Peeter Joot's (OLD) Blog.

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Posts Tagged ‘state multiplicity’

An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

In A compilation of notes, so far, for ‘PHY452H1S Basic Statistical Mechanics’ I posted a link this compilation of statistical mechanics course notes.

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

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PHY452H1S Basic Statistical Mechanics. Lecture 7: Ideal gas and SHO phase space volume calculations. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on February 5, 2013

[Click here for a PDF of this post with nicer formatting]

Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

Review. Classical phase space calculation

\begin{aligned}E_{\mathrm{ideal}} = \sum_i \frac{\mathbf{p}_i^2}{2 m}\end{aligned} \hspace{\stretch{1}}(1.2.1)

From this we calculated \gamma(E), and

\begin{aligned}\frac{d\gamma(E)}{dE} = \Omega_{\mathrm{classical}}(E)\end{aligned} \hspace{\stretch{1}}(1.2.2)

Fudging with a requirement that \Delta x \Delta p \sim h, we corrected this as

\begin{aligned}\Omega_{\mathrm{quantum}}(E) = \frac{\Omega_{\mathrm{classical}}(E)}{N! h^{3N}}\end{aligned} \hspace{\stretch{1}}(1.2.3)

Now let’s do the quantum calculation.

Quantum calculation

Recall that for the solutions of the Quantum free particle in a box, as in (Fig 1), our solutions are

Fig 1: 1D Quantum free particle in a box

 

\begin{aligned}\Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left( \frac{ n \pi x}{L} \right),\end{aligned} \hspace{\stretch{1}}(1.2.4)

where n = 1, 2, \cdots, and

\begin{aligned}\epsilon_n = \frac{n^2 h^2}{8 m L^2}\end{aligned} \hspace{\stretch{1}}(1.2.5)

.

In three dimensions, with n_i = 1, 2, \cdots we have

\begin{aligned}\Psi_{n_1, n_2, n_3}(x, y, z) = \left( \frac{2}{L} \right)^{3/2} \sin\left( \frac{ n_1 \pi x}{L} \right)\sin\left( \frac{ n_2 \pi x}{L} \right)\sin\left( \frac{ n_3 \pi x}{L} \right)\end{aligned} \hspace{\stretch{1}}(1.2.6)

and

\begin{aligned}\epsilon_{n_1, n_2, n_3} = \frac{h^2}{8 m L^2} \left( n_1^2 + n_2^2 + n_3^2 \right)\end{aligned} \hspace{\stretch{1}}(1.2.7)

\begin{aligned}\gamma^{3d}_{\mathrm{classical}}(E) = \underbrace{V}_{L^3}\int d^3 p \theta \left( E - \frac{\mathbf{p}^2}{2m} \right)= V \frac{4 \pi}{3} (2 m E)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.2.8)

so that

\begin{aligned}\gamma^{3d}_{\mathrm{corrected}}(E) = V \frac{4 \pi}{3} \frac{(2 m E)^{3/2}}{h^3}\end{aligned} \hspace{\stretch{1}}(1.2.9)

\begin{aligned}\gamma^{3d}_{\mathrm{quantum}}(E) = \sum_{n_1, n_2, n_3} \Theta(E - \epsilon_{n_1, n_2, n_3} ).\end{aligned} \hspace{\stretch{1}}(1.2.10)

How do the multiplicities scale by energy? We’ll have expect something like (Fig 2).

Fig 2: Multiplicities for free quantum particle in a 3D box

 

Provided the energies E \gg 3h^2/(8 m L) are large enough, we can approximate the sum with

\begin{aligned}\sum_{n_1, n_2, n_3} \sim \int_0^\infty dn_1 dn_2 dn_3\end{aligned} \hspace{\stretch{1}}(1.2.11)

So

\begin{aligned}\gamma^{3d}_{\mathrm{quantum}} \left( E \gg \frac{h^2}{8 m L^2} \right) \approx\int_0^\infty dn_1 dn_2 dn_3 \Theta \left( E - \frac{h^2}{8 m L^2} \left( n_1^2 + n_2^2 + n_3^2 \right) \right)=\frac{1}{{8}}\frac{4 \pi}{3} \left( \frac{8 m L^2 E}{h^2} \right)^{3/2}=L^3\frac{4 \pi}{3} \frac{\left( 2 m E \right)^{3/2}}{h^3}\end{aligned} \hspace{\stretch{1}}(1.2.12)

Harmonic oscillator in 1D.

Our phase space is of the form (Fig 3).

