A) Calculate the entropy S and energy U for a quantum...

a) Calculate the entropy S and energy U for a quantum harmonic oscillator and hence determine the heat capacity at constant volume, CV. b) i) Consider a 1-dimensional classical simple harmonic oscillator, which has Hamiltonian H = p^2 / 2m + 1/2 ?^2 x^2. Obtain Hamilton's equations and then solve them to find x and p as a function of time. Peform a temporal average of the Hamiltonian to obtain the average energy. ii) Calculate the entropy S and energy U for a classical harmonic oscillator. Using your result for U and comparing to the expression you found in part i), determine the amplitude of the motion at temperature T. Also using your result for U, determine the heat capacity at constant volume, CV. Compare your answer to the classical limit of the result you found in part a).

Transcript

00:01 Hello students, for part a we want to find the entropy, energy and heat capacity for a quantum harmonic oscillator.

00:10 We know that energy of a harmonic oscillator en is equal to n plus 1 by 2 h -cut omega where n is equal to 0, 1, 2, etc.

00:23 Infinity.

00:24 Now we can write the partition function for the energy particle.

00:29 That is z1 equal to summation n equal to 0 to infinity e raised to minus beta en.

00:37 This can be written as summation n equal to 0 to infinity e raised to minus n plus 1 by 2 h -cut omega by kt.

00:48 Which is equal to summation e raised to minus n plus 1 by 2 x.

00:55 Here we put x equal to h -cut omega by kt.

01:01 Now z1 equal to e raised to minus 1 by 2 x summation n equal to 0 to infinity e raised to minus nx.

01:11 That is equal to e raised to minus x by 2 1 plus e raised to minus x plus e raised to minus 2x plus etc.

01:25 That is equal to e raised to minus x by 2, 1 by 1minus e raised to minus x.

01:31 That is z1 equal to e raised to minus x by 2 1 minus e raised to minus x.

01:37 Now substitute back for x that is z equal to e raised to minus beta h cut omega by 2 by 1 minus e raised to minus beta h cut omega by 2.

01:51 That is equal to 1 by 2 sin h beta h cut omega by 2.

01:59 Taking the log we get log z equal to log of 1 by 2 sin h beta h cut omega by 2 which is equal to minus log 2 sin h beta h cut omega by 2.

02:22 Now for part a its first part is free energy.

02:27 That is f equal to minus k b t log z.

02:32 That is f equal to k b t log 2 sin h beta h cut omega by 2.

02:45 That is the free energy f equal to k b t log 2 sin h h cut omega by 2 k b t.

03:00 Now we want to find the entropy.

03:06 That is entropy s equal to minus dou f by dou t at constant volume.

03:14 That is equal to on calculation we get the entropy s equal to h cut omega by 2 t cot h h cut omega by 2 k b t minus k b log 2 sin h h cut omega by 2 k b t...

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