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Diffstat (limited to 'content/know')
| -rw-r--r-- | content/know/concept/fourier-transform/index.pdc | 150 | ||||
| -rw-r--r-- | content/know/concept/lagrangian-mechanics/index.pdc | 131 | ||||
| -rw-r--r-- | content/know/concept/landau-quantization/index.pdc | 127 | ||||
| -rw-r--r-- | content/know/concept/pulay-mixing/index.pdc | 13 |
4 files changed, 402 insertions, 19 deletions
diff --git a/content/know/concept/fourier-transform/index.pdc b/content/know/concept/fourier-transform/index.pdc index 3be47ff..1d1e27d 100644 --- a/content/know/concept/fourier-transform/index.pdc +++ b/content/know/concept/fourier-transform/index.pdc @@ -25,8 +25,8 @@ The **forward** FT is defined as follows, where $A$, $B$, and $s$ are unspecifie $$\begin{aligned} \boxed{ \tilde{f}(k) - = \hat{\mathcal{F}}\{f(x)\} - = A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x} + \equiv \hat{\mathcal{F}}\{f(x)\} + \equiv A \int_{-\infty}^\infty f(x) \exp\!(i s k x) \dd{x} } \end{aligned}$$ @@ -35,8 +35,8 @@ The **inverse Fourier transform** (iFT) undoes the forward FT operation: $$\begin{aligned} \boxed{ f(x) - = \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\} - = B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k x) \dd{k} + \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\} + \equiv B \int_{-\infty}^\infty \tilde{f}(k) \exp\!(- i s k x) \dd{k} } \end{aligned}$$ @@ -46,9 +46,9 @@ again. Let us verify this, by rearranging the integrals to get the $$\begin{aligned} \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\} - &= A B \int_{-\infty}^\infty \exp(-i s k x) \int_{-\infty}^\infty f(x') \exp(i s k x') \dd{x'} \dd{k} + &= A B \int_{-\infty}^\infty \exp\!(-i s k x) \int_{-\infty}^\infty f(x') \exp\!(i s k x') \dd{x'} \dd{k} \\ - &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp(i s k (x' - x)) \dd{k} \Big) \dd{x'} + &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp\!(i s k (x' - x)) \dd{k} \Big) \dd{x'} \\ &= 2 \pi A B \int_{-\infty}^\infty f(x') \: \delta(s(x' - x)) \dd{x'} = \frac{2 \pi A B}{|s|} f(x) @@ -74,15 +74,15 @@ on whether the analysis is for forward ($s > 0$) or backward-propagating ## Derivatives -The FT of a derivative has a very interesting property. +The FT of a derivative has a very useful property. Below, after integrating by parts, we remove the boundary term by assuming that $f(x)$ is localized, i.e. $f(x) \to 0$ for $x \to \pm \infty$: $$\begin{aligned} \hat{\mathcal{F}}\{f'(x)\} - &= A \int_{-\infty}^\infty f'(x) \exp(i s k x) \dd{x} + &= A \int_{-\infty}^\infty f'(x) \exp\!(i s k x) \dd{x} \\ - &= A \big[ f(x) \exp(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x} + &= A \big[ f(x) \exp\!(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp\!(i s k x) \dd{x} \\ &= (- i s k) \tilde{f}(k) \end{aligned}$$ @@ -110,13 +110,139 @@ $$\begin{aligned} Derivatives in the frequency domain have an analogous property: $$\begin{aligned} + \dv[n]{\tilde{f}}{k} + &= A \dv[n]{k} \int_{-\infty}^\infty f(x) \exp\!(i s k x) \dd{x} + \\ + &= A \int_{-\infty}^\infty (i s x)^n f(x) \exp\!(i s k x) \dd{x} + = \hat{\mathcal{F}}\{ (i s x)^n f(x) \} +\end{aligned}$$ + + +## Multiple dimensions + +The Fourier transform is straightforward to generalize to $N$ dimensions. +Given a scalar field $f(\vb{x})$ with $\vb{x} = (x_1, ..., x_N)$, +its FT $\tilde{f}(\vb{k})$ is defined as follows: + +$$\begin{aligned} + \boxed{ + \tilde{f}(\vb{k}) + \equiv \hat{\mathcal{F}}\{f(\vb{x})\} + \equiv A \int_{-\infty}^\infty f(\vb{x}) \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}} + } +\end{aligned}$$ + +Where the wavevector $\vb{k} = (k_1, ..., k_N)$. +Likewise, the inverse FT is given by: + +$$\begin{aligned} \boxed{ - \dv[n]{\tilde{f}}{k} - = A \int_{-\infty}^\infty (i s x)^n f(x) \exp(i s k x) \dd{x} - = \hat{\mathcal{F}}\{ (i s x)^n f(x) \} + f(\vb{x}) + \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(\vb{k})\} + \equiv B \int_{-\infty}^\infty \tilde{f}(\vb{k}) \exp\!