From 922a0bbeb81f9a0297c6a728d243cbec75cf9c3b Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Mon, 29 Mar 2021 09:15:42 +0200
Subject: Expand knowledge base, move WKB approximation
---
content/know/concept/bells-theorem/index.pdc | 378 +++++++++++++++++++++
.../index.pdc | 207 -----------
content/know/concept/wkb-approximation/index.pdc | 207 +++++++++++
.../know/concept/young-dupre-relation/index.pdc | 102 ++++++
4 files changed, 687 insertions(+), 207 deletions(-)
create mode 100644 content/know/concept/bells-theorem/index.pdc
delete mode 100644 content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc
create mode 100644 content/know/concept/wkb-approximation/index.pdc
create mode 100644 content/know/concept/young-dupre-relation/index.pdc
(limited to 'content')
diff --git a/content/know/concept/bells-theorem/index.pdc b/content/know/concept/bells-theorem/index.pdc
new file mode 100644
index 0000000..8d35c84
--- /dev/null
+++ b/content/know/concept/bells-theorem/index.pdc
@@ -0,0 +1,378 @@
+---
+title: "Bell's theorem"
+firstLetter: "B"
+publishDate: 2021-03-28
+categories:
+- Physics
+- Quantum mechanics
+- Quantum information
+
+date: 2021-03-28T14:41:32+02:00
+draft: false
+markup: pandoc
+---
+
+# Bell's theorem
+
+**Bell's theorem** states that the laws of quantum mechanics
+cannot be explained by theories built on
+so-called **local hidden variables** (LHVs).
+
+Suppose that we have two spin-1/2 particles, called $A$ and $B$,
+in an entangled [Bell state](/know/concept/bell-state/):
+
+$$\begin{aligned}
+ \ket{\Psi^{-}}
+ = \frac{1}{\sqrt{2}} \Big( \ket{\uparrow \downarrow} - \ket{\downarrow \uparrow} \Big)
+\end{aligned}$$
+
+Since they are entangled,
+if we measure the $z$-spin of particle $A$, and find e.g. $\ket{\uparrow}$,
+then particle $B$ immediately takes the opposite state $\ket{\downarrow}$.
+The point is that this collapse is instant,
+regardless of the distance between $A$ and $B$.
+
+Einstein called this effect "action-at-a-distance",
+and used it as evidence that quantum mechanics is an incomplete theory.
+He said that there must be some **hidden variable** $\lambda$
+that determines the outcome of measurements of $A$ and $B$
+from the moment the entangled pair is created.
+However, according to Bell's theorem, he was wrong.
+
+To prove this, let us assume that Einstein was right, and some $\lambda$,
+which we cannot understand, let alone calculate or measure, controls the results.
+We want to know the spins of the entangled pair
+along arbitrary directions $\vec{a}$ and $\vec{b}$,
+so the outcomes for particles $A$ and $B$ are:
+
+$$\begin{aligned}
+ A(\vec{a}, \lambda) = \pm 1
+ \qquad \quad
+ B(\vec{b}, \lambda) = \pm 1
+\end{aligned}$$
+
+Where $\pm 1$ are the eigenvalues of the Pauli matrices
+in the chosen directions $\vec{a}$ and $\vec{b}$:
+
+$$\begin{aligned}
+ \hat{\sigma}_a
+ &= \vec{a} \cdot \vec{\sigma}
+ = a_x \hat{\sigma}_x + a_y \hat{\sigma}_y + a_z \hat{\sigma}_z
+ \\
+ \hat{\sigma}_b
+ &= \vec{b} \cdot \vec{\sigma}
+ = b_x \hat{\sigma}_x + b_y \hat{\sigma}_y + b_z \hat{\sigma}_z
+\end{aligned}$$
+
+Whether $\lambda$ is a scalar or a vector does not matter;
+we simply demand that it follows an unknown probability distribution $\rho(\lambda)$:
+
+$$\begin{aligned}
+ \int \rho(\lambda) \dd{\lambda} = 1
+ \qquad \quad
+ \rho(\lambda) \ge 0
+\end{aligned}$$
+
+The product of the outcomes of $A$ and $B$ then has the following expectation value.
