Expand knowledge base

4 files changed, 507 insertions, 5 deletions

diff --git a/content/know/concept/beltrami-identity/index.pdc b/content/know/concept/beltrami-identity/index.pdc
new file mode 100644
index 0000000..a466a63
--- /dev/null
+++ b/content/know/concept/beltrami-identity/index.pdc
@@ -0,0 +1,138 @@
+---
+title: "Beltrami identity"
+firstLetter: "B"
+publishDate: 2022-09-17
+categories:
+- Physics
+- Mathematics
+
+date: 2022-09-10T13:39:06+02:00
+draft: false
+markup: pandoc
+---
+
+# Beltrami identity
+
+Consider a general functional $J[f]$ of the following form,
+with $f(x)$ an unknown function:
+
+$$\begin{aligned}
+    J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x}
+\end{aligned}$$
+
+Where $L$ is the Lagrangian.
+To find the $f$ that maximizes or minimizes $J[f]$,
+the [calculus of variations](/know/concept/calculus-of-variations/)
+states that the Euler-Lagrange equation must be solved for $f$:
+
+$$\begin{aligned}
+    0 = \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f'} \Big)
+\end{aligned}$$
+
+We now want to know exactly how $L$ depends on the free variable $x$,
+since it is a function of $x$, $f(x)$ and $f'(x)$.
+Using the chain rule:
+
+$$\begin{aligned}
+    \dv{L}{x}
+    = \pdv{L}{f} \dv{f}{x} + \pdv{L}{f'} \dv{f'}{x} + \pdv{L}{x}
+\end{aligned}$$
+
+Substituting the Euler-Lagrange equation into the first term gives us:
+
+$$\begin{aligned}
+    \dv{L}{x}
+    &= f' \dv{x} \Big( \pdv{L}{f'} \Big) + \dv{f'}{x} \pdv{L}{f'} + \pdv{L}{x}
+    \\
+    &= \dv{x} \bigg( f' \pdv{L}{f'} \bigg) + \pdv{L}{x}
+\end{aligned}$$
+
+Although we started from the "hard" derivative $\dv*{L}{x}$,
+we arrive at an expression for the "soft" derivative $\pdv*{L}{x}$,
+describing the *explicit* dependence of $L$ on $x$:
+
+$$\begin{aligned}
+    - \pdv{L}{x}
+    = \dv{x} \bigg( f' \pdv{L}{f'} - L \bigg)
+\end{aligned}$$
+
+What if $L$ does not explicitly depend on $x$, i.e. $\pdv*{L}{x} = 0$?
+In that case, the equation can be integrated to give the **Beltrami identity**:
+
+$$\begin{aligned}
+    \boxed{
+        f' \pdv{L}{f'} - L
+        = C
+    }
+\end{aligned}$$
+
+Where $C$ is a constant.
+This says that the left-hand side is a conserved quantity in $x$,
+which could be useful to know.
+If we insert a concrete expression for $L$,
+the Beltrami identity might be easier to solve for $f$ than the full Euler-Lagrange equation.
+The assumption $\pdv*{L}{x} = 0$ is justified;
+for example, if $x$ is time, it means that the potential is time-independent.
+
+
+## Higher dimensions
+
+Above, a 1D problem was considered, i.e. $f$ depended only on a single variable $x$.
+Consider now a 2D problem, such that $J[f]$ is given by:
+
+$$\begin{aligned}
+    J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
+\end{aligned}$$
+
+In which case the Euler-Lagrange equation takes the following form:
+
+$$\begin{aligned}
+    0 = \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f_x} \Big) - \dv{y} \Big( \pdv{L}{f_y} \Big)
+\end{aligned}$$
+
+Once again, we calculate the hard $x$-derivative of $L$ (the $y$-derivative is analogous):
+
+$$\begin{aligned}
+    \dv{L}{x}
+    &= \pdv{L}{f} \dv{f}{x} + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_y} \dv{f_y}{x} + \pdv{L}{x}
+    \\
+    &= \dv{f}{x} \bigg( \dv{x} \Big( \pdv{L}{f_x} \Big) + \dv{y} \Big( \pdv{L}{f_y} \Big) \bigg)
+    + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_y} \dv{f_y}{x} + \pdv{L}{x}
+    \\
+    &= \dv{x} \Big( f_x \pdv{L}{f_x} \Big) + \dv{y} \Big( f_x \pdv{L}{f_y} \Big) + \pdv{L}{x}
+\end{aligned}$$
+
+This time, we arrive at the following expression for the soft derivative $\pdv*{L}{x}$:
+
+$$\begin{aligned}
+    - \pdv{L}{x}
+    &= \dv{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{y} \Big( f_x \pdv{L}{f_y} \Big)
+\end{aligned}$$
+
+Due to the derivatives, this cannot be cleanly turned into an analogue of the 1D Beltrami identity,
+and therefore we use that name only in the 1D case.
