Expand knowledge base

8 files changed, 720 insertions, 7 deletions

diff --git a/content/know/concept/convolution-theorem/index.pdc b/content/know/concept/convolution-theorem/index.pdc
index 1454cc0..2712c21 100644
--- a/content/know/concept/convolution-theorem/index.pdc
+++ b/content/know/concept/convolution-theorem/index.pdc
@@ -69,7 +69,8 @@ $$\begin{aligned}
 ## Laplace transform
 
 For functions $f(t)$ and $g(t)$ which are only defined for $t \ge 0$,
-the convolution theorem can also be stated using the Laplace transform:
+the convolution theorem can also be stated using
+the [Laplace transform](/know/concept/laplace-transform/):
 
 $$\begin{aligned}
     \boxed{(f * g)(t) = \hat{\mathcal{L}}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
diff --git a/content/know/concept/hamiltonian-mechanics/index.pdc b/content/know/concept/hamiltonian-mechanics/index.pdc
new file mode 100644
index 0000000..0a7306e
--- /dev/null
+++ b/content/know/concept/hamiltonian-mechanics/index.pdc
@@ -0,0 +1,312 @@
+---
+title: "Hamiltonian mechanics"
+firstLetter: "H"
+publishDate: 2021-07-03
+categories:
+- Physics
+- Classical mechanics
+
+date: 2021-07-03T14:39:14+02:00
+draft: false
+markup: pandoc
+---
+
+# Hamiltonian mechanics
+
+**Hamiltonian mechanics** is an alternative formulation of classical mechanics,
+which equivalent to Newton's laws,
+but often mathematically advantageous.
+It is built on the shoulders of [Lagrangian mechanics](/know/concept/lagrangian-mechanics/),
+which is in turn built on [variational calculus](/know/concept/calculus-of-variations/).
+
+
+## Definitions
+
+In Lagrangian mechanics, use a Lagrangian $L$,
+which depends on position $q(t)$ and velocity $\dot{q}(t)$,
+to define the momentum $p(t)$ as a derived quantity.
+Hamiltonian mechanics switches the roles of $\dot{q}$ and $p$:
+the **Hamiltonian** $H$ is a function of $q$ and $p$,
+and the velocity $\dot{q}$ is derived from it:
+
+$$\begin{aligned}
+    \pdv{L(q, \dot{q})}{\dot{q}} = p
+    \qquad \quad
+    \pdv{H(q, p)}{p} \equiv \dot{q}
+\end{aligned}$$
+
+Conveniently, this switch turns out to be
+[Legendre transformation](/know/concept/legendre-transform/):
+$H$ is the Legendre transform of $L$,
+with $p = \partial L / \partial \dot{q}$ taken as
+the coordinate to replace $\dot{q}$.
+Therefore:
+
+$$\begin{aligned}
+    \boxed{
+        H(q, p) \equiv \dot{q} \: p - L(q, \dot{q})
+    }
+\end{aligned}$$
+
+This almost always works,
+because $L$ is usually a second-order polynomial of $\dot{q}$,
+and thus convex as required for Legendre transformation.
+In the above expression,
+$\dot{q}$ must be rewritten in terms of $p$ and $q$,
+which is trivial, since $p$ is proportional to $\dot{q}$ by definition.
+
+The Hamiltonian $H$ also has a direct physical meaning:
+for a mass $m$, and for $L = T - V$,
+it is straightforward to show that $H$ represents the total energy $T + V$:
+
+$$\begin{aligned}
+    H
+    = \dot{q} \: p - L
+    = m \dot{q}^2 - L
+    = 2 T - (T - V)
+    = T + V
+\end{aligned}$$
+
+Just as Lagrangian mechanics,
+Hamiltonian mechanics scales well for large systems.
+Its definition is generalized as follows to $N$ objects,
+where $p$ is shorthand for $p_1, ..., p_N$:
+
+$$\begin{aligned}
+    \boxed{
+        H(q, p)
+        \equiv \bigg( \sum_{n = 1}^N \dot{q}_n \: p_n \bigg) - L(q, \dot{q})
+    }
+\end{aligned}$$
+
+The positions and momenta $(q, p)$ form a phase space,
+i.e. they fully describe the state.