+Note that we only multiply $A$ and $B$ for shared $\lambda$-values:
+this is what makes it a **local** hidden variable:
+
+$$\begin{aligned}
+ \expval{A_a B_b}
+ = \int \rho(\lambda) \: A(\vec{a}, \lambda) \: B(\vec{b}, \lambda) \dd{\lambda}
+\end{aligned}$$
+
+From this, two inequalities can be derived,
+which both prove Bell's theorem.
+
+
+## Bell inequality
+
+If $\vec{a} = \vec{b}$, then we know that $A$ and $B$ always have opposite spins:
+
+$$\begin{aligned}
+ A(\vec{a}, \lambda)
+ = A(\vec{b}, \lambda)
+ = - B(\vec{b}, \lambda)
+\end{aligned}$$
+
+The expectation value of the product can therefore be rewritten as follows:
+
+$$\begin{aligned}
+ \expval{A_a B_b}
+ = - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda}
+\end{aligned}$$
+
+Next, we introduce an arbitrary third direction $\vec{c}$,
+and use the fact that $( A(\vec{b}, \lambda) )^2 = 1$:
+
+$$\begin{aligned}
+ \expval{A_a B_b} - \expval{A_a B_c}
+ &= - \int \rho(\lambda) \Big( A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) - A(\vec{a}, \lambda) \: A(\vec{c}, \lambda) \Big) \dd{\lambda}
+ \\
+ &= - \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big) A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda}
+\end{aligned}$$
+
+Inside the integral, the only factors that can be negative
+are the last two, and their product is $\pm 1$.
+Taking the absolute value of the whole left,
+and of the integrand on the right, we thus get:
+
+$$\begin{aligned}
+ \Big| \expval{A_a B_b} - \expval{A_a B_c} \Big|
+ &\le \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big)
+ \: \Big| A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \Big| \dd{\lambda}
+ \\
+ &\le \int \rho(\lambda) \dd{\lambda} - \int \rho(\lambda) A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \dd{\lambda}
+\end{aligned}$$
+
+Since $\rho(\lambda)$ is a normalized probability density function,
+we arrive at the **Bell inequality**:
+
+$$\begin{aligned}
+ \boxed{
+ \Big| \expval{A_a B_b} - \expval{A_a B_c} \Big|
+ \le 1 + \expval{A_b B_c}
+ }
+\end{aligned}$$
+
+Any theory involving an LHV $\lambda$ must obey this inequality.
+The problem, however, is that quantum mechanics dictates the expectation values
+for the state $\ket{\Psi^{-}}$:
+
+$$\begin{aligned}
+ \expval{A_a B_b} = - \vec{a} \cdot \vec{b}
+\end{aligned}$$
+
+Finding directions which violate the Bell inequality is easy:
+for example, if $\vec{a}$ and $\vec{b}$ are orthogonal,
+and $\vec{c}$ is at a $\pi/4$ angle to both of them,
+then the left becomes $0.707$ and the right $0.293$,
+which clearly disagrees with the inequality,
+meaning that LHVs are impossible.
+
+
+## CHSH inequality
+
+The **Clauser-Horne-Shimony-Holt** or simply **CHSH inequality**
+takes a slightly different approach, and is more useful in practice.
+
+Consider four spin directions, two for $A$ called $\vec{a}_1$ and $\vec{a}_2$,
+and two for $B$ called $\vec{b}_1$ and $\vec{b}_2$.
