A simple harmonic one-dimensional quantum oscillator has energy levels given by

Question

A simple harmonic one-dimensional quantum oscillator has energy levels given by En = (n+1/2)nw where w is the characteristic frequency of the oscillator and where the quantum number n can assume the possible integral values n = 0,1.2,.... Derive an expression for the canonical partition function of a single quantum oscillator. Using the result of (a), find the average energy E for the system of N oscillators. Sketch E/(Nhw) a function of kT/hw. This is the Planck result for the thermal average energy of photons in a single mode of frequency w. Find the Helmholtz free energy, specific heat and entropy of the N-oscillator system. Sketch A/(Nhw), C/Nk and S/Nk as functions of kT /hw. What is the high-temperature limit of the heat capacity? Show that the high and low temperature limits make sense.

Explanation / Answer

Z = exp(-En)    sum over n (0to ) , where En = (n + 0.5)hbar*    Z = exp(-*[0 + 0.5]hbar*) + exp(-*[1 +0.5]hbar*) + exp(-*[2 + 0.5]hbar*) +...+exp(-*[n + 0.5]hbar*) Z = exp(-*hbar*/2)* [ exp(-hbar**0) +exp(-hbar**1) + exp(-hbar**2) + .... +exp(-*hbar**n)] Z = exp(-*hbar*/2)* [ 1 + exp(-*hbar*) +exp(-*hbar*)2 + .... +exp(-*hbar*)n] recall that a*rn = a/(1 -r)           when summed from n = 0 to n => So Z = exp(-*hbar*/2) * { 1 / [1 -exp(-*hbar*)] } Z = exp(-*hbar*/2)/ [1 - exp(-*hbar*)] b) ln(Z) = -*hbar*/2 - ln[1 -exp(-*hbar*)] = -d ln(Z)/d = hbar*/2 + hbar**exp(-*hbar*)/[1 - exp(-*hbar*)] = (hbar*)*[1/2 + 1/(exp(*hbar*) - 1)] c) A = -lnZ / = hbar*/2 + ln[1 -exp(-*hbar*)] / A = E - TS S = (E - A)/T = kb**(E - A) E and A from b and c, respectively) Cv = dE/dT = (1/kT2)*dE/d =(/T)*d/d recall = (hbar*)*[1/2 + 1/(exp(*hbar*)- 1) ] d/d =hbar**hbar**exp(*hbar*)/[exp(*hbar*)- 1] 2 Cv = (/T)*(hbar*)2*exp(*hbar*)/[exp(*hbar*) - 1]2 As -> 0 (T -> ) Cv should approach kB/2

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