To determine the total number of microstates, we need to determine the density of states in one dimension. The density of states function (g(E)) is the number of microstates per unit length per unit energy range.
In one dimension for a particle of mass 'm' in a box of length 'L' we have the energy given
E: hk2 2
From the boundary conditions of the cuboid at x=0 and x=L, we deduce that 'k' can take on discrete values in units of 2π/L. Consequently, the quantum states are in units of L/2π. The number of states 'N' for momentum 'k' will be equal to,
2KL N = TT
And the energy E at any momentum "k" is given by
E: hk2 2
The density of states in 1D is given by
9(E) 1 IN L DE = 1 DN dk L dk de
we know
NP 27 dk 1
AND,
hk DE Ε dk dk dЕ = m hk m
Notation in terms of energy E,
dk DE -m 1/2 E-1/2 -1/2
The density of states will be
9(E) 1 dN 1 dN dk IDE = I dk DE L L dE V2 m1/2E-12 πh
The total number of microstates at energy 'E' is,
2m Z(E) = [” 9(E)E = Z dE = E-1/2 DE 6.* -12dE = 22m VE ић пh
(i) A two-particle system will have the number of microstates will be,
Z(E) = 21(E1Z2E - E1)
where Z1 and Z2 are the microstates due to particle 1 and particle 2 calculated above. E1 is the energy of particle 1, so the energy of particle 2 is E-E1.
(ii) For indistinguishable particles, the number of microstates will be
Z(E) 21(E1)Z2(E – E1) 2!
where Z1 and Z2 are the microstates caused by particle 1 and particle 2, respectively, calculated as above. E1 is the energy of particle 1 and (E-E1) is the energy of particle 2. 2! it comes as a result of the fact that if we exchange two particles, due to their indistinguishability, the result will be the same.
E is the total energy of the system for both the above cases.
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