Lecture 28-Statistics of Fermions and Bosons

[@damian.gamlath]

1:00 Agenda of lecture
1:35 *Boson statistics*
3:40 Trying to calculate the Partition Function for a system of Bosons using Canonical Ensemble (point: it doesn't work)
12:55 Switch to Grand Canonical Ensemble to find the Partition Function (this shows why we need to use GCE for quantum stats)
22:10 Deriving statistics for Bosons
32:25 Bose Einstein statistics for average occupancy
32:45 *Fermion statistics*
37:25 Quick summary of the difference in the even/odd parity for Bosons and Fermions and how it affects statistics for such particles
44:05 *Occupancy for both B.E. and F.D. stats*
*NOTES*
12:55 The reason we can't do this restricted sum is because the number of particles is limited, but number of possible energies for Bosons is infinite. So we are trying to do a sum within a sum, but one of those sums runs to infinity, but the other runs to a limited number of particles.

[@perfectlyimperfect6849]

Excellent lecture... thankyou so much sir for this crystal clear concept

[@岡安一壽-g2y]

Thank you sir.
Another method
ni:number of particles with energy εi
gi:number of states in energy level εi
N=Σni
E=Σniεi
(n1,n2,･･･,nk):distribution
W(n1,n2,･･･,nk):the number of states that the distribution is (n1,n2,･･･,nk)
P(n1,n2,･･･,nk):the probability that the distribution is (n1,n2,･･･,nk)
(,,･･･,):disrribution in equilibrium
P(n1,n2,･･･,nk)=W(n1,n2,･･･,nk)exp{-(α1+α2･･･+αN)}･exp(-βE)
αN,β:undetermined multipliers of Lagrange's multiplier method
Bosons
W(n1,n2,･･･,nk)=(1/Πni!)Πgi(gi+1)･･･(gi+ni-1)
P(n1,n2,･･･,nk)=(1/Πni!)Πgi(gi+1)･･･(gi+ni-1)･exp{-(α1+α2･･･+αN)}･Πexp(-βεi)^ni
αN:N is the subscript of α
In equilibrium, P(,,･･･,)=Pmax
Since P(･･･,+1,･･･)≒P(･･･,,･･･)=Pmax
P(･･･,+1,･･･)/P(･･･,,･･･)≒1
Therefore
P(･･･,+1,･･･)/P(･･･,,･･･)=(gi+)exp(-αN-βεi)/(+1)≒1

+1≒ →　
(gi+)exp(-αN-βεi)/≒1 (1)
Accordingly　≒(gi+)exp(-αN-βεi)
And　N=Σ=exp(-αN)Σ(gi+)exp(-βεi)
∴ exp(αN)=(1/N)Σ(gi+)exp(-βεi)=ZN/N　　
where ZN≡Σ(gi+)exp(-βεi)
Therefore　αN=ln(ZN/N)
By using (1)　/gi≒1/{exp(αN+βεi)-1}=1/{(ZN/N)exp(βεi)-1}
And P(n1,n2,･･･,nk)=(N!/Πni!)Πgi(gi+1)･･･(gi+ni-1)･1/(Z1･Z2･･･ZN)･exp(-βE)
Fermions
W(n1,n2,･･･,nk)=(1/Πni!)Πgi(gi-1)･･･(gi-ni+1)
P(n1,n2,･･･,nk)=(1/Πni!)Πgi(gi-1)･･･(gi-ni+1)･exp{-(α1+α2･･･+αN)}･Πexp(-βεi)^ni
αN:N is the subscript of α
In equilibrium, P(,,･･･,)=Pmax
Since P(･･･,+1,･･･)≒P(･･･,,･･･)=Pmax
P(･･･,+1,･･･)/P(･･･,,･･･)≒1
Therefore
P(･･･,+1,･･･)/P(･･･,,･･･)=(gi-)exp(-αN-βεi)/(+1)≒1

+1≒ →　
(gi-)exp(-αN-βεi)/≒1 (2)
Accordingly　≒(gi-)exp(-αN-βεi)
And　N=Σ=exp(-αN)Σ(gi-)exp(-βεi)
∴ exp(αN)=(1/N)Σ(gi-)exp(-βεi)=ZN/N　　
where ZN≡Σ(gi-)exp(-βεi)
Therefore　αN=ln(ZN/N)
By using (2)　/gi≒1/{exp(αN+βεi)+1}=1/{(ZN/N)exp(βεi)+1}
And P(n1,n2,･･･,nk)=(N!/Πni!)Πgi(gi-1)･･･(gi-ni+1)･1/(Z1･Z2･･･ZN)･exp(-βE)

[@ghgft]

Thank you sir.. wonderful lecture series

[@srinugadde4676]

Very good and nicely explained

[@ranjitchoudhury4977]

Thanks a lot.

[@ShailendraSingh-f7u]

Too much class sir

[@mirmohsin2986]

Wariyah asal

[@ankitmishra2723]

Thanks sir