Skip to content

Changes

Page history
Update Optical transitions: Move Hamiltonian to separate page and remove... authored by Florian Bayer's avatar Florian Bayer 
Update Optical transitions: Move Hamiltonian to separate page and remove equation environment (https://gitlab.com/gitlab-org/gitlab-foss/-/issues/61410).
Show whitespace changes
Inline Side-by-side
physics/Optical-transitions.md
View page @ bd63aaae
...@@ -15,12 +15,10 @@ There are two variants for this ansatz common in literature. ...@@ -15,12 +15,10 @@ There are two variants for this ansatz common in literature.
We derive the perturbation Hamiltonian by considering the EM wave as an additional potential for the atom core and the electron, such that $`H = H_0 +H_\text{EM}`$ with We derive the perturbation Hamiltonian by considering the EM wave as an additional potential for the atom core and the electron, such that $`H = H_0 +H_\text{EM}`$ with
```math ```math
\begin{equation}
\begin{split} \begin{split}
H_\text{EM} &= - e \phi(\vec{r}_e) + q \phi(\vec{r}_c) \\ H_\text{EM} &= - e \phi(\vec{r}_e) + q \phi(\vec{r}_c) \\
&= e\left[\phi(\vec{r}_c) - \phi(\vec{r}_e)\right] + (q-e)\phi(\vec{r}_c) &= e\left[\phi(\vec{r}_c) - \phi(\vec{r}_e)\right] + (q-e)\phi(\vec{r}_c)
\end{split} \end{split}
\end{equation}
``` ```
where $`\phi`$ is the electric potential of the EM wave. where $`\phi`$ is the electric potential of the EM wave.
...@@ -28,25 +26,21 @@ Since the last term only describes the atomic core, which is not of interesst in ...@@ -28,25 +26,21 @@ Since the last term only describes the atomic core, which is not of interesst in
For the first term we use a Taylor series up to first order in the electron-core distance $`\vec{r}`$ for the electric potential. As the EM wave potential varies on length scales much larger then the electron-core distance, we can neglect higher orders. This is known as the dipol approximation. Thus we can write For the first term we use a Taylor series up to first order in the electron-core distance $`\vec{r}`$ for the electric potential. As the EM wave potential varies on length scales much larger then the electron-core distance, we can neglect higher orders. This is known as the dipol approximation. Thus we can write
```math ```math
\begin{equation}
\begin{split} \begin{split}
H_\text{EM} &= - e \vec{r} \cdot \vec{\nabla}\phi(\vec{r}_c) \\ H_\text{EM} &= - e \vec{r} \cdot \vec{\nabla}\phi(\vec{r}_c) \\
&= - \vec{\mu}_e \cdot \vec{E}(\vec{r}_c) &= - \vec{\mu}_e \cdot \vec{E}(\vec{r}_c)
\end{split} \end{split}
\end{equation}
``` ```
where $`\vec{\mu}_e`$ is an effective electric dipole moment and $`\vec{E}(t) = \vec{E} \cos(\omega t)`$. where $`\vec{\mu}_e`$ is an effective electric dipole moment and $`\vec{E}(t) = \vec{E} \cos(\omega t)`$.
Since the commutator $`[H_0,H_\text{EM}]`$ does not vanish, both Hamiltonians do not have the same eigenstates and $`H_\text{EM}`$ will introduce transitions between our known eigenstates of $`H_0`$. The strength of such a transition is given by the transition matrix element Since the commutator $`[H_0,H_\text{EM}]`$ does not vanish, both Hamiltonians do not have the same eigenstates and $`H_\text{EM}`$ will introduce transitions between our known eigenstates of $`H_0`$. The strength of such a transition is given by the transition matrix element
```math ```math
\begin{equation}
\begin{split} \begin{split}
v_{fi} &= \langle \psi_f | H_\text{EM} | \psi_i \rangle \\ v_{fi} &= \langle \psi_f | H_\text{EM} | \psi_i \rangle \\
&= e \vec{E} \langle \psi_f | \vec{r} | \psi_i \rangle \\ &= e \vec{E} \langle \psi_f | \vec{r} | \psi_i \rangle \\
&= \frac{-i \hbar e \vec{E}}{(E_f - E_i)m} \langle \psi_f | \vec{p} | \psi_i \rangle &= \frac{-i \hbar e \vec{E}}{(E_f - E_i)m} \langle \psi_f | \vec{p} | \psi_i \rangle
\end{split} \end{split}
\end{equation}
``` ```
We can use the above commutator relation to [show](https://en.wikipedia.org/wiki/Transition_dipole_moment#In_terms_of_momentum) the last identity. Note that this identity is only true, as long as there is only the $`p^2/2m`$ term in the Hamiltonian. Upon inclusion of a vector potential, there will also be a term linear in $`\vec{A} \cdot \vec{p}`$. In our case this is negelectable, as this will in the end not contribute to transitions but just act as homogenoues energy offset for all states. We can use the above commutator relation to [show](https://en.wikipedia.org/wiki/Transition_dipole_moment#In_terms_of_momentum) the last identity. Note that this identity is only true, as long as there is only the $`p^2/2m`$ term in the Hamiltonian. Upon inclusion of a vector potential, there will also be a term linear in $`\vec{A} \cdot \vec{p}`$. In our case this is negelectable, as this will in the end not contribute to transitions but just act as homogenoues energy offset for all states.