+
+However, if $\pdv*{L}{x} = 0$, this equation is still useful.
+For an off-topic demonstration of this fact,
+let us choose $x$ as the transverse coordinate, and integrate over it to get:
+
+$$\begin{aligned}
+    0
+    &= - \int_{x_0}^{x_1} \pdv{L}{x} \dd{x}
+    \\
+    &= \int_{x_0}^{x_1} \dv{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{y} \Big( f_x \pdv{L}{f_y} \Big) \dd{x}
+    \\
+    &= \Big[ f_x \pdv{L}{f_x} - L \Big]_{x_0}^{x_1} + \dv{y} \int_{x_0}^{x_1} \Big( f_x \pdv{L}{f_y} \Big) \dd{x}
+\end{aligned}$$
+
+If our boundary conditions cause the boundary term to vanish (as is often the case),
+then the integral on the right is a conserved quantity with respect to $y$.
+While not as elegant as the 1D Beltrami identity,
+the above 2D counterpart still fulfills the same role.
+
+
+
+## References
+1.  O. Bang,
+    *Nonlinear mathematical physics: lecture notes*, 2020,
+    unpublished.
diff --git a/content/know/concept/calculus-of-variations/index.pdc b/content/know/concept/calculus-of-variations/index.pdc
index 26c5753..9cae283 100644
--- a/content/know/concept/calculus-of-variations/index.pdc
+++ b/content/know/concept/calculus-of-variations/index.pdc
@@ -251,7 +251,7 @@ meaning they do not depend on any derivatives of any $f_n(x)$:
 
 $$\begin{aligned}
     \phi_m(f_1, ..., f_N, x) = 0
-    \qquad
+    \qquad \qquad
     \int_{x_0}^{x_1} \phi_m(f_1, ..., f_N, x) \dd{x} = C_m
 \end{aligned}$$
 
@@ -261,12 +261,12 @@ by simply redefining the constraint as $\phi_m^0 = \phi_m - C_m = 0$.
 
 To solve this constrained optimization problem for $f_n(x)$,
 we introduce [Lagrange multipliers](/know/concept/lagrange-multiplier/) $\lambda_m$.
-In the former case $\lambda_m(x)$ is a function of all $x$, while in the
+In the former case $\lambda_m(x)$ is a function of $x$, while in the
 latter case $\lambda_m$ is constant:
 
 $$\begin{aligned}
-    \int \lambda_m(x_i) \: \phi_m(\{f_n\}, x) \dd{x} = 0
-    \qquad
+    \int \lambda_m(x) \: \phi_m(\{f_n\}, x) \dd{x} = 0
+    \qquad \qquad
     \lambda_m \int \phi_m(\{f_n\}, x) \dd{x} = \lambda_m C_m
 \end{aligned}$$
 
diff --git a/content/know/concept/ritz-method/index.pdc b/content/know/concept/ritz-method/index.pdc
new file mode 100644
index 0000000..320771a
--- /dev/null
+++ b/content/know/concept/ritz-method/index.pdc
@@ -0,0 +1,364 @@
+---
+title: "Ritz method"
+firstLetter: "R"
+publishDate: 2022-09-18
+categories:
+- Physics
+- Mathematics
+- Perturbation
+- Quantum mechanics
+- Numerical methods
+
+date: 2022-09-10T13:33:07+02:00
+draft: false
+markup: pandoc
+---
+
+# Ritz method
+
+In various branches of physics,
+the **Ritz method** is a technique to approximately find the lowest solutions to an eigenvalue problem.
+Some call it the **Rayleigh-Ritz method**, the **Ritz-Galerkin method**,
+or simply the **variational method**.
+
+
+
+## Background
+
+In the context of [variational calculus](/know/concept/calculus-of-variations/),
+consider the following functional to be optimized:
+
+$$\begin{aligned}
+    R[u]
+    = \frac{1}{S} \int_a^b p(x) \big|u_x(x)\big|^2 - q(x) \big|u(x)\big|^2 \dd{x}
+\end{aligned}$$
+
+Where $u(x) \in \mathbb{C}$ is the unknown function,
+and $p(x), q(x) \in \mathbb{R}$ are given.