+Let us introduce the following abbreviations:
+
+$$\begin{aligned}
+ A_1 &= A(\vec{a}_1, \lambda)
+ \qquad \quad
+ A_2 = A(\vec{a}_2, \lambda)
+ \\
+ B_1 &= B(\vec{b}_1, \lambda)
+ \qquad \quad
+ B_2 = B(\vec{b}_2, \lambda)
+\end{aligned}$$
+
+From the definition of the expectation value,
+we know that the difference is given by:
+
+$$\begin{aligned}
+ \expval{A_1 B_1} - \expval{A_1 B_2}
+ = \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \Big) \dd{\lambda}
+\end{aligned}$$
+
+We introduce some new terms and rearrange the resulting expression:
+
+$$\begin{aligned}
+ \expval{A_1 B_1} - \expval{A_1 B_2}
+ &= \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \pm A_1 B_1 A_2 B_2 \mp A_1 B_1 A_2 B_2 \Big) \dd{\lambda}
+ \\
+ &= \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
+ - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
+\end{aligned}$$
+
+Taking the absolute value of both sides
+and invoking the triangle inequality then yields:
+
+$$\begin{aligned}
+ \Big| \expval{A_1 B_1} - \expval{A_1 B_2} \Big|
+ &= \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
+ - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg|
+ \\
+ &\le \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} \!\bigg|
+ + \bigg|\! \int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg|
+\end{aligned}$$
+
+Using the fact that the product of $A$ and $B$ is always either $-1$ or $+1$,
+we can reduce this to:
+
+$$\begin{aligned}
+ \Big| \expval{A_1 B_1} - \expval{A_1 B_2} \Big|
+ &\le \int \rho(\lambda) \Big| A_1 B_1 \Big| \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
+ + \!\int \rho(\lambda) \Big| A_1 B_2 \Big| \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
+ \\
+ &\le \int \rho(\lambda) \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
+ + \!\int \rho(\lambda) \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
+\end{aligned}$$
+
+Evaluating these integrals gives us the following inequality,
+which holds for both choices of $\pm$:
+
+$$\begin{aligned}
+ \Big| \expval{A_1 B_1} - \expval{A_1 B_2} \Big|
+ &\le 2 \pm \expval{A_2 B_2} \pm \expval{A_2 B_1}
+\end{aligned}$$
+
+We should choose the signs such that the right-hand side is as small as possible, that is:
+
+$$\begin{aligned}
+ \Big| \expval{A_1 B_1} - \expval{A_1 B_2} \Big|
+ &\le 2 \pm \Big( \expval{A_2 B_2} + \expval{A_2 B_1} \Big)
+ \\
+ &\le 2 - \Big| \expval{A_2 B_2} + \expval{A_2 B_1} \Big|
+\end{aligned}$$
+
+Rearranging this and once again using the triangle inequality,
+we get the CHSH inequality:
+
+$$\begin{aligned}
+ 2
+ &\ge \Big| \expval{A_1 B_1} - \expval{A_1 B_2} \Big| + \Big| \expval{A_2 B_2} + \expval{A_2 B_1} \Big|
+ \\
+ &\ge \Big| \expval{A_1 B_1} - \expval{A_1 B_2} + \expval{A_2 B_2} + \expval{A_2 B_1} \Big|
+\end{aligned}$$
+
+The quantity on the right-hand side is sometimes called the **CHSH quantity** $S$,
+and measures the correlation between the spins of $A$ and $B$:
+
+$$\begin{aligned}
+ \boxed{
+ S \equiv \expval{A_2 B_1} + \expval{A_2 B_2} + \expval{A_1 B_1} - \expval{A_1 B_2}
+ }
+\end{aligned}$$
+
+The CHSH inequality places an upper bound on the magnitude of $S$
+for LHV-based theories:
+
+$$\begin{aligned}
+ \boxed{
+ |S| \le 2
+ }
+\end{aligned}$$
+
+
+## Tsirelson's bound
+
+Quantum physics can violate the CHSH inequality, but by how much?
+Consider the following two-particle operator,
+whose expectation value is the CHSH quantity, i.e. $S = \expval*{\hat{S}}$:
+
+$$\begin{aligned}
+ \hat{S}
+ = \hat{A}_2 \otimes \hat{B}_1 + \hat{A}_2 \otimes \hat{B}_2 + \hat{A}_1 \otimes \hat{B}_1 - \hat{A}_1 \otimes \hat{B}_2
+\end{aligned}$$
+
+Where $\otimes$ is the tensor product,
+and e.g. $\hat{A}_1$ is the Pauli matrix for the $\vec{a}_1$-direction.
