Free energy of a harmonic oscillator. The 1D quantum harmonic oscillator has equ

Question

Free energy of a harmonic oscillator. The 1D quantum harmonic oscillator has equally spaced energy states with energy E_n = (n + 1/2)h omega, where n is a positive integer or zero, and a) is the frequency of the oscillator. Calculate a) the partition function for a single harmonic oscillator for a constant temperature tau. Write the final result as a single term instead of a sum over states. b) the Helmholtz free energy, c) the entropy, d) Show that in the high temperature limit, where h omega/tau

Explanation / Answer

In this case the harmonic oscillator is 1-D and we consider a constant temperature

now from the Hamiltonian of 1-D oscillator is

H= P^2/2m +1/2m *omega^2X^2

from this we get the energy states or the energy eigen function

En= (n+1/2)hcut omega

From this we get the generl form of cannonical partition function

Q(beta) =summation of exp(-beta(n+1/2)hcut omega)

using Taylor expansion we get that

Q(beta) = exp(-beta*hcut*omega/2)/1-exp(-beta*hcut*omega/2) ..

= 1/exp(beta*hcut*omega/2) -exp(-beta*hcut*omega/2)......a)

Again free energy means

A= -1/betalnQ(beta) =hcut*omega/2 +1/beta*ln(1-exp(beta*hcut*omega)) b)

Again average energy E= (1/2+<n>)hcut*omega

Now entropy S= klnQ(beta) +E/T = kln(1-exp(-beta*hcut*omega)) +hcut*omega/T *exp(-beta*hcut*omega/1-exp(-beta*hcut*omega)) c)

d) at high temperature hcut tends to zero thus giving the said eexpression

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