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Diffstat (limited to 'source/know/concept')
| -rw-r--r-- | source/know/concept/boltzmann-equation/index.md | 4 | ||||
| -rw-r--r-- | source/know/concept/boussinesq-wave-theory/index.md | 2 | ||||
| -rw-r--r-- | source/know/concept/convolution-theorem/index.md | 77 | ||||
| -rw-r--r-- | source/know/concept/electric-dipole-approximation/index.md | 89 | ||||
| -rw-r--r-- | source/know/concept/ghz-paradox/index.md | 10 | ||||
| -rw-r--r-- | source/know/concept/hydrogen-atom/index.md | 4 | ||||
| -rw-r--r-- | source/know/concept/laser-rate-equations/index.md | 14 | ||||
| -rw-r--r-- | source/know/concept/legendre-transform/index.md | 31 | ||||
| -rw-r--r-- | source/know/concept/parsevals-theorem/index.md | 10 | ||||
| -rw-r--r-- | source/know/concept/salt-equation/index.md | 7 | ||||
| -rw-r--r-- | source/know/concept/sturm-liouville-theory/index.md | 4 |
11 files changed, 143 insertions, 109 deletions
diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md index 9cb3bcd..5f4add0 100644 --- a/source/know/concept/boltzmann-equation/index.md +++ b/source/know/concept/boltzmann-equation/index.md @@ -65,7 +65,7 @@ But what about the collision term? Expressions for it exist, which are almost exact in many cases, but unfortunately also quite difficult to work with. In addition, $$f$$ is a 7-dimensional function, -so the BTE is already hard to solve without collisions. +so the BTE is already hard to solve without collisions! We only present the simplest case, known as the **Bhatnagar-Gross-Krook approximation**: if the equilibrium state $$f_0(\vb{r}, \vb{v})$$ is known, @@ -314,7 +314,7 @@ For the sake of clarity, we write out the pressure term, including the outer div $$\begin{aligned} \nabla \cdot (\vb{V} \cdot \hat{P}) - &= (\nabla \cdot \hat{P}{}^{\mathrm{T}}) \cdot \vb{V} + &= (\nabla \cdot \hat{P}{}^\top) \cdot \vb{V} = \nabla \cdot \begin{bmatrix} P_{xx} & P_{xy} & P_{xz} \\ diff --git a/source/know/concept/boussinesq-wave-theory/index.md b/source/know/concept/boussinesq-wave-theory/index.md index ad2fe4c..e5fd433 100644 --- a/source/know/concept/boussinesq-wave-theory/index.md +++ b/source/know/concept/boussinesq-wave-theory/index.md @@ -574,7 +574,7 @@ Because $$\phi_\xi$$ is real, we need the right-hand side to be positive, so $$w > 2 \phi$$; for $$\phi \to 0$$, this means that $$w > 0$$. This equation is similar to the one encountered when solving the [Korteweg-de Vries equation](/know/concept/korteweg-de-vries-equation/) -and is integrated in the same way; refer there for details. +and is integrated in the same way; look there for details. The result is: $$\begin{aligned} diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md index d10d85d..3f9eafb 100644 --- a/source/know/concept/convolution-theorem/index.md +++ b/source/know/concept/convolution-theorem/index.md @@ -7,54 +7,59 @@ categories: layout: "concept" --- -The **convolution theorem** states that a convolution in the direct domain -is equal to a product in the frequency domain. This is especially useful -for computation, replacing an $$\mathcal{O}(n^2)$$ convolution with an -$$\mathcal{O}(n \log(n))$$ transform and product. +The **convolution theorem** states that a convolution in the real domain +is equal to a product in the frequency domain. +This fact is especially useful for computation, +as it allows replacing an $$\mathcal{O}(n^2)$$ convolution +with an $$\mathcal{O}(n \log(n))$$ transform and product. ## Fourier