From 9d9693af6fb94ef4404a3c2399cb38842e5ca822 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Fri, 12 May 2023 21:19:19 +0200
Subject: Improve knowledge base

---
 .../know/concept/blasius-boundary-layer/index.md   | 82 +++++++++-------------
 source/know/concept/bloch-sphere/index.md          | 58 +++++++++------
 source/know/concept/dirac-notation/index.md        | 76 ++++++++++----------
 source/know/concept/gram-schmidt-method/index.md   | 59 +++++++++++-----
 source/know/concept/matsubara-sum/index.md         |  2 +-
 5 files changed, 157 insertions(+), 120 deletions(-)

(limited to 'source/know')

diff --git a/source/know/concept/blasius-boundary-layer/index.md b/source/know/concept/blasius-boundary-layer/index.md
index de80f96..bca933f 100644
--- a/source/know/concept/blasius-boundary-layer/index.md
+++ b/source/know/concept/blasius-boundary-layer/index.md
@@ -12,88 +12,76 @@ layout: "concept"
 In fluid dynamics, the **Blasius boundary layer** is an application of
 the [Prandtl equations](/know/concept/prandtl-equations/),
 which govern the flow of a fluid
-at large Reynolds number $$\mathrm{Re} \gg 1$$
-close to a surface.
+at large [Reynolds number](/know/concept/reynolds-number/)
+$$\mathrm{Re} \gg 1$$ close to a surface.
 Specifically, the Blasius layer is the solution
 for a half-plane approached from the edge by a fluid.
 
-A fluid with velocity field $$\va{v} = U \vu{e}_x$$ flows to the plane,
-which starts at $$y = 0$$ and exists for $$x \ge 0$$.
-To describe this, we make an ansatz
-for the *slip-flow* region's $$x$$-velocity $$v_x(x, y)$$:
+Let the half-plane lie in the $$(x,z)$$-plane (i.e. at $$y = 0$$)
+and exist for all $$x \ge 0$$, such that its edge lies on the $$z$$-axis.
+A fluid with velocity field $$\va{v} = U \vu{e}_x$$
+approaches the half-plane's edge head-on.
+To describe the fluid's movements around the plane,
+we make an ansatz for the so-called **slip-flow region**'s $$x$$-velocity $$v_x(x, y)$$:
 
 $$\begin{aligned}
     v_x
     = U f'(s)
-    \qquad \quad
+    \qquad \qquad
     s
     \equiv \frac{y}{\delta(x)}
 \end{aligned}$$
 
+Where $$\delta(x) \equiv \sqrt{\nu x / U}$$ is the boundary layer thickness
+estimate that was used to derive the Prandtl equations.
 Note that $$f'(s)$$ is the derivative of an unknown $$f(s)$$,
-and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$.
-Furthermore, $$\delta(x)$$ is the thickness of the stationary boundary layer at the surface.
-To derive the Prandtl equations,
-the estimate $$\delta(x) = \sqrt{\nu x / U}$$ was used,
-which we will stick with.
-For later use, it is worth writing the derivatives of $$s$$:
+and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$,
+i.e. the fluid is stationary at the half-plane's surface $$s = 0$$,
+and has velocity $$U$$ far away at $$s \to \infty$$.
 
-$$\begin{aligned}
-    \pdv{s}{x}
-    = - y \frac{\delta'}{\delta^2}
-    = - s \frac{\delta'}{\delta}
-    \qquad \quad
-    \pdv{s}{y}
-    = \frac{1}{\delta}
-\end{aligned}$$
-
-Inserting the ansatz for $$v_x$$ into the incompressibility condition then yields:
+Inserting the ansatz into the incompressibility condition $$\nabla \cdot \va{v} = 0$$ yields:
 
 $$\begin{aligned}
     \pdv{v_y}{y}
     = - \pdv{v_x}{x}
-    = U s f'' \frac{\delta'}{\delta}
+    = - \pdv{v_x}{s} \pdv{s}{x}
+    = U \frac{\delta'}{\delta} s f''
 \end{aligned}$$
 
