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Posts Tagged ‘hole’

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

Posted by peeterjoot on March 27, 2013

Here’s my second update of my notes compilation for this course, including all of the following:

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

plus everything detailed in the description of my first update and before.

PHY452H1S Basic Statistical Mechanics. Lecture 18: 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.)]

Disclaimer

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

Review

Last time we found that the low temperature behaviour or the chemical potential was quadratic as in fig. 1.1. \begin{aligned}\mu =\mu(0) - a \frac{T^2}{T_{\mathrm{F}}}\end{aligned} \hspace{\stretch{1}}(1.1.1)

Specific heat \begin{aligned}E = \sum_\mathbf{k} n_{\mathrm{F}}(\epsilon_\mathbf{k}, T) \epsilon_\mathbf{k}\end{aligned} \hspace{\stretch{1}}(1.1.2) \begin{aligned}\frac{E}{V} &= \frac{1}{{(2\pi)^3}} \int d^3 \mathbf{k} n_{\mathrm{F}}(\epsilon_\mathbf{k}, T) \epsilon_\mathbf{k} \\ &= \int d\epsilon N(\epsilon) n_{\mathrm{F}}(\epsilon, T) \epsilon,\end{aligned} \hspace{\stretch{1}}(1.1.3)

where \begin{aligned}N(\epsilon) = \frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)^{3/2}\sqrt{\epsilon}.\end{aligned} \hspace{\stretch{1}}(1.1.4)

Low temperature $C_{\mathrm{V}}$ \begin{aligned}\frac{\Delta E(T)}{V}=\int_0^\infty d\epsilon N(\epsilon)\left( n_{\mathrm{F}}(\epsilon, T) - n_{\mathrm{F}}(\epsilon, 0) \right)\end{aligned} \hspace{\stretch{1}}(1.1.5)

The only change in the distribution fig. 1.2, that is of interest is over the step portion of the distribution, and over this range of interest $N(\epsilon)$ is approximately constant as in fig. 1.3. \begin{aligned}N(\epsilon) \approx N(\mu)\end{aligned} \hspace{\stretch{1}}(1.0.6a) \begin{aligned}\mu \approx \epsilon_{\mathrm{F}},\end{aligned} \hspace{\stretch{1}}(1.0.6b)

so that \begin{aligned}\Delta e \equiv\frac{\Delta E(T)}{V}\approx N(\epsilon_{\mathrm{F}})\int_0^\infty d\epsilon\left( n_{\mathrm{F}}(\epsilon, T) - n_{\mathrm{F}}(\epsilon, 0) \right)=N(\epsilon_{\mathrm{F}})\int_{-\epsilon_{\mathrm{F}}}^\infty d x (\epsilon_{\mathrm{F}} + x)\left( n_{\mathrm{F}}(\epsilon + x, T) - n_{\mathrm{F}}(\epsilon_{\mathrm{F}} + x, 0) \right).\end{aligned} \hspace{\stretch{1}}(1.0.7)

Here we’ve made a change of variables $\epsilon = \epsilon_{\mathrm{F}} + x$, so that we have near cancelation of the $\epsilon_{\mathrm{F}}$ factor \begin{aligned}\Delta e &= N(\epsilon_{\mathrm{F}})\epsilon_{\mathrm{F}}\int_{-\epsilon_{\mathrm{F}}}^\infty d x \underbrace{\left( n_{\mathrm{F}}(\epsilon + x, T) - n_{\mathrm{F}}(\epsilon_{\mathrm{F}} + x, 0) \right)}_{\text{almost equal everywhere}}+N(\epsilon_{\mathrm{F}})\int_{-\epsilon_{\mathrm{F}}}^\infty d x x\left( n_{\mathrm{F}}(\epsilon + x, T) - n_{\mathrm{F}}(\epsilon_{\mathrm{F}} + x, 0) \right) \\ &\approx N(\epsilon_{\mathrm{F}})\int_{-\infty}^\infty d x x\left( \frac{1}{{ e^{\beta x} +1 }} - {\left.{{\frac{1}{{ e^{\beta x} +1 }}}}\right\vert}_{{T \rightarrow 0}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.8)

Here we’ve extended the integration range to $-\infty$ since this doesn’t change much. FIXME: justify this to myself? Taking derivatives with respect to temperature we have \begin{aligned}\frac{\delta e}{T} &= -N(\epsilon_{\mathrm{F}})\int_{-\infty}^\infty d x x\frac{1}{{(e^{\beta x} + 1)^2}}\frac{d}{dT} e^{\beta x} \\ &= N(\epsilon_{\mathrm{F}})\int_{-\infty}^\infty d x x\frac{1}{{(e^{\beta x} + 1)^2}}e^{\beta x}\frac{x}{k_{\mathrm{B}} T^2}\end{aligned} \hspace{\stretch{1}}(1.0.9)