Fig 3: 1D classical SHO phase space

 

Where the number of states in this classical picture are found with:

\begin{aligned}\gamma^{\mathrm{classical}}(E) = \int dx dp \theta\left( E - \left( \frac{1}{{2}} k x^2 + \frac{1}{{2m }} p^2 \right) \right).\end{aligned} \hspace{\stretch{1}}(1.2.13)

Rescale

\begin{aligned}\tilde{x} = x \sqrt{ \frac{k}{2}}\end{aligned} \hspace{\stretch{1}}(1.0.14a)

\begin{aligned}\tilde{p} = \frac{p}{\sqrt{2m}}\end{aligned} \hspace{\stretch{1}}(1.0.14b)

so that we have

\begin{aligned}\gamma^{\mathrm{classical}}(E) = \int d\tilde{x} d \tilde{p} \sqrt{\frac{2 \times 2 m}{k}} \theta\left( E - \tilde{x}^2 - \tilde{p}^2 \right)=2 \sqrt{\frac{m}{k}} \pi E= 2 \pi \sqrt{\frac{m}{k}} E.\end{aligned} \hspace{\stretch{1}}(1.0.15)

\begin{aligned}\gamma^{\mathrm{SHO}}_{\mathrm{corrected}}(E) = 2 \pi \sqrt{\frac{m}{k}} \frac{E}{h}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

How about the quantum calculation?

We have for the energy

\begin{aligned}E_n^{\mathrm{SHO}} = \left( n + \frac{1}{{2}} \right) \hbar \omega\end{aligned} \hspace{\stretch{1}}(1.0.17a)

\begin{aligned}\omega = \sqrt{\frac{k}{m}}\end{aligned} \hspace{\stretch{1}}(1.0.17b)

\begin{aligned}\hbar = \frac{h}{2 \pi}\end{aligned} \hspace{\stretch{1}}(1.0.17c)

graphing the counts (Fig 4), we again have stepping as a function of energy, but no multiplicities this time

Fig 4: 1D quantum SHO states per energy level

\begin{aligned}\gamma_{\mathrm{quantum}}(E) = \sum_{n = 0}^\infty \Theta\left( E - \left( n + \frac{1}{{2}} \hbar \omega \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.18)

we make the continuous approximation for the summation again, and throwing away the zero point energy, we have

\begin{aligned}\gamma_{\mathrm{quantum}}(E \gg \hbar \omega) \approx\int_{0}^\infty dn \Theta\left( E - n \hbar \omega \right)= 2 \pi \frac{E}{h} \sqrt{\frac{m}{k}}\end{aligned} \hspace{\stretch{1}}(1.0.19)

Why N!?

We have a problem with out counting here. Consider some particles in a box as in (Fig 5).

Fig 5: Three particles in a box

 

  1. particle 1 at \mathbf{x}_1
  2. particle 2 at \mathbf{x}_2
  3. particle 3 at \mathbf{x}_3

or

  1. particle 1 at \mathbf{x}_2
  2. particle 2 at \mathbf{x}_3
  3. particle 3 at \mathbf{x}_1

This is fine in the classical picture, but in the quantum picture with an assumption of indistinguishability, no two particles (say electrons) cannot be labelled in this fashion.

\paragraphAndIndex{Gibbs paradox}

\begin{aligned}\underbrace{S_{\mathrm{ideal}}^{(\mathrm{E}, \mathrm{N}, \mathrm{V})}}_{\text{Statistical entropy}}= k_{\mathrm{B}} \ln \left( \frac{\Omega_{\mathrm{classical}}}{h^{3N}} \right)\underbrace{\approx}_{N \gg 1} k_{\mathrm{B}} \left( N \ln V + \frac{3 N}{2} \ln \left( \frac{4 \pi m E }{3 N h^2} \right) + \frac{3 N}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.20)

Suppose we double the volume as in (Fig 6), then our total entropy for the bigger system would be

Fig 6: Gibbs volume doubling argument. Two identical systems allowed to mix

 

\begin{aligned}S_{\mathrm{total}}^{(\mathrm{E}, \mathrm{N}, \mathrm{V})}= k_{\mathrm{B}} \ln \left( \frac{\Omega_{\mathrm{classical}}}{h^{3N}} \right)\approx k_{\mathrm{B}} \left( (2 N) \ln (2 V) + \frac{3 (2 N)}{2} \ln \left( \frac{4 \pi m (2 E) }{2 ( 2 N) h^2} \right) + \frac{3 (2 N)}{2} \right).\end{aligned} \hspace{\stretch{1}}(1.0.21)

We have

\begin{aligned}S_{\mathrm{total}} = S_1 + S_2 + k_{\mathrm{B}} (2 N) \ln 2= S_1 + S_2 + k_{\mathrm{B}} \ln 2^{2 N}.\end{aligned} \hspace{\stretch{1}}(1.0.22)

This is telling us that each particle could be in either the left or the right side, but we know that this uncertainty shouldn’t be in the final answer. We must drop this k_{\mathrm{B}} term.

So, if we assume that these particles are identical, and divide \Omega by N!, then we find

\begin{aligned}S_{\mathrm{ideal}} = k_{\mathrm{B}} \left( N \ln \frac{V}{N} + \frac{3 N}{2} \ln \left( \frac{4 \pi m E }{3 N h^2} \right) + \frac{5 N}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.23)

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