-Which we integrate to get an expression for the $$y$$-velocity $$v_y$$, namely:
+Which we integrate by parts to get an expression for the $$y$$-velocity $$v_y$$, namely:
 
 $$\begin{aligned}
     v_y
     = U \frac{\delta'}{\delta} \int s f'' \dd{y}
+    = U \delta' \bigg( s f' - \int f' \dd{s} \bigg)
     = U \delta' \: (s f' - f)
 \end{aligned}$$
 
 Now, consider the main Prandtl equation,
-assuming that the attack velocity $$U$$ is constant:
-
-$$\begin{aligned}
-    v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
-    = \nu \pdvn{2}{v_x}{y}
-\end{aligned}$$
-
-Inserting our expressions for $$v_x$$ and $$v_y$$ into this leads us to:
-
-$$\begin{aligned}
-    - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f)
-    = \nu U \frac{1}{\delta^2} f'''
-\end{aligned}$$
-
-After  multiplying it by $$\delta^2 / U$$ and cancelling out some terms,
-it reduces to:
+assuming the attack velocity $$U$$ is constant.
+Inserting our expressions for $$v_x$$ and $$v_y$$ into it gives:
 
 $$\begin{aligned}
-    \nu f''' + U \delta' \delta f'' f
-    = 0
+    \nu \pdvn{2}{v_x}{y}
+    &= v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
+    \\
+    \nu \pdvn{2}{v_x}{s} \pdvn{2}{s}{y}
+    &= v_x \pdv{v_x}{s} \pdv{s}{x} + v_y \pdv{v_x}{s} \pdv{s}{y}
+    \\
+    \nu U \frac{1}{\delta^2} f'''
+    &= - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f)
 \end{aligned}$$
 
-Then, substituting $$\delta(x) = \sqrt{\nu x / U}$$ and $$\delta'(x) = (1/2) \sqrt{\nu / (U x)}$$ yields:
+We multiply by $$\delta^2 / U$$, cancel out some terms,
+and substitute $$\delta(x) \equiv \sqrt{\nu x / U}$$, leaving:
 
 $$\begin{aligned}
-    \nu f''' + U \frac{\nu}{2 U} f'' f
-    = 0
+    \nu f'''
+    &= - U \delta' \delta f'' f
+    = - U \frac{\nu}{2 U} f'' f
 \end{aligned}$$
 
-Simplifying this leads us to the **Blasius equation**,
+This leads us to the **Blasius equation**,
 which is a nonlinear ODE for $$f(s)$$:
 
 $$\begin{aligned}
diff --git a/source/know/concept/bloch-sphere/index.md b/source/know/concept/bloch-sphere/index.md
index 99ac45d..b348d22 100644
--- a/source/know/concept/bloch-sphere/index.md
+++ b/source/know/concept/bloch-sphere/index.md
@@ -21,23 +21,23 @@ and their extremes are the eigenstates of the Pauli matrices:
 
 $$\begin{aligned}
     \hat{\sigma}_z
-    \to \{\Ket{0}, \Ket{1}\}
+    \to \{\ket{0}, \ket{1}\}
     \qquad
     \hat{\sigma}_x
-    \to \{\Ket{+}, \Ket{-}\}
+    \to \{\ket{+}, \ket{-}\}
     \qquad
     \hat{\sigma}_y
-    \to \{\Ket{+i}, \Ket{-i}\}
+    \to \{\ket{+i}, \ket{-i}\}
 \end{aligned}$$
 
-Where the latter two states are expressed as follows in the conventional $$z$$-basis:
+Where the latter two pairs are expressed as follows in the conventional $$z$$-basis:
 
 $$\begin{aligned}
-    \Ket{\pm}
-    = \frac{\Ket{0} \pm \Ket{1}}{\sqrt{2}}
-    \qquad \quad
-    \Ket{\pm i}
-    = \frac{\Ket{0} \pm i \Ket{1}}{\sqrt{2}}
+    \ket{\pm}
+    = \frac{\ket{0} \pm \ket{1}}{\sqrt{2}}
+    \qquad \qquad
+    \ket{\pm i}
+    = \frac{\ket{0} \pm i \ket{1}}{\sqrt{2}}
 \end{aligned}$$
 
