From dedb366c3a78f61c64f6be627ea091e71e009f7d Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 2 Oct 2021 15:40:20 +0200
Subject: Expand knowledge base
---
.../know/concept/einstein-coefficients/index.pdc | 16 +-
content/know/concept/fabry-perot-cavity/index.pdc | 6 +
.../know/concept/maxwell-bloch-equations/index.pdc | 427 +++++++++++++++++++++
content/know/concept/rabi-oscillation/index.pdc | 24 +-
.../know/concept/rutherford-scattering/index.pdc | 242 ++++++++++++
.../concept/rutherford-scattering/one-body.png | Bin 0 -> 66663 bytes
.../concept/rutherford-scattering/two-body.png | Bin 0 -> 40242 bytes
7 files changed, 699 insertions(+), 16 deletions(-)
create mode 100644 content/know/concept/maxwell-bloch-equations/index.pdc
create mode 100644 content/know/concept/rutherford-scattering/index.pdc
create mode 100644 content/know/concept/rutherford-scattering/one-body.png
create mode 100644 content/know/concept/rutherford-scattering/two-body.png
(limited to 'content/know/concept')
diff --git a/content/know/concept/einstein-coefficients/index.pdc b/content/know/concept/einstein-coefficients/index.pdc
index 80707c6..9feaf8c 100644
--- a/content/know/concept/einstein-coefficients/index.pdc
+++ b/content/know/concept/einstein-coefficients/index.pdc
@@ -170,19 +170,21 @@ $$\begin{aligned}
= \frac{\big|\!\matrixel{a}{H_1}{b}\!\big|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ba} - \omega) t / 2 \big)}{(\omega_{ba} - \omega)^2}
\end{aligned}$$
-If the location of the nucleus of the atom has $z = 0$,
+If the nucleus is at $z = 0$,
then generally $\ket{1}$ and $\ket{2}$ will be even or odd functions of $z$,
-such that $\matrixel{1}{z}{1} = \matrixel{2}{z}{2} = 0$, leading to:
+meaning that $\matrixel{1}{z}{1} = \matrixel{2}{z}{2} = 0$
+(see also [Laporte's selection rule](/know/concept/selection-rules/)),
+leading to:
$$\begin{gathered}
- \matrixel{1}{H_1}{2} = - E_0 d
+ \matrixel{1}{H_1}{2} = - E_0 d^*
\qquad
- \matrixel{2}{H_1}{1} = - E_0 d^*
+ \matrixel{2}{H_1}{1} = - E_0 d
\\
\matrixel{1}{H_1}{1} = \matrixel{2}{H_1}{2} = 0
\end{gathered}$$
-Where $d \equiv q \matrixel{1}{z}{2}$ is a constant,
+Where $d \equiv q \matrixel{2}{z}{1}$ is a constant,
namely the $z$-component of the **transition dipole moment**.
The chance of an upward jump (i.e. absorption) is:
@@ -284,12 +286,12 @@ Let us return to the matrix elements of the perturbation $\hat{H}_1$,
and define the polarization unit vector $\vec{n}$:
$$\begin{aligned}
- \matrixel{1}{\hat{H}_1}{2}
+ \matrixel{2}{\hat{H}_1}{1}
= - \vec{d} \cdot \vec{E}_0
= - E_0 (\vec{d} \cdot \vec{n})
\end{aligned}$$
-Where $\vec{d} \equiv q \matrixel{1}{\vec{r}}{2}$ is
+Where $\vec{d} \equiv q \matrixel{2}{\vec{r}}{1}$ is
the full **transition dipole moment** vector, which is usually complex.
The goal is to calculate the average of $|\vec{d} \cdot \vec{n}|^2$.
diff --git a/content/know/concept/fabry-perot-cavity/index.pdc b/content/know/concept/fabry-perot-cavity/index.pdc
index 2f1f84f..d40852f 100644
--- a/content/know/concept/fabry-perot-cavity/index.pdc
+++ b/content/know/concept/fabry-perot-cavity/index.pdc
@@ -120,6 +120,12 @@ $$\begin{aligned}
&= (1 - r_R) B_m \exp\!\big( i (\tilde{n}_C \!-\! \tilde{n}_R) \tilde{k}_m L/2 \big)
\end{aligned}$$
+Note that we have not demanded continuity of the electric field.
+This is because the mirrors are infinitely thin "magic" planes;
+if we had instead used a full physical mirror structure,
+then the we would have demanded continuity,
+as you might have expected.