(- i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{k}} } \end{aligned}$$ +In practice, in $N$D, there is not as much disagreement about +the constants $A$, $B$ and $s$ as in 1D: +typically $A = 1$ and $B = 1 / (2 \pi)^N$, with $s = \pm 1$. +Any choice will do, as long as: + +$$\begin{aligned} + \boxed{ + A B + = \bigg( \frac{|s|}{2 \pi} \bigg)^{\!N} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-constants-ND"/> +<label for="proof-constants-ND">Proof</label> +<div class="hidden"> +<label for="proof-constants-ND">Proof.</label> +The inverse FT of the forward FT of $f(\vb{x})$ must be equal to $f(\vb{x})$ again, so: + +$$\begin{aligned} + \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\} + &= A B \int \exp\!(- i s \vb{k} \cdot \vb{x}) + \int f(\vb{x}') \exp\!(i s \vb{k} \cdot \vb{x}') \dd[N]{\vb{x}'} \dd[N]{\vb{k}} + \\ + &= (2 \pi)^N A B \int f(\vb{x}') + \Big( \frac{1}{(2 \pi)^N} \int \exp\!(i s \vb{k} \cdot (\vb{x}' - \vb{x})) \dd[N]{\vb{k}} \Big) \dd[N]{\vb{x}'} + \\ + &= (2 \pi)^N A B \int f(\vb{x}') + \Big( \prod_{n = 1}^N \frac{1}{2 \pi} \int \exp\!(i s k_n (x_n' - x_n)) \dd{k_n} \Big) \dd[N]{\vb{x}'} +\end{aligned}$$ + +Here, we recognize the definition of the Dirac delta function again, +leading to: + +$$\begin{aligned} + \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\} + &= (2 \pi)^N A B \int f(\vb{x}') + \Big( \prod_{n = 1}^N \delta(s(x_n' - x_n)) \Big) \dd[N]{\vb{x}'} + \\ + &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \dd[N]{\vb{x}'} + = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x}) +\end{aligned}$$ +</div> +</div> + +Differentiation is more complicated for $N > 1$, +but the FT is still useful, +notably for the Laplacian $\nabla^2 \equiv \dv*[2]{x_1} + ... + \dv*[2]{x_N}$. +Let $|\vb{k}|$ be the norm of $\vb{k}$, +then for a localized $f$: + +$$\begin{aligned} + \boxed{ + \hat{\mathcal{F}}\{\nabla^2 f(\vb{x})\} + = - s^2 |\vb{k}|^2 \tilde{f}(\vb{k}) + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-laplacian"/> +<label for="proof-laplacian">Proof</label> +<div class="hidden"> +<label for="proof-laplacian">Proof.</label> +We insert $\nabla^2 f$ into the FT, +decompose the exponential and the Laplacian, +and then integrate by parts (limits $\pm \infty$ omitted): + +$$\begin{aligned} + \hat{\mathcal{F}}\{\nabla^2 f\} + &= A \int \big( \nabla^2 f \big) \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}} + \\ + &= A \int \Big( \sum_{n = 1}^N \pdv[2]{f}{x_n} \Big) \Big( \prod_{m = 1}^N \exp\!(i s k_m x_m) \Big) \dd[N]{\vb{x}} + \\ + &= A \sum_{n = 1}^N \bigg[ \pdv{f}{x_n} \exp\!(i s \vb{k} \cdot \vb{x}) \bigg] + - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}} +\end{aligned}$$ + +Just like in 1D, we get rid of the boundary term +by assuming that all derivatives $\dv*{f}{x_n}$ are nicely localized. +To proceed, we then integrate by parts again: + +$$\begin{aligned} + \hat{\mathcal{F}}\{\nabla^2 f\} + &= - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \Big( \prod_{m = 1}^N \exp\!(i s k_m x_m) \Big) \dd[N]{\vb{x}} + \\ + &= - A \sum_{n = 1}^N i s k_n \bigg[ f \exp\!