 More generally, every point on the surface of the sphere
@@ -45,11 +45,12 @@ describes a pure qubit state in terms of the angles $$\theta$$ and $$\varphi$$,
 respectively the elevation and azimuth:
 
 $$\begin{aligned}
-    \Ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \Ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \Ket{1}
+    \ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \ket{1}
 \end{aligned}$$
 
-We can generalize this further by describing points using the **Bloch vector** $$\vec{r}$$,
-with radius $$r \le 1$$:
+Another way to describe states is the **Bloch vector** $$\vec{r}$$,
+which is simply the $$(x,y,z)$$-coordinates of a point on the sphere.
+Let the radius $$r \le 1$$:
 
 $$\begin{aligned}
     \boxed{
@@ -60,8 +61,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Note that $$\vec{r}$$ is not actually a qubit state,
-but rather an implicit description of one,
-meaning that it does not need to be normalized.
+but rather a description of one.
 The main point of the Bloch vector is that it allows us
 to describe the qubit using a [density operator](/know/concept/density-operator/):
 
@@ -73,7 +73,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where $$\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$$ is the Pauli "vector".
-Now, we know that $$\hat{\rho}$$ represents a pure ensemble
+Now, we know that a density matrix represents a pure ensemble
 if and only if it is idempotent, i.e. $$\hat{\rho}^2 = \hat{\rho}$$:
 
 $$\begin{aligned}
@@ -82,8 +82,8 @@ $$\begin{aligned}
     = \frac{1}{4} \Big( \hat{I} + 2 (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
 \end{aligned}$$
 
-You can easily convince yourself that if $$(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$$,
-then we get $$\hat{\rho}$$ again, and the state is pure:
+You can easily convince yourself that, if $$(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$$,
+we get $$\hat{\rho}$$ again, so the state is pure:
 
 $$\begin{aligned}
     (\vec{r} \cdot \vec{\sigma})^2
@@ -109,11 +109,28 @@ else if $$r < 1$$ it is mixed.
 
 Another useful property of the Bloch vector
 is that the expectation value of the Pauli matrices
-are given by the corresponding component of $$\vec{r}$$,
-for example for $$\hat{\sigma}_z$$:
+are given by the corresponding component of $$\vec{r}$$:
 
 $$\begin{aligned}
-    \Expval{\hat{\sigma}_z}
+    \boxed{
+        \begin{aligned}
+            \expval{\hat{\sigma}_{x}}
+            &= r_{x}
+            \\
+            \expval{\hat{\sigma}_{y}}
+            &= r_{y}
+            \\
+            \expval{\hat{\sigma}_{z}}
+            &= r_{z}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+This is a consequence of the above form of the density operator $$\hat{\rho}$$.
+For example for $$\hat{\sigma}_z$$:
+
+$$\begin{aligned}
+    \expval{\hat{\sigma}_z}
     &= \Tr(\hat{\rho} \hat{\sigma}_z)
     = \frac{1}{2} \Tr\!\big(\hat{\sigma}_z + (\vec{r} \cdot \vec{\sigma}) \hat{\sigma}_z \big)
     = \frac{1}{2} \Tr\!\big( (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z) \hat{\sigma}_z \big)
@@ -124,6 +141,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## References
 1.  N. Brunner,
     *Quantum information theory: lecture notes*,
diff --git a/source/know/concept/dirac-notation/index.md b/source/know/concept/dirac-notation/index.md
index 46cc325..2830a33 100644
--- a/source/know/concept/dirac-notation/index.md
+++ b/source/know/concept/dirac-notation/index.md
@@ -8,47 +8,49 @@ categories:
 layout: "concept"
 ---
 