+
## References
diff --git a/content/know/concept/maxwell-bloch-equations/index.pdc b/content/know/concept/maxwell-bloch-equations/index.pdc
new file mode 100644
index 0000000..3a0df1b
--- /dev/null
+++ b/content/know/concept/maxwell-bloch-equations/index.pdc
@@ -0,0 +1,427 @@
+---
+title: "Maxwell-Bloch equations"
+firstLetter: "M"
+publishDate: 2021-10-02
+categories:
+- Physics
+- Quantum mechanics
+- Electromagnetism
+
+date: 2021-09-09T21:17:52+02:00
+draft: false
+markup: pandoc
+---
+
+# Maxwell-Bloch equations
+
+For an electron in a two level system with time-independent states
+$\ket{g}$ (ground) and $\ket{e}$ (excited),
+consider the following general solution
+to the full Schrödinger equation:
+
+$$\begin{aligned}
+ \ket{\Psi}
+ &= c_g \: \ket{g} \exp\!(-i E_g t / \hbar) + c_e \: \ket{e} \exp\!(-i E_e t / \hbar)
+\end{aligned}$$
+
+Perturbing this system with
+an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
+introduces a time-dependent sinusoidal term $\hat{H}_1$ to the Hamiltonian.
+In the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
+$\hat{H}_1$ is given by:
+
+$$\begin{aligned}
+ \hat{H}_1(t)
+ = - \hat{\vb{p}} \cdot \vb{E}(t)
+ \qquad \quad
+ \hat{\vb{p}}
+ \equiv q \hat{\vb{x}}
+ \qquad \quad
+ \vb{E}(t)
+ = \vb{E}_0 \cos\!(\omega t)
+\end{aligned}$$
+
+Where $\vb{E}$ is an [electric field](/know/concept/electric-field/),
+and $\hat{\vb{p}}$ is the dipole moment operator.
+From [Rabi oscillation](/know/concept/rabi-oscillation/),
+we know that the time-varying coefficients $c_g$ and $c_e$
+can then be described by:
+
+$$\begin{aligned}
+ \dv{c_g}{t}
+ &= i \frac{q \matrixel{g}{\hat{\vb{x}}}{e} \cdot \vb{E}_0}{2 \hbar} \exp\!\big( i \omega t \!-\! i \omega_0 t \big) \: c_e
+ \\
+ \dv{c_e}{t}
+ &= i \frac{q \matrixel{e}{\hat{\vb{x}}}{g} \cdot \vb{E}_0}{2 \hbar} \exp\!\big(\!-\! i \omega t \!+\! i \omega_0 t \big) \: c_g
+\end{aligned}$$
+
+We want to rearrange these equations a bit.
+Therefore, we split the electric field $\vb{E}$ like so,
+where the amplitudes $\vb{E}_0^{-}$ and $\vb{E}_0^{+}$ may be slowly varying:
+
+$$\begin{aligned}
+ \vb{E}(t)
+ = \vb{E}^{-}(t) + \vb{E}^{+}(t)
+ = \vb{E}_0^{-} \exp\!(i \omega t) + \vb{E}_0^{+} \exp\!(-i \omega t)
+\end{aligned}$$
+
+Since $\vb{E}$ is real, $\vb{E}_0^{+} = (\vb{E}_0^{-})^*$.
+Similarly, we define the transition dipole moment $\vb{p}_0^{-}$:
+
+$$\begin{aligned}
+ \vb{p}_0^{-}
+ \equiv q \matrixel{e}{\vb{x}}{g}
+ \qquad \quad
+ \vb{p}_0^{+}
+ \equiv (\vb{p}_0^{-})^*
+ = q \matrixel{g}{\vb{x}}{e}
+\end{aligned}$$
+
+With these, the equations for $c_g$ and $c_e$ can be rewritten as shown below.
+Note that $\vb{E}^{-}$ and $\vb{E}^{+}$ include the driving plane wave,
+and the *rotating wave approximation* is still made:
+
+$$\begin{aligned}
+ \dv{c_g}{t}
+ &= \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \exp\!(- i \omega_0 t) \: c_e
+ \\
+ \dv{c_e}{t}
+ &= \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \exp\!(i \omega_0 t) \: c_g
+\end{aligned}$$
+
+
+## Optical Bloch equations
+
+For $\ket{\Psi}$ as defined above,
+the corresponding pure [density operator](/know/concept/density-operator/)
+$\hat{\rho}$ is as follows:
+
+$$\begin{aligned}
+ \hat{\rho}
+ = \ket{\Psi} \bra{\Psi}
+ =
+ \begin{bmatrix}
+ c_e c_e^* & c_e c_g^* \exp\!(-i \omega_0 t) \\
+ c_g c_e^* \exp\!(i \omega_0 t) & c_g c_g^*
+ \end{bmatrix}
+ \equiv
+ \begin{bmatrix}
+ \rho_{ee} & \rho_{eg} \\
+ \rho_{ge} & \rho_{gg}
+ \end{bmatrix}
+\end{aligned}$$
+
+Where $\omega_0 \equiv (E_e \!-\! E_g) / \hbar$ is the resonance frequency.
