[[Saito Group (Open)]]

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


* 1st week [Electron-Photon Interactions] [#te145562]
--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.

--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.

--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.

--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] [#te145562]

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