-**Dirac notation** is a notation to do calculations in a [Hilbert space](/know/concept/hilbert-space/)
-without needing to worry about the space's representation. It is
-basically the *lingua franca* of quantum mechanics.
-
-In Dirac notation there are **kets** $$\Ket{V}$$ from the Hilbert space
-$$\mathbb{H}$$ and **bras** $$\Bra{V}$$ from a dual $$\mathbb{H}'$$ of the
-former. Crucially, the bras and kets are from different Hilbert spaces
-and therefore cannot be added, but every bra has a corresponding ket and
-vice versa.
+**Dirac notation** enables us to do calculations
+in a general [Hilbert space](/know/concept/hilbert-space/)
+without needing to worry about the space's representation.
+It is the *lingua franca* of quantum mechanics.
+
+In Dirac notation there are
+**kets** $$\ket{V}$$ from the Hilbert space $$\mathbb{H}$$
+and **bras** $$\bra{V}$$ from its dual space $$\mathbb{H}'$$.
+Crucially, the bras and kets are from different Hilbert spaces
+and therefore cannot be added,
+but every bra has a corresponding ket and vice versa.
 
 Bras and kets can be combined in two ways: the **inner product**
-$$\Inprod{V}{W}$$, which returns a scalar, and the **outer product**
-$$\Ket{V} \Bra{W}$$, which returns a mapping $$\hat{L}$$ from kets $$\Ket{V}$$
-to other kets $$\Ket{V'}$$, i.e. a linear operator. Recall that the
-Hilbert inner product must satisfy:
+$$\inprod{V}{W}$$, which returns a scalar, and the **outer product**
+$$\ket{V} \bra{W}$$, which returns a linear operator
+that maps kets $$\ket{V}$$ to other kets $$\ket{V'}$$.
+Recall that by definition the Hilbert inner product must satisfy:
 
 $$\begin{aligned}
-    \Inprod{V}{W} = \Inprod{W}{V}^*
+    \inprod{V}{W} = \inprod{W}{V}^*
 \end{aligned}$$
 
-So far, nothing has been said about the actual representation of bras or
-kets. If we represent kets as $$N$$-dimensional columns vectors, the
-corresponding bras are given by the kets' adjoints, i.e. their transpose
-conjugates:
+So far, nothing has been said about the actual representation of bras or kets.
+If we represent kets as $$N$$-dimensional columns vectors,
+the corresponding bras are given by the kets' adjoints,
+i.e. their transpose conjugates:
 
 $$\begin{aligned}
-    \Ket{V} =
+    \ket{V} =
     \begin{bmatrix}
         v_1 \\ \vdots \\ v_N
     \end{bmatrix}
     \quad \implies \quad
-    \Bra{V} =
+    \bra{V} =
     \begin{bmatrix}
         v_1^* & \cdots & v_N^*
     \end{bmatrix}
 \end{aligned}$$
 
-The inner product $$\Inprod{V}{W}$$ is then just the familiar dot product $$V \cdot W$$:
+The inner product $$\inprod{V}{W}$$ is then just the familiar dot product $$V \cdot W$$:
 
 $$\begin{gathered}
-    \Inprod{V}{W}
+    \inprod{V}{W}
     =
     \begin{bmatrix}
         v_1^* & \cdots & v_N^*
@@ -60,10 +62,11 @@ $$\begin{gathered}
     = v_1^* w_1 + ... + v_N^* w_N
 \end{gathered}$$
 
-Meanwhile, the outer product $$\Ket{V} \Bra{W}$$ creates an $$N \cross N$$ matrix:
+Meanwhile, the outer product $$\ket{V} \bra{W}$$ creates an $$N \cross N$$ matrix,
+which can be thought of as applying an operation to any vector it multiplies:
 
 $$\begin{gathered}
-    \Ket{V} \Bra{W}
+    \ket{V} \bra{W}
     =
     \begin{bmatrix}
         v_1 \\ \vdots \\ v_N
@@ -80,15 +83,14 @@ $$\begin{gathered}
     \end{bmatrix}
 \end{gathered}$$
 
