Minor rewrites and corrections

1 files changed, 74 insertions, 52 deletions

diff --git a/content/know/concept/electric-dipole-approximation/index.pdc b/content/know/concept/electric-dipole-approximation/index.pdc
index 265babf..67c73ee 100644
--- a/content/know/concept/electric-dipole-approximation/index.pdc
+++ b/content/know/concept/electric-dipole-approximation/index.pdc
@@ -20,120 +20,142 @@ The general Hamiltonian of an electron in such a wave is given by:
 
 $$\begin{aligned}
     \hat{H}
-    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{2 m} (\vec{A} \cdot \vec{P} + \vec{P} \cdot \vec{A}) + \frac{q^2 \vec{A}{}^2}{2m} + V
+    &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \varphi
+    \\
+    &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \varphi
+\end{aligned}$$
+
+With charge $q = - e$,
+canonical momentum operator $\vu{P} = - i \hbar \nabla$,
+and magnetic vector potential $\vb{A}(\vb{x}, t)$.
+We reduce this by fixing the Coulomb gauge $\nabla \cdot \vb{A} = 0$,
+so that $\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$:
+
+$$\begin{aligned}
+    \comm*{\vb{A}}{\vu{P}} \psi
+    &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi)
+    \\
+    &= i \hbar (\nabla \cdot \vb{A}) \psi
+    = 0
 \end{aligned}$$
 
-With charge $q = - e$
-and electromagnetic vector potential $\vec{A}(\vec{r}, t)$.
-We reduce this by fixing the Coulomb gauge $\nabla \cdot \vec{A} = 0$,
-so that $\vec{A} \cdot \vec{P} = \vec{P} \cdot \vec{A}$,
-and assume that $\vec{A}{}^2$ is negligible:
+Where $\psi$ is an arbitrary test function.
+Assuming $\vb{A}$ is so small that $\vb{A}{}^2$ is negligible, we split $\hat{H}$ as follows,
+where $\hat{H}_1$ can be regarded as a perturbation to $\hat{H}_0$:
 
 $$\begin{aligned}
     \hat{H}
     = \hat{H}_0 + \hat{H}_1
     \qquad \quad
     \hat{H}_0
-    \equiv \frac{\vec{P}{}^2}{2 m} + V
+    \equiv \frac{\vu{P}{}^2}{2 m} + q \varphi
     \qquad \quad
     \hat{H}_1
-    \equiv - \frac{q}{m} \vec{P} \cdot \vec{A}
+    \equiv - \frac{q}{m} \vu{P} \cdot \vb{A}
 \end{aligned}$$
 
-We have split $\hat{H}$ into $\hat{H}_0$
-and a perturbation $\hat{H}_1$, since $\vec{A}$ is small.
-In an electromagnetic wave,
-$\vec{A}$ is oscillating sinusoidally in time and space as follows:
+In an electromagnetic wave, $\vb{A}$ is oscillating sinusoidally in time and space:
 
 $$\begin{aligned}
-    \vec{A}(\vec{r}, t) = - i \vec{A}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
+    \vb{A}(\vb{x}, t) = \vb{A}_0 \sin\!(\vb{k} \cdot \vb{x} - \omega t)
 \end{aligned}$$
 
-The corresponding perturbative
-[electric field](/know/concept/electric-field/) $\vec{E}$
-points in the same direction:
+Mathematically, it is more convenient to represent this with a complex exponential,
+whose real part should be taken at the end of the calculation:
 
 $$\begin{aligned}
-    \vec{E}(\vec{r}, t)
-    = - \pdv{\vec{A}}{t}
-    = \vec{E}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
+    \vb{A}(\vb{x}, t) = - i \vb{A}_0 \exp\!(i \vb{k} \cdot \vb{x} - i \omega t)
 \end{aligned}$$
 
-Where $\vec{E}_0 = \omega \vec{A}_0$.
+The corresponding perturbative [electric field](/know/concept/electric-field/) $\vb{E}$ is then given by:
+
+$$\begin{aligned}
+    \vb{E}(\vb{x}, t)
+    = - \pdv{\vb{A}}{t}
+    = \vb{E}_0 \exp\!(i \vb{k} \cdot \vb{x} - i \omega t)
+\end{aligned}$$
+
+Where $\vb{E}_0 = \omega \vb{A}_0$.
 Let us restrict ourselves to visible light,
-whose wavelength $2 \pi / k \approx 10^{-6} \:\mathrm{m}$.
-Meanwhile, an atomic orbital is on the order of $10^{-10} \:\mathrm{m}$,
-so $\vec{k} \cdot \vec{r}$ is negligible:
+whose wavelength $2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$.
+Meanwhile, an atomic orbital is several Bohr $\sim 10^{-10} \:\mathrm{m}$,
+so $\vb{k} \cdot \vb{x}$ is negligible:
 
 $$\begin{aligned}
     \boxed{
-        \vec{E}(\vec{r}, t)
-        \approx \vec{E}_0 \exp\!(- i \omega t)
+        \vb{E}(\vb{x}, t)
+        \approx \vb{E}_0 \exp\!(- i \omega t)
     }
 \end{aligned}$$
 
 This is the **electric dipole approximation**:
-we ignore all spatial variation of $\vec{E}$,
+we ignore all spatial variation of $\vb{E}$,
 and only consider its temporal oscillation.
 Also, since we have not used the word "photon",
 we are implicitly treating the radiation classically,
 and the electron quantum-mechanically.
 