+We take the $t$-derivative of the matrix elements,
+and insert the equations for $c_g$ and $c_e$:
+
+$$\begin{aligned}
+ \dv{\rho_{gg}}{t}
+ &= \dv{c_g}{t} c_g^* + c_g \dv{c_g^*}{t}
+ \\
+ &= \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \exp\!(- i \omega_0 t) \: c_e c_g^*
+ - \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \exp\!(i \omega_0 t) \: c_g c_e^*
+ \\
+ \dv{\rho_{ee}}{t}
+ &= \dv{c_e}{t} c_e^* + c_e \dv{c_e^*}{t}
+ \\
+ &= \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \exp\!(i \omega_0 t) \: c_g c_e^*
+ - \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \exp\!(- i \omega_0 t) \: c_e c_g^*
+ \\
+ \dv{\rho_{ge}}{t}
+ &= \dv{c_g}{t} c_e^* \exp\!(i \omega_0 t) + c_g \dv{c_e^*}{t} \exp\!(i \omega_0 t) + i \omega_0 c_g c_e^* \exp\!(i \omega_0 t)
+ \\
+ &= \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \: c_e c_e^*
+ - \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \: c_g c_g^*
+ + i \omega_0 c_g c_e^* \exp\!(i \omega_0 t)
+ \\
+ \dv{\rho_{eg}}{t}
+ &= \dv{c_e}{t} c_g^* \exp\!(-i \omega_0 t) + c_e \dv{c_g^*}{t} \exp\!(-i \omega_0 t) - i \omega_0 c_e c_g^* \exp\!(- i \omega_0 t)
+ \\
+ &= \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \: c_g c_g^*
+ - \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \: c_e c_e^*
+ - i \omega_0 c_e c_g^* \: \exp\!(- i \omega_0 t)
+\end{aligned}$$
+
+Recognizing the density matrix elements allows us
+to reduce these equations to:
+
+$$\begin{aligned}
+ \dv{\rho_{gg}}{t}
+ &= \frac{i}{\hbar} \Big( \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} - \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} \Big)
+ \\
+ \dv{\rho_{ee}}{t}
+ &= \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} - \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} \Big)
+ \\
+ \dv{\rho_{ge}}{t}
+ &= i \omega_0 \rho_{ge} + \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \big( \rho_{ee} - \rho_{gg} \big)
+ \\
+ \dv{\rho_{eg}}{t}
+ &= - i \omega_0 \rho_{eg} + \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \big( \rho_{gg} - \rho_{ee} \big)
+\end{aligned}$$
+
+These equations are correct if nothing else is affecting $\hat{\rho}$.
+But in practice, these quantities decay due to various processes,
+e.g. spontaneous emission (see [Einstein coefficients](/know/concept/einstein-coefficients/)).
+
+Let $\rho_{ee}$ decays with rate $\gamma_e$.
+Since the total probability $\rho_{ee} + \rho_{gg} = 1$,
+we thus have:
+
+$$\begin{aligned}
+ \Big( \dv{\rho_{ee}}{t} \Big)_{e}
+ = - \gamma_e \rho_{ee}
+ \quad \implies \quad
+ \Big( \dv{\rho_{gg}}{t} \Big)_{e}
+ = \gamma_e \rho_{ee}
+\end{aligned}$$
+
+Meanwhile, for whatever reason,
+let $\rho_{gg}$ decay into $\rho_{ee}$ with rate $\gamma_g$:
+
+$$\begin{aligned}
+ \Big( \dv{\rho_{gg}}{t} \Big)_{g}
+ = - \gamma_g \rho_{gg}
+ \quad \implies \quad
+ \Big( \dv{\rho_{gg}}{t} \Big)_{g}
+ = \gamma_g \rho_{gg}
+\end{aligned}$$
+
+And finally, let the diagonal (perpendicular) matrix elements
+both decay with rate $\gamma_\perp$:
+
+$$\begin{aligned}
+ \Big( \dv{\rho_{eg}}{t} \Big)_{\perp}
+ = - \gamma_\perp \rho_{eg}
+ \qquad \quad
+ \Big( \dv{\rho_{ge}}{t} \Big)_{\perp}
+ = - \gamma_\perp \rho_{ge}
+\end{aligned}$$
+
+Putting everything together,
+we arrive at the **optical Bloch equations** governing $\hat{\rho}$:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ \dv{\rho_{gg}}{t}
+ &= \gamma_e \rho_{ee} - \gamma_g \rho_{gg}
+ + \frac{i}{\hbar} \Big( \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} - \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} \Big)
+ \\
+ \dv{\rho_{ee}}{t}
+ &= \gamma_g \rho_{gg} - \gamma_e \rho_{ee}
+ + \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} - \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} \Big)
+ \\
+ \dv{\rho_{ge}}{t}
+ &= - \Big( \gamma_\perp - i \omega_0 \Big) \rho_{ge}
+ + \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \Big( \rho_{ee} - \rho_{gg} \Big)
+ \\
+ \dv{\rho_{eg}}{t}
+ &= - \Big( \gamma_\perp + i \omega_0 \Big) \rho_{eg}
+ + \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \Big( \rho_{gg} - \rho_{ee} \Big)
+ \end{aligned}
+ }
+\end{aligned}$$
+
+Many authors simplify these equations a bit by choosing
+$\gamma_g = 0$ and $\gamma_\perp = \gamma_e / 2$.