+
+An extremely useful concept in Hamiltonian mechanics
+is the **Poisson bracket** (PB),
+which is a binary operation on two quantities $A(q, p)$ and $B(q, p)$,
+denoted by $\{A, B\}$:
+
+$$\begin{aligned}
+    \boxed{
+        \{ A, B \}
+        \equiv \sum_{n = 1}^N \Big( \pdv{A}{q_n} \pdv{B}{p_n} - \pdv{A}{p_n} \pdv{B}{q_n} \Big)
+    }
+\end{aligned}$$
+
+
+## Canonical equations
+
+Lagrangian mechanics has a single Euler-Lagrange equation per object,
+yielding $N$ second-order equations of motion in total.
+In contrast, Hamiltonian mechanics has $2 N$ first-order equations of motion,
+known as **Hamilton's canonical equations**:
+
+$$\begin{aligned}
+    \boxed{
+        - \pdv{H}{q_n} = \dot{p}_n
+        \qquad
+        \pdv{H}{p_n} = \dot{q}_n
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-canoneq"/>
+<label for="proof-canoneq">Proof</label>
+<div class="hidden">
+<label for="proof-canoneq">Proof.</label>
+For the first equation,
+we differentiate $H$ with respect to $q_n$,
+and use the chain rule:
+$$\begin{aligned}
+    \pdv{H}{q_n}
+    &= \pdv{q_n} \Big( \sum_{j} \dot{q}_j \: p_j - L \Big)
+    \\
+    &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{q_n} + p_j \pdv{\dot{q}_j}{q_n} \Big)
+    - \Big( \pdv{L}{q_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{q_n} \Big) \bigg)
+    \\
+    &= \sum_{j} \Big( p_j \pdv{\dot{q}_j}{q_n} - \pdv{L}{q_n} - p_j \pdv{\dot{q}_j}{q_n} \Big)
+    = - \pdv{L}{q_n}
+\end{aligned}$$
+
+We use the Euler-Lagrange equation here,
+leading to the desired equation:
+
+$$\begin{aligned}
+    - \pdv{L}{q_n} = - \dv{t} \Big( \pdv{L}{\dot{q}_n} \Big) = - \dv{p_n}{t} = - \dot{p}_n
+\end{aligned}$$
+
+The second equation is somewhat trivial,
+since $H$ is defined to satisfy it in the first place.
+Nevertheless, we can prove it by brute force,
+using the same approach as above:
+$$\begin{aligned}
+    \pdv{H}{p_n}
+    &= \pdv{p_n} \Big( \sum_{j} \dot{q}_j \: p_j - L \Big)
+    \\
+    &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{p_n} + p_j \pdv{\dot{q}_j}{p_n} \Big)
+    - \Big( \pdv{L}{q_j} \pdv{q_j}{p_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{p_n} \Big) \bigg)
+    \\
+    &= \dot{q}_n + \sum_{j} \Big( p_j \pdv{\dot{q}_j}{p_n}
+    - 0 \pdv{L}{q_j} - p_j \pdv{\dot{q}_j}{p_n} \Big)
+    = \dot{q}_n
+\end{aligned}$$
+</div>
+</div>
+
+Just like in Lagrangian mechanics, if $H$ does not explicitly contain $q_n$,
+then $q_n$ is called a **cyclic coordinate**, and leads to the conservation of $p_n$:
+
+$$\begin{aligned}
+    \dot{p}_n = - \pdv{H}{q_n} = 0
+    \quad \implies \quad
+    p_n = \mathrm{conserved}
+\end{aligned}$$
+
+Of course, there may be other conserved quantities.