+In addition, $S$ is the norm of $u$, which we demand be constant
+with respect to a weight function $w(x) \in \mathbb{R}$:
+
+$$\begin{aligned}
+    S
+    = \int_a^b w(x) \big|u(x)\big|^2 \dd{x}
+\end{aligned}$$
+
+To handle this normalization requirement,
+we introduce a [Lagrange multiplier](/know/concept/lagrange-multiplier/) $\lambda$,
+and define the Lagrangian $\Lambda$ for the full constrained optimization problem as:
+
+$$\begin{aligned}
+    \Lambda
+    \equiv \frac{1}{S} \bigg( \big( p |u_x|^2 - q |u|^2 \big) - \lambda \big( w |u|^2 \big) \bigg)
+\end{aligned}$$
+
+The resulting Euler-Lagrange equation is then calculated in the standard way, yielding:
+
+$$\begin{aligned}
+    0
+    &= \pdv{\Lambda}{u^*} - \dv{x} \Big( \pdv{\Lambda}{u_x^*} \Big)
+    \\
+    &= - \frac{1}{S} \bigg( q u + \lambda w u + \dv{x} \big( p u_x \big) \bigg)
+\end{aligned}$$
+
+Which is clearly satisfied if and only if the following equation is fulfilled:
+
+$$\begin{aligned}
+    \dv{x} \big( p u_x \big) + q u
+    = - \lambda w u
+\end{aligned}$$
+
+This has the familiar form of a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/) (SLP),
+with $\lambda$ representing an eigenvalue.
+SLPs have useful properties, but before we can take advantage of those,
+we need to handle an important detail: the boundary conditions (BCs) on $u$.
+The above equation is only a valid SLP for certain BCs,
+as seen in the derivation of Sturm-Liouville theory.
+
+Let us return to the definition of $R[u]$,
+and integrate it by parts:
+
+$$\begin{aligned}
+    R[u]
+    &= \frac{1}{S} \int_a^b p u_x u_x^* - q u u^* \dd{x}
+    \\
+    &= \frac{1}{S} \Big[ p u_x u^* \Big]_a^b - \frac{1}{S} \int_a^b \dv{x} \Big(p u_x\Big) u^* + q u u^* \dd{x}
+\end{aligned}$$
+
+The boundary term vanishes for a subset of the BCs that make a valid SLP,
+including Dirichlet BCs $u(a) = u(b) = 0$, Neumann BCs $u_x(a) = u_x(b) = 0$, and periodic BCs.
+Therefore, we assume that this term does indeed vanish,
+such that we can use Sturm-Liouville theory later:
+
+$$\begin{aligned}
+    R[u]
+    &= - \frac{1}{S} \int_a^b \bigg( \dv{x} \Big(p u_x\Big) + q u \bigg) u^* \dd{x}
+    \equiv - \frac{1}{S} \int_a^b u^* \hat{H} u \dd{x}
+\end{aligned}$$
+
+Where $\hat{H}$ is the self-adjoint Sturm-Liouville operator.
+Because the constrained Euler-Lagrange equation is now an SLP,
+we know that it has an infinite number of real discrete eigenvalues $\lambda_n$ with a lower bound,
+corresponding to mutually orthogonal eigenfunctions $u_n(x)$.
+
+To understand the significance of this result,
+suppose we have solved the SLP,
+and now insert one of the eigenfunctions $u_n$ into $R$:
+
+$$\begin{aligned}
+    R[u_n]
+    &= - \frac{1}{S_n} \int_a^b u_n^* \hat{H} u_n \dd{x}
+    = \frac{1}{S_n} \int_a^b u_n^* \lambda_n w u_n \dd{x}
+    \\
+    &= \frac{1}{S_n} \lambda_n \int_a^b w |u_n|^2 \dd{x}
+    = \frac{S_n}{S_n} \lambda_n
+\end{aligned}$$
+
+Where $S_n$ is the normalization of $u_n$.
+In other words, when given $u_n$,
+the functional $R$ yields the corresponding eigenvalue $\lambda_n$:
+
+$$\begin{aligned}
+    \boxed{
+        R[u_n]
+        = \lambda_n
+    }
+\end{aligned}$$
+
+This powerful result was not at all clear from $R$'s initial definition.