With $\beta x = y$, we have for $T \ll T_{\mathrm{F}}$ \begin{aligned}\frac{C}{V} &= N(\epsilon_{\mathrm{F}})\int_{-\infty}^\infty \frac{ dy y^2 e^y }{ (e^y + 1)^2 k_{\mathrm{B}} T^2} (k_{\mathrm{B}} T)^3 \\ &= N(\epsilon_{\mathrm{F}}) k_{\mathrm{B}}^2 T\underbrace{\int_{-\infty}^\infty \frac{ dy y^2 e^y }{ (e^y + 1)^2 } }_{\pi^2/3} \\ &= \frac{\pi^2}{3} N(\epsilon_{\mathrm{F}}) k_{\mathrm{B}} (k_{\mathrm{B}} T).\end{aligned} \hspace{\stretch{1}}(1.0.10)

Using eq. 1.1.4 at the Fermi energy and \begin{aligned}\frac{N}{V} = \rho\end{aligned} \hspace{\stretch{1}}(1.0.11a) \begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2 m}\end{aligned} \hspace{\stretch{1}}(1.0.11b) \begin{aligned}k_{\mathrm{F}} = \left( 6 \pi^2 \rho \right)^{1/3},\end{aligned} \hspace{\stretch{1}}(1.0.11c)

we have \begin{aligned}N(\epsilon_{\mathrm{F}}) &= \frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)^{3/2}\sqrt{\epsilon_{\mathrm{F}}} \\ &= \frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)^{3/2}\frac{\hbar k_{\mathrm{F}}}{\sqrt{2m}} \\ &= \frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)^{3/2}\frac{\hbar }{\sqrt{2m}} \left( 6 \pi^2 \rho \right)^{1/3} \\ &= \frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)\left( 6 \pi^2 \frac{N}{V} \right)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.0.12)

Giving \begin{aligned}\frac{C}{N} &= \frac{\pi^2}{3} \frac{V}{N}\frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)\left( 6 \pi^2 \frac{N}{V} \right)^{1/3}k_{\mathrm{B}} (k_{\mathrm{B}} T) \\ &= \left( \frac{m}{6 \hbar^2} \right)\left( \frac{V}{N} \right)^{2/3}\left( 6 \pi^2 \right)^{1/3}k_{\mathrm{B}} (k_{\mathrm{B}} T) \\ &= \left( \frac{ \pi^2 m}{3 \hbar^2} \right)\left( \frac{V}{\pi^2 N} \right)^{2/3}k_{\mathrm{B}} (k_{\mathrm{B}} T) \\ &= \left( \frac{ \pi^2 m}{\hbar^2} \right)\frac{\hbar^2}{2 m \epsilon_{\mathrm{F}}}k_{\mathrm{B}} (k_{\mathrm{B}} T),\end{aligned} \hspace{\stretch{1}}(1.0.13)

or \begin{aligned}\boxed{\frac{C}{N} = \frac{\pi^2}{2} k_{\mathrm{B}} \frac{ k_{\mathrm{B}} T}{\epsilon_{\mathrm{F}}}.}\end{aligned} \hspace{\stretch{1}}(1.0.14)

This is illustrated in fig. 1.4.

Relativisitic gas

1. Relativisitic gas \begin{aligned}\epsilon_\mathbf{k} = \pm \hbar v \left\lvert {\mathbf{k}} \right\rvert.\end{aligned} \hspace{\stretch{1}}(1.0.15) \begin{aligned}\epsilon = \sqrt{(m_0 c^2)^2 + c^2 (\hbar \mathbf{k})^2}\end{aligned} \hspace{\stretch{1}}(1.0.16)

2. graphene
3. massless Dirac Fermion

We can think of this state distribution in a condensed matter view, where we can have a hole to electron state transition by supplying energy to the system (i.e. shining light on the substrate). This can also be thought of in a relativisitic particle view where the same state transition can be thought of as a positron electron pair transition. A round trip transition will have to supply energy like $2 m_0 c^2$ as illustrated in fig. 1.6.

Graphene

Consider graphene, a 2D system. We want to determine the density of states $N(\epsilon)$, \begin{aligned}\int \frac{d^2 \mathbf{k}}{(2 \pi)^2} \rightarrow \int_{-\infty}^\infty d\epsilon N(\epsilon),\end{aligned} \hspace{\stretch{1}}(1.0.17)

We’ll find a density of states distribution like fig. 1.7. \begin{aligned}N(\epsilon) = \text{constant factor} \frac{\left\lvert {\epsilon} \right\rvert}{v},\end{aligned} \hspace{\stretch{1}}(1.0.18) \begin{aligned}C \sim \frac{d}{dT} \int N(\epsilon) n_{\mathrm{F}}(\epsilon) \epsilon d\epsilon,\end{aligned} \hspace{\stretch{1}}(1.0.19) \begin{aligned}\Delta E \sim \underbrace{T}_{\text{window}}\times\underbrace{T}_{\text{energy}}\times\underbrace{T}_{\text{number of states}}\sim T^3\end{aligned} \hspace{\stretch{1}}(1.0.20)

so that \begin{aligned}C_{\mathrm{Dimensionless}} \sim T^2\end{aligned} \hspace{\stretch{1}}(1.0.21)