+Generally speaking, the $t$-derivative of an arbitrary quantity $A(q, p, t)$ is as follows,
+where $\pdv*{t}$ is a "soft" derivative
+(only affects explicit occurrences of $t$),
+and $\dv*{t}$ is a "hard" derivative
+(also affects implicit $t$ inside $q$ and $p$):
+
+$$\begin{aligned}
+    \boxed{
+        \dv{A}{t}
+        = \{ A, H \} + \pdv{A}{t}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-diff-t"/>
+<label for="proof-diff-t">Proof</label>
+<div class="hidden">
+<label for="proof-diff-t">Proof.</label>
+We differentiate via the multivariate chain rule,
+insert the canonical equations,
+and eventually recognize the PB definition:
+
+$$\begin{aligned}
+    \dv{A}{t}
+    &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{q_n}{t} + \pdv{A}{p_n} \pdv{p_n}{t} \Big) + \pdv{A}{t}
+    \\
+    &= \sum_{n} \Big( \pdv{A}{q_n} \dot{q}_n + \pdv{A}{p_n} \dot{p}_n \Big) + \pdv{A}{t}
+    \\
+    &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{H}{p_n} - \pdv{A}{p_n} \pdv{H}{q_n} \Big) + \pdv{A}{t}
+\end{aligned}$$
+</div>
+</div>
+
+Assuming that $H$ does not explicitly depend on $t$,
+the above property naturally leads us to an alternative
+way of writing Hamilton's canonical equations:
+
+$$\begin{aligned}
+    \dot{q}_n = \{ q_n, H \}
+    \qquad \quad
+    \dot{p}_n = \{ p_n, H \}
+\end{aligned}$$
+
+
+
+## Canonical coordinates
+
+So far, we have assumed that the phase space coordinates $(q, p)$
+are the *positions* and *canonical momenta*, respectively,
+and that led us to Hamilton's canonical equations.
+
+In theory, we could make a transformation of the following general form:
+
+$$\begin{aligned}
+    q \to Q(q, p)
+    \qquad \quad
+    p \to P(q, p)
+\end{aligned}$$
+
+However, most choices of $(Q, P)$ would not preserve Hamilton's equations.
+Any $(Q, P)$ that do keep this form
+are known as **canonical coordinates**,
+and the corresponding transformation is a **canonical transformation**.
+That is, any $(Q, P)$ that satisfy:
+
+$$\begin{aligned}
+    - \pdv{H}{Q_n} = \dot{P}_n
+    \qquad \quad
+    \pdv{H}{P_n} = \dot{Q}_n
+\end{aligned}$$
+
+Then we might as well write $H(q, p)$ as $H(Q, P)$.
+So, which $(Q, P)$ fulfill this?
+It turns out that the following must be satisfied for all $n, j$,
+where $\delta_{nj}$ is the Kronecker delta:
+
+$$\begin{aligned}
+    \boxed{
+        \{ Q_n, Q_j \} = \{ P_n, P_j \} = 0
+        \qquad
+        \{ Q_n, P_j \} = \delta_{nj}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-cantrans"/>
+<label for="proof-cantrans">Proof</label>
+<div class="hidden">
+<label for="proof-cantrans">Proof.</label>
+Assuming that $Q_n$, $P_n$ and $H$ do not explicitly depend on $t$,
+we use our expression for the $t$-derivative of an arbitrary quantity,
+and apply the multivariate chain rule to it:
+
+$$\begin{aligned}
+    \dot{Q}_n
+    &= \{Q_n, H\}
+    = \sum_{n} \bigg( \pdv{Q_n}{q_n} \pdv{H}{p_n} - \pdv{Q_n}{p_n} \pdv{H}{q_n} \bigg)
+    \\
+    &= \sum_{n, j} \bigg( \pdv{Q_n}{q_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{p_n} + \pdv{H}{P_j} \pdv{P_j}{p_n} \Big)
+    - \pdv{Q_n}{p_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{q_n} + \pdv{H}{P_j} \pdv{P_j}{q_n} \Big) \bigg)
+    \\
+    &= \sum_{n, j} \bigg( \pdv{H}{Q_j} \Big( \pdv{Q_n}{q_n} \pdv{Q_j}{p_n} - \pdv{Q_n}{p_n} \pdv{Q_j}{q_n} \Big)
+    + \pdv{H}{P_j} \Big( \pdv{Q_n}{q_n} \pdv{P_j}{p_n} - \pdv{Q_n}{p_n} \pdv{P_j}{q_n} \Big) \bigg)
+    \\
+    &= \sum_{j} \bigg( \pdv{H}{Q_j} \{Q_n, Q_j\} + \pdv{H}{P_j} \{Q_n, P_j\} \bigg)
+\end{aligned}$$
+
+This is equivalent to Hamilton's equation $\dot{Q}_n = \pdv*{H}{P_n}$
+if and only if $\{Q_n, Q_j\} = 0$ for all $n$ and $j$,
+and if $\{Q_n, P_j\} = \delta_{nj}$.
