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cepheid

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## Homework Statement

**Cyclotron resonance for a spheroidal energy surface.**Consider the energy surface

[tex] \epsilon(\mathbf{k}) = \hbar^2 \left( \frac{k_x^2 + k_y^2}{2m_t} + \frac{k_z^2}{2m_l} \right) [/tex]

where

*m*is the transverse mass parameter and

_{t}*m*is the longitudinal mass parameter. A surface on which [itex] \epsilon(\mathbf{k}) [/itex] is constant will be a spheroid. Use the equation of motion with [itex] \mathbf{v} = \hbar^{-1} \nabla_{\mathbf{k}} \epsilon [/itex] to show that [itex] \omega_c = eB/(m_l m_t)^{1/2}c [/itex] when the static magnetic field

_{l}**B**lies in the

*xy*plane.

## Homework Equations

__Dynamics of Bloch Electrons__

The equation of motion for an electron subject to the periodic potential of a crystal lattice is

[tex] \hbar \frac{d\mathbf{k}}{dt} = -\frac{e}{c}\mathbf{v} \times \mathbf{B} \ \ \ \textrm{cgs} [/tex]

[tex] \hbar \frac{d\mathbf{k}}{dt} = -e\mathbf{v} \times \mathbf{B} \ \ \ \textrm{SI} [/tex]

[tex] \hbar \frac{d\mathbf{k}}{dt} = -e\mathbf{v} \times \mathbf{B} \ \ \ \textrm{SI} [/tex]

## The Attempt at a Solution

[tex] \mathbf{v} = \hbar^{-1} \nabla_{\mathbf{k}} \epsilon(\mathbf{k}) = \hbar^{-1} \left( \hat{k}_x \frac{\partial}{\partial k_x} + \hat{k}_y \frac{\partial}{\partial k_y} + \hat{k}_z \frac{\partial}{\partial k_z} \right) \hbar^2 \left( \frac{k_x^2 + k_y^2}{2m_t} + \frac{k_z^2}{2m_l} \right) [/tex]

[tex] = \hbar \left( \hat{k}_x \frac{k_x}{m_t} + \hat{k}_y \frac{k_y}{m_t} + \hat{k}_z \frac{k_z}{m_l} \right) [/tex]

Apply the equation of motion with

[tex] \mathbf{B} = B_x \hat{x} + B_y \hat{y} [/tex]

[tex] \frac{d\mathbf{k}}{dt} = -\frac{e}{c}\left( \hat{k}_x \frac{k_x}{m_t} + \hat{k}_y \frac{k_y}{m_t} + \hat{k}_z \frac{k_z}{m_l} \right) \times (B_x \hat{x} + B_y \hat{y}) [/tex]

[tex] \frac{d\mathbf{k}}{dt} = -\frac{e}{c}\left( \hat{k}_x \frac{k_x}{m_t} + \hat{k}_y \frac{k_y}{m_t} + \hat{k}_z \frac{k_z}{m_l} \right) \times (B_x \hat{x} + B_y \hat{y}) [/tex]

Right so, um how am I supposed to proceed to compute such a cross product? :uhh:

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