Quantum Statistical

aQUANTUM STATISTICS PHYSICSNEED FOR QUANTUM STATISTICS In classical statistics, we assume that it is possible to simultaneously determine the position and momentum coordinates of a gaseous particle as precisely as we like. It means that these particles are distinguishable and can be labelled. But this is not true in practice. Heisenberg’s uncertainty principle forbids determination of the position (q) and the momentum (p) of a particle simultaneously with infinite precision. If the uncertainties in the measurements of q and p are Δq and Δp, respectively, the product ΔqΔp cannot be made less than ℏ/2. i.e ΔqΔp⩾ℏ

2where ℏ=h

2𝜋 and h=6.62 ⨯10-34 Js is Planck’s constant. It implies that when we study the behaviour of an assembly of identical particles statistically,we should treat it as a collection of indistinguishable particles. Further, blackbody radiation deserves a unique place in physics because it gave birth to quantum theory. BOSE EINSTEIN STATISTICSFollowing are the conditions of Bose-Einstein statistics(1) Particles are identical and indistinguishable.(2) Particles don't obey Pauli exclusion principle.(3) The sum of particles in each quantum state is the total number of particles.(4) The sum of energy of each particle in the quantum state is the total energy.Let us consider N particles in system. Let particles are in first, second, ... ith cell with average energy 𝜀1,𝜀2,…, 𝜀i for each particle. Let g is the degenency factor. Suppose that ni particles are arranged in a row and distributed among gi quantum states with (gi-1) partitions in between. The total number of possible arrangements of particles and partitions is equal to the total number of permutations of (ni+gi-1) objects in a row. Therefore the total possible ways of arranging ni particles with (gi-1) partitions =(ni+gi-1)!As the particles are identical and indistinguishable the possible number of distinct arrangements =(ni+gi-1)!

ni!(gi-1)! The total number of different and distinguishable ways of arranging N paricles in all the available energy states is given by,a

Ω	=(n1+g1-1) n1!(g1-1)×(n2+g2-1) n2!(g2-1)×…
∴Ω	= ∏i(ni+gi-1)! |ni!(gi-1)! ...............(i)

Also we have, logx!=xlogx-x, for large x∴ From (i), |logΩ= ∑ilog(ni+gi-1)!-logni!-log(gi-1)! = ∑i(ni+gi-1)log(n i+gi-1)-(ni+gi-1)-|| ||[(nilogni-ni)-(gi-1)log(gi-1)-(gi-1)]∴logΩ= ∑i(ni+gi-1)log(ni+gi-1)-nilogni-(gi-1)|log(gi-1)]For Maxima 𝛿logΩ = 0i.e 𝛿[ ∑i(ni+gi-1)log(ni+gi-1)-nilogni-(gi-1 )|log(gi-1)]=0 ∑i𝛿(ni+gi-1)log(ni+gi-1)-𝛿[nilogni]-𝛿(gi-1)|log(gi-1)=0 ∑ilog(ni+gi-1).𝛿ni+(ni+gi-1).𝛿ni

(ni+gi-1)- logni𝛿ni-ni.𝛿ni

ni=0 [ indipendent on gi] ∑ilog(ni+gi-1).𝛿ni+𝛿ni-logni𝛿ni-𝛿ni=0∴𝛿logΩ= ∑i[log(ni+gi-1)𝛿ni-logni𝛿ni]=0 ∴ ∑ i[-log(ni+gi-1)+logni]𝛿ni=0.............(ii)Also, ∑i𝛿ni=0 ..............(iii)and ∑i𝜀i𝛿n=0 ................(iv)Multiplying (iii) by 𝛼,(iv) by 𝛽 and adding to (ii) we get, ∑i[-log(ni+gi-1)+logni+𝛼+𝛽𝜀i]𝛿n i=0 or, -loge(ni+gi-1)+logni+𝛼+𝛽𝜀i=0 ∴-log(ni+gi)+logni+𝛼+𝛽𝜀i=0 {1 can be neglected compared with |(ni+gi)}∴log(ni

ni+gi)=-𝛼-𝛽𝜀ior.-log(ni +gi

ni)=-(𝛼+𝛽𝜀i)ni+gi

ni=e𝛼+𝛽𝜖ior.1+gi/ni=e𝛼+𝛽𝜖i or.a

gi ni	=(ea+𝛽𝜖i-1)
	[ni=gi (e𝛼+𝛽𝜖i-1)]