+
+Next, we do the exact same thing with $P_n$ instead of $Q_n$,
+giving an analogous result:
+
+$$\begin{aligned}
+    \dot{P}_n
+    &= \{P_n, H\}
+    = \sum_{n} \bigg( \pdv{P_n}{q_n} \pdv{H}{p_n} - \pdv{P_n}{p_n} \pdv{H}{q_n} \bigg)
+    \\
+    &= \sum_{n, j} \bigg( \pdv{P_n}{q_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{p_n} + \pdv{H}{P_j} \pdv{P_j}{p_n} \Big)
+    - \pdv{P_n}{p_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{q_n} + \pdv{H}{P_j} \pdv{P_j}{q_n} \Big) \bigg)
+    \\
+    &= \sum_{n, j} \bigg( \pdv{H}{Q_j} \Big( \pdv{P_n}{q_n} \pdv{Q_j}{p_n} - \pdv{P_n}{p_n} \pdv{Q_j}{q_n} \Big)
+    + \pdv{H}{P_j} \Big( \pdv{P_n}{q_n} \pdv{P_j}{p_n} - \pdv{P_n}{p_n} \pdv{P_j}{q_n} \Big) \bigg)
+    \\
+    &= \sum_{j} \bigg( \pdv{H}{Q_j} \{P_n, Q_j\} + \pdv{H}{P_j} \{P_n, P_j\} \bigg)
+\end{aligned}$$
+
+Which is equivalent to Hamilton's equation $\dot{P}_n = -\pdv*{H}{Q_n}$
+if and only if $\{P_n, P_j\} = 0$,
+and $\{Q_n, P_j\} = - \delta_{nj}$.
+The PB is anticommutative,
+i.e. $\{A, B\} = - \{B, A\}$.
+</div>
+</div>
+
+If you have experience with quantum mechanics,
+the latter equation should look suspiciously similar
+to the *canonical commutation relation* $[\hat{Q}, \hat{P}] = i \hbar$.
+
+
+
+## References
+1.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.
diff --git a/content/know/concept/harmonic-oscillator/index.pdc b/content/know/concept/harmonic-oscillator/index.pdc
index 5d97a24..4219602 100644
--- a/content/know/concept/harmonic-oscillator/index.pdc
+++ b/content/know/concept/harmonic-oscillator/index.pdc
@@ -68,7 +68,8 @@ $$\begin{aligned}
     x(t) = \cos\!(\omega_0 t)
 \end{aligned}$$
 
-When using Lagrangian mechanics or Hamiltonian mechanics,
+When using [Lagrangian](/know/concept/lagrangian-mechanics/)
+or Hamiltonian mechanics,
 we need to know the potential energy $V(x)$
 added to the system by a displacement to $x$.
 This equals the work done by the displacement,
diff --git a/content/know/concept/lagrangian-mechanics/index.pdc b/content/know/concept/lagrangian-mechanics/index.pdc
index 24e1f93..dcf555b 100644
--- a/content/know/concept/lagrangian-mechanics/index.pdc
+++ b/content/know/concept/lagrangian-mechanics/index.pdc
@@ -4,6 +4,7 @@ firstLetter: "L"
 publishDate: 2021-07-01
 categories:
 - Physics
+- Classical mechanics
 
 date: 2021-07-01T18:44:43+02:00
 draft: false
@@ -16,7 +17,10 @@ markup: pandoc
 which is equivalent to Newton's laws,
 but offers some advantages.
 Its mathematical backbone is the
-[calculus of variations](/know/concept/calculus-of-variations/).
+[calculus of variations](/know/concept/calculus-of-variations/),
+and hence it is built on the **principle of least action**,
+which states that the path taken by a system
+will be a minimum of the **action** (i.e. energy cost) of that path.
