Expand knowledge base

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!