...@@ -56,22 +50,18 @@ We can use the above commutator relation to [show](https://en.wikipedia.org/wiki ...@@ -56,22 +50,18 @@ We can use the above commutator relation to [show](https://en.wikipedia.org/wiki
Let's write our Hamiltonian as Let's write our Hamiltonian as
```math ```math
\begin{equation}
\begin{split} \begin{split}
H &= \frac{(\vec{p} +e \vec{A})^2}{2m_0} +V_c\\ H &= \frac{(\vec{p} +e \vec{A})^2}{2m_0} +V_c\\
&= \frac{(\vec{p} +e \vec{A}_0 +e \vec{A}_\text{EM})^2}{2m_0} +V_c\\ &= \frac{(\vec{p} +e \vec{A}_0 +e \vec{A}_\text{EM})^2}{2m_0} +V_c\\
=H_0 + H_\text{EM} &= \left[\frac{(\vec{p} +e \vec{A}_0)^2}{2m_0} +V_c\right] + \left[\frac{e \vec{A}_\text{EM} \cdot\vec{p}}{m_0}\right] =H_0 + H_\text{EM} &= \left[\frac{(\vec{p} +e \vec{A}_0)^2}{2m_0} +V_c\right] + \left[\frac{e \vec{A}_\text{EM} \cdot\vec{p}}{m_0}\right]
\end{split} \end{split}
\end{equation}
``` ```
where we have split the vector potential $`A`$ into the static part $`A_0`$ of the electric and magnetic fields applied to the `kdotpy` state calculation and $`\vec{A}_\text{EM} = \frac{\vec{E}}{\omega} \cos(\omega t - \vec{q}\cdot\vec{r})`$ for the EM wave. We used the identity $`\vec{p} \cdot \vec{A} = \vec{A} \cdot \vec{p}`$ valid for divergence free fields of EM waves and negelected the energy offset terms $`\vec{A}^2_\text{EM} + 2 \vec{A}_\text{0}\cdot\vec{A}_\text{EM}`$. The latter term is by the way the same term missing in the $`\vec{r} \leftrightarrow\vec{p}`$ relation above in [variant 1](#variant-1-atomistic-approach). where we have split the vector potential $`A`$ into the static part $`A_0`$ of the electric and magnetic fields applied to the `kdotpy` state calculation and $`\vec{A}_\text{EM} = \frac{\vec{E}}{\omega} \cos(\omega t - \vec{q}\cdot\vec{r})`$ for the EM wave. We used the identity $`\vec{p} \cdot \vec{A} = \vec{A} \cdot \vec{p}`$ valid for divergence free fields of EM waves and negelected the energy offset terms $`\vec{A}^2_\text{EM} + 2 \vec{A}_\text{0}\cdot\vec{A}_\text{EM}`$. The latter term is by the way the same term missing in the $`\vec{r} \leftrightarrow\vec{p}`$ relation above in [variant 1](#variant-1-atomistic-approach).