-If the kets are instead represented by functions $$f(x)$$ of
-$$x \in [a, b]$$, then the bras represent *functionals* $$F[u(x)]$$ which
-take an unknown function $$u(x)$$ as an argument and turn it into a scalar
-using integration:
+If the kets are instead represented by continuous functions $$f(x)$$ of $$x \in [a, b]$$,
+then the bras are *functionals* $$F[u(x)]$$
+that take an arbitrary function $$u(x)$$ as an argument and return a scalar:
 
 $$\begin{aligned}
-    \Ket{f} = f(x)
+    \ket{f} = f(x)
     \quad \implies \quad
-    \Bra{f}
+    \bra{f}
     = F[u(x)]
     = \int_a^b f^*(x) \: u(x) \dd{x}
 \end{aligned}$$
@@ -96,23 +98,25 @@ $$\begin{aligned}
 Consequently, the inner product is simply the following familiar integral:
 
 $$\begin{gathered}
-    \Inprod{f}{g}
+    \inprod{f}{g}
     = F[g(x)]
     = \int_a^b f^*(x) \: g(x) \dd{x}
 \end{gathered}$$
 
-However, the outer product becomes something rather abstract:
+However, the outer product is then rather abstract:
+a continuous analogue of a matrix:
 
 $$\begin{gathered}
-    \Ket{f} \Bra{g}
+    \ket{f} \bra{g}
     = f(x) \: G[u(x)]
     = f(x) \int_a^b g^*(\xi) \: u(\xi) \dd{\xi}
 \end{gathered}$$
 
-This result makes more sense if we surround it by a bra and a ket:
+This maybe makes more sense if we surround it
+by a bra $$\bra{u}$$ and a ket $$\ket{w}$$ and rearrange:
 
 $$\begin{aligned}
-    \Bra{u} \!\Big(\!\Ket{f} \Bra{g}\!\Big)\! \Ket{w}
+    \bra{u} \!\Big(\!\ket{f} \bra{g}\!\Big)\! \ket{w}
     &= U\big[f(x) \: G[w(x)]\big]
     = U\Big[ f(x) \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big]
     \\
@@ -120,7 +124,7 @@ $$\begin{aligned}
     \\
     &= \Big( \int_a^b u^*(x) \: f(x) \dd{x} \Big) \Big( \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big)
     \\
-    &= \Inprod{u}{f} \Inprod{g}{w}
+    &= \inprod{u}{f} \inprod{g}{w}
 \end{aligned}$$
 
 
diff --git a/source/know/concept/gram-schmidt-method/index.md b/source/know/concept/gram-schmidt-method/index.md
index 70ad512..a62522e 100644
--- a/source/know/concept/gram-schmidt-method/index.md
+++ b/source/know/concept/gram-schmidt-method/index.md
@@ -8,39 +8,66 @@ categories:
 layout: "concept"
 ---
 
-Given a set of linearly independent non-orthonormal vectors
-$$\ket{V_1}, \ket{V_2}, ...$$ from a [Hilbert space](/know/concept/hilbert-space/),
-the **Gram-Schmidt method**
-turns them into an orthonormal set $$\ket{n_1}, \ket{n_2}, ...$$ as follows:
+Given a set of $$N$$ linearly independent vectors $$\ket{V_1}, \ket{V_2}, ...$$
+from a [Hilbert space](/know/concept/hilbert-space/),
+the **Gram-Schmidt method** is an algorithm that turns them
+into an orthonormal set $$\ket{n_1}, \ket{n_2}, ...$$ as follows
+(in [Dirac notation](/know/concept/dirac-notation/)):
 
 1.  Take the first vector $$\ket{V_1}$$ and normalize it to get $$\ket{n_1}$$:
     
     $$\begin{aligned}
-        \ket{n_1} = \frac{\ket{V_1}}{\sqrt{\inprod{V_1}{V_1}}}
+        \ket{n_1}
+        = \frac{\ket{V_1}}{\sqrt{\inprod{V_1}{V_1}}}
     \end{aligned}$$
     