This is Bose Einstein distribution law. Fermi Dirac Statistics The condition for Fermi Dirac statistics are(1) The sub levels consists only ni= 0 or 1. That is particle obey Pauli exclusion principle.(2) The particles are identical and indistinguishable.(3) The sum of energy possessed by a particle in each sub-levels of different energy levels is the total energy. Let g be the degeneracy factor. Let ni be the number of particles for energy level 𝜀i. The first particle can be placed in any one of the available gi states i.e., this particle can be assigned to any of gi sets of quantum numbers. Thus first particle can be distributed in gi different ways and the process continues. Thus the total number of different ways of arranging ni particles among the gi states with energy level 𝜀i is=gi(gi-1)…[gi-(ni-1)]=gi!

(g i-ni)!Further, if the particles are taken to be indistinguishable, it will not possible to detect any difference when ni particles are reshuffled in to different states occupied by them in the energy level 𝜀i.∴ The total number of different and distinguishable ways isgi!

ni!(gi-ni)!∴ The total number of different and distinguishable ways of getting the distribution n1,n2,n3 etc. among the various energy levels 𝜀1,𝜀2,…𝜀i ete can be calculated by multiplying the various factors.a

∴Ω	=g1! n1!(g1-n1)!⋅g2! n2!(g2-n2)!⋯= ∏igi! ni!(gi-ni)!
∴ logΩ	= ∑i[loggi!-logni!-log(gi-ni)!]
∴logΩ	=∑(( giloggi-gi)-(nilogni-ni)-(gi-ni)log(gi-ni)-(gi-ni)]

a

logΩ	= ∑i[gi-loggi-nilogni-(gi-ni)log(gi-ni) ]

a

𝛿logΩ

= ∑i[𝛿gi-𝛿loggi-𝛿(nilogni)-𝛿(gi-ni)log(gi-ni)]

= ∑i-ni.1

ni𝛿ni- logni𝛿ni-[(gi-n i).1

gi-ni)𝛿(-ni)+log(gi-ni)𝛿(-ni)}a

∴𝛿logΩ

= ∑i[logni-log(gi-ni)]𝛿ni

For maximum 𝛿(logΩ)=0|⇒ ∑i[logn i-log(gi-ni)]𝛿ni=0....................(i)But ∑i𝛿ni=0 ...........................(2)and ∑i𝜖i𝛿ni=0............................(3) Multiply (2) by 𝛼 and (3) by 𝛽 and add to (1)[∑logni-log(gi-ni)+𝛼+𝛽𝜖i] 𝛿ni=0|here, 𝛿ni≠0so, or logni-log(gi-ni)+𝛼+𝛽𝜀i=0 logni-log(gi-ni)=-𝛼-𝛽𝜀i log(gi-ni)-logni=𝛼+𝛽𝜀ior, logg i-ni

ni=𝛼+𝛽𝜖i gi-ni

ni=e𝛼+𝛽𝜖ior, gi

ni-1=e𝛼+𝛽𝜖ior, gi

ni=1+e(𝛼+𝛽𝜖i)∴ni=gi

e𝛼+𝛽𝜀i+1 This is Fermi Dirac distribution law. Comparision of THREE STATISTICSThe expressions for the most probable distributions in the thres statistics are:(1) Maxwell's Boltzmann statistics:n1=gi

e𝛼+𝛽𝜀i⇒gi

ni=e𝛼+𝛽𝜖iwhere 𝛼,𝛽 are constant, gi is the degeneracy of ith level and ni are the number of particles for energy 𝜀i.(2) Bose Einstein statistics:a

