diff options
| -rw-r--r-- | content/know/concept/einstein-coefficients/index.pdc | 16 | ||||
| -rw-r--r-- | content/know/concept/fabry-perot-cavity/index.pdc | 6 | ||||
| -rw-r--r-- | content/know/concept/maxwell-bloch-equations/index.pdc | 427 | ||||
| -rw-r--r-- | content/know/concept/rabi-oscillation/index.pdc | 24 | ||||
| -rw-r--r-- | content/know/concept/rutherford-scattering/index.pdc | 242 | ||||
| -rw-r--r-- | content/know/concept/rutherford-scattering/one-body.png | bin | 0 -> 66663 bytes | |||
| -rw-r--r-- | content/know/concept/rutherford-scattering/two-body.png | bin | 0 -> 40242 bytes | |||
| -rw-r--r-- | sources/know/concept/rutherford-scattering/one-body.svg | 275 | ||||
| -rw-r--r-- | sources/know/concept/rutherford-scattering/two-body.svg | 358 |
9 files changed, 1332 insertions, 16 deletions
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. diff --git a/content/know/concept/rutherford-scattering/one-body.png b/content/know/concept/rutherford-scattering/one-body.png Binary files differnew file mode 100644 index 0000000..2f837bb --- /dev/null +++ b/content/know/concept/rutherford-scattering/one-body.png diff --git a/content/know/concept/rutherford-scattering/two-body.png b/content/know/concept/rutherford-scattering/two-body.png Binary files differnew file mode 100644 index 0000000..b67b6d1 --- /dev/null +++ b/content/know/concept/rutherford-scattering/two-body.png diff --git a/sources/know/concept/rutherford-scattering/one-body.svg b/sources/know/concept/rutherford-scattering/one-body.svg new file mode 100644 index 0000000..67fcff4 --- /dev/null +++ b/sources/know/concept/rutherford-scattering/one-body.svg @@ -0,0 +1,275 @@ +<?xml version="1.0" encoding="UTF-8" standalone="no"?> +<!-- Created with Inkscape (http://www.inkscape.org/) --> + +<svg + width="799.08087" + height="490.07523" + viewBox="0 0 211.42348 129.66574" + version="1.1" + id="svg5" + inkscape:version="1.1.1 (3bf5ae0d25, 2021-09-20)" + sodipodi:docname="one-body.svg" + inkscape:export-filename="/home/user/git/prefetch/content/know/concept/rutherford-scattering/one-body.png" + inkscape:export-xdpi="128" + inkscape:export-ydpi="128" + xmlns:inkscape="http://www.inkscape.org/namespaces/inkscape" + xmlns:sodipodi="http://sodipodi.sourceforge.net/DTD/sodipodi-0.dtd" + xmlns="http://www.w3.org/2000/svg" + xmlns:svg="http://www.w3.org/2000/svg"> + <sodipodi:namedview + id="namedview7" + pagecolor="#ffffff" + bordercolor="#999999" + borderopacity="1" + inkscape:pageshadow="0" + inkscape:pageopacity="1" + inkscape:pagecheckerboard="0" + inkscape:document-units="mm" + showgrid="false" + units="px" + inkscape:zoom="1.4142136" + inkscape:cx="198.34345" + inkscape:cy="243.95183" + inkscape:window-width="2560" + inkscape:window-height="1440" + inkscape:window-x="0" + inkscape:window-y="0" + inkscape:window-maximized="1" + inkscape:current-layer="layer1" + width="1280px" /> + <defs + id="defs2"> + <marker + style="overflow:visible" + id="Arrow2Sstart" + refX="0" + refY="0" + orient="auto" + inkscape:stockid="Arrow2Sstart" + inkscape:isstock="true"> + <path + transform="matrix(0.3,0,0,0.3,-0.69,0)" + d="M 8.7185878,4.0337352 -2.2072895,0.01601326 8.7185884,-4.0017078 c -1.7454984,2.3720609 -1.7354408,5.6174519 -6e-7,8.035443 z" + style="fill:context-stroke;fill-rule:evenodd;stroke:context-stroke;stroke-width:0.625;stroke-linejoin:round" + id="path1012" /> + </marker> + <marker + style="overflow:visible" + id="Arrow2Send" + refX="0" + refY="0" + orient="auto" + inkscape:stockid="Arrow2Send" + inkscape:isstock="true"> + <path + transform="matrix(-0.3,0,0,-0.3,0.69,0)" + d="M 8.7185878,4.0337352 -2.2072895,0.01601326 8.7185884,-4.0017078 c -1.7454984,2.3720609 -1.7354408,5.6174519 -6e-7,8.035443 z" + style="fill:context-stroke;fill-rule:evenodd;stroke:context-stroke;stroke-width:0.625;stroke-linejoin:round" + id="path1015" /> + </marker> + <marker + style="overflow:visible" + id="Arrow1Send" + refX="0" + refY="0" + orient="auto" + inkscape:stockid="Arrow1Send" + inkscape:isstock="true"> + <path + transform="matrix(-0.2,0,0,-0.2,-1.2,0)" + style="fill:context-stroke;fill-rule:evenodd;stroke:context-stroke;stroke-width:1pt" + d="M 0,0 5,-5 -12.5,0 5,5 Z" + id="path997" /> + </marker> + <inkscape:path-effect + effect="bspline" + id="path-effect1802" + is_visible="true" + lpeversion="1" + weight="33.333333" + steps="2" + helper_size="0" + apply_no_weight="true" + apply_with_weight="true" + only_selected="false" /> + <marker + style="overflow:visible" + id="Arrow2Mend" + refX="0" + refY="0" + orient="auto" + inkscape:stockid="Arrow2Mend" + inkscape:isstock="true"> + <path + transform="scale(-0.6)" + d="M 8.7185878,4.0337352 -2.2072895,0.01601326 8.7185884,-4.0017078 c -1.7454984,2.3720609 -1.7354408,5.6174519 -6e-7,8.035443 z" + style="fill:context-stroke;fill-rule:evenodd;stroke:context-stroke;stroke-width:0.625;stroke-linejoin:round" + id="path1009" /> + </marker> + <marker + style="overflow:visible" + id="Arrow1Mend" + refX="0" + refY="0" + orient="auto" + inkscape:stockid="Arrow1Mend" + inkscape:isstock="true"> + <path + transform="matrix(-0.4,0,0,-0.4,-4,0)" + style="fill:context-stroke;fill-rule:evenodd;stroke:context-stroke;stroke-width:1pt" + d="M 0,0 5,-5 -12.5,0 5,5 Z" + id="path991" /> + </marker> + <marker + style="overflow:visible" + id="Arrow2Lend" + refX="0" + refY="0" + orient="auto" + inkscape:stockid="Arrow2Lend" + inkscape:isstock="true"> + <path + transform="matrix(-1.1,0,0,-1.1,-1.1,0)" + d="M 8.7185878,4.0337352 -2.2072895,0.01601326 8.7185884,-4.0017078 c -1.7454984,2.3720609 -1.7354408,5.6174519 -6e-7,8.035443 z" + style="f |