(i s \vb{k} \cdot \vb{x}) \bigg] + + A \sum_{n = 1}^N (i s k_n)^2 \int f \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}} +\end{aligned}$$ + +Once again, we remove the boundary term +by assuming that $f$ is localized, yielding: + +$$\begin{aligned} + \hat{\mathcal{F}}\{\nabla^2 f\} + &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}} + = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f} +\end{aligned}$$ +</div> +</div> + ## References diff --git a/content/know/concept/lagrangian-mechanics/index.pdc b/content/know/concept/lagrangian-mechanics/index.pdc new file mode 100644 index 0000000..24e1f93 --- /dev/null +++ b/content/know/concept/lagrangian-mechanics/index.pdc @@ -0,0 +1,131 @@ +--- +title: "Lagrangian mechanics" +firstLetter: "L" +publishDate: 2021-07-01 +categories: +- Physics + +date: 2021-07-01T18:44:43+02:00 +draft: false +markup: pandoc +--- + +# Lagrangian mechanics + +**Lagrangian mechanics** is a formulation of classical mechanics, +which is equivalent to Newton's laws, +but offers some advantages. +Its mathematical backbone is the +[calculus of variations](/know/concept/calculus-of-variations/). + +For a moving object with position $x(t)$ and velocity $\dot{x}(t)$, +we define the Lagrangian $L$ as the difference +between its kinetic and potential energies: + +$$\begin{aligned} + \boxed{ + L(x, \dot{x}, t) \equiv T - V = \frac{1}{2} m \dot{x}^2 - V(x) + } +\end{aligned}$$ + +From variational calculus we then get the Euler-Lagrange equation, +which in this case turns out to just be Newton's second law: + +$$\begin{aligned} + \dv{t} \Big( \pdv{L}{\dot{x}} \Big) = \pdv{L}{x} + \qquad \implies \qquad + m \ddot{x} = - \pdv{V}{x} = F +\end{aligned}$$ + +But compared to Newtonian mechanics, +Lagrangian mechanics scales better for large systems. +For example, to describe the dynamics of $N$ objects $x_1(t), ..., x_N(t)$, +we only need a single $L$ +from which the equations of motion can easily be derived. +Getting these equations directly from Newton's laws could get messy. + +At no point have we assumed Cartesian coordinates: +the Euler-Lagrange equations keep their form +for any independent coordinates $q_1(t), ..., q_N(t)$: + +$$\begin{aligned} + \dv{t} \Big( \pdv{L}{\dot{q_n}} \Big) = \pdv{L}{q_n} +\end{aligned}$$ + +We define the **canonical momentum conjugate** $p_n(t)$ +and the **generalized force conjugate** $F_n(t)$ as follows, +such that we can always get Newton's second law: + +$$\begin{aligned} + \boxed{ + p_n \equiv \pdv{L}{\dot{q}_n} \qquad F_n \equiv \pdv{L}{q_n} + } + \qquad \implies \qquad + \dv{p_n}{t} = F_n +\end{aligned}$$ + +But this is actually a bit misleading, +since $p_n$ need not be a momentum, nor $F_n$ a force, +although often they are. +For example, $p_n$ could be angular momentum, and $F_n$ torque. + +Another advantage of Lagrangian mechanics is that +the conserved quantities can be extracted from $L$ using Noether's theorem. +In the simplest case, if $L$ does not depend on $q_n$ +(then known as a **cyclic coordinate**), +then we know that the "momentum" $p_n$ is a conserved quantity: + +$$\begin{aligned} + F_n = \pdv{L}{q_n} = 0 + \qquad \implies \qquad + \dv{p_n}{t} = 0 +\end{aligned}$$ + +Now, as the number of particles $N$ increases to infinity, +variational calculus will give infinitely many coupled equations, +which is obviously impractical. + +Such a system can be regarded as continuous, so the $N$ functions $q_n$ +can be replaced by a single density function $u(x,t)$. +This approach can also be used for continuous fields, +in which case the complex conjugate $u^*$ is often included. +The Lagrangian $L$ then becomes: + +$$\begin{aligned} + L(u, u^*, u_x, u_x^*, u_t, u_t^*, x, t) + = \int_{-\infty}^\infty \mathcal{L}(u, u^*, u_x, u_x^*, u_t, u_t^*, x, t) \dd{x} +\end{aligned}$$ + +Where $\mathcal{L}$ is known as the **Lagrangian density**. +By inserting this into the functional $J$ +used for the derivation of the Euler-Lagrange equations, we get: + +$$\begin{aligned} + J[u] + = \int_{t_0}^{t_1} L \dd{t} + = \int_{t_0}^{t_1} \int_{-\infty}^\infty \mathcal{L} \dd{x} \dd{t} +\end{aligned}$$ + +This is simply 2D variational problem, +so the Euler-Lagrange equations will be two PDEs: + +$$\begin{aligned} + 0 &= \pdv{\mathcal{L}}{u} - \pdv{x} \Big( \pdv{\mathcal{L}}{u_x} \Big) - \pdv{t} \Big( \pdv{\mathcal{L}}{u_t} \Big) + \\ + 0 &= \pdv{\mathcal{L}}{u^*} - \pdv{x} \Big( \pdv{\mathcal{L}}{u_x^*} \Big) - \pdv{t} \Big( \pdv{\mathcal{L}}{u_t^*} \Big) +\end{aligned}$$ + +If $\mathcal{L}$ is real, +then these two Euler-Lagrange equations will in fact be identical. + +Finally, note that for abstract fields, +the Lagrangian density $\mathcal{L}$ rarely has +a physical interpretation, and is not unique. +Instead, it must be reverse-engineered from a relevant equation. + + + +## References +1. R. Shankar, + *Principles of quantum mechanics*, 2nd edition, + Springer. diff --git a/content/know/concept/landau-quantization/index.pdc b/content/know/concept/landau-quantization/index.pdc new file mode 100644 index 0000000..4212078 --- /dev/null +++ b/content/know/concept/landau-quantization/index.pdc @@ -0,0 +1,127 @@ +--- +title: "Landau quantization" +firstLetter: "L" +publishDate: 2021-07-01 +categories: +- Physics +- Quantum mechanics + +date: 2021-07-01T18:44:30+02:00 +draft: false +markup: pandoc +--- + +# Landau quantization + +When a particle with charge $q$ is moving in a homogeneous magnetic field, +quantum mechanics decrees that its allowed energies split +into degenerate discrete **Landau levels**, +a phenomenon known as **Landau quantization**. + +Starting from the Hamiltonian $\hat{H}$ for a particle with mass $m$ +in a vector potential $\vec{A}(\hat{Q})$: + +$$\begin{aligned} + \hat{H} + &= \frac{1}{2 m} \big( \hat{p} - q \vec{A} \big)^2 +\end{aligned}$$ + +We choose $\vec{A} = (- \hat{y} B, 0, 0)$, +yielding a magnetic field $\vec{B} = \nabla \times \vec{A}$ +pointing in the $z$-direction with strength $B$. +The Hamiltonian becomes: + +$$\begin{aligned} + \hat{H} + &= \frac{\big( \hat{p}_x - q B \hat{y} \big)^2}{2 m} + \frac{\hat{p}_y^2}{2 m} + \frac{\hat{p}_z^2}{2 m} +\end{aligned}$$ + +The only position operator occurring in $\hat{H}$ is $\hat{y}$, +so $[\hat{H}, \hat{p}_x] = [\hat{H}, \hat{p}_z] = 0$. +Because $\hat{p}_z$ appears in an unmodified kinetic energy term, +and the corresponding $\hat{z}$ does not occur at all, +the particle has completely free motion in the $z$-direction. +Likewise, because $\hat{x}$ does not occur in $\hat{H}$, +we can replace $\hat{p}_x$ by its eigenvalue $\hbar k_x$, +although the motion is not free, due to $q B \hat{y}$. + +Based on the absence of $\hat{x}$ and $\hat{z}$, +we make the following ansatz for the wavefunction $\Psi$: +a plane wave in the $x$ and $z$ directions, multiplied by an unknown $\phi(y)$: + +$$\begin{aligned} + \Psi(x, y, z) + = \phi(y) \exp(i k_x x + i k_z z) +\end{aligned}$$ + +Inserting this into the time-independent Schrödinger equation gives, +after dividing out the plane wave exponential $\exp(i k_x x + i k_z z)$: + +$$\begin{aligned} + E \phi + &= \frac{1}{2 m} \Big( (\hbar k_x - q B y)^2 + \hat{p}_y^2 + \hbar^2 k_z^2 \Big) \phi +\end{aligned}$$ + +By defining the cyclotron frequency $\omega_c \equiv q B / m$ and rearranging, +we can turn this into a 1D quantum harmonic oscillator in $y$, +with a couple of extra terms: + +$$\begin{aligned} + \Big( E - \frac{\hbar^2 k_z^2}{2 m} \Big) \phi + &= \bigg( \frac{1}{2} m \omega_c^2 \Big( y - \frac{\hbar k_x}{m \omega_c} \Big)^2 + \frac{\hat{p}_y^2}{2 m} \bigg) \phi +\end{aligned}$$ + +The potential minimum is shifted by $y_0 = \hbar k_x / (m \omega_c)$, +and a plane wave in $z$ contributes to the energy $E$. +In any case, the energy levels of this type of system are well-known: + +$$\begin{aligned} + \boxed{ + E_n = \hbar \omega_c \Big(n + \frac{1}{2}\Big) + \frac{\hbar^2 k_z^2}{2 m} + } +\end{aligned}$$ + +And $\Psi_n$ is then as follows, +where $\phi$ is the known quantum harmonic oscillator solution: + +$$\begin{aligned} + \Psi_n(x, y, z) + = \phi_n(y - y_0) \exp(i k_x x + i k_z z) +\end{aligned}$$ + +Note that this wave function contains $k_x$ (also inside $y_0$), +but $k_x$ is absent from the energy $E_n$. +This implies degeneracy: +assuming periodic boundary conditions $\Psi(x\!