n i	=gi e𝛼+𝛽𝜖i-1
or, gi ni+1	=e𝛼+𝛽𝜖i

(3) Fermi Dirac statistics:ni=gi

e𝛼+𝛽𝜀i+1or. gi

ni-1=e𝛼+𝛽iIt is to be noted that if g1

n i is very large in comparison to unity we may write, gi

ni≈gi

ni+1≈gi

ni-1i.e., For large values of gi

ni Bose Einstein and Fermi Dirac distributions approach the Maxwell-Boltzmann distribution. This is the case for normat existence of gases when the temperature is not too law and pressure is not too high.ELECTRON GASElectrons in a metal belong to a most characteristics system of fermions because electrons obey the exclusion principle. For electrons in a metal, the energy levels are grouped in bands. Practically at all temperature, the lower level energy bands are filled with electrons. The upper level energy bands are partially filled with electrons. The distribution of electrons is to be considered only in the upper bands called conduction bands. The zero energy level is taken at the lowest level of conduction band. It is assumed that the electrons have frce movements within the conductor, provided the energy associated with the electrons is of the order of upper energy bands.We know from Fermi-Diarc distribution law,ni=gi

e𝛼+𝛽𝜀i+1where n i is the number of particles in ith cell and gi is the degeneracy factor for ith cell. As the energy of electron in the conduction band is continuous the degeneracy factor gi is replaced by g(E)dE and ni by n(E)dE i.e., n(E)dE=g(E)dE

eaeE/KT+1Here n(E) refers to number of electrons, g(E) refers number of phase cells.But g(E)dE in terms of momentum is,g(p)dp=8𝜋v

h3 P2dP=8𝜋v

h3P.PdPBut E=1

2mv2 and P=mvaP=(2mE)1/2∴P2=2mE⇒2PdP=2mdE⇒PdP=mdEa∴g(E)dE=8𝜋v

h3(2mE)1/2mdE =82𝜋v

h3m3/2 E1/2dE∴n(E)dE=82𝜋v

h3⋅m3/2E1/2dE

e𝛼eE/KT+1But for clectron gas 𝛼=-EF

KT,EF is Fermi energy∴n(E)dE=82𝜋v

h3m3/2E1/2dE

e(E-EF) /KT+1This is exact form of Fermi Dirac law of energy distribution electrons. At the temperature T=0K, number of electrons is equal to the total number of energy states occupied by the electrons from zero to EF.a

∴n	=EF∫0g(E)dE=82𝜋v h3m3/2EF∫0E1/2dE
	=82𝜋Vm3/2 h3[2 3E3/2]EF0
	=162𝜋Vm3/2 3h3⋅E3/2F
or, E3/2F	=3nh3 162𝜋Vm3/2
or, EF=	h2 2m(3n 8𝜋V)2/3

FERMI LEVEL AND FERMI ENERGYThe topmost filled energy level at absolute zero temperature is called Fermi level and corresponding energy at topmost level (Fermi level) is called Fermi energy. i.e., the energy values upto which all the states are full at 0K and above which all the energy states are empty is known as Fermi energy.

Numerical Suppose that we have two state system, in which the first state has energy 𝜀 and second state 3𝜀. Give the ratio of probability of occupancy of first level to the probability of occupancy of second level, and simplify it.[TU 2075]Solution:There are two energy states 𝜀1 and 𝜀2 are 𝜀 and 3𝜀 for n1 amd n2 level with probability of occurrence P1 for state 1st and P2 for that of 2nd . Then probability P1 and P2 for two staes are,P1=Cexp(-𝜀1

kBT)And P2=Cexp (-𝜀2

kBT)Dividing (ii) by (i)Now, P1

P2=exp(𝜀2-𝜀1

kBT)[Taking room temperature]or, P1

P2=exp( 3𝜀-𝜀

kBT)or, P1

P2=exp(2𝜀

kBT)∴ The ratio of probability of occupancy of first level to the probability of occupancy of second level is exp(2𝜀

kBT).