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PHY452H1S Basic Statistical Mechanics. Lecture 17: Fermi gas thermodynamics. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 26, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]


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

Fermi gas thermodynamics

  • Energy was found to be

    \begin{aligned}\frac{E}{N} = \frac{3}{5} \epsilon_{\mathrm{F}}\qquad \text{where} \quad T = 0.\end{aligned} \hspace{\stretch{1}}(1.2.1)

  • Pressure was found to have the form fig. 1.1

    Fig 1.1: Pressure in Fermi gas

  • The chemical potential was found to have the form fig. 1.2.

    \begin{aligned}e^{\beta \mu} = \rho \lambda_{\mathrm{T}}^3\end{aligned} \hspace{\stretch{1}}(1.0.2a)

    \begin{aligned}\lambda_{\mathrm{T}} = \frac{h}{\sqrt{ 2 \pi m k_{\mathrm{B}} T}},\end{aligned} \hspace{\stretch{1}}(1.0.2b)

    so that the zero crossing is approximately when

    \begin{aligned}e^{\beta \times 0} = 1 = \rho \lambda_{\mathrm{T}}^3.\end{aligned} \hspace{\stretch{1}}(1.0.3)

    That last identification provides the relation T \sim  T_{\mathrm{F}}. FIXME: that bit wasn’t clear to me.

    Fig 1.2: Chemical potential in Fermi gas

How about at other temperatures?

  • \mu(T) = ?
  • E(T) = ?
  • C_{\mathrm{V}}(T) = ?

We had

\begin{aligned}N = \sum_k \frac{1}{{e^{\beta (\epsilon_k - \mu)} + 1}} = \sum_{\mathbf{k}} n_{\mathrm{F}}(\epsilon_\mathbf{k})\end{aligned} \hspace{\stretch{1}}(1.0.4)

\begin{aligned}E(T) =\sum_k \epsilon_\mathbf{k} n_{\mathrm{F}}(\epsilon_\mathbf{k}).\end{aligned} \hspace{\stretch{1}}(1.0.5)

FIXME: references to earlier sections where these were derived.

We can define a density of states

\begin{aligned}\sum_\mathbf{k} &= \sum_\mathbf{k} \int_{-\infty}^\infty d\epsilon  \delta(\epsilon  - \epsilon_\mathbf{k}) \\ &= \int_{-\infty}^\infty d\epsilon \sum_\mathbf{k}\delta(\epsilon  - \epsilon_\mathbf{k}),\end{aligned} \hspace{\stretch{1}}(1.0.6)

where the liberty to informally switch the order of differentiation and integration has been used. This construction allows us to write a more general sum

\begin{aligned}\sum_\mathbf{k} f(\epsilon_\mathbf{k}) &= \sum_\mathbf{k} \int_{-\infty}^\infty d\epsilon  \delta(\epsilon  - \epsilon_\mathbf{k}) f(\epsilon_\mathbf{k}) \\ &= \sum_\mathbf{k}\int_{-\infty}^\infty d\epsilon \delta(\epsilon  - \epsilon_\mathbf{k})f(\epsilon_\mathbf{k}) \\ &=\int_{-\infty}^\infty d\epsilon  f(\epsilon_\mathbf{k})\left( \sum_\mathbf{k} \delta(\epsilon  - \epsilon_\mathbf{k}) \right).\end{aligned} \hspace{\stretch{1}}(1.0.7)

This sum, evaluated using a continuum approximation, is

\begin{aligned}N(\epsilon ) &\equiv \sum_\mathbf{k}\delta(\epsilon  - \epsilon_\mathbf{k}) \\ &= \frac{V}{(2 \pi)^3} \int d^3 \mathbf{k} \delta\left( \epsilon  - \frac{\hbar^2 k^2}{2 m} \right) \\ &= \frac{V}{(2 \pi)^3} 4 \pi \int_0^\infty k^2 dk \delta\left( \epsilon  - \frac{\hbar^2 k^2}{2 m} \right)\end{aligned} \hspace{\stretch{1}}(1.0.8)


\begin{aligned}\delta(g(x)) = \sum_{x_0} \frac{\delta(x - x_0)}{\left\lvert {g'(x_0)} \right\rvert},\end{aligned} \hspace{\stretch{1}}(1.0.9)

where the roots of g(x) are x_0, we have

\begin{aligned}N(\epsilon ) &= \frac{V}{(2 \pi)^3} 4 \pi \int_0^\infty k^2 dk \delta\left( k - \frac{\sqrt{2 m \epsilon }}{\hbar} \right)\frac{m \hbar }{ \hbar^2 \sqrt{2 m \epsilon }} \\ &= \frac{V}{(2 \pi)^3} 2 \pi \frac{2 m \epsilon }{\hbar^2}\frac{2 m \hbar }{ \hbar^2 \sqrt{2 m \epsilon }} \\ &= V \left( \frac{2 m}{\hbar^2} \right)^{3/2} \frac{1}{{4 \pi^2}} \sqrt{\epsilon }.\end{aligned} \hspace{\stretch{1}}(1.0.10)

In 2D this would be

\begin{aligned}N(\epsilon ) \sim  V \int dk k \delta \left( \epsilon  - \frac{\hbar^2 k^2}{2m} \right) = V \frac{\sqrt{2 m \epsilon }}{\hbar} \frac{m \hbar}{\hbar^2 \sqrt{ 2 m \epsilon }} \sim  V\end{aligned} \hspace{\stretch{1}}(1.0.11)

and in 1D

\begin{aligned}N(\epsilon ) &\sim  V \int dk \delta \left( \epsilon  - \frac{\hbar^2 k^2}{2m} \right) \\ &= V \frac{m \hbar}{\hbar^2 \sqrt{ 2 m \epsilon }} \\ &\sim  \frac{1}{{\sqrt{\epsilon }}}.\end{aligned} \hspace{\stretch{1}}(1.0.12)

What happens when we have linear energy momentum relationships?