-2.  Begin loop. Take the next non-orthonormal vector $$\ket{V_j}$$, and
+2.  Begin loop. Take the next input vector $$\ket{V_j}$$, and
     subtract from it its projection onto every already-processed vector:
     
     $$\begin{aligned}
-        \ket{n_j'} = \ket{V_j} - \ket{n_1} \inprod{n_1}{V_j} - \ket{n_2} \inprod{n_2}{V_j} - ... - \ket{n_{j-1}} \inprod{n_{j-1}}{V_{j-1}}
+        \ket{g_j}
+        = \ket{V_j} - \ket{n_1} \inprod{n_1}{V_j} - \ket{n_2} \inprod{n_2}{V_j} - ... - \ket{n_{j-1}} \inprod{n_{j-1}}{V_j}
     \end{aligned}$$
     
-    This leaves only the part of $$\ket{V_j}$$ which is orthogonal to
-    $$\ket{n_1}$$, $$\ket{n_2}$$, etc. This why the input vectors must be
-    linearly independent; otherwise $$\Ket{n_j'}$$ may become zero at some
-    point.
+    This leaves only the part of $$\ket{V_j}$$
+    that is orthogonal to all previous $$\ket{n_k}$$.
+    This why the input vectors must be linearly independent;
+    otherwise $$\ket{g_j}$$ could become zero.
     
-3.  Normalize the resulting ortho*gonal* vector $$\ket{n_j'}$$ to make it
+    On a computer, the resulting $$\ket{g_j}$$ will
+    not be perfectly orthogonal due to rounding errors.
+    The above description of step #2 is particularly bad.
+    A better approach is:
+    
+    $$\begin{aligned}
+        \ket{g_j^{(1)}}
+        &= \ket{V_j} - \ket{n_1} \inprod{n_1}{V_j}
+        \\
+        \ket{g_j^{(2)}}
+        &= \ket{g_j^{(1)}} - \ket{n_2} \inprod{n_2}{g_j^{(1)}}
+        \\
+        \vdots
+        \\
+        \ket{g_j}
+        = \ket{g_j^{(j-1)}}
+        &= \ket{g_j^{(j-2)}} - \ket{n_{j-2}} \inprod{n_{j-2}}{g_j^{(j-2)}}
+    \end{aligned}$$
+    
+    In other words, instead of projecting $$\ket{V_j}$$ directly onto all $$\ket{n_k}$$,
+    we instead project only the part of $$\ket{V_j}$$ that has already been made orthogonal
+    to all previous $$\ket{n_m}$$ with $$m < k$$.
+    This is known as the **modified Gram-Schmidt method**.
+    
+3.  Normalize the resulting ortho*gonal* vector $$\ket{g_j}$$ to make it
     ortho*normal*:
     
     $$\begin{aligned}
-                \ket{n_j} = \frac{\ket{n_j'}}{\sqrt{\inprod{n_j'}{n_j'}}}
+        \ket{n_j}
+        = \frac{\ket{g_j}}{\sqrt{\inprod{g_j}{g_j}}}
     \end{aligned}$$
     
-4.  Loop back to step 2, taking the next vector $$\ket{V_{j+1}}$$.
-
-If you are unfamiliar with this notation, take a look at [Dirac notation](/know/concept/dirac-notation/).
+4.  Loop back to step 2, taking the next vector $$\ket{V_{j+1}}$$,
+    until all $$N$$ have been processed.
 
 
 
diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-sum/index.md
index 8b903d4..aef8379 100644
--- a/source/know/concept/matsubara-sum/index.md
+++ b/source/know/concept/matsubara-sum/index.md
@@ -36,7 +36,7 @@ $$\begin{aligned}
     = \sum_{z_p} e^{z_p \tau} \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) h(z) \big)
 \end{aligned}$$
 
-Where we have applied the residue theorem
+Where we have applied the [residue theorem](/know/concept/residue-theorem/)
 to get a sum over all simple poles $$z_p$$
 of either $$g$$ or $$h$$ (but not both) enclosed by $$C$$.
 Clearly, we could make this look like a Matsubara sum,
-- 
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