 
 For a moving object with position $x(t)$ and velocity $\dot{x}(t)$,
 we define the Lagrangian $L$ as the difference
diff --git a/content/know/concept/laplace-transform/index.pdc b/content/know/concept/laplace-transform/index.pdc
new file mode 100644
index 0000000..bd7673b
--- /dev/null
+++ b/content/know/concept/laplace-transform/index.pdc
@@ -0,0 +1,125 @@
+---
+title: "Laplace transform"
+firstLetter: "L"
+publishDate: 2021-07-02
+categories:
+- Mathematics
+- Physics
+
+date: 2021-07-02T15:48:30+02:00
+draft: false
+markup: pandoc
+---
+
+# Laplace transform
+
+The **Laplace transform** is an integral transform
+that losslessly converts a function $f(t)$ of a real variable $t$,
+into a function $\tilde{f}(s)$ of a complex variable $s$,
+where $s$ is sometimes called the **complex frequency**,
+analogously to the [Fourier transform](/know/concept/fourier-transform/).
+The transform is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{f}(s)
+        \equiv \hat{\mathcal{L}}\{f(t)\}
+        \equiv \int_0^\infty f(t) \exp\!(- s t) \dd{t}
+    }
+\end{aligned}$$
+
+Depending on $f(t)$, this integral may diverge.
+This is solved by restricting the domain of $\tilde{f}(s)$
+to $s$ where $\mathrm{Re}\{s\} > s_0$,
+for an $s_0$ large enough to compensate for the growth of $f(t)$.
+
+
+## Derivatives
+
+The derivative of a transformed function is the transform
+of the original mutliplied by its variable.
+This is especially useful for transforming ODEs with variable coefficients:
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{f}'(s) = - \hat{\mathcal{L}}\{t f(t)\}
+    }
+\end{aligned}$$
+
+This property generalizes nicely to higher-order derivatives of $s$, so:
+
+$$\begin{aligned}
+    \boxed{
+        \dv[n]{\tilde{f}}{s} = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-dv-s"/>
+<label for="proof-dv-s">Proof</label>
+<div class="hidden">
+<label for="proof-dv-s">Proof.</label>
+The exponential $\exp\!(- s t)$ is the only thing that depends on $s$ here:
+
+$$\begin{aligned}
+    \dv[n]{\tilde{f}}{s}
+    &= \dv[n]{s} \int_0^\infty f(t) \exp\!(- s t) \dd{t}
+    \\
+    &= \int_0^\infty (-t)^n f(t) \exp\!(- s t) \dd{t}
+    = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
+\end{aligned}$$
+</div>
+</div>
+
+The Laplace transform of a derivative introduces the initial conditions into the result.
+Notice that $f(0)$ is the initial value in the original $t$-domain:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\mathcal{L}}\{ f'(t) \} = - f(0) + s \tilde{f}(s)
+    }
+\end{aligned}$$
+
+This property generalizes to higher-order derivatives,
+although it gets messy quickly.
+Once again, the initial values of the lower derivatives appear:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\}
+        = - \sum_{j = 0}^{n - 1} s^j f^{(n - 1 - j)}(0) + s^n \tilde{f}(s)
+    }
+\end{aligned}$$
+
+Where $f^{(n)}(t)$ is shorthand for the $n$th derivative of $f(t)$,
+and $f^{(0)}(t) = f(t)$.
+As an example, $\hat{\mathcal{L}}\{f'''(t)\}$ becomes
+$- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s)$.
+
+<div class="accordion">
+<input type="checkbox" id="proof-dv-t"/>
+<label for="proof-dv-t">Proof</label>
+<div class="hidden">
+<label for="proof-dv-t">Proof.</label>
+We integrate by parts and use the fact that $\lim_{x \to \infty} \exp\!(-x) = 0$:
+
+$$\begin{aligned}
+    \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\}
+    &= \int_0^\infty f^{(n)}(t) \exp\!(- s t) \dd{t}
+    \\
+    &= \big[ f^{(n - 1)}(t) \exp\!(- s t) \big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t}
+    \\
+    &= - f^{(n - 1)}(0) + s \big[ f^{(n - 2)}(t) \exp\!(- s t) \big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t}
+\end{aligned}$$
+
+And so on.
+By partially integrating $n$ times in total we arrive at the conclusion.