+
+
+## Including Maxwell's equations
+
+This two-level system has a dipole moment $\vb{p}$ as follows,
+where we use [Laporte's selection rule](/know/concept/selection-rules/)
+to remove diagonal terms, by assuming that
+the electron's orbitals are odd or even:
+
+$$\begin{aligned}
+ \vb{p}
+ &= \matrixel{\Psi}{\hat{\vb{p}}}{\Psi}
+ \\
+ &= q \Big( c_g c_g^* \matrixel{g}{\hat{\vb{x}}}{g} + c_e c_e^* \matrixel{e}{\hat{\vb{x}}}{e}
+ + c_g c_e^* \matrixel{e}{\hat{\vb{x}}}{g} \exp\!(i \omega_0 t) + c_e c_g^* \matrixel{g}{\hat{\vb{x}}}{e} \exp\!(-i \omega_0 t) \Big)
+ \\
+ &= q \Big( \rho_{ge} \matrixel{e}{\hat{\vb{x}}}{g} + \rho_{eg} \matrixel{g}{\hat{\vb{x}}}{e} \Big)
+ = \vb{p}_0^{-} \rho_{ge}(t) + \vb{p}_0^{+} \rho_{eg}(t)
+ \equiv \vb{p}^{-}(t) + \vb{p}^{+}(t)
+\end{aligned}$$
+
+Where we have split $\vb{p}$ analogously to $\vb{E}$
+by defining $\vb{p}^{+} \equiv \vb{p}_0^{+} \rho_{eg}$.
+Its equation of motion can then be found from the optical Bloch equations:
+
+$$\begin{aligned}
+ \dv{\vb{p}^{+}}{t}
+ = \vb{p}_0^{+} \dv{\rho_{eg}}{t}
+ = - \vb{p}_0^{+} \Big( \gamma_\perp + i \omega_0 \Big) \rho_{eg}
+ + \frac{i}{\hbar} \vb{p}_0^{+} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \Big( \rho_{gg} - \rho_{ee} \Big)
+\end{aligned}$$
+
+Some authors do not bother multiplying $\rho_{ge}$ by $\vb{p}_0^{+}$.
+In any case, we arrive at:
+
+$$\begin{aligned}
+ \boxed{
+ \dv{\vb{p}^{+}}{t}
+ = - \Big( \gamma_\perp + i \omega_0 \Big) \vb{p}^{+}
+ - \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \vb{p}_0^{+} d
+ }
+\end{aligned}$$
+
+Where we have defined the **population inversion** $d \in [-1, 1]$ as follows,
+which quantifies the electron's excitedness:
+
+$$\begin{aligned}
+ d
+ \equiv \rho_{ee} - \rho_{gg}
+\end{aligned}$$
+
+From the optical Bloch equations,
+we find its equation of motion to be:
+
+$$\begin{aligned}
+ \dv{d}{t}
+ &= \dv{\rho_{ee}}{t} - \dv{\rho_{gg}}{t}
+ = 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ + \frac{i 2}{\hbar} \Big( \vb{p}^{-} \cdot \vb{E}^{+} - \vb{p}^{+} \cdot \vb{E}^{-} \Big)
+\end{aligned}$$
+
+We can rewrite the first two terms in the following intuitive form,
+which describes a decay with
+rate $\gamma_\parallel \equiv \gamma_g + \gamma_e$
+towards an equilbrium $d_0$:
+
+$$\begin{aligned}
+ 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ = \gamma_\parallel (d_0 - d)
+ \qquad \quad
+ d_0
+ \equiv \frac{\gamma_g - \gamma_e}{\gamma_g + \gamma_e}
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-inversion-decay"/>
+<label for="proof-inversion-decay">Proof</label>
+<div class="hidden">
+<label for="proof-inversion-decay">Proof.</label>
+We introduce some new terms, and reorganize the expression:
+
+$$\begin{aligned}
+ 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ &= 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ + \gamma_g \rho_{ee} - \gamma_g \rho_{ee}