+The square of this operator is then given by:
+
+$$\begin{aligned}
+ \hat{S}^2
+ = \quad &\hat{A}_2^2 \otimes \hat{B}_1^2 + \hat{A}_2^2 \otimes \hat{B}_1 \hat{B}_2
+ + \hat{A}_2 \hat{A}_1 \otimes \hat{B}_1^2 - \hat{A}_2 \hat{A}_1 \otimes \hat{B}_1 \hat{B}_2
+ \\
+ + &\hat{A}_2^2 \otimes \hat{B}_2 \hat{B}_1 + \hat{A}_2^2 \otimes \hat{B}_2^2
+ + \hat{A}_2 \hat{A}_1 \otimes \hat{B}_2 \hat{B}_1 - \hat{A}_2 \hat{A}_1 \otimes \hat{B}_2^2
+ \\
+ + &\hat{A}_1 \hat{A}_2 \otimes \hat{B}_1^2 + \hat{A}_1 \hat{A}_2 \otimes \hat{B}_1 \hat{B}_2
+ + \hat{A}_1^2 \otimes \hat{B}_1^2 - \hat{A}_1^2 \otimes \hat{B}_1 \hat{B}_2
+ \\
+ - &\hat{A}_1 \hat{A}_2 \otimes \hat{B}_2 \hat{B}_1 - \hat{A}_1 \hat{A}_2 \otimes \hat{B}_2^2
+ - \hat{A}_1^2 \otimes \hat{B}_2 \hat{B}_1 + \hat{A}_1^2 \otimes \hat{B}_2^2
+ \\
+ = \quad &\hat{A}_2^2 \otimes \hat{B}_1^2 + \hat{A}_2^2 \otimes \hat{B}_2^2 + \hat{A}_1^2 \otimes \hat{B}_1^2 + \hat{A}_1^2 \otimes \hat{B}_2^2
+ \\
+ + &\hat{A}_2^2 \otimes \acomm*{\hat{B}_1}{\hat{B}_2} - \hat{A}_1^2 \otimes \acomm*{\hat{B}_1}{\hat{B}_2}
+ + \acomm*{\hat{A}_1}{\hat{A}_2} \otimes \hat{B}_1^2 - \acomm*{\hat{A}_1}{\hat{A}_2} \otimes \hat{B}_2^2
+ \\
+ + &\hat{A}_1 \hat{A}_2 \otimes \comm*{\hat{B}_1}{\hat{B}_2} - \hat{A}_2 \hat{A}_1 \otimes \comm*{\hat{B}_1}{\hat{B}_2}
+\end{aligned}$$
+
+Spin operators are unitary, so their square is the identity,
+e.g. $\hat{A}_1^2 = \hat{I}$. Therefore $\hat{S}^2$ reduces to:
+
+$$\begin{aligned}
+ \hat{S}^2
+ &= 4 \: (\hat{I} \otimes \hat{I}) + \comm*{\hat{A}_1}{\hat{A}_2} \otimes \comm*{\hat{B}_1}{\hat{B}_2}
+\end{aligned}$$
+
+The *norm* $\norm*{\hat{S}^2}$ of this operator
+is the largest possible expectation value $\expval*{\hat{S}^2}$,
+which is the same as its largest eigenvalue.
+It is given by:
+
+$$\begin{aligned}
+ \norm{\hat{S}^2}
+ &= 4 + \norm{\comm*{\hat{A}_1}{\hat{A}_2} \otimes \comm*{\hat{B}_1}{\hat{B}_2}}
+ \\
+ &\le 4 + \norm{\comm*{\hat{A}_1}{\hat{A}_2}} \norm{\comm*{\hat{B}_1}{\hat{B}_2}}
+\end{aligned}$$
+
+We find a bound for the norm of the commutators by using the triangle inequality, such that:
+
+$$\begin{aligned}
+ \norm{\comm*{\hat{A}_1}{\hat{A}_2}}
+ = \norm{\hat{A}_1 \hat{A}_2 - \hat{A}_2 \hat{A}_1}
+ \le \norm{\hat{A}_1 \hat{A}_2} + \norm{\hat{A}_2 \hat{A}_1}
+ \le 2 \norm{\hat{A}_1 \hat{A}_2}
+ \le 2
+\end{aligned}$$
+
+And $\norm*{\comm*{\hat{B}_1}{\hat{B}_2}} \le 2$ for the same reason.