+</div>
+</div>
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/content/know/concept/legendre-transform/index.pdc b/content/know/concept/legendre-transform/index.pdc
index 290a89a..9cb6824 100644
--- a/content/know/concept/legendre-transform/index.pdc
+++ b/content/know/concept/legendre-transform/index.pdc
@@ -27,9 +27,9 @@ $$\begin{aligned}
     y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C
 \end{aligned}$$
 
-The Legendre transform $L(f')$ is defined such that $L(f'(x_0)) = C$ (or
-sometimes $-C$ instead) for all $x_0 \in [a, b]$, where $C$ is the
-constant corresponding to the tangent line at $x = x_0$. This yields:
+The Legendre transform $L(f')$ is defined such that $L(f'(x_0)) = C$
+(or sometimes $-C$) for all $x_0 \in [a, b]$,
+where $C$ corresponds to the tangent line at $x = x_0$. This yields:
 
 $$\begin{aligned}
     L(f'(x)) = f'(x) \: x - f(x)
@@ -85,7 +85,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Legendre transformation is important in physics,
-since it connects Lagrangian and Hamiltonian mechanics to each other.
+since it connects [Lagrangian](/know/concept/lagrangian-mechanics/)
+and Hamiltonian mechanics to each other.
 It is also used to convert between thermodynamic potentials.
 
 
diff --git a/content/know/concept/path-integral-formulation/index.pdc b/content/know/concept/path-integral-formulation/index.pdc
new file mode 100644
index 0000000..c66aed8
--- /dev/null
+++ b/content/know/concept/path-integral-formulation/index.pdc
@@ -0,0 +1,188 @@
+---
+title: "Path integral formulation"
+firstLetter: "P"
+publishDate: 2021-07-03
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-07-03T14:39:50+02:00
+draft: false
+markup: pandoc
+---
+
+# Path integral formulation
+
+In quantum mechanics, the **path integral formulation**
+is an alternative description of quantum mechanics,
+which is equivalent to the "traditional" Schrödinger equation.
+Whereas the latter is based on [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/),
+the former comes from [Lagrangian mechanics](/know/concept/lagrangian-mechanics/).
+
+It expresses the [propagator](/know/concept/propagator/) $K$
+using the following sum over all possible paths $x(t)$,
+which all go from the initial position $x_0$ at time $t_0$
+to the destination $x_N$ at time $t_N$:
+
+$$\begin{aligned}
+    \boxed{
+        K(x_N, t_N; x_0, t_0)
+        = A \sum_{\mathrm{all}\:x(t)} \exp(i S[x] / \hbar)
+    }
+\end{aligned}$$
+
+Where $A$ normalizes.
+$S[x]$ is the classical action of the path $x$, whose minimization yields
+the Euler-Lagrange equation from Lagrangian mechanics.
+Note that each path is given an equal weight,
+even unrealistic paths that make big detours.
+
+This apparent problem solves itself,
+thanks to the fact that paths close to the classical optimum $x_c(t)$
+have an action close to $S_c = S[x_c]$,
+while the paths far away have very different actions.
+Since $S[x]$ is inside a complex exponential,
+this means that paths close to $x_c$ add contructively,
+and the others add destructively and cancel out.
+
+An interesting way too look at it is by varying $\hbar$:
+as its value decreases, minor action differences yield big phase differences,
+which make the quantum wave function stay closer to $x_c$.
+In the limit $\hbar \to 0$, quantum mechanics thus turns into classical mechanics.