+ + \gamma_e \rho_{gg} - \gamma_e \rho_{gg}
+ \\
+ &= \gamma_g (\rho_{gg} + \rho_{ee}) - \gamma_e (\rho_{gg} + \rho_{ee})
+ + \gamma_g (\rho_{gg} - \rho_{ee}) + \gamma_e (\rho_{gg} - \rho_{ee})
+\end{aligned}$$
+
+Since the total probability $\rho_{gg} + \rho_{ee} = 1$,
+and $d \equiv \rho_{ee} - \rho_{gg}$, this reduces to:
+
+$$\begin{aligned}
+ 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ &= \gamma_g - \gamma_e - (\gamma_g + \gamma_e) d
+ \\
+ &= (\gamma_g + \gamma_e) \Big( \frac{\gamma_g - \gamma_e}{\gamma_g + \gamma_e} - d \Big)
+ \\
+ &= \gamma_\parallel ( d_0 - d )
+\end{aligned}$$
+</div>
+</div>
+
+With this, the equation for the population inversion $d$
+takes the following final form:
+
+$$\begin{aligned}
+ \boxed{
+ \dv{d}{t}
+ = \gamma_\parallel (d_0 - d) + \frac{i 2}{\hbar} \Big( \vb{p}^{-} \cdot \vb{E}^{+} - \vb{p}^{+} \cdot \vb{E}^{-} \Big)
+ }
+\end{aligned}$$
+
+Finally, we would like a relation between the polarization
+and the electric field $\vb{E}$,
+for which we turn to [Maxwell's equations](/know/concept/maxwells-equations/).
+We start from Faraday's law,
+and split $\vb{B} = \mu_0 (\vb{H} + \vb{M})$:
+
+$$\begin{aligned}
+ \nabla \cross \vb{E}
+ = - \pdv{\vb{B}}{t}
+ = - \mu_0 \pdv{\vb{H}}{t} - \mu_0 \pdv{\vb{M}}{t}
+\end{aligned}$$
+
+We assume that there is no magnetization $\vb{M} = 0$.
+Then we we take the curl of both sides,
+and replace $\nabla \cross \vb{H}$ with Ampère's circuital law:
+
+$$\begin{aligned}
+ \nabla \cross \big( \nabla \cross \vb{E} \big)
+ = - \mu_0 \pdv{}{t} \big( \nabla \cross \vb{H} \big)
+ = - \mu_0 \pdv{}{t} \Big( \vb{J}_\mathrm{free} + \pdv{\vb{D}}{t} \Big)
+\end{aligned}$$
+
+Inserting the definition $\vb{D} = \varepsilon_0 \vb{E} + \vb{P}$
+together with Ohm's law $\vb{J}_\mathrm{free} = \sigma \vb{E}$ yields:
+
+$$\begin{aligned}
+ \nabla \cross \big( \nabla \cross \vb{E} \big)
+ = - \mu_0 \sigma \pdv{\vb{E}}{t} - \mu_0 \varepsilon_0 \pdv[2]{\vb{E}}{t} - \mu_0 \pdv[2]{\vb{P}}{t}
+\end{aligned}$$
+
+Where $\sigma$ is the medium's conductivity, if any;
+many authors assume $\sigma = 0$.
+Next, we rewrite the left side using a vector identity,
+and assume no net charge $\nabla \cdot \vb{E} = 0$:
+
+$$\begin{aligned}
+ \nabla^2 \vb{E} - \nabla \big( \nabla \cdot \vb{E} \big)
+ = \nabla^2 \vb{E}
+ = \mu_0 \sigma \pdv{\vb{E}}{t} + \mu_0 \varepsilon_0 \pdv[2]{\vb{E}}{t} + \mu_0 \pdv[2]{\vb{P}}{t}
+\end{aligned}$$
+
+After some rearranging,
+we arrive at a variant of the electromagnetic wave equation:
+
+$$\begin{aligned}
+ \nabla^2 \vb{E} - \mu_0 \sigma \pdv{\vb{E}}{t} - \mu_0 \varepsilon_0 \pdv[2]{\vb{E}}{t}
+ &= \mu_0 \pdv[2]{\vb{P}}{t}
+\end{aligned}$$
+
+It is trivial to show that $\vb{E}$ and $\vb{P}$
+can be replaced by $\vb{E}^{+}$ and $\vb{P}^{+}$.