Using the dipole approximation ($`\lambda \gg r`$ or $`\vec{q} \to 0`$) and Euler's formula we can rewrite Using the dipole approximation ($`\lambda \gg r`$ or $`\vec{q} \to 0`$) and Euler's formula we can rewrite
```math ```math
\begin{equation}
\vec{A}_\text{EM} = \frac{\vec{E}}{2\omega} \left(e^{-i\omega t} + \text{c.c.}\right) \vec{A}_\text{EM} = \frac{\vec{E}}{2\omega} \left(e^{-i\omega t} + \text{c.c.}\right)
\end{equation}
``` ```
As in variant 1 we can derive the transition matrix element $`v_{fi}`$. As in variant 1 we can derive the transition matrix element $`v_{fi}`$.
The time varying phase factor and its complex conjugate can be identified as the two terms responsible for absorption and stimulated emission (see Fermi's Golden Rule). To fullfill energy conservation, we identify $`\hbar \omega = E_f - E_i`$. The time varying phase factor and its complex conjugate can be identified as the two terms responsible for absorption and stimulated emission (see Fermi's Golden Rule). To fullfill energy conservation, we identify $`\hbar \omega = E_f - E_i`$.
...@@ -81,46 +71,38 @@ The time varying phase factor and its complex conjugate can be identified as the ...@@ -81,46 +71,38 @@ The time varying phase factor and its complex conjugate can be identified as the
Both variants above yield the transition matrix element Both variants above yield the transition matrix element
```math ```math
\begin{equation}
\begin{split} \begin{split}
v_{fi} &= \langle \psi_f | H_\text{EM} | \psi_i \rangle \\ v_{fi} &= \langle \psi_f | H_\text{EM} | \psi_i \rangle \\
&= \frac{ e \vec{E}}{2m\omega} \langle \psi_f | \vec{p} | \psi_i \rangle &= \frac{ e \vec{E}}{2m\omega} \langle \psi_f | \vec{p} | \psi_i \rangle
\end{split} \end{split}
\end{equation}
``` ```
up to a global phase factor ($`-i`$), which can be neglected, as we will only deal with absolute squares $`|v_{fi}|^2`$ later on. up to a global phase factor ($`-i`$), which can be neglected, as we will only deal with absolute squares $`|v_{fi}|^2`$ later on.
Let's split the components Let's split the components
```math ```math
\begin{equation}
\begin{split} \begin{split}
\vec{E}\cdot\vec{p} &= E (\textbf{e}_x p_x +\textbf{e}_y p_y +\textbf{e}_z p_z) \\ \vec{E}\cdot\vec{p} &= E (\textbf{e}_x p_x +\textbf{e}_y p_y +\textbf{e}_z p_z) \\
&= \frac{E}{\sqrt{2}}(\textbf{e}_x p_x +\textbf{e}_y p_y) + E\textbf{e}_z p_z &= \frac{E}{\sqrt{2}}(\textbf{e}_x p_x +\textbf{e}_y p_y) + E\textbf{e}_z p_z
\end{split} \end{split}
\end{equation}
``` ```
with linear and circular polarization unit vectors $`\textbf{e}_i`$, $`\textbf{e}_\pm = \frac{1}{\sqrt{2}} (\textbf{e}_x \pm i\textbf{e}_y)`$ and $`p_\pm = p_x \pm ip_y`$. with linear and circular polarization unit vectors $`\textbf{e}_i`$, $`\textbf{e}_\pm = \frac{1}{\sqrt{2}} (\textbf{e}_x \pm i\textbf{e}_y)`$ and $`p_\pm = p_x \pm ip_y`$.
A typical experimental setup of magneto-optical investigations of layered sample structures is the Faraday configuration, where the EM wave is incident onto the sample parallel to a static magnetic field vector. These two vectors are usually oriented perpendicular to the sample surface, i.e. along the layer growth direction ($`z`$). The $`\textbf{e}_z`$ component of an EM wave traveling in $`z`$ direction is zero and we only need to consider the transition matrices A typical experimental setup of magneto-optical investigations of layered sample structures is the Faraday configuration, where the EM wave is incident onto the sample parallel to a static magnetic field vector. These two vectors are usually oriented perpendicular to the sample surface, i.e. along the layer growth direction ($`z`$). The $`\textbf{e}_z`$ component of an EM wave traveling in $`z`$ direction is zero and we only need to consider the transition matrices
```math ```math
\begin{equation}
\begin{split} \begin{split}
v_x \pm i v_y = v_\pm = \frac{ e E}{2\sqrt{2}m\omega} p_\pm v_x \pm i v_y = v_\pm = \frac{ e E}{2\sqrt{2}m\omega} p_\pm
\end{split} \end{split}
\end{equation}
``` ```
describing the interaction of the system with left and right circular polarized EM waves. We will see later on, that the circular basis is the most convenient polarization basis choice for our system. describing the interaction of the system with left and right circular polarized EM waves. We will see later on, that the circular basis is the most convenient polarization basis choice for our system.