transform -The convolution theorem is usually expressed as follows, where -$$\hat{\mathcal{F}}$$ is the [Fourier transform](/know/concept/fourier-transform/), -and $$A$$ and $$B$$ are constants from its definition: +The convolution theorem is usually expressed as follows, +where $$\hat{\mathcal{F}}$$ is the [Fourier transform](/know/concept/fourier-transform/), +and $$A$$ and $$B$$ are the constants from its definition: $$\begin{aligned} \boxed{ \begin{aligned} - A \cdot (f * g)(x) &= \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\ - B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\} + A \cdot (f * g)(x) + &= \hat{\mathcal{F}}{}^{-1}\Big\{ \tilde{f}(k) \: \tilde{g}(k) \Big\} + \\ + B \cdot (\tilde{f} * \tilde{g})(k) + &= \hat{\mathcal{F}}\Big\{ f(x) \: g(x) \Big\} \end{aligned} } \end{aligned}$$ {% include proof/start.html id="proof-fourier" -%} -We expand the right-hand side of the theorem and -rearrange the integrals: +We expand the right-hand side of the theorem and rearrange the integrals: $$\begin{aligned} - \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} - &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \: e^{i s k x'} \dd{x'} \Big) e^{-i s k x} \dd{k} + \hat{\mathcal{F}}{}^{-1}\Big\{ \tilde{f}(k) \: \tilde{g}(k) \Big\} + &= B \int_{-\infty}^\infty \tilde{f}(k) \bigg( A \int_{-\infty}^\infty g(x') \: e^{i s k x'} \dd{x'} \bigg) e^{-i s k x} \dd{k} \\ - &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \: e^{-i s k (x - x')} \dd{k} \Big) \dd{x'} + &= A \int_{-\infty}^\infty g(x') \bigg( B \int_{-\infty}^\infty \tilde{f}(k) \: e^{-i s k (x - x')} \dd{k} \bigg) \dd{x'} \\ &= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'} - = A \cdot (f * g)(x) + \\ + &= A \cdot (f * g)(x) \end{aligned}$$ Then we do the same again, this time starting from a product in the $$x$$-domain: $$\begin{aligned} - \hat{\mathcal{F}}\{f(x) \: g(x)\} - &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \: e^{-i s x k'} \dd{k'} \Big) e^{i s k x} \dd{x} + \hat{\mathcal{F}}\Big\{ f(x) \: g(x) \Big\} + &= A \int_{-\infty}^\infty f(x) \bigg( B \int_{-\infty}^\infty \tilde{g}(k') \: e^{-i s x k'} \dd{k'} \bigg) e^{i s k x} \dd{x} \\ - &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \: e^{i s x (k - k')} \dd{x} \Big) \dd{k'} + &= B \int_{-\infty}^\infty \tilde{g}(k') \bigg( A \int_{-\infty}^\infty f(x) \: e^{i s x (k - k')} \dd{x} \bigg) \dd{k'} \\ &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'} - = B \cdot (\tilde{f} * \tilde{g})(k) + \\ + &= B \cdot (\tilde{f} * \tilde{g})(k) \end{aligned}$$ {% include proof/end.html id="proof-fourier" %} @@ -62,45 +67,51 @@ $$\begin{aligned} ## Laplace transform -For functions $$f(t)$$ and $$g(t)$$ which are only defined for $$t \ge 0$$, +For functions $$f(t)$$ and $$g(t)$$ that are only defined for $$t \ge 0$$, the convolution theorem can also be stated using the [Laplace transform](/know/concept/laplace-transform/): $$\begin{aligned} - \boxed{(f * g)(t) = \hat{\mathcal{L}}{}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}} + \boxed{ + (f * g)(t) + = \hat{\mathcal{L}}{}^{-1}\Big\{ \tilde{f}(s) \: \tilde{g}(s) \Big\} + } \end{aligned}$$ -Because the inverse Laplace transform $$\hat{\mathcal{L}}{}^{-1}$$ is -unpleasant, the theorem is often stated using the forward transform -instead: +Because the inverse Laplace transform $$\hat{\mathcal{L}}{}^{-1}$$ is usually difficult, +the theorem is often stated using the forward transform instead: $$\begin{aligned} - \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)} + \boxed{ + \hat{\mathcal{L}}\Big\{ (f * g)(t) \Big\} + = \tilde{f}(s) \: \tilde{g}(s) + } \end{aligned}$$ {% include proof/start.html id="proof-laplace" -%} We expand the left-hand side. Note that the lower integration limit is 0 instead of $$-\infty$$, -because we set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$: +because we choose to set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$: $$\begin{aligned} - \hat{\mathcal{L}}\{(f * g)(t)\} - &= \int_0^\infty \Big( \int_0^\infty g(t') \: f(t - t') \dd{t'} \Big) e^{-s t} \dd{t} + \hat{\mathcal{L}}\Big\{ (f * g)(t) \Big\} + &= \int_0^\infty \bigg( \int_0^\infty g(t') \: f(t - t') \dd{t'} \bigg) e^{-s t} \dd{t} \\ - &= \int_0^\infty \Big( \int_0^\infty f(t - t') \: e^{-s t} \dd{t} \Big) g(t') \dd{t'} + &= \int_0^\infty \bigg( \int_0^\infty f(t - t') \: e^{-s t} \dd{t} \bigg) \: g(t') \dd{t'} \end{aligned}$$ Then we define a new integration variable $$\tau = t - t'$$, yielding: $$\begin{aligned} - \hat{\mathcal{L}}\{(f * g)(t)\} - &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s (\tau + t')} \dd{\tau} \Big) g(t') \dd{t'} + \hat{\mathcal{L}}\Big\{ (f * g)(t) \Big\} + &= \int_0^\infty \bigg( \int_0^\infty f(\tau) \: e^{-s (\tau + t')} \dd{\tau} \bigg) \: g(t') \dd{t'} \\ - &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s \tau} \dd{\tau} \Big) g(t') \: e^{-s t'} \dd{t'} + &= \int_0^\infty \bigg( \int_0^\infty f(\tau) \: e^{-s \tau} \dd{\tau} \bigg) \: g(t') \: e^{-s t'} \dd{t'} \\ &= \int_0^\infty \tilde{f}(s) \: g(t') \: e^{-s t'} \dd{t'} - = \tilde{f}(s) \: \tilde{g}(s) + \\ + &= \tilde{f}(s) \: \tilde{g}(s) \end{aligned}$$ {% include proof/end.html id="proof-laplace" %} diff --git a/source/know/concept/electric-dipole-approximation/index.md b/source/know/concept/electric-dipole-approximation/index.md index 35cf00c..06f0f45 100644 --- a/source/know/concept/electric-dipole-approximation/index.md +++ b/source/know/concept/electric-dipole-approximation/index.md @@ -13,20 +13,22 @@ layout: "concept" Suppose that an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/) is travelling through an atom, and affecting the electrons. -The general Hamiltonian of an electron in such a wave is given by: +The general Hamiltonian of an electron in an electromagnetic field is: $$\begin{aligned} \hat{H} - &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \varphi + &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \Phi \\ - &= \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 + &= \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 \Phi \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}$$: +Where $$q < 0$$ is the electron's charge, +$$\vu{P} = - i \hbar \nabla$$ is the canonical momentum operator, +$$\vb{A}$$ is the magnetic vector potential, +and $$\Phi$$ is the electric scalar potential. +We start by fixing the Coulomb gauge $$\nabla \cdot \vb{A} = 0$$ +such that $$\vu{P}$$ and $$\vb{A}$$ commute; +let $$\psi$$ be an arbitrary test function: $$\begin{aligned} \comm{\vb{A}}{\vu{P}} \psi @@ -35,20 +37,29 @@ $$\begin{aligned} &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi) \\ &= i \hbar (\nabla \cdot \vb{A}) \psi - = 0 + \\ + &= 0 +\end{aligned}$$ + +Meaning $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$. +Furthermore, we assume that $$\vb{A}$$ is so small that $$\vb{A}{}^2$$ is negligible, +so the Hamiltonian is reduced to: + +$$\begin{aligned} + \hat{H} + &\approx \frac{\vu{P}{}^2}{2 m} - \frac{q}{m} \vu{P} \cdot \vb{A} + q \Phi \end{aligned}$$ -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 $$\hat{H}_0$$: +We now split $$\hat{H}$$ like so, +where $$\hat{H}_1$$ can be regarded as a perturbation to the "base" $$\hat{H}_0$$: $$\begin{aligned} \hat{H} = \hat{H}_0 + \hat{H}_1 - \qquad \quad + \qquad\qquad \hat{H}_0 - \equiv \frac{\vu{P}{}^2}{2 m} + q \varphi - \qquad \quad + \equiv \frac{\vu{P}{}^2}{2 m} + q \Phi + \qquad\qquad \hat{H}_1 \equiv - \frac{q}{m} \vu{P} \cdot \vb{A} \end{aligned}$$ @@ -56,14 +67,16 @@ $$\begin{aligned} In an electromagnetic wave, $$\vb{A}$$ is oscillating sinusoidally in time and space: $$\begin{aligned} - \vb{A}(\vb{x}, t) = \vb{A}_0 \sin(\vb{k} \cdot \vb{x} - \omega t) + \vb{A}(\vb{x}, t) + = \vb{A}_0 \sin(\vb{k} \cdot \vb{x} - \omega t) \end{aligned}$$ Mathematically, it is more convenient to represent this with a complex exponential, whose real part should be taken at the end of the calculation: $$\begin{aligned} - \vb{A}(\vb{x}, t) = - i \vb{A}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t) + \vb{A}(\vb{x}, t) + = - i \vb{A}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t) \end{aligned}$$ The corresponding perturbative [electric field](/know/concept/electric-field/) $$\vb{E}$$ is then given by: @@ -75,10 +88,10 @@ $$\begin{aligned} \end{aligned}$$ Where $$\vb{E}_0 = \omega \vb{A}_0$$. -Let us restrict ourselves to visible light, -whose wavelength $$2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$$. -Meanwhile, an atomic orbital is several Bohr radii $$\sim 10^{-10} \:\mathrm{m}$$, -so $$\vb{k} \cdot \vb{x}$$ is negligible: +Light in and around the visible spectrum +has a wavelength $$2 \pi / |\vb{k}| \sim 10^{-7} \:\mathrm{m}$$, +while an atomic orbital is several Bohr radii $$\sim 10^{-10} \:\mathrm{m}$$, +so $$\vb{k} \cdot \vb{x}$$ is very small. Therefore: $$\begin{aligned} \boxed{ @@ -96,14 +109,13 @@ and the electron quantum-mechanically. 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 $$\vu{P} = m \: \idv{\vu{x}}{t}$$ +To do so, we use that momentum $$\vu{P} \equiv m \: \idv{\vu{x}}{t}$$ and evaluate this in the [interaction picture](/know/concept/interaction-picture/): $$\begin{aligned} \vu{P} - = m \idv{\vu{x}}{t} + &= m \dv{\vu{x}}{t} = m \frac{i}{\hbar} \comm{\hat{H}_0}{\vu{x}} - = m \frac{i}{\hbar} (\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0) \end{aligned}$$ Taking the off-diagonal inner product with @@ -111,34 +123,37 @@ the two-level system's states $$\Ket{1}$$ and $$\Ket{2}$$ gives: $$\begin{aligned} \matrixel{2}{\vu{P}}{1} - = m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1} - = m i \omega_0 \matrixel{2}{\vu{x}}{1} + &= m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1} + \\ + &= m i \omega_0 \matrixel{2}{\vu{x}}{1} \end{aligned}$$ -Therefore, $$\vu{P} / m = i \omega_0 \vu{x}$$, -where $$\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$$ is the resonance of the energy gap, +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: +Therefore, $$\vu{P} / m = i \omega_0 \vu{x}$$, so we get: $$\begin{aligned} \hat{H}_1(t) &= - \frac{q}{m} \vu{P} \cdot \vb{A} - = - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp(- i \omega t) \\ - &\approx - q \vu{x} \cdot \vb{E}_0 \exp(- i \omega t) - = - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t) + &= - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp(- i \omega t) + \\ + &\approx - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t) \end{aligned}$$ -Where $$\vu{d} \equiv q \vu{x} = - e \vu{x}$$ is +Where $$\vu{d} \equiv q \vu{x}$$ is the **transition dipole moment operator** of the electron, -hence the name **electric dipole approximation**. +hence the name *electric dipole approximation*. Finally, we take the real part, yielding: $$\begin{aligned} \boxed{ - \hat{H}_1(t) - = - \vu{d} \cdot \vb{E}(t) - = - q \vu{x} \cdot \vb{E}_0 \cos(\omega t) + \begin{aligned} + \hat{H}_1(t) + &= - \vu{d} \cdot \vb{E}(t) + \\ + &= - q \vu{x} \cdot \vb{E}_0 \cos(\omega t) + \end{aligned} } \end{aligned}$$ diff --git a/source/know/concept/ghz-paradox/index.md b/source/know/concept/ghz-paradox/index.md index 9951883..758e12f 100644 --- a/source/know/concept/ghz-paradox/index.md +++ b/source/know/concept/ghz-paradox/index.md @@ -11,13 +11,13 @@ layout: "concept" The **Greenberger-Horne-Zeilinger** or **GHZ paradox** is an alternative proof of [Bell's theorem](/know/concept/bells-theorem/) -that does not use inequalities, -but the three-particle entangled **GHZ state** $$\ket{\mathrm{GHZ}}$$ instead, +that does not use inequalities, but instead +the three-particle entangled **GHZ state** $$\ket{\mathrm{GHZ}}$$: $$\begin{aligned} \boxed{ \ket{\mathrm{GHZ}} - = \frac{1}{\sqrt{2}} \Big( \ket{000} + \ket{111} \Big) + \equiv \frac{1}{\sqrt{2}} \Big( \ket{000} + \ket{111} \Big) } \end{aligned}$$ @@ -49,8 +49,8 @@ $$\begin{aligned} In other words, the GHZ state is a simultaneous eigenstate of these composite operators, with eigenvalues $$+1$$ and $$-1$$, respectively. -Let us introduce two more operators in the same way, -so that we have a set of four observables, +Let us do the same for two more operators, +so that we have a set of four observables for which $$\ket{\mathrm{GHZ}}$$ gives these eigenvalues: $$\begin{aligned} diff --git a/source/know/concept/hydrogen-atom/index.md b/source/know/concept/hydrogen-atom/index.md index a8443dd..9692d10 100644 --- a/source/know/concept/hydrogen-atom/index.md +++ b/source/know/concept/hydrogen-atom/index.md @@ -525,7 +525,7 @@ $$\begin{aligned} \boxed{ \frac{1}{\lambda_0} = \frac{\omega}{2 \pi c} - = \mathcal{R}\Big( \frac{1}{n_i^2} - \frac{1}{n_f^2} \Big) + = R_\mathrm{H} \bigg( \frac{1}{n_i^2} - \frac{1}{n_f^2} \bigg) } \end{aligned}$$ @@ -554,7 +554,7 @@ and is sometimes used as a unit in calculations: $$\begin{aligned} \mathrm{Ry} - = 2 \pi \hbar c R_\mathrm{H} + \equiv 2 \pi \hbar c R_\mathrm{H} = |E_1| = \frac{\mu}{2 \hbar^2} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2 \approx 13.61 \:\mathrm{eV} diff --git a/source/know/concept/laser-rate-equations/index.md b/source/know/concept/laser-rate-equations/index.md index 1f42f73..c81f02b 100644 --- a/source/know/concept/laser-rate-equations/index.md +++ b/source/know/concept/laser-rate-equations/index.md @@ -97,14 +97,14 @@ $$\begin{aligned} \end{aligned}$$ Typically, $$\gamma_\perp$$ is much larger than the rate of any other decay process, -in which case $$\ipdv{}{\vb{P}0^{+}\!}{t}$$ is negligible compared to $$\gamma_\perp \vb{P}_0^{+}$$. +in which case $$\ipdv{\vb{P}_0^{+}\!}{t}$$ is negligible compared to $$\gamma_\perp \vb{P}_0^{+}$$. Effectively, this means that the polarization $$\vb{P}_0^{+}$$ near-instantly follows the electric field $$\vb{E}^{+}\!$$. -Setting $$\ipdv{}{\vb{P}0^{+}\!}{t} \approx 0$$, the second MBE becomes: +Setting $$\ipdv{\vb{P}_0^{+}\!}{t} \approx 0$$, the second MBE becomes: $$\begin{aligned} \vb{P}^{+} - = -\frac{i |g|^2}{\hbar (\gamma_\perp + i (\omega_0 \!