+The norm is the largest eigenvalue, therefore:
+
+$$\begin{aligned}
+ \norm{\hat{S}^2}
+ \le 4 + 2 \cdot 2
+ = 8
+ \quad \implies \quad
+ \norm{\hat{S}}
+ \le \sqrt{8}
+ = 2 \sqrt{2}
+\end{aligned}$$
+
+We thus arrive at **Tsirelson's bound**,
+which states that quantum mechanics can violate
+the CHSH inequality by a factor of $\sqrt{2}$:
+
+$$\begin{aligned}
+ \boxed{
+ |S|
+ \le 2 \sqrt{2}
+ }
+\end{aligned}$$
+
+Importantly, this is a *tight* bound,
+meaning that there exist certain spin measurement directions
+for which Tsirelson's bound becomes an equality, for example:
+
+$$\begin{aligned}
+ \hat{A}_1 = \hat{\sigma}_z
+ \qquad
+ \hat{A}_2 = \hat{\sigma}_x
+ \qquad
+ \hat{B}_1 = \frac{\hat{\sigma}_z + \hat{\sigma}_x}{\sqrt{2}}
+ \qquad
+ \hat{B}_2 = \frac{\hat{\sigma}_z - \hat{\sigma}_x}{\sqrt{2}}
+\end{aligned}$$
+
+Using the fact that $\expval{A_a B_b} = - \vec{a} \cdot \vec{b}$,
+it can then be shown that $S = 2 \sqrt{2}$ in this case.
+
+
+
+## References
+1. D.J. Griffiths, D.F. Schroeter,
+ *Introduction to quantum mechanics*, 3rd edition,
+ Cambridge.
+2. J.B. Brask,
+ *Quantum information: lecture notes*,
+ 2021, unpublished.
diff --git a/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc b/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc
deleted file mode 100644
index cf44fc8..0000000
--- a/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc
+++ /dev/null
@@ -1,207 +0,0 @@
----
-title: "Wentzel-Kramers-Brillouin approximation"
-firstLetter: "W"
-publishDate: 2021-02-22
-categories:
-- Quantum mechanics
-- Physics
-
-date: 2021-02-22T21:38:35+01:00
-draft: false
-markup: pandoc
----
-
-# Wentzel-Kramers-Brillouin approximation
-
-In quantum mechanics, the **Wentzel-Kramers-Brillouin** or simply the **WKB
-approximation** is a technique to approximate the wave function $\psi(x)$ of
-the one-dimensional time-independent Schrödinger equation. It is an example
-of a **semiclassical approximation**, because it tries to find a
-balance between classical and quantum physics.
-
-In classical mechanics, a particle travelling in a potential $V(x)$
-along a path $x(t)$ has a total energy $E$ as follows, which we
-rearrange:
-
-$$\begin{aligned}
- E = \frac{1}{2} m \dot{x}^2 + V(x)
- \quad \implies \quad
- m^2 (x')^2 = 2 m (E - V(x))
-\end{aligned}$$
-
-The left-hand side of the rearrangement is simply the momentum squared,
-so we define the magnitude of the momentum $p(x)$ accordingly:
-
-$$\begin{aligned}
- p(x) = \sqrt{2 m (E - V(x))}
-\end{aligned}$$
-
-Note that this is under the assumption that $E > V$, which is always the
-case in classical mechanics, but not necessarily so in quantum
-mechanics, but we stick with it for now. We rewrite the Schrödinger
-equation:
-
-$$\begin{aligned}
- 0
- = \dv[2]{\psi}{x} + \frac{2 m}{\hbar^2} (E - V) \psi
- = \dv[2]{\psi}{x} + \frac{p^2}{\hbar^2} \psi
-\end{aligned}$$
-
-If $V(x)$ were constant, and by extension $p(x)$ too, then the solution
-is easy:
-
-$$\begin{aligned}
- \psi(x)
- = \psi(0) \exp(\pm i p x / \hbar)
-\end{aligned}$$
-
-This form is reminiscent of the generator of translations. In practice,
-$V(x)$ and $p(x)$ vary with $x$, but we can still salvage this solution
-by assuming that $V(x)$ varies slowly compared to the wavelength
-$\lambda(x) = 2 \pi / k(x)$, where $k(x) = p(x) / \hbar$ is the
-wavenumber. The solution then takes the following form:
-
-$$\begin{aligned}
- \psi(x)
- = \psi(0) \exp\!\Big(\!\pm\! \frac{i}{\hbar} \int_0^x \chi(\xi) \dd{\xi} \Big)
-\end{aligned}$$
-
-$\chi(\xi)$ is an unknown function, which intuitively should be related
-to $p(x)$. The purpose of the integral is to accumulate the change of
-$\chi$ from the initial point $0$ to the current position $x$.