We can derive the matrix elements $`v_{fi}`$ of the $`v`$ matrices by using the commutator relation $`[x,H_0] = \frac{i\hbar}{m} p_x `$ (and analogeously for other spatial coordinates) again. For any Hamiltonian the commutator acts like a derivation with respect to the momentum in the same spatial direction. This follows from the $`[x_i,p_j] \propto \delta_{ij}`$ commutator relations and the product rule for the commutator of operator products. We can derive the matrix elements $`v_{fi}`$ of the $`v`$ matrices by using the commutator relation $`[x,H_0] = \frac{i\hbar}{m} p_x `$ (and analogeously for other spatial coordinates) again. For any Hamiltonian the commutator acts like a derivation with respect to the momentum in the same spatial direction. This follows from the $`[x_i,p_j] \propto \delta_{ij}`$ commutator relations and the product rule for the commutator of operator products.
```math ```math
\begin{equation}
\begin{split} \begin{split}
[x,H_0] = \frac{\text{d}H_0}{\text{d}p_x} [x,H_0] = \frac{\text{d}H_0}{\text{d}p_x}
\end{split} \end{split}
\end{equation}
``` ```
This means that we can get $`v_{x,fi}`$ directly from the matrix reprentation of the $`k \cdot p`$ matrix $`H_{0,fi}`$ by taking the derivation to the respective $`k_x`$ momentum. Note that in the 2D periodic case only $`p_z`$ is an operator (central finite difference) in the Hamiltonian matrix, while $`p_{x,y}`$ just act on the plane wave parts $`e^{i(k_x x + k_y y)}`$ of the wave function and are replaced by $`k_{x,y}`$ momentum values. This means that we can get $`v_{x,fi}`$ directly from the matrix reprentation of the $`k \cdot p`$ matrix $`H_{0,fi}`$ by taking the derivation to the respective $`k_x`$ momentum. Note that in the 2D periodic case only $`p_z`$ is an operator (central finite difference) in the Hamiltonian matrix, while $`p_{x,y}`$ just act on the plane wave parts $`e^{i(k_x x + k_y y)}`$ of the wave function and are replaced by $`k_{x,y}`$ momentum values.
...@@ -128,48 +110,15 @@ This means that we can get $`v_{x,fi}`$ directly from the matrix reprentation of ...@@ -128,48 +110,15 @@ This means that we can get $`v_{x,fi}`$ directly from the matrix reprentation of
Therefore, we get Therefore, we get
```math ```math
\begin{equation}
\begin{split} \begin{split}
v_{x,fi} &= \frac{\text{d}H_{0,fi}}{\text{d}k_x}\\ v_{x,fi} &= \frac{\text{d}H_{0,fi}}{\text{d}k_x}\\
v_{\pm,fi} &= 2 \frac{\text{d}H_{0,fi}}{\text{d}k_\mp} = \frac{ e E}{\sqrt{2}m\omega_{fi}} \frac{\text{d}H_{0,fi}}{\text{d}k_\mp} v_{\pm,fi} &= 2 \frac{\text{d}H_{0,fi}}{\text{d}k_\mp} = \frac{ e E}{\sqrt{2}m\omega_{fi}} \frac{\text{d}H_{0,fi}}{\text{d}k_\mp}
\end{split} \end{split}
\end{equation}
``` ```
where we have used $`\frac{\text{d}}{\text{d}k_\pm} = \frac{\text{d} k_x}{\text{d}k_\pm} \frac{\text{d}}{\text{d}k_x} + \frac{\text{d} k_y}{\text{d}k_\pm} \frac{\text{d}}{\text{d}k_y} = \frac{1}{2}\left(\frac{\text{d}}{\text{d}k_x} \mp i\frac{\text{d}}{\text{d}k_y} \right)`$. where we have used $`\frac{\text{d}}{\text{d}k_\pm} = \frac{\text{d} k_x}{\text{d}k_\pm} \frac{\text{d}}{\text{d}k_x} + \frac{\text{d} k_y}{\text{d}k_\pm} \frac{\text{d}}{\text{d}k_y} = \frac{1}{2}\left(\frac{\text{d}}{\text{d}k_x} \mp i\frac{\text{d}}{\text{d}k_y} \right)`$.
