Question

PROBLEM 1 5 points] In classical statistical mechanics, the canonical partition function for a single harmonic oscillator is of the form d dp e Δ ΔΊΔ ) is the regulating spatial and momentum resolution cutoffs, which are often Chosen to be at the scale of the atoms (and n) and are important for making entropy dimensionless but they drop out in parts (b) and (c). Moreover, Z factorizes as Z ZzZp with Z. 3 Calculate the partition function and the Helmholtz Free energy, F =-kBT log(Z). This involves Gaussian integrals which you can find everywhere, e.g., in your Thermodynamics Phys224 text book. a. b. [2 Find the internal energy of the harmonic oscillator as an expectation value, using that the probability density to find the oscillator in position z with momentum p is equal to c. [2 Evaluate the specific heat and show that it obeys the so-called equipartition theorem, i.е., that C-2 × per oscillator. This proves the empirical Law of Dulong-Petit which says that the vibrations of the solid contribute to the specific heat an amount C- DNkB, with D 3 the dimensions of space three directions to oscillate) and N the number of vibrating units

Answer 1

For simplicity use,

$\beta=\frac{1}{k_BT}$

Gaussian integral

$\int_{-\infty}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$

Using the above

$Z=\frac{1}{\Delta}\sqrt{\frac{\pi}{\beta K/2}}\sqrt{\frac{\pi}{2m\beta}}$

$Z=\frac{1}{\Delta}\frac{\pi}{\beta}\frac{1}{\sqrt{mK}}$

So that

$F=-\frac{1}{\beta}\log Z=-\frac{1}{\beta}\log\left (\frac{\pi}{\Delta\beta}\frac{1}{\sqrt{mK}} \right )$

(b)

Average energy definition

$\langle E\rangle=\sum E(x,p)P(x,p)$

Or more appropriately in this case the sum is rather an integral,

$\langle E\rangle=\frac{1}{Z}\int E(x,p)e^{-\beta E(x,p)}dxdp$

The following trick is the general result for average internal energy

$\langle E\rangle=-\frac{1}{Z}\int \frac{\partial }{\partial \beta}e^{-\beta E(x,p)}dxdp$

$\langle E\rangle=-\frac{1}{Z} \frac{\partial }{\partial \beta}\int e^{-\beta E(x,p)}dxdp$

$\langle E\rangle=-\frac{1}{Z} \frac{\partial Z}{\partial \beta}=-\frac{\partial \log Z}{\partial \beta}$

Using the expression for Z from part(a)

$\langle E\rangle=-\frac{\partial }{\partial \beta}\left ( \textup{contants}-\log\beta \right )$

$\langle E\rangle=\frac{1}{\beta}=k_BT$

(c)

Specific heat

$C=\frac{\partial\langle E\rangle }{\partial T}=k_B$