+
+## Time-slicing derivation
+
+The most popular way to derive the path integral formulation proceeds as follows:
+starting from the definition of the propagator $K$,
+we divide the time interval $t_N - t_0$ into $N$ "slices"
+of equal width $\Delta t = (t_N - t_0) / N$,
+where $N$ is large:
+
+$$\begin{aligned}
+    K(x_N, t_N; x_0, t_0)
+    &= \matrixel{x_N}{e^{- i \hat{H} (t_N - t_0) / \hbar}}{x_0}
+    = \matrixel{x_N}{e^{- i \hat{H} \Delta t / \hbar} \cdots e^{- i \hat{H} \Delta t / \hbar}}{x_0}
+\end{aligned}$$
+
+Between the exponentials we insert $N\!-\!1$ identity operators
+$\hat{I} = \int \ket{x} \bra{x} \dd{x}$,
+and define $x_j = x(t_j)$ for an arbitrary path $x(t)$:
+
+$$\begin{aligned}
+    K
+    &= \idotsint \matrixel{x_N}{e^{- i \hat{H} \Delta t / \hbar}}{x_{N-1}} \cdots \matrixel{x_1}{e^{- i \hat{H} \Delta t / \hbar}}{x_0}
+    \dd{x_1} \cdots \dd{x_{N - 1}}
+\end{aligned}$$
+
+For sufficiently small time steps $\Delta t$ (i.e. large $N$
+we make the following approximation
+(which would be exact, were it not for the fact that
+$\hat{T}$ and $\hat{V}$ are operators):
+
+$$\begin{aligned}
+    e^{- i \hat{H} \Delta t / \hbar}
+    = e^{- i (\hat{T} + \hat{V}) \Delta t / \hbar}
+    \approx e^{- i \hat{T} \Delta t / \hbar} e^{- i \hat{V} \Delta t / \hbar}
+\end{aligned}$$
+
+Since $\hat{V} = V(x_j)$,
+we can take it out of the inner product as a constant factor:
+
+$$\begin{aligned}
+    \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar} e^{- i \hat{V} \Delta t / \hbar}}{x_j}
+    = e^{- i V(x_j) \Delta t / \hbar} \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j}
+\end{aligned}$$
+
+Here we insert the identity operator
+expanded in the momentum basis $\hat{I} = \int \ket{p} \bra{p} \dd{p}$,
+and commute it with the kinetic energy $\hat{T} = \hat{p}^2 / (2m)$ to get:
+
+$$\begin{aligned}
+    \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j}
+    = \int_{-\infty}^\infty \braket{x_{j+1}}{p} \exp\!\Big(\!-\! i \frac{p^2 \Delta t}{2 m \hbar}\Big) \braket{p}{x_j} \dd{p}
+\end{aligned}$$
+
+In the momentum basis $\ket{p}$,
+the position basis vectors
+are represented by plane waves:
+
+$$\begin{aligned}
+    \braket{p}{x_j}
+    = \frac{1}{\sqrt{2 \pi \hbar}} \exp\!\Big( \!-\! i \frac{x_j p}{\hbar} \Big)
+    \qquad
+    \braket{x_{j+1}}{p}
+    = \frac{1}{\sqrt{2 \pi \hbar}} \exp\!\Big( i \frac{x_{j+1} p}{\hbar} \Big)
+\end{aligned}$$
+
+With this, we return to the inner product and further evaluate the integral:
+
+$$\begin{aligned}
+    \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j}
+    &= \frac{1}{2 \pi \hbar} \int_{-\infty}^\infty
+    \exp\!\Big(\!-\! i \frac{p^2 \Delta t}{2 m \hbar}\Big) \exp\!\Big(i \frac{(x_{j+1} - x_j) p}{\hbar}\Big) \:dp
+    \\
+    &= \frac{1}{2 \pi \hbar} \sqrt{\frac{2 \pi m \hbar}{i \Delta t}} \exp\!\Big( i \frac{m (x_{j+1} - x_j)^2}{2 \hbar \Delta t} \Big)
+\end{aligned}$$
+
+Inserting this back into the definition of the propagator $K(x_N, t_N; x_0, t_0)$ yields:
+
+$$\begin{aligned}
+    K
+    = \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2}
+    \idotsint
+    \exp\!\bigg(\! \sum_{j = 0}^{N - 1} i \Big( \frac{m (x_{j+1} \!-\! x_j)^2}{2 \hbar \Delta t} - \frac{V(x_j) \Delta t}{\hbar} \Big) \!\bigg)
+    \dd{x_1} \cdots \dd{x_{N-1}}
+\end{aligned}$$
+
+For large $N$ and small $\Delta t$, the sum in the exponent becomes an integral:
+
+$$\begin{aligned}
+    \frac{i}{\hbar} \sum_{j = 0}^{N - 1} \Big( \frac{m (x_{j+1} \!-\! x_j)^2}{2 \Delta t^2} - V(x_j) \Big) \Delta t
+    \quad \to \quad
+    \frac{i}{\hbar} \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau}
+\end{aligned}$$
+
+Upon closer inspection, this integral turns out to be the classical action $S[x]$,
+with the integrand being the Lagrangian $L$:
+
+$$\begin{aligned}
+    S[x(t)]
+    = \int_{t_0}^{t_N} L(x, \dot{x}, \tau) \dd{\tau}
+    = \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau}
+\end{aligned}$$
+
+The definition of the propagator $K$ is then further reduced to the following:
+
+$$\begin{aligned}
+    K
+    = \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2}
+    \idotsint \exp(i S[x] / \hbar) \dd{x_1} \cdots \dd{x_{N-1}}
+\end{aligned}$$
+
+Finally, for the purpose of normalization,
+we define the integral over all paths $x(t)$ as follows,
+where we write $D[x]$ instead of $\dd{x}$:
+
+$$\begin{aligned}
+    \int D[x]
+    \equiv \lim_{N \to \infty} \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2} \idotsint \dd{x_1} \cdots \dd{x_{N-1}}
+\end{aligned}$$
+
+We thus arrive at **Feynman's path integral**,
+which sums over all possible paths $x(t)$:
+
+$$\begin{aligned}
+    K
+    = \int \exp(i S[x] / \hbar) \:D[x]
+    = A \sum_{\mathrm{all}\:x(t)} \exp(i S[x] / \hbar)
+\end{aligned}$$
+
+
+
+## References
+1.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.