+It is equally trivial to convert
+the dipole $\vb{p}^{+}$ and inversion $d$
+into their macroscopic versions $\vb{P}^{+}$ and $D$,
+simply by summing over all atoms in the medium.
+We thus arrive at the **Maxwell-Bloch equations**:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ \mu_0 \pdv[2]{\vb{P}^{+}}{t}
+ &= \nabla^2 \vb{E}^{+} - \mu_0 \sigma \pdv{\vb{E}^{+}}{t} - \mu_0 \varepsilon_0 \pdv[2]{\vb{E}^{+}}{t}
+ \\
+ \pdv{\vb{P}^{+}}{t}
+ &= - \Big( \gamma_\perp + i \omega_0 \Big) \vb{P}^{+}
+ - \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \vb{p}_0^{+} D
+ \\
+ \pdv{D}{t}
+ &= \gamma_\parallel (D_0 - D) + \frac{i 2}{\hbar} \Big( \vb{P}^{-} \cdot \vb{E}^{+} - \vb{P}^{+} \cdot \vb{E}^{-} \Big)
+ \end{aligned}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. F. Kärtner,
+ [Ultrafast optics: lecture notes](https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-977-ultrafast-optics-spring-2005/lecture-notes/),
+ 2005, MIT.
+2. H. Haken,
+ *Light: volume 2: laser light dynamics*,
+ 1985, North-Holland.
+3. H.J. Metcalf, P. van der Straten,
+ *Laser cooling and trapping*,
+ 1999, Springer.
diff --git a/content/know/concept/rabi-oscillation/index.pdc b/content/know/concept/rabi-oscillation/index.pdc
index cf393a4..a488de0 100644
--- a/content/know/concept/rabi-oscillation/index.pdc
+++ b/content/know/concept/rabi-oscillation/index.pdc
@@ -87,11 +87,15 @@ while $\exp\!(i (\omega \!-\! \omega_0) t)$ does not.
Dropping the respective terms thus leaves us with:
$$\begin{aligned}
- \dv{c_a}{t}
- = - i \frac{V_{ab}}{2 \hbar} \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b
- \qquad \quad
- \dv{c_b}{t}
- = - i \frac{V_{ba}}{2 \hbar} \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t \big) \: c_a
+ \boxed{
+ \begin{aligned}
+ \dv{c_a}{t}
+ &= - i \frac{V_{ab}}{2 \hbar} \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b
+ \\
+ \dv{c_b}{t}
+ &= - i \frac{V_{ba}}{2 \hbar} \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t \big) \: c_a
+ \end{aligned}
+ }
\end{aligned}$$
Now we can solve this system of coupled equations exactly.
@@ -186,7 +190,7 @@ the special case of exact resonance $\omega = \omega_0$:
$$\begin{aligned}
\Omega
- \equiv \frac{V_{ab}}{\hbar}
+ \equiv \frac{V_{ba}}{\hbar}
\end{aligned}$$
As an example, Rabi oscillation arises
@@ -195,7 +199,7 @@ where $\hat{H}_1$ is:
$$\begin{aligned}
\hat{H}_1(t)
- = - q \vec{r} \cdot \vec{E}_0 \cos(\omega t)
+ = - q \vec{r} \cdot \vec{E}_0 \cos\!(\omega t)
\end{aligned}$$
After making the rotating wave approximation,
@@ -203,12 +207,14 @@ the resulting Rabi frequency is given by:
$$\begin{aligned}
\Omega
- = \frac{\vec{d} \cdot \vec{E}_0}{\hbar}
+ = - \frac{\vec{d} \cdot \vec{E}_0}{\hbar}
\end{aligned}$$
Where $\vec{E}_0$ is the [electric field](/know/concept/electric-field/) amplitude,
-and $\vec{d} \equiv q \matrixel{a}{\vec{r}}{b}$ is the transition dipole moment
+and $\vec{d} \equiv q \matrixel{b}{\vec{r}}{a}$ is the transition dipole moment
of the electron between orbitals $\ket{a}$ and $\ket{b}$.
+Apparently, some authors define $\vec{d}$ with the opposite sign,
+thereby departing from its classical interpretation.
diff --git a/content/know/concept/rutherford-scattering/index.pdc b/content/know/concept/rutherford-scattering/index.pdc
new file mode 100644
index 0000000..481e4d1
--- /dev/null
+++ b/content/know/concept/rutherford-scattering/index.pdc
@@ -0,0 +1,242 @@
+---
+title: "Rutherford scattering"
+firstLetter: "R"
+publishDate: 2021-10-02
+categories:
+- Physics
+
+date: 2021-09-23T16:22:07+02:00
+draft: false
+markup: pandoc
+---
+
+# Rutherford scattering
+
+**Rutherford scattering** or **Coulomb scattering**
+is an elastic pseudo-collision of two electrically charged particles.