-\! \omega))} \vb{E}^{+} D + = -\frac{i |g|^2}{\hbar (\gamma_\perp + i (\omega_0 - \omega))} \vb{E}^{+} D = \frac{|g|^2 \gamma(\omega)}{\hbar \gamma_\perp} \vb{E}^{+} D \end{aligned}$$ @@ -137,7 +137,7 @@ $$\begin{aligned} &= i (\omega - \Omega) \vb{E}_0^{+} + i \frac{|g|^2 \omega \gamma(\omega)}{2 \hbar \varepsilon_0 \gamma_\perp n^2} \vb{E}_0^{+} D \end{aligned}$$ -Next, we insert our ansatz for $$\vb{E}^{+}\!$$ and $$\vb{P}^{+}\!$$ +Next, we insert our ansatz for $$\vb{E}^{+}$$ and $$\vb{P}^{+}$$ into the third MBE, and rewrite $$\vb{P}_0^{+}$$ as above. Using our identity for $$\gamma(\omega)$$, and the fact that $$\vb{E}_0^{+} \cdot \vb{E}_0^{-} = |\vb{E}|^2$$, we find: @@ -293,11 +293,13 @@ $$\begin{aligned} \boxed{ \begin{aligned} \pdv{N_p}{t} - &= - (\gamma_\mathrm{out} + \gamma_\mathrm{abs} + \gamma_\mathrm{loss}) N_p + \gamma_\mathrm{spon} N_e + G_\mathrm{stim} N_p N_e + &= - (\gamma_\mathrm{out} + \gamma_\mathrm{abs} + \gamma_\mathrm{loss}) N_p + + \gamma_\mathrm{spon} N_e + G_\mathrm{stim} N_p N_e \\ \pdv{N_e}{t} &= R_\mathrm{pump} + \gamma_\mathrm{abs} N_p - - (\gamma_\mathrm{spon} + \gamma_\mathrm{n.r.} + \gamma_\mathrm{leak}) N_e - G_\mathrm{stim} N_p N_e + - (\gamma_\mathrm{spon} + \gamma_\mathrm{n.r.} + \gamma_\mathrm{leak}) N_e + - G_\mathrm{stim} N_p N_e \end{aligned} } \end{aligned}$$ diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md index c4fdeb4..d09613f 100644 --- a/source/know/concept/legendre-transform/index.md +++ b/source/know/concept/legendre-transform/index.md @@ -10,48 +10,49 @@ layout: "concept" The **Legendre transform** of a function $$f(x)$$ is a new function $$L(f')$$, which depends only on the derivative $$f'(x)$$ of $$f(x)$$, -and from which the original function $$f(x)$$ can be reconstructed. +and from which the original $$f(x)$$ can be reconstructed. The point is that $$L(f')$$ contains the same information as $$f(x)$$, just in a different form, analogously to e.g. the [Fourier transform](/know/concept/fourier-transform/). Let us choose an arbitrary point $$x_0 \in [a, b]$$ in the domain of $$f(x)$$. Consider a line $$y(x)$$ tangent to $$f(x)$$ at $$x = x_0$$, -which must have slope $$f'(x_0)$$, and intersects the $$y$$-axis at $$-C$$: +which has slope $$f'(x_0)$$ and intersects the $$y$$-axis at $$y = -C$$: $$\begin{aligned} y(x) - = f'(x_0) (x - x_0) + f(x_0) - = f'(x_0) \: x - C + &= f'(x_0) (x - x_0) + f(x_0) + \\ + &= f'(x_0) \: x - C \end{aligned}$$ Where $$C \equiv f'(x_0) \: x_0 - f(x_0)$$. -We now define the Legendre transform $$L(f')$$ such that -for all $$x_0 \in [a, b]$$ we have $$L(f'(x_0)) = C$$ +We now define the *Legendre transform* $$L(f')$$, +such that for all $$x_0 \in [a, b]$$ we have $$L(f'(x_0)) = C$$ (some authors use $$-C$$ instead). Renaming $$x_0$$ to $$x$$: $$\begin{aligned} L(f'(x)) - = f'(x) \: x - f(x) + &= f'(x) \: x - f(x) \end{aligned}$$ We want this function to depend only on the derivative $$f'$$, but currently $$x$$ still appears here as a variable. -We fix this problem in the easiest possible way: +We solve this problem in the easiest possible way: by assuming that $$f'(x)$$ is invertible for all $$x \in [a, b]$$. If $$x(f')$$ is the inverse of $$f'(x)$$, then $$L(f')$$ is given by: $$\begin{aligned} \boxed{ L(f') - = f' \: x(f') - f(x(f')) + = f' \: x(f') - f\big(x(f')\big) } \end{aligned}$$ The only requirement for the existence of the Legendre transform is thus the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$, -which can only