+2.  L.E. Ballentine,
+    *Quantum mechanics: a modern development*, 2nd edition,
+    World Scientific.
diff --git a/content/know/concept/propagator/index.pdc b/content/know/concept/propagator/index.pdc
new file mode 100644
index 0000000..2f18c4d
--- /dev/null
+++ b/content/know/concept/propagator/index.pdc
@@ -0,0 +1,81 @@
+---
+title: "Propagator"
+firstLetter: "P"
+publishDate: 2021-07-04
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-07-04T10:46:47+02:00
+draft: false
+markup: pandoc
+---
+
+# Propagator
+
+In quantum mechanics, the **propagator** $K(x_f, t_f; x_i, t_i)$
+gives the probability amplitude that a particle
+starting at $x_i$ at $t_i$ ends up at position $x_f$ at $t_f$.
+It is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        K(x_f, t_f; x_i, t_i)
+        \equiv \matrixel{x_f}{\hat{U}(t_f, t_i)}{x_i}
+    }
+\end{aligned}$$
+
+Where $\hat{U} \equiv \exp\!(- i t \hat{H} / \hbar)$ is the time-evolution operator.
+The probability that a particle travels
+from $(x_i, t_i)$ to $(x_f, t_f)$ is then given by:
+
+$$\begin{aligned}
+    P
+    &= \big| K(x_f, t_f; x_i, t_i) \big|^2
+\end{aligned}$$
+
+Given a general (i.e. non-collapsed) initial state $\psi_i(x) \equiv \psi(x, t_i)$,
+we must integrate over $x_i$:
+
+$$\begin{aligned}
+    P
+    &= \bigg| \int_{-\infty}^\infty K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \bigg|^2
+\end{aligned}$$
+
+And if the final state $\psi_f(x) \equiv \psi(x, t_f)$
+is not a basis vector either, then we integrate twice:
+
+$$\begin{aligned}
+    P
+    &= \bigg| \iint_{-\infty}^\infty \psi_f^*(x_f) \: K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \dd{x_f} \bigg|^2
+\end{aligned}$$
+
+Given a $\psi_i(x)$, the propagator can also be used
+to find the full final wave function:
+
+$$\begin{aligned}
+    \boxed{
+        \psi(x_f, t_f)
+        = \int_{-\infty}^\infty \psi_i(x_i) K(x_f, t_f; x_i, t_i) \:dx_i
+    }
+\end{aligned}$$
+
+Sometimes the name "propagator" is also used to refer to
+the so-called *fundamental solution* or *Green's function* $G$
+of the time-dependent Schrödinger equation,
+which is related to $K$ by:
+
+$$\begin{aligned}
+    \boxed{
+        G(x_f, t_f; x_i, t_i)
+        = - \frac{i}{\hbar} \: \Theta(t_f - t_i) \: K(x_f, t_f; x_i, t_i)
+    }
+\end{aligned}$$
+
+Where $\Theta(t)$ is the [Heaviside step function](/know/concept/heaviside-step-function/).
+The definition of $G$ is that it satisfies the following equation,
+where $\delta$ is the [Dirac delta function](/know/concept/dirac-delta-function/):
+
+$$\begin{aligned}
+    \Big( i \hbar \pdv{t_f} - \hat{H} \Big) G = \delta(x_f - x_i) \: \delta(t_f - t_i)
+\end{aligned}$$