+It is not a true collision, and is caused by Coulomb repulsion.
+
+The general idea is illustrated below.
+Consider two particles 1 and 2, with the same charge sign.
+Let 2 be initially at rest, and 1 approach it with velocity $\vb{v}_1$.
+Coulomb repulsion causes 1 to deflect by an angle $\theta$,
+and pushes 2 away in the process:
+
+<a href="two-body.png">
+<img src="two-body.png" style="width:50%;display:block;margin:auto;">
+</a>
+
+Here, $b$ is called the **impact parameter**.
+Intuitively, we expect $\theta$ to be larger for smaller $b$.
+
+By combining Coulomb's law with Newton's laws,
+these particles' equations of motion are found to be as follows,
+where $r = |\vb{r}_1 - \vb{r}_2|$ is the distance between 1 and 2:
+
+$$\begin{aligned}
+ m_1 \dv{\vb{v}_1}{t}
+ = \vb{F}_1
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}_1 - \vb{r}_2}{r^3}
+ \qquad \quad
+ m_2 \dv{\vb{v}_2}{t}
+ = \vb{F}_2
+ = - \vb{F}_1
+\end{aligned}$$
+
+Using the [reduced mass](/know/concept/reduced-mass/)
+$\mu \equiv m_1 m_2 / (m_1 \!+\! m_2)$,
+we turn this into a one-body problem:
+
+$$\begin{aligned}
+ \mu \dv{\vb{v}}{t}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3}
+\end{aligned}$$
+
+Where $\vb{v} \equiv \vb{v}_1 \!-\! \vb{v}_2$ is the relative velocity,
+and $\vb{r} \equiv \vb{r}_1 \!-\! \vb{r}_2$ is the relative position.
+The latter is as follows in
+[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/)
+$(r, \varphi, z)$:
+
+$$\begin{aligned}
+ \vb{r}
+ = r \cos{\varphi} \:\vu{e}_x + r \sin{\varphi} \:\vu{e}_y + z \:\vu{e}_z
+ = r \:\vu{e}_r + z \:\vu{e}_z
+\end{aligned}$$
+
+These new coordinates are sketched below,
+where the origin represents $\vb{r}_1 = \vb{r}_2$.
+Crucially, note the symmetry:
+if the "collision" occurs at $t = 0$,
+then by comparing $t > 0$ and $t < 0$
+we can see that $v_x$ is unchanged for any given $\pm t$,
+while $v_y$ simply changes sign:
+
+<a href="one-body.png">
+<img src="one-body.png" style="width:60%;display:block;margin:auto;">
+</a>
+
+From our expression for $\vb{r}$,
+we can find $\vb{v}$ by differentiating with respect to time:
+
+$$\begin{aligned}
+ \vb{v}
+ &= \big( r' \cos{\varphi} - r \varphi' \sin{\varphi} \big) \:\vu{e}_x
+ + \big( r' \sin{\varphi} + r \varphi' \cos{\varphi} \big) \:\vu{e}_y + z' \:\vu{e}_z
+ \\
+ &= r' \: \big( \cos{\varphi} \:\vu{e}_x + \sin{\varphi} \:\vu{e}_y \big)
+ + r \varphi' \: \big( \!-\! \sin{\varphi} \:\vu{e}_x + \cos{\varphi} \:\vu{e}_y \big) + z' \:\vu{e}_z
+ \\
+ &= r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi + z' \:\vu{e}_z
+\end{aligned}$$
+
+Where we have recognized the basis vectors $\vu{e}_r$ and $\vu{e}_\varphi$.
