excited states quickly turns into an ion + free e. ) SingletTriplet Spatial part of parahelium is symmetric  higher e-e interaction  higher energy than orthohelium counterpart Do Prob 5.10

The Periodic Table n \ l0s1p2d3f4g (2l+1) Filling order of the periodic table. l – degeneracy lifted by e-e interaction (screening).

Ground State Electron Configuration Spectral Term : Ground state spectral terms are determined by Hund’s rules (see Prob 5.13 ) : 1.Highest S. 2.Highest L. 3.No more than half filled : J = | L  S |. More than half-filled : J = L+S. See R.Eisberg,R.Resnick, “Quantum Physics”, 2 nd ed., §10-3. E.g., C = (2p) 2. m 11 01 szsz   S = 1, L = 1, J = 0 Do Prob 5.14

5.3. Solids 1.The Free Electron Gas 2.Band Structure

The Free Electron Gas Solid modelled as a rectangular infinite well with dimensions { l i, i = 1, 2, 3 }.  Let for   Set  i =1, 2, 3 unless stated otherwise

Boundary Conditions   Normalization :  

k-Space Density   Volume occupied by one allowed (stationary) state in the 1 st quadrant of the k-space is = Volume of solid   k is over allowed k in 1 st quadrant of k-space.  d 3 k is over all k-space. By symmetry, all 8 quadrants are equivalent. = k-space density = density of allowed states with  n i  1.

Fermi Energy Consider a solid of N atoms, each contributing q free electrons. In the ground state, these qN electrons will occupy the lowest qN states. Since these states occupy a sphere centered at the origin of the k-space. Let the radius of this sphere be k F.  ( Factor 2 comes from spin degeneracy. ) = free electron density  Fermi energy

Total Energy Total energy of the ground state :   Compressing the solid increases its engergy. Work need be done to compress it. Electrons exerts outward pressure on the solid boundary. degeneracy pressure ( Caused by Pauli exclusion )

Miscellaneous  Energy per e. Ideal gas :  e gas : Do Prob 5.16 and for Al

Band Structure 1-D crystal with periodic potentiala = lattice constant Simplest model :1-D Dirac comb For a finite solid with N atoms, strict mathematical periodicity can be achieved by imposing the periodic boundary condition

Bloch’s Theorem Bloch’s theorem : K = const Proof :Let D be the displacement operator :   i.e. or   D & H can share the same eigenstates. Let  Since  0, setting completes the proof.

Periodic Boundary Condition  For a 1-D solid with N atoms, imposing the periodic BC gives  K is real  As expected

Dirac Comb 1-D Dirac comb : For x  ja :  

Periodic BC Impose periodic BC : Bloch’s theorem  Let  x  ja  continuous at x = 0   0 < x < a   a < x < 0  

 Condition for consistency is

Let  Sincethere’s no solution for | f (z) | > 1 ( band gaps ).

  10 Do Prob 5.19 with   20

5.4. Quantum Statistical Mechanics 1.An Example 2.The General Case 3.The Most Probable Configuration 4.Physical Significance of  and  5.The Blackbody Spectrum  Fundamental assumption of statistical mechanics : In thermal equilibrium, each distinct system configuration of the same E is equally likely to occur. ( Can also be stated in terms of ensemble. )

An Example Consider 3 non-interacting particles ( all of mass m ) in an 1-D infinite square well. Let There’re 13 combinations of (n A, n B, n C ) that can satisfy the condition : (11, 11, 11), (13, 13, 5), (13, 5, 13), (5, 13, 13), (1, 1, 19), (1, 19, 1), (19, 1, 1), (5, 7, 17), (5, 17, 7), (7, 5, 17), (7, 17, 5), (17, 5, 7), (17, 7, 5).

Occupation Number Specification of system configuration: (n A, n B, n C )occupation number (11, 11, 11)( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... )N 11 = 3 (13, 13, 5), (13, 5, 13), (5, 13, 13)( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... )N 5 = 1 N 13 = 2 (1, 1, 19), (1, 19, 1), (19, 1, 1),( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... )N 1 = 2 N 19 = 1 (5, 7, 17), (5, 17, 7), (7, 5, 17), (7, 17, 5), (17, 5, 7), (17, 7, 5). ( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... )N 5 = 1 N 7 = 1 N 17 = 1

P n : Particles Distinguishable P n = Probability of finding a particle with energy E n. P1P1 (3/13) (2/3) = 2/13 P5P5 (9/13) (1/3) = 3/13 P7P7 (6/13) (1/3) = 2/13 P 11 (1/13) (3/3) = 1/13 P 13 (3/13) (2/3) = 2/13 P 17 (6/13) (1/3) = 2/13 P 19 (3/13) (1/3) = 1/13 Sum= 1 Particles are distinguishable  each configurations is distinct.  Each is equally likely when system is in equilibrium. (n A, n B, n C ) (11, 11, 11) (13, 13, 5), (13, 5, 13), (5, 13, 13) (1, 1, 19), (1, 19, 1), (19, 1, 1), (5, 7, 17), (5, 17, 7), (7, 5, 17), (7, 17, 5), (17, 5, 7), (17, 7, 5). Total = 13 distinct configurations

