Title: ETH_Sem5 / Solid_State_Intro / Woche_06

Q: Formula for DOS of acoustic phonons
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What is illogical about this formula?
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A: \(\nu(\omega) = \frac{1}{2\pi^2} \big( \frac{2}{v_T^3} + \frac{1}{v_L^3} \big) \omega^2\); DOS cannot grow indefinitely with \(\omega\) because the number of normal modes is fixed.
The Debye approximation solves this problem.

Q: What is the idea behind the Debye approximation?
A: - DOS only given by the three acoustic phonon branches.
- The Debye frequency \(\omega_D\) acts as a cutoff that prevents the DOS to grow indefinitely.
\(\nu(\omega) = \frac{9n}{V_0w_D^3} \;\omega^2 \;\Theta(\omega_D - \omega)\)
- \(\omega_D\) is choosen to yield the total number of normal modes upon integration
\(V \int_{0}^{\omega_D} \nu(\omega)\text{d}\omega = 3nV/V_0\)

Q: How does the specific heat derived by the Debye approximation behave at \(T \rightarrow 0\) and \(T \rightarrow \infty\)
\(C_V = \frac{9nVk_B}{V_0} \;\left( \frac{T}{T_D} \right)^3 \int_{0}^{T_D/T} \text{d}y \;\frac{y^4\exp(y)}{(\exp(y)-1)^2} \) where \(y = \frac{\hbar\omega}{k_BT}\)
A: As \(T \rightarrow 0\) the integral goes from 0 to infinity and thus is constant. At low temperatures the specific heat of a solid goes as the 3rd power of temperature.
As \(T \rightarrow \infty\) (assuming the solid wouldn't melt) the classical result \(C_V = \frac{3nVk_B}{V_0}\) is revocered.

Q: How is the Debye temperature defined? What are its typical values?
A: \(T_D = \hbar\omega_D / k_B\) is typically a few hundered Kelvin.

Q: Explain the setup for a relaxation calorimeter.
A: Whole setup is placed in a vacuum.
Small wires with known thermal conductivity connect thermal bath and platform.

Q: How is a specific heat measurement done with a relaxation caloriemeter?
A: - Set thermal bath to desired temperature and wait for bath, platform and sample to reach thermal equilibrium.
- Turn on heater.
- Sample warms up and wires conduct heat away.
- Saturation is achieved were heat input of heater and heat drain through the wires are equal.
- Turn off heater. Relaxation occurs.
- Calculate specific heat through measured temperature, heat input, thermal conductivity of wire and observed ramp and relaxation time.

Q: What is the dispersion relation for a free electron.
A: \(\varepsilon(\vec{k}) = \frac{\hbar^2}{2m_e}k^2 = \frac{p^2}{2m_e}\)

Q: What is the wavefunction for an electron in a box?
What are the allowed wave vectors \(\vec{k}\)  due to imposed  periodic boundry conditions.
A: \(\psi(\vec{r}) = \frac{1}{\sqrt{V}}\exp(i\vec{k}\vec{r})\)
\(k_x = 0, \pm \frac{2\pi}{L_x}, \pm \frac{4\pi}{L_x}, ...\)
\(k_y = 0, \pm \frac{2\pi}{L_y}, \pm \frac{4\pi}{L_y}, ...\)
\(k_z = 0, \pm \frac{2\pi}{L_z}, \pm \frac{4\pi}{L_z}, ...\)

Q: What is the DOS for a free electron? How is it different from the DOS calculation for Phonons?
A: The DOS for the free electron is calculated similarly by counting the number of states in a spherical shell of radius \(k\) and thickness \(\text{d}k\).
\(\nu(\varepsilon) = \frac{m_e^{3/2}}{\sqrt{2}\pi^2\hbar^3}\sqrt{\varepsilon} \;\text{d}\varepsilon\)
But an electron is characterized not only by its momentum but also by its intrinsic spin.
This leads to an additional factor of two.
\(\nu(\varepsilon) = \frac{\sqrt{2}m_e^{3/2}}{\pi^2\hbar^3}\sqrt{\varepsilon} \;\text{d}\varepsilon\)

List the differences of an ideal gas of phonons vs an ideal gas of electrons
phonons are  bosons
electrons are fermions
the number of phonons is dependent on  temperature alone = not externally controlled
the number of electrons is dependent on temperature, charge and potential = externally controlled

Q: Explain the chemical potential \(\mu\)
A: The chemical energy \(\mu\) is the energy required to move the electron from the box out to infinity.
So box is a potential well of \(-\mu\). This potential energy is added to the kinetic energy of the electron \(\varepsilon(\vec{k})\) for the calculation of Fermi-Dirac distribution

Q: Fermi-Dirac distribution
A: \(\langle n \rangle = f(\varepsilon) = \frac{1}{\exp(\frac{\varepsilon - \mu}{k_B T})+1}\)

Q: Consider Fermi-Dirac distribution at \(T \rightarrow 0\)
\(f(\varepsilon) = \frac{1}{\exp(\frac{\varepsilon - \mu}{k_B T})+1}\)
A: It approaches a step function
\(f(\varepsilon) \rightarrow \Theta(\mu -\varepsilon)\) (blue graph)

Q: How is the Fermi energy defined?
A: The value of the chemical potential \(\mu\) as \(T \rightarrow 0\)