+If we choose the coordinate system such that all dynamics are in the $(x,y)$-plane,
+i.e. $z(t) = 0$, we have:
+
+$$\begin{aligned}
+ \vb{r}
+ = r \: \vu{e}_r
+ \qquad \qquad
+ \vb{v}
+ = r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi
+\end{aligned}$$
+
+Consequently, the angular momentum $\vb{L}$ is as follows,
+pointing purely in the $z$-direction:
+
+$$\begin{aligned}
+ \vb{L}(t)
+ = \mu \vb{r} \cross \vb{v}
+ = \mu \big( r \vu{e}_r \cross r \varphi' \vu{e}_\varphi \big)
+ = \mu r^2 \varphi' \:\vu{e}_z
+\end{aligned}$$
+
+Now, from the figure above,
+we can argue geometrically that at infinity $t = \pm \infty$,
+the ratio $b/r$ is related to the angle $\chi$ between $\vb{v}$ and $\vb{r}$ like so:
+
+$$\begin{aligned}
+ \frac{b}{r(\pm \infty)}
+ = \sin{\chi(\pm \infty)}
+ \qquad \quad
+ \chi(t)
+ \equiv \measuredangle(\vb{r}, \vb{v})
+\end{aligned}$$
+
+With this, we can rewrite
+the magnitude of the angular momentum $\vb{L}$ as follows,
+where the total velocity $|\vb{v}|$ is a constant,
+thanks to conservation of energy:
+
+$$\begin{aligned}
+ \big| \vb{L}(\pm \infty) \big|
+ = \mu \big| \vb{r} \cross \vb{v} \big|
+ = \mu r |\vb{v}| \sin{\chi}
+ = \mu b |\vb{v}|
+\end{aligned}$$
+
+However, conveniently,
+angular momentum is also conserved, i.e. $\vb{L}$ is constant in time:
+
+$$\begin{aligned}
+ \vb{L}'(t)
+ &= \mu \big( \vb{r} \cross \vb{v}' + \vb{v} \cross \vb{v} \big)
+ = \vb{r} \cross (\mu \vb{v}')
+ = \vb{r} \cross \Big( \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3} \Big)
+ = 0
+\end{aligned}$$
+
+Where we have replaced $\mu \vb{v}'$ with the equation of motion.
+Thanks to this, we can equate the two preceding expressions for $\vb{L}$,
+leading to the relation below.
+Note the appearance of a new minus,
+because the sketch shows that $\varphi' < 0$,
+i.e. $\varphi$ decreases with increasing $t$:
+
+$$\begin{aligned}
+ - \mu r^2 \dv{\varphi}{t}
+ = \mu b |\vb{v}|
+ \quad \implies \quad
+ \dd{t}
+ = - \frac{r^2}{b |\vb{v}|} \dd{\varphi}
+\end{aligned}$$
+
+Now, at last, we turn to the main equation of motion.
+Its $y$-component is given by:
+
+$$\begin{aligned}
+ \mu \dv{v_y}{t}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{y}{r^3}
+ \quad \implies \quad
+ \mu \dd{v_y}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{y}{r^3} \dd{t}
+\end{aligned}$$
+
+We replace $\dd{t}$ with our earlier relation,
+and recognize geometrically that $y/r = \sin{\varphi}$:
+
+$$\begin{aligned}
+ \mu \dd{v_y}
+ = - \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \frac{y}{r} \dd{\varphi}
+ = - \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \sin{\varphi} \dd{\varphi}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \dd{(\cos{\varphi})}
+\end{aligned}$$
+
+Integrating this from the initial state $i$ at $t = -\infty$
+to the final state $f$ at $t = \infty$ yields:
+
+$$\begin{aligned}
+ \Delta v_y
+ \equiv \int_{i}^{f} \dd{v_y}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos{\varphi_i} \big)
+\end{aligned}$$
+
+From symmetry, we see that $\varphi_i = \pi \!-\! \varphi_f$,
+and that $\Delta v_y = v_{y,f} \!-\! v_{y,i} = 2 v_{y,f}$, such that:
+
+$$\begin{aligned}
+ 2 v_{y,f}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos\!(\pi \!-\! \varphi_f) \big)
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( 2 \cos{\varphi_f} \big)
+\end{aligned}$$
+
+Furthermore, geometrically, at $t = \infty$
+we notice that $v_{y,f} = |\vb{v}| \sin{\varphi_f}$,
+leading to:
+
+$$\begin{aligned}
+ 2 |\vb{v}| \sin{\varphi_f}
+ = \frac{q_1 q_2}{2 \pi \varepsilon_0 b |\vb{v}| \mu} \cos{\varphi_f}
+\end{aligned}$$
+
+Rearranging this yields the following equation
+for the final polar angle $\varphi_f \equiv \varphi(\infty)$:
+
+$$\begin{aligned}
+ \tan{\varphi_f}
+ = \frac{\sin{\varphi_f}}{\cos{\varphi_f}}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu}
+\end{aligned}$$
+
+However, we want $\theta$, not $\varphi_f$.
+One last use of symmetry and geometry
+tells us that $\theta = 2 \varphi_f$,
+and we thus arrive at the celebrated **Rutherford scattering formula**:
+
+$$\begin{aligned}
+ \boxed{
+ \tan\!\Big( \frac{\theta}{2} \Big)
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. P.M. Bellan,
+ *Fundamentals of plasma physics*,
+ 1st edition, Cambridge.
+2. M. Salewski, A.H. Nielsen,
+ *Plasma physics: lecture notes*,
+ 2021, unpublished.
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