Saito Group (Open)

Started by Thomas.2013.09.02 This is Thomas's logbook for September of 2013.

1st week [Electron-Photon Interactions]

  • 09.02
    • The Hamiltonian for the electron-photon interaction is given by H = \frac{1}{2m} \left(\mathbf{p}-\frac{q}{c}\mathbf{A}\right)\left(\mathbf{p}-\frac{q}{c}\mathbf{A}\right) +q\phi, in this equation \mathbf{p} is the linear momentum, q is the charge of the particle, \mathbf{A} is the vector potential, m is the particle's mass, c is the speed of light and \phi is the scalar potential. It is possible to check this by using the canonical equations from Hamilton, \dot{\mathbf{r}}=\frac{\partial H}{\partial \mathbf{p}} and \dot{\mathbf{p}}=-\nabla H.
  • From those last equations together with the following identities  \frac{d \mathbf{A}}{d t} = \frac{\partial \mathbf{A}}{\partial t} + \left(\frac{d \mathbf{r}}{d t} \cdot \nabla \right)\mathbf{A}=\frac{\partial \mathbf{A}}{\partial t} + (\mathbf{v} \cdot \nabla)\mathbf{A} and (\mathbf{v} \cdot \nabla) \mathbf{A} = -\mathbf{v} \times (\nabla \times \mathbf{A}) + \nabla(\mathbf{v}\cdot \mathbf{A}) With all this considered it was possible to arrive at the equation of motion of a particle under an electromagnetic field, m \ddot{\mathbf{r}}=\frac{q}{c}(\mathbf{v} \times \mathbf{B}) + q \mathbf{E}.
  • Next steps:
  • Consider the wave functions from quantum well, hydrogen atom, and solve with the electromagnetic(electron-photon interaction) perturbation.
  • 09.03
    • By looking at the previous Hamiltonian in a different shape, \mathcal{H}=\left[\frac{p^2}{2m} + V(\mathbf{r})\right] -\frac{q}{cm} \mathbf{p} \cdot \mathbf{A} + \frac{q^2}{c^2} \frac{\mathbf{A}^2}{2m} where the term inside the brackets is actually the unperturbed hamiltonian, and the other terms are the perturbations that arise from the electomagnetic field influence.
  • As the electromagnetic field is just a perturbation, it is not so strong, therefore it is possible to disregard \mathbf{A}^2, and consider the perturbation Hamiltonian as \mathcal{H}_{\mbox{eR}}=-\frac{q}{cm} \mathbf{p} \cdot \mathbf{A}.
  • Taking a monochromatic wave over a quantum well of length R it is possible to find the new wavefunction after taking the \left< \phi_i |\mathcal{H}_{\mbox{eR}}|\phi_f \right>.
  • Next Steps:
  • Find the actual new wavefunctions.
  • 09.04
    • Following the condition for finding the perturbation result. The wave equations for the states n=1, initial (i), and n=2, final (f), are considered. Also for the perturbation, the Coulomb gauge will be chosen, \nabla \cdot \mathbf{A}=0, as a result, the perturbation Hamiltonian will be \mathcal{H}_{\mbox{eR}}=\frac{e \hbar}{m \omega} \mathbf{E(\mathbf{r},t)} \cdot \nabla .
  • So, \left< \phi_1 |\frac{e \hbar}{m \omega} \mathbf{E(\mathbf{r},t)} \cdot \nabla  |\phi_2 \right> with \phi_1=\frac{1}{\sqrt{R}} \cos{\frac{\pi x}{R}} and \phi_2=\frac{1}{\sqrt{R}} \cos{\frac{2\pi x}{R}} which would be -\frac{8 e \hbar \mathbf{E} }{3 R m \omega}.
  • Next Steps:
  • Evaluate other transitions, in order to get the linear combination which it is the new wavefunction.
  • 09.05
    • After evaluating the \left< \phi_2 |\frac{e \hbar}{m \omega} \mathbf{E(\mathbf{r},t)} \cdot \nabla  |\phi_1 \right> the result was -\frac{10 e \hbar \mathbf{E} }{3 R m \omega}. The \left< \phi_1 |\frac{e \hbar}{m \omega} \mathbf{E(\mathbf{r},t)} \cdot \nabla  |\phi_1 \right> and \left< \phi_2 |\frac{e \hbar}{m \omega} \mathbf{E(\mathbf{r},t)} \cdot \nabla  |\phi_2 \right> are null.

2nd week [Raman Spectroscopy Book, Chapter 5 Exercises]

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Last-modified: 2013-09-13 (Fri) 02:00:11 (2391d)