-Next, we want to convert $\hat{H}_1$
-to use the electric field $\vec{E}$ instead of the potential $\vec{A}$.
-To do so, we rewrite the momemtum $\vec{P} = m \: \dv*{\vec{r}}{t}$
+Next, we want to rewrite $\hat{H}_1$
+to use the electric field $\vb{E}$ instead of the potential $\vb{A}$.
+To do so, we use that $\vu{P} = m \: \dv*{\vu{x}}{t}$
 and evaluate this in the [interaction picture](/know/concept/interaction-picture/):
 
 $$\begin{aligned}
-    \matrixel{2}{\dv*{\vec{r}}{t}}{1}
-    &= \frac{i}{\hbar} \matrixel{2}{[\hat{H}_0, \vec{r}]}{1}
-    = \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vec{r} - \vec{r} \hat{H}_0}{1}
-    \\
-    &= \frac{i}{\hbar} (E_2 - E_1) \matrixel{2}{\vec{r}}{1}
-    = i \omega_0 \matrixel{2}{\vec{r}}{1}
+    \vu{P}
+    = m \dv*{\vu{x}}{t}
+    = m \frac{i}{\hbar} \comm*{\hat{H}_0}{\vu{x}}
+    = m \frac{i}{\hbar} (\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0)
+\end{aligned}$$
+
+Taking the off-diagonal inner product with
+the two-level system's states $\ket{1}$ and $\ket{2}$ gives:
+
+$$\begin{aligned}
+    \matrixel{2}{\vu{P}}{1}
+    = m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1}
+    = m i \omega_0 \matrixel{2}{\vu{x}}{1}
 \end{aligned}$$
 
-Therefore, $\vec{P} / m = i \omega_0 \vec{r}$,
-where $\omega_0 \equiv (E_2 - E_1) / \hbar$ is the resonance frequency of the transition,
-close to which we assume that $\vec{A}$ and $\vec{E}$ are oscillating, i.e. $\omega \approx \omega_0$.
+Therefore, $\vu{P} / m = i \omega_0 \vu{x}$,
+where $\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$ is the resonance of the energy gap,
+close to which we assume that $\vb{A}$ and $\vb{E}$ are oscillating, i.e. $\omega \approx \omega_0$.
 We thus get:
 
 $$\begin{aligned}
     \hat{H}_1(t)
-    &= - \frac{q}{m} \vec{P} \cdot \vec{A}
-    = - (- i i) q \omega_0 \vec{r} \cdot \vec{A}_0 \exp\!(- i \omega t)
+    &= - \frac{q}{m} \vu{P} \cdot \vb{A}
+    = - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp\!(- i \omega t)
     \\
-    &\approx - q \vec{r} \cdot \vec{E}_0 \exp\!(- i \omega t)
-    = - \vec{d} \cdot \vec{E}_0 \exp\!(- i \omega t)
+    &\approx - q \vu{x} \cdot \vb{E}_0 \exp\!(- i \omega t)
+    = - \vu{d} \cdot \vb{E}_0 \exp\!(- i \omega t)
 \end{aligned}$$
 
-Where $\vec{d} \equiv q \vec{r} = - e \vec{r}$ is
+Where $\vu{d} \equiv q \vu{x} = - e \vu{x}$ is
 the **transition dipole moment operator** of the electron,
 hence the name **electric dipole approximation**.
-Finally, since electric fields are actually real
-(we let it be complex for mathematical convenience),
-we take the real part, yielding:
+Finally, we take the real part, yielding:
 
 $$\begin{aligned}
     \boxed{
         \hat{H}_1(t)
-        = - q \vec{r} \cdot \vec{E}_0 \cos\!(\omega t)
+        = - \vu{d} \cdot \vb{E}(t)
+        = - q \vu{x} \cdot \vb{E}_0 \cos\!(\omega t)
     }
 \end{aligned}$$
 
 If this approximation is too rough,
-$\vec{E}$ can always be Taylor-expanded in $(i \vec{k} \cdot \vec{r})$:
+$\vb{E}$ can always be Taylor-expanded in $(i \vb{k} \cdot \vb{x})$:
 
 $$\begin{aligned}
-    \vec{E}(\vec{r}, t)
-    = \vec{E}_0 \Big( 1 + (i \vec{k} \cdot \vec{r}) + \frac{1}{2} (i \vec{k} \cdot \vec{r})^2 + \: ... \Big) \exp\!(- i \omega t)
+    \vb{E}(\vb{x}, t)
+    = \vb{E}_0 \Big( 1 + (i \vb{k} \cdot \vb{x}) + \frac{1}{2} (i \vb{k} \cdot \vb{x})^2 + \: ... \Big) \exp\!(- i \omega t)
 \end{aligned}$$
 
 Taking the real part then yields the following series of higher-order correction terms:
 
 $$\begin{aligned}
-    \vec{E}(\vec{r}, t)
-    = \vec{E}_0 \Big( \cos\!(\omega t) + (\vec{k} \cdot \vec{r}) \sin\!(\omega t) - \frac{1}{2} (\vec{k} \cdot \vec{r})^2 \cos\!(\omega t) + \: ... \Big)
+    \vb{E}(\vb{x}, t)
+    = \vb{E}_0 \Big( \cos\!(\omega t) + (\vb{k} \cdot \vb{x}) \sin\!(\omega t) - \frac{1}{2} (\vb{k} \cdot \vb{x})^2 \cos\!(\omega t) + \: ... \Big)
 \end{aligned}$$