Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Posts Tagged ‘hole’

Final update of notes for PHY487 (condensed matter physics)

Posted by peeterjoot on January 20, 2014

Here is what will likely be the final update of my class notes from Winter 2013, University of Toronto Condensed Matter Physics course (PHY487H1F), taught by Prof. Stephen Julian.

Official course description: “Introduction to the concepts used in the modern treatment of solids. The student is assumed to be familiar with elementary quantum mechanics. Topics include: bonding in solids, crystal structures, lattice vibrations, free electron model of metals, band structure, thermal properties, magnetism and superconductivity (time permitting)”

This document contains:

• Plain old lecture notes. These mirror what was covered in class, possibly augmented with additional details.
• Personal notes exploring details that were not clear to me from the lectures, or from the texts associated with the lecture material.
• Assigned problems. Like anything else take these as is.
• Some worked problems attempted as course prep, for fun, or for test preparation, or post test reflection.
• Links to Mathematica workbooks associated with this course.
My thanks go to Professor Julian for teaching this course.

NOTE: This v.5 update of these notes is still really big (~18M).  Some of my mathematica generated 3D images result in very large pdfs.

Changelog for this update (relative to the first, and second, and third, and the last pre-exam Changelogs).

January 19, 2014 Quadratic Deybe

January 19, 2014 One atom basis phonons in 2D

January 07, 2014 Two body harmonic oscillator in 3D
Figure out a general solution for two interacting harmonic oscillators, then use the result to calculate the matrix required for a 2D two atom diamond lattice with horizontal, vertical and diagonal nearest neighbour coupling.

December 04, 2013 Lecture 24: Superconductivity (cont.)

December 04, 2013 Problem Set 10: Drude conductivity and doped semiconductors.

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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.

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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)

Fig 1.1: Fermi gas chemical potential

 

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.

Fig 1.2: Fermi distribution

Fig 1.3: Fermi gas density of states

\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.

Fig 1.4: Specific heat per Fermion

 

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

    Fig 1.5: Relativisitic gas energy distribution

     

    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.

    Fig 1.6: Hole to electron round trip transition energy requirement

     

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.

Fig 1.7: Density of states for 2D linear energy momentum distribution

 

\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)

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