P n : Fermions P n = Probability of finding a particle with energy E n. P1P1 0 P5P5 (1/1) (1/3) = 1/3 P7P7 P 11 0 P 13 0 P 17 (1/1) (1/3) = 1/3 P 19 0 Sum= 1 No state can be doubly occupied. Allowed distinct configurations are : ( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... )N 5 = 1, N 7 = 1, N 17 = 1 occupation number ( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... )N 11 = 3 ( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... )N 5 = 1 N 13 = 2 ( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... )N 1 = 2 N 19 = 1 ( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... )N 5 = 1 N 7 = 1 N 17 = 1 Total = 1 distinct configurations

P n : Bosons P n = Probability of finding a particle with energy E n. P1P1 (1/4) (2/3) = 1/6 P5P5 (1/4) (1/3) + (1/4) (1/3) = 1/6 P7P7 (1/4) (1/3) = 1/12 P 11 (1/4) (3/3) = 1/4 P 13 (1/4) (2/3) = 1/6 P 17 (1/4) (1/3) = 1/12 P 19 (1/4) (1/3) = 1/12 Sum= 1 Allowed distinct configurations are : ( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... )N 11 = 3 ( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... )N 5 = 1, N 13 = 2 ( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... )N 1 = 2, N 19 = 1 ( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... )N 5 = 1, N 7 = 1, N 17 = 1 occupation number ( 0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,... )N 11 = 3 ( 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,0,0,... )N 5 = 1 N 13 = 2 ( 2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,... )N 1 = 2 N 19 = 1 ( 0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,... )N 5 = 1 N 7 = 1 N 17 = 1 Total = 4 distinct configurations

The General Case Consider a system whose 1-particle energies are E i with degeneracy d i, i = 1,2,3,... Now, N particles are put into the system such that N i particles have energy E i. Question : For a given configuration (N 1, N 2, N 3,... ), what is the number, Q(N 1, N 2, N 3,... ), of distinct states allowed?

Distinguishable Particles Each energy level E i corresponds to a bin with d i compartments. The problem is equivalent to finding the number of distinct ways to put N particles, with N i of them going into the d i compartments of the ith bin. Ans.: 1. There’re ways, without regard of picking order, to choose N 1 particles from the whole N particles. 2. When putting a particle into the 1 st bin, there’re d 1 choices of compartments.  the number of ways to put N 1 particles into the 1 st bin is  3. For the 2 nd bin, one starts with N  N 1 particles so one gets

Fermions 1. Fermions are indistinguishable, so it doesn’t matter which ones are going to which bin. 2. Each bin compartment can accept atmost one fermion. So the number of way to place N i fermions into the i th bin is if 

Bosons 1. Bosons are indistinguishable, so it doesn’t matter which ones are going to which bin. 2. Each bin compartment can accept any number of bosons. Consider placing N i bosons into the d i compartments of the i th bin. Let the bosons be represented by N i dots on a line. By inserting d i  1 partitions, the dots are “placed” into d i compartments. If the dots & partitions are all distinct, the number of all possible arrangement is However, the dots & partitions are indistinguishable among themselves. Hence

The Most Probable Configuration An isolated system in thermal equilibrium has a fixed total energy E, and fixed number of particle N, i.e., Since each possible way to share E among N particles has the same probability to exist, the most probable configuration (N 1, N 2, N 3,... ) is the one with the maximum Q(N 1, N 2, N 3,... ). Since Q(N 1, N 2, N 3,... ) involves a lot of factorials, it’s easier to work with lnQ.

Lagrange Multipliers Problem is to maximize ln Q, subject to constraintsand Using Lagrange’s multiplier method, we maximize, without constraint, i.e., we set Note : are just the original constraints.

Distinguishable Particles  Stirling’s approximation :for z >>1 

Fermions  d i, N i >>1 : 

Bosons  d i, N i >>1 :  d i  1  d i Do Prob 5.26, 5.27

Physical Significance of  and  Ideal gas in 3-D infinite square well. Taking the “bin” as the spherical shell between k and k+dk, the degeneracy d k is just the number of states in the shell :

Distinguishable Particles    cf. kinetic theory result :  universal

Distributions Set DFBDFB  Maxwell – Boltzmann Fermi – Dirac Bose - Einstein   Chemical potential n(  )  Most probable number of particles in a state with energy .  occupation number

Fermi-Dirac as T  0 T  0 :  (0)  E F T  0 T > 0

Ideal Gas Distinguishable particles  Indistinguishable particles : FBFB FBFB fixes  gives Read Prob Do Prob 5.28

The Blackbody Spectrum Photon  quantum of EM field  spin 1, massless boson with v  c. Only m   1 occurs. Number not conserved    0. Energy density in range d   Black body spectrum

Blackbody Spectrum 6000 K 4000 K 2000 K Visible region Do Prob 5.31 Read Prob 5.30