+
+
+
+## Justification
+
+But what if we do not know the eigenfunctions? Is $R$ still useful?
+Yes, as we shall see. Suppose we make an educated guess $u(x)$
+for the ground state (i.e. lowest-eigenvalue) solution $u_0(x)$:
+
+$$\begin{aligned}
+    u(x)
+    = u_0(x) + \sum_{n = 1}^\infty c_n u_n(x)
+\end{aligned}$$
+
+Here, we are using the fact that the eigenfunctions of an SLP form a complete set,
+so our (known) guess $u$ can be expanded in the true (unknown) eigenfunctions $u_n$.
+We are assuming that $u$ is already quite close to its target $u_0$,
+such that the (unknown) expansion coefficients $c_n$ are small;
+specifically $|c_n|^2 \ll 1$.
+Let us start from what we know:
+
+$$\begin{aligned}
+    \boxed{
+        R[u]
+        = - \frac{\displaystyle\int u^* \hat{H} u \dd{x}}{\displaystyle\int u^* w u \dd{x}}
+    }
+\end{aligned}$$
+
+This quantity is known as the **Rayleigh quotient**.
+Inserting our ansatz $u$,
+and using that the true $u_n$ have corresponding eigenvalues $\lambda_n$:
+
+$$\begin{aligned}
+    R[u]
+    &= - \frac{\displaystyle\int \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \: \hat{H} \Big\{ u_0 + \sum_n c_n u_n \Big\} \dd{x}}
+    {\displaystyle\int w \Big( u_0 + \sum_n c_n u_n \Big) \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \dd{x}}
+    \\
+    &= - \frac{\displaystyle\int \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \Big( \!-\! \lambda_0 w u_0 - \sum_n c_n \lambda_n w u_n \Big) \dd{x}}
+    {\displaystyle\int w  \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \Big( u_0 + \sum_n c_n u_n \Big) \dd{x}}
+\end{aligned}$$
+
+For convenience, we switch to [Dirac notation](/know/concept/dirac-notation/)
+before evaluating further.
+
+$$\begin{aligned}
+    R
+    &= \frac{\displaystyle \Big( \bra{u_0} + \sum_n c_n^* \bra{u_n} \Big) \cdot \Big( \lambda_0 \ket{w u_0} + \sum_n c_n \lambda_n \ket{w u_n} \Big)}
+    {\displaystyle \Big( \bra{u_0} + \sum_n c_n^* \bra{u_n} \Big) \cdot \Big( \ket{w u_0} + \sum_n c_n \ket{w u_n} \Big)}
+    \\
+    &= \frac{\displaystyle \lambda_0 \braket{u_0}{w u_0} + \lambda_0 \sum_{n = 1}^\infty c_n^* \braket{u_n}{w u_0}
+    + \sum_{n = 1}^\infty c_n \lambda_n \braket{u_0}{w u_n} + \sum_{m n} c_n c_m^* \lambda_n \braket{u_m}{w u_n}}
+    {\displaystyle \braket{u_0}{w u_0} + \sum_{n = 1}^\infty c_n^* \braket{u_n}{w u_0}
+    + \sum_{n = 1}^\infty c_n \braket{u_0}{w u_n} + \sum_{m n} c_n c_m^* \braket{u_m}{w u_n}}
+\end{aligned}$$
+
+Using orthogonality $\braket{u_m}{w u_n} = S_n \delta_{mn}$,
+and the fact that $n \neq 0$ by definition, we find:
+
+$$\begin{aligned}
+    R
+    &= \frac{\displaystyle \lambda_0 S_0 + \lambda_0 \sum_n c_n^* S_n \delta_{n0}
+    + \sum_n c_n \lambda_n S_n \delta_{n0} + \sum_{m n} c_n c_m^* \lambda_n S_n \delta_{mn}}
+    {\displaystyle S_0 + \sum_n c_n^* S_n \delta_{n0} + \sum_n c_n S_n \delta_{n0} + \sum_{m n} c_n c_m^* S_n \delta_{mn}}
+    \\
+    &= \frac{\displaystyle \lambda_0 S_0 + 0 + 0 + \sum_{n} c_n c_n^* \lambda_n S_n}
+    {\displaystyle S_0 + 0 + 0 + \sum_{n} c_n c_n^* S_n}
+    = \frac{\displaystyle \lambda_0 S_0 + \sum_{n} |c_n|^2 \lambda_n S_n}
+    {\displaystyle S_0 + \sum_{n} |c_n|^2 S_n}
+\end{aligned}$$
+
+It is always possible to choose our normalizations such that $S_n = S$ for all $u_n$, leaving:
+
+$$\begin{aligned}
+    R