+\!L_x) = \Psi(x)$, +then $k_x$ can take values of the form $2 \pi n / L_x$, for $n \in \mathbb{Z}$. + +However, $k_x$ also occurs in the definition of $y_0$, so the degeneracy +is finite, since $y_0$ must still lie inside the system, +or, more formally, $y_0 \in [0, L_y]$: + +$$\begin{aligned} + 0 \le y_0 = \frac{\hbar k_x}{m \omega_c} = \frac{\hbar 2 \pi n}{q B L_x} \le L_y +\end{aligned}$$ + +Isolating this for $n$, we find the following upper bound of the degeneracy: + +$$\begin{aligned} + \boxed{ + n \le + \frac{q B L_x L_y}{2 \pi \hbar} = \frac{q B A}{h} + } +\end{aligned}$$ + +Where $A \equiv L_x L_y$ is the area of the confinement in the $(x,y)$-plane. +Evidently, the degeneracy of each level increases with larger $B$, +but since $\omega_c = q B / m$, the energy gap between each level increases too. +In other words: the [density of states](/know/concept/density-of-states/) +is a constant with respect to the energy, +but the states get distributed across the $E_n$ differently depending on $B$. + + + +## References +1. L.E. Ballentine, + *Quantum mechanics: a modern development*, 2nd edition, + World Scientific. diff --git a/content/know/concept/pulay-mixing/index.pdc b/content/know/concept/pulay-mixing/index.pdc index 8daa54f..4e7a411 100644 --- a/content/know/concept/pulay-mixing/index.pdc +++ b/content/know/concept/pulay-mixing/index.pdc @@ -16,8 +16,8 @@ by generating a series $\rho_1$, $\rho_2$, etc. converging towards the desired solution $\rho_*$. **Pulay mixing**, also often called **direct inversion in the iterative subspace** (DIIS), -is an effective method to speed up convergence, -which also helps to avoid periodic divergences. +can speed up the convergence for some types of problems, +and also helps to avoid periodic divergences. The key concept it relies on is the **residual vector** $R_n$ of the $n$th iteration, which in some way measures the error of the current $\rho_n$. @@ -113,17 +113,16 @@ $\lambda = - \braket{R_{n+1}}{R_{n+1}}$, where $R_{n+1}$ is the *predicted* residual of the next iteration, subject to the two assumptions. -This method is very effective. However, in practice, the earlier inputs $\rho_1$, $\rho_2$, etc. are much further from $\rho_*$ than $\rho_n$, -so usually only the most recent $N$ inputs $\rho_{n - N}$, ..., $\rho_n$ are used: +so usually only the most recent $N\!+\!1$ inputs $\rho_{n - N}$, ..., $\rho_n$ are used: $$\begin{aligned} \rho_{n+1} - = \sum_{m = N}^n \alpha_m \rho_m + = \sum_{m = n-N}^n \alpha_m \rho_m \end{aligned}$$ -You might be confused by the absence of all $\rho_m^\mathrm{new}$ +You might be confused by the absence of any $\rho_m^\mathrm{new}$ in the creation of $\rho_{n+1}$, as if the iteration's outputs are being ignored. This is due to the first assumption, which states that $\rho_n^\mathrm{new}$ are $\rho_n$ are already similar, @@ -155,7 +154,7 @@ while still giving more weight to iterations with smaller residuals. Pulay mixing is very effective for certain types of problems, e.g. density functional theory, -where it can accelerate convergence by up to one order of magnitude! +where it can accelerate convergence by up to two orders of magnitude! |