Suppose that we have a linear energy momentum relationship like

\begin{aligned}\epsilon_\mathbf{k} = v \left\lvert {\mathbf{k}} \right\rvert.\end{aligned} \hspace{\stretch{1}}(1.0.13)

An example of such a relationship is the high velocity relation between the energy and momentum of a particle

\begin{aligned}\epsilon_\mathbf{k} = \sqrt{ m_0^2 c^4 + p^2 c^2 } \sim  \left\lvert {\mathbf{p}} \right\rvert c.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Another example is graphene, a carbon structure of the form fig. 1.3. The energy and momentum for such a structure is related in roughly as shown in fig. 1.4, where

Fig 1.3: Graphene bond structure


Fig 1.4: Graphene energy momentum dependence


\begin{aligned}\epsilon_\mathbf{k} = \pm v_{\mathrm{F}} \left\lvert {\mathbf{k}} \right\rvert.\end{aligned} \hspace{\stretch{1}}(1.0.15)

Continuing with the 3D case we have

FIXME: Is this (or how is this) related to the linear energy momentum relationships for Graphene like substances?

\begin{aligned}N = V \int_0^\infty\underbrace{n_{\mathrm{F}}(\epsilon )}_{1/(e^{\beta (\epsilon  - \mu)} + 1)}\underbrace{N(\epsilon )}_{\epsilon ^{1/2}}\end{aligned} \hspace{\stretch{1}}(1.0.16)

\begin{aligned}\rho &= \frac{N}{V} \\ &= \left( \frac{2m}{\hbar^2 } \right)^{3/2} \frac{1}{{ 4 \pi^2}}\int_0^\infty d\epsilon  \frac{\epsilon ^{1/2}}{z^{-1} e^{\beta \epsilon } + 1} \\ &= \left( \frac{2m}{\hbar^2 } \right)^{3/2} \frac{1}{{ 4 \pi^2}}\left( k_{\mathrm{B}} T \right)^{3/2}\int_0^\infty dx \frac{x^{1/2}}{z^{-1} e^{x} + 1}\end{aligned} \hspace{\stretch{1}}(1.0.17)

where z = e^{\beta \mu} as usual, and we write x = \beta \epsilon . For the low temperature asymptotic behavior see [1] appendix section E. For z large it can be shown that this is

\begin{aligned}\int_0^\infty dx \frac{x^{1/2}}{z^{-1} e^{x} + 1}\approx \frac{2}{3}\left( \ln z \right)^{3/2}\left( 1 + \frac{\pi^2}{8} \frac{1}{{(\ln z)^2}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.18)

so that

\begin{aligned}\rho &\approx  \left( \frac{2m}{\hbar^2 } \right)^{3/2} \frac{1}{{ 4 \pi^2}}\left( k_{\mathrm{B}} T \right)^{3/2}\frac{2}{3}\left( \ln z \right)^{3/2}\left( 1 + \frac{\pi^2}{8} \frac{1}{{(\ln z)^2}} \right) \\ &= \left( \frac{2m}{\hbar^2 } \right)^{3/2} \frac{1}{{ 4 \pi^2}}\frac{2}{3}\mu^{3/2}\left( 1 + \frac{\pi^2}{8} \frac{1}{{(\beta \mu)^2}} \right) \\ &= \left( \frac{2m}{\hbar^2 } \right)^{3/2} \frac{1}{{ 4 \pi^2}}\frac{2}{3}\mu^{3/2}\left( 1 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{\mu} \right)^2 \right) \\ &= \rho_{T = 0}\left( \frac{\mu}{ \epsilon_{\mathrm{F}} } \right)^{3/2}\left( 1 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{\mu} \right)^2 \right)\end{aligned} \hspace{\stretch{1}}(1.0.19)

Assuming a quadratic form for the chemical potential at low temperature as in fig. 1.5, we have

Fig 1.5: Assumed quadratic form for low temperature chemical potential


\begin{aligned}1 &= \left( \frac{\mu}{ \epsilon_{\mathrm{F}} } \right)^{3/2}\left( 1 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{\mu} \right)^2 \right) \\ &= \left( \frac{\epsilon_{\mathrm{F}} - a T^2}{ \epsilon_{\mathrm{F}} } \right)^{3/2}\left( 1 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{\epsilon_{\mathrm{F}} - a T^2} \right)^2 \right) \\ &\approx  \left( 1 - \frac{3}{2} a \frac{T^2}{\epsilon_{\mathrm{F}}} \right)\left( 1 + \frac{\pi^2}{8} \frac{(k_{\mathrm{B}} T)^2}{\epsilon_{\mathrm{F}}^2} \right) \\ &\approx  1 - \frac{3}{2} a \frac{T^2}{\epsilon_{\mathrm{F}}} + \frac{\pi^2}{8} \frac{(k_{\mathrm{B}} T)^2}{\epsilon_{\mathrm{F}}^2},\end{aligned} \hspace{\stretch{1}}(1.0.20)


\begin{aligned}a = \frac{\pi^2}{12} \frac{k_{\mathrm{B}}^2}{\epsilon_{\mathrm{F}}},\end{aligned} \hspace{\stretch{1}}(1.0.21)

We have used a Taylor expansion (1 + x)^n \approx  1 + n x for small x, for an end result of

\begin{aligned}\mu = \epsilon_{\mathrm{F}} - \frac{\pi^2}{12} \frac{(k_{\mathrm{B}} T)^2}{\epsilon_{\mathrm{F}}}.\end{aligned} \hspace{\stretch{1}}(1.0.22)


[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.


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