+    &= \frac{\displaystyle \lambda_0 S + \sum_{n} |c_n|^2 \lambda_n S}
+    {\displaystyle S + \sum_{n} |c_n|^2 S}
+    = \frac{\displaystyle \lambda_0 + \sum_{n} |c_n|^2 \lambda_n}
+    {\displaystyle 1 + \sum_{n} |c_n|^2}
+\end{aligned}$$
+
+And finally, after rearranging the numerator, we arrive at the following relation:
+
+$$\begin{aligned}
+    R
+    &= \frac{\displaystyle \lambda_0 + \sum_{n} |c_n|^2 \lambda_0 + \sum_{n} |c_n|^2 (\lambda_n - \lambda_0)}
+    {\displaystyle 1 + \sum_{n} |c_n|^2}
+    = \lambda_0 + \frac{\displaystyle \sum_{n} |c_n|^2 (\lambda_n - \lambda_0)}
+    {\displaystyle 1 + \sum_{n} |c_n|^2}
+\end{aligned}$$
+
+Thus, if we improve our guess $u$,
+then $R[u]$ approaches the true eigenvalue $\lambda_0$.
+For numerically finding $u_0$ and $\lambda_0$, this gives us a clear goal: minimize $R$, because:
+
+$$\begin{aligned}
+    \boxed{
+        R[u]
+        = \lambda_0 + \frac{\displaystyle \sum_{n = 1}^\infty |c_n|^2 (\lambda_n - \lambda_0)}
+        {\displaystyle 1 + \sum_{n = 1}^\infty |c_n|^2}
+        \ge \lambda_0
+    }
+\end{aligned}$$
+
+Note that the convergence to $\lambda_0$ goes as $|c_n|^2$,
+while $u$ converges to $u_0$ as $|c_n|$ by definition,
+so even a fairly bad guess $u$ will give a decent estimate for $\lambda_0$.
+
+
+
+## The method
+
+In the following, we stick to Dirac notation,
+since the results hold for both continuous functions $u(x)$ and discrete vectors $\vb{u}$,
+as long as the operator $\hat{H}$ is self-adjoint.
+Suppose we express our guess $\ket{u}$ as a linear combination
+of *known* basis vectors $\ket{f_n}$ with weights $a_n \in \mathbb{C}$:
+
+$$\begin{aligned}
+    \ket{u}
+    = \sum_{n = 0}^\infty c_n \ket{u_n}
+    = \sum_{n = 0}^\infty a_n \ket{f_n}
+    \approx \sum_{n = 0}^{N - 1} a_n \ket{f_n}
+\end{aligned}$$
+
+For numerical tractability, we truncate the sum at $N$ terms,
+and for generality, we allow $\ket{f_n}$ to be non-orthogonal,
+as described by an *overlap matrix* with elements $S_{mn}$:
+
+$$\begin{aligned}
+    \braket{f_m}{w f_n} = S_{m n}
+\end{aligned}$$
+
+From the discussion above,
+we know that the ground-state eigenvalue $\lambda_0$ is estimated by:
+
+$$\begin{aligned}
+    \lambda_0
+    \approx \lambda
+    = R[u]
+    = \frac{\braket*{u}{\hat{H} u}}{\braket{u}{w u}}
+    = \frac{\displaystyle \sum_{m n} a_m^* a_n \braket*{f_m}{\hat{H} f_n}}{\displaystyle \sum_{m n} a_m^* a_n \braket{f_m}{w f_n}}
+    \equiv \frac{\displaystyle \sum_{m n} a_m^* a_n H_{m n}}{\displaystyle \sum_{m n} a_m^* a_n S_{mn}}
+\end{aligned}$$
+
+And we also know that our goal is to minimize $R[u]$,
+so we vary $a_k^*$ to find its extremum:
+
+$$\begin{aligned}
+    0
+    = \pdv{R}{a_k^*}
+    &= \frac{\displaystyle \Big( \sum_{n} a_n H_{k n} \Big) \Big( \sum_{m n} a_n a_m^* S_{mn} \Big)
+    - \Big( \sum_{n} a_n S_{k n} \Big) \Big( \sum_{m n} a_n a_m^* H_{mn} \Big)}
+    {\Big( \displaystyle \sum_{m n} a_n a_m^* S_{mn} \Big)^2}
+    \\
+    &= \frac{\displaystyle \Big( \sum_{n} a_n H_{k n} \Big) - R[u] \Big( \sum_{n} a_n S_{k n}\Big)}{\braket{u}{w u}}
+    = \frac{\displaystyle \sum_{n} a_n \big(H_{k n} - \lambda S_{k n}\big)}{\braket{u}{w u}}
+\end{aligned}$$
+
+Clearly, this is only satisfied if the following holds for all $k = 0, 1, ..., N\!-\!1$:
+
+$$\begin{aligned}
+    0