-Let us write this as an indefinite integral for convenience:
-
-$$\begin{aligned}
- \psi(x)
- = \psi(0) \exp\!\bigg( \!\pm\! \frac{i}{\hbar} \Big( \int \chi(x) \dd{x} - C \Big) \bigg)
-\end{aligned}$$
-
-Where $C = \int \chi(x) \dd{x} |_{x = 0}$ is the initial point of the definite integral.
-For simplicity, we absorb the constant $C$ into $\psi(0)$.
-We can now clearly see that:
-
-$$\begin{aligned}
- \psi'(x) = \pm \frac{i}{\hbar} \chi(x) \psi(x)
- \quad \implies \quad
- \chi(x) = \pm \frac{\hbar}{i} \frac{\psi'(x)}{\psi(x)}
-\end{aligned}$$
-
-Next, we insert this ansatz for $\psi(x)$ into the Schrödinger equation
-to get:
-
-$$\begin{aligned}
- 0
- &= \pm \frac{i}{\hbar} \dv{(\chi \psi)}{x} + \frac{p^2}{\hbar^2} \psi
- = \pm \frac{i}{\hbar} \chi' \psi \pm \frac{i}{\hbar} \chi \psi' + \frac{p^2}{\hbar^2} \psi
- = \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi
-\end{aligned}$$
-
-Dividing out $\psi$ and rearranging gives us the following, which is
-still exact:
-
-$$\begin{aligned}
- \pm \frac{\hbar}{i} \chi'
- = p^2 - \chi^2
-\end{aligned}$$
-
-Next, we expand this as a power series of $\hbar$. This is why it is
-called *semiclassical*: so far we have been using full quantum mechanics,
-but now we are treating $\hbar$ as a parameter which controls the
-strength of quantum effects:
-
-$$\begin{aligned}
- \chi(x) = \chi_0(x) + \frac{\hbar}{i} \chi_1(x) + \frac{\hbar^2}{i^2} \chi_2(x) + ...
-\end{aligned}$$
-
-The heart of the WKB approximation is its assumption that quantum effects are
-sufficiently weak (i.e. $\hbar$ is small enough) that we only need to
-consider the first two terms, or, more specifically, that we only go up to
-$\hbar$, not $\hbar^2$ or higher. Inserting the first two terms of this
-expansion into the equation:
-
-$$\begin{aligned}
- \pm \frac{\hbar}{i} \chi_0'
- &= p^2 - \chi_0^2 - 2 \frac{\hbar}{i} \chi_0 \chi_1
-\end{aligned}$$
-
-Where we have discarded all terms containing $\hbar^2$. At order
-$\hbar^0$, we then get the expected classical result for $\chi_0(x)$:
-
-$$\begin{aligned}
- 0 = p^2 - \chi_0^2
- \quad \implies \quad
- \chi_0(x) = p(x)
-\end{aligned}$$
-
-While at order $\hbar$, we get the following quantum-mechanical
-correction:
-
-$$\begin{aligned}
- \pm \frac{\hbar}{i} \chi_0'
- = - 2 \frac{\hbar}{i} \chi_0 \chi_1
- \quad \implies \quad
- \chi_1(x) = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)}
-\end{aligned}$$
-
-Therefore, our approximated wave function $\psi(x)$ currently looks like
-this:
-
-$$\begin{aligned}
- \psi(x)
- &\approx \psi(0) \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int \chi_0(x) \dd{x} \Big) \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big)
-\end{aligned}$$
-
-We can reduce the latter exponential using integration by substitution:
-
-$$\begin{aligned}
- \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big)
- &= \exp\!\Big( \!-\! \frac{1}{2} \int \frac{\chi_0'(x)}{\chi_0(x)} \dd{x} \Big)
- = \exp\!\Big( \!-\! \frac{1}{2} \int \frac{1}{\chi_0}\:d\chi_0 \Big)
- \\
- &= \exp\!\Big( \!-\! \frac{1}{2} \ln\!\big(\chi_0(x)\big) \Big)
- = \frac{1}{\sqrt{\chi_0(x)}}
- = \frac{1}{\sqrt{p(x)}}
-\end{aligned}$$
-
-In the WKB approximation for $E > V$, the solution $\psi(x)$ is thus
-given by:
-
-$$\begin{aligned}
- \boxed{
- \psi(x) \approx \frac{A}{\sqrt{p(x)}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big)
- }
-\end{aligned}$$
-
-What if $E < V$? In classical mechanics, this is not allowed; a ball
-cannot simply go through a potential bump without the necessary energy.