As we already have a symbolic Hamiltonian defintion in `kdotpy`, we can use the method `deriv(to)` to construct suitable symbolic transition matrices which just need to be evaluated for certain Landau Levels (when using `kdotpy-ll`). As we already have a symbolic [Hamiltonian definition](physics/Kane-Hamiltonian) in `kdotpy`, we can use the method `deriv(to)` to construct suitable symbolic transition matrices which just need to be evaluated for certain Landau Levels (when using `kdotpy-ll`).
With the $`\Gamma`$-point orbital basis
```math
\begin{equation}
\begin{split}
|1\rangle &= |\Gamma_6, +\tfrac{1}{2}\rangle = |S, \uparrow\rangle\\
|2\rangle &= |\Gamma_6, -\tfrac{1}{2}\rangle = |S, \downarrow\rangle\\
|3\rangle &= |\Gamma_8, +\tfrac{3}{2}\rangle = \tfrac{1}{\sqrt{2}}|X +iY, \uparrow\rangle\\
|4\rangle &= |\Gamma_8, +\tfrac{1}{2}\rangle = \tfrac{1}{\sqrt{6}}\left[|X +iY, \downarrow\rangle - 2|Z, \uparrow\rangle\right]\\
|5\rangle &= |\Gamma_8, -\tfrac{1}{2}\rangle = -\tfrac{1}{\sqrt{6}}\left[|X -iY, \uparrow\rangle + 2|Z, \downarrow\rangle\right]\\
|6\rangle &= |\Gamma_8, -\tfrac{3}{2}\rangle = -\tfrac{1}{\sqrt{2}}|X -iY, \downarrow\rangle\\
|7\rangle &= |\Gamma_7, +\tfrac{1}{2}\rangle = \tfrac{1}{\sqrt{3}}\left[|X +iY, \downarrow\rangle + |Z, \uparrow\rangle\right]\\
|8\rangle &= |\Gamma_7, -\tfrac{1}{2}\rangle = \tfrac{1}{\sqrt{3}}\left[|X -iY, \uparrow\rangle - |Z, \downarrow\rangle\right]
\end{split}
\end{equation}
```
the $`8 \times 8`$ Kane Hamiltonian matrix is
```math
\begin{equation}
H =
\begin{pmatrix}
\textcolor{blue}{T} & 0 & \textcolor{green}{-\tfrac{1}{\sqrt{2}}P k_+} & & \textcolor{green}{\tfrac{1}{\sqrt{6}}P k_-} & 0 & -\tfrac{1}{\sqrt{3}}P k_z & \textcolor{green}{-\tfrac{1}{\sqrt{3}}P k_-} \\
0 & \textcolor{blue}{T} & 0 & \textcolor{green}{-\tfrac{1}{\sqrt{6}}P k_+} & \sqrt{\tfrac{2}{3}}P k_z & \textcolor{green}{\tfrac{1}{\sqrt{2}}P k_+} & \textcolor{green}{-\tfrac{1}{\sqrt{3}}P k_+} & \tfrac{1}{\sqrt{3}}P k_z \\
\textcolor{green}{-\tfrac{1}{\sqrt{2}}P k_-} & 0 & \textcolor{blue}{U+V} & \textcolor{green}{-S_-} & \textcolor{red}{R} & 0 & \textcolor{green}{\tfrac{1}{\sqrt{2}} S_-} & \textcolor{red}{-\sqrt{2} R} \\
\sqrt{\tfrac{2}{3}}P k_z & \textcolor{green}{-\tfrac{1}{\sqrt{6}}P k_-} & \textcolor{green}{-S^\dag_-} & \textcolor{blue}{U-V} & \textcolor{green}{C} & \textcolor{red}{R} & \textcolor{blue}{\sqrt{2} V} & -\sqrt{\tfrac{3}{2}} S_-
\end{pmatrix}
\end{equation}
```
ToDo: Unfinished Hamiltonian. For some reason this does not render correctly in some cases.
## Spectra ## Spectra
......