be true if $$f(x)$$ is either convex or concave, +which is only satisfied if $$f(x)$$ is either convex or concave, meaning its derivative $$f'(x)$$ is monotonic. The derivative of $$L(f')$$ with respect to $$f'$$ is simply $$x(f')$$. @@ -72,9 +73,11 @@ To show this, let $$g(L')$$ be the Legendre transform of $$L(f')$$: $$\begin{aligned} g(L') - = L' \: f'(L') - L(f'(L')) - = x(f') \: f' - f' \: x(f') + f(x(f')) - = f(x) + &= L' \: f'(L') - L(f'(L')) + \\ + &= x(f') \: f' - f' \: x(f') + f(x(f')) + \\ + &= f(x) \end{aligned}$$ Moreover, a Legendre transform is always invertible, @@ -84,7 +87,7 @@ so a proof is: $$\begin{aligned} L''(f') - = \dv{}{f'} \Big( \dv{L}{f'} \Big) + &= \dv{}{f'} \Big( \dv{L}{f'} \Big) = \dv{x(f')}{f'} = \dv{x}{f'(x)} = \frac{1}{f''(x)} diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md index a7ce0bf..b2db490 100644 --- a/source/know/concept/parsevals-theorem/index.md +++ b/source/know/concept/parsevals-theorem/index.md @@ -17,9 +17,11 @@ where $$A$$, $$B$$, and $$s$$ are constants from the FT's definition: $$\begin{aligned} \boxed{ \begin{aligned} - \inprod{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)} + \inprod{f(x)}{g(x)} + &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)} \\ - \inprod{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \Inprod{f(x)}{g(x)} + \inprod{\tilde{f}(k)}{\tilde{g}(k)} + &= \frac{2 \pi A^2}{|s|} \inprod{f(x)}{g(x)} \end{aligned} } \end{aligned}$$ @@ -71,8 +73,8 @@ $$\begin{aligned} For this reason, physicists like to define the Fourier transform -with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$, because then it nicely -conserves the functions' normalization. +with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$, +because then it nicely preserves the functions' normalization. diff --git a/source/know/concept/salt-equation/index.md b/source/know/concept/salt-equation/index.md index 77f4755..d7f8ef3 100644 --- a/source/know/concept/salt-equation/index.md +++ b/source/know/concept/salt-equation/index.md @@ -268,9 +268,10 @@ it emits photons without any significant light amplification taking place. Upon gradually increasing the pump $$D_0$$ in the active region, all $$\Imag(k_n)$$ become less negative, until one hits the real axis $$\Imag(k_n) = 0$$, -at which point that mode starts lasing, -i.e. the Light gets Amplified by [Stimulated Emission](/know/concept/einstein-coefficients/) (LASE). -After that, $$D_0$$ can be increased even further until some other $$k_n$$ become real. +at which point that mode starts *lasing*: +its Light gets Amplified by [Stimulated Emission](/know/concept/einstein-coefficients/) of Radiation (LASER). +After that, $$D_0$$ can be increased even further until some other $$k_n$$ become real, +so there are multiple active modes competing for charge carriers. Below threshold (i.e. before any mode is lasing), the problem is linear in $$\Psi_n$$, but above threshold it is nonlinear via $$h(\vb{x})$$. diff --git a/source/know/concept/sturm-liouville-theory/index.md b/source/know/concept/sturm-liouville-theory/index.md index d7984b5..117d893 100644 --- a/source/know/concept/sturm-liouville-theory/index.md +++ b/source/know/concept/sturm-liouville-theory/index.md @@ -151,9 +151,9 @@ $$\begin{aligned} \boxed{ \begin{aligned} \hat{L}_\mathrm{SL} \{u(x)\} - &= \hat{L}_\mathrm{SL}^\dagger \{u(x)\} - \\ &= \Big( p(x) \: u'(x) \Big)' + q(x) \: u(x) + \\ + &= \hat{L}_\mathrm{SL}^\dagger \{u(x)\} \end{aligned} } \end{aligned}$$ |