+    = \sum_{n = 0}^{N - 1} a_n \big(H_{k n} - \lambda S_{k n}\big)
+\end{aligned}$$
+
+For illustrative purposes,
+we can write this as a matrix equation
+with $M_{k n} \equiv H_{k n} - \lambda S_{k n}$:
+
+$$\begin{aligned}
+    \begin{bmatrix}
+        M_{0,0} & M_{0,1} & \cdots & M_{0,N-1} \\
+        M_{1,0} & \ddots & & \vdots \\
+        \vdots & & \ddots & \vdots \\
+        M_{N-1,0} & \cdots & \cdots & M_{N-1,N-1}
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        a_0 \\ a_1 \\ \vdots \\ a_{N-1}
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        0 \\ 0 \\ \vdots \\ 0
+    \end{bmatrix}
+\end{aligned}$$
+
+Note that this looks like an eigenvalue problem for $\lambda$.
+Indeed, demanding that $\overline{M}$ cannot simply be inverted
+(i.e. the solution is non-trivial)
+yields a characteristic polynomial for $\lambda$:
+
+$$\begin{aligned}
+    0
+    = \det\!\Big[ \overline{M} \Big]
+    = \det\!\Big[ \overline{H} - \lambda \overline{S} \Big]
+\end{aligned}$$
+
+This gives a set of $\lambda$,
+which are the exact eigenvalues of $\overline{H}$,
+and the estimated eigenvalues of $\hat{H}$
+(recall that $\overline{H}$ is $\hat{H}$ expressed in a truncated basis).
+The eigenvector $\big[ a_0, a_1, ..., a_{N-1} \big]$ of the lowest $\lambda$
+gives the optimal weights to approximate $\ket{u_0}$ in the basis $\{\ket{f_n}\}$.
+Likewise, the higher $\lambda$'s eigenvectors approximate
+excited (i.e. non-ground) eigenstates of $\hat{H}$,
+although in practice the results are less accurate the higher we go.
+
+The overall accuracy is determined by how good our truncated basis is,
+i.e. how large a subspace it spans
+of the [Hilbert space](/know/concept/hilbert-space/) in which the true $\ket{u_0}$ resides.
+Clearly, adding more basis vectors will improve the results,
+at the cost of computation.
+For example, if $\hat{H}$ represents a helium atom,
+a good choice for $\{\ket{f_n}\}$ would be hydrogen orbitals,
+since those are qualitatively similar.
+
+You may find this result unsurprising;
+it makes some intuitive sense that approximating $\hat{H}$
+in a limited basis would yield a matrix $\overline{H}$ giving rough eigenvalues.
+The point of this discussion is to rigorously show
+the validity of this approach.
+
+Nowadays, there exist many other methods to calculate eigenvalues
+of complicated operators $\hat{H}$,
+but an attractive feature of the Ritz method is that it is single-step,
+whereas its competitors tend to be iterative.
+
+
+
+## References
+1.  G.B. Arfken, H.J. Weber,
+    *Mathematical methods for physicists*, 6th edition, 2005,
+    Elsevier.
+2.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/content/know/concept/sturm-liouville-theory/index.pdc b/content/know/concept/sturm-liouville-theory/index.pdc
index e530b89..ef41f19 100644
--- a/content/know/concept/sturm-liouville-theory/index.pdc
+++ b/content/know/concept/sturm-liouville-theory/index.pdc
@@ -253,7 +253,7 @@ $$\begin{aligned}
 
 When you're solving a differential eigenvalue problem,
 knowing that all eigenvalues are real is a *huge* simplification,
-so it is always worth checking whether you're dealing with an SLP.
+so it is always worth checking whether you are dealing with an SLP.
 
 Another useful fact of SLPs is that they always
 have an infinite number of discrete eigenvalues.