-However, in quantum mechanics, particles can **tunnel** through barriers.
-
-Conveniently, all we need to change for the WKB approximation is to let
-the momentum take imaginary values:
-
-$$\begin{aligned}
- p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)}
-\end{aligned}$$
-
-And then take the absolute value in the appropriate place in front of
-$\psi(x)$:
-
-$$\begin{aligned}
- \boxed{
- \psi(x) \approx \frac{A}{\sqrt{|p(x)|}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big)
- }
-\end{aligned}$$
-
-In the classical region ($E > V$), the wave function oscillates, and
-in the quantum-mechanical region ($E < V$) it is exponential. Note that for
-$E \approx V$ the approximation breaks down, due to the appearance of
-$p(x)$ in the denominator.
-
-
-## References
-1. D.J. Griffiths, D.F. Schroeter,
- *Introduction to quantum mechanics*, 3rd edition,
- Cambridge.
-2. R. Shankar,
- *Principles of quantum mechanics*, 2nd edition,
- Springer.
diff --git a/content/know/concept/wkb-approximation/index.pdc b/content/know/concept/wkb-approximation/index.pdc
new file mode 100644
index 0000000..985bcec
--- /dev/null
+++ b/content/know/concept/wkb-approximation/index.pdc
@@ -0,0 +1,207 @@
+---
+title: "WKB approximation"
+firstLetter: "W"
+publishDate: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+
+date: 2021-02-22T21:38:35+01:00
+draft: false
+markup: pandoc
+---
+
+# WKB approximation
+
+In quantum mechanics, the **Wentzel-Kramers-Brillouin** or simply the **WKB
+approximation** is a technique to approximate the wave function $\psi(x)$ of
+the one-dimensional time-independent Schrödinger equation. It is an example
+of a **semiclassical approximation**, because it tries to find a
+balance between classical and quantum physics.
+
+In classical mechanics, a particle travelling in a potential $V(x)$
+along a path $x(t)$ has a total energy $E$ as follows, which we
+rearrange:
+
+$$\begin{aligned}
+ E = \frac{1}{2} m \dot{x}^2 + V(x)
+ \quad \implies \quad
+ m^2 (x')^2 = 2 m (E - V(x))
+\end{aligned}$$
+
+The left-hand side of the rearrangement is simply the momentum squared,
+so we define the magnitude of the momentum $p(x)$ accordingly:
+
+$$\begin{aligned}
+ p(x) = \sqrt{2 m (E - V(x))}
+\end{aligned}$$
+
+Note that this is under the assumption that $E > V$, which is always the
+case in classical mechanics, but not necessarily so in quantum
+mechanics, but we stick with it for now. We rewrite the Schrödinger
+equation:
+
+$$\begin{aligned}
+ 0
+ = \dv[2]{\psi}{x} + \frac{2 m}{\hbar^2} (E - V) \psi
+ = \dv[2]{\psi}{x} + \frac{p^2}{\hbar^2} \psi
+\end{aligned}$$
+
+If $V(x)$ were constant, and by extension $p(x)$ too, then the solution
+is easy:
+
+$$\begin{aligned}
+ \psi(x)
+ = \psi(0) \exp(\pm i p x / \hbar)
+\end{aligned}$$
+
+This form is reminiscent of the generator of translations. In practice,
+$V(x)$ and $p(x)$ vary with $x$, but we can still salvage this solution
+by assuming that $V(x)$ varies slowly compared to the wavelength
+$\lambda(x) = 2 \pi / k(x)$, where $k(x) = p(x) / \hbar$ is the
+wavenumber. The solution then takes the following form:
+
+$$\begin{aligned}
+ \psi(x)
+ = \psi(0) \exp\!\Big(\!\pm\! \frac{i}{\hbar} \int_0^x \chi(\xi) \dd{\xi} \Big)
+\end{aligned}$$
+
+$\chi(\xi)$ is an unknown function, which intuitively should be related
+to $p(x)$. The purpose of the integral is to accumulate the change of
+$\chi$ from the initial point $0$ to the current position $x$.
+Let us write this as an indefinite integral for convenience:
+
+$$\begin{aligned}
+ \psi(x)
+ = \psi(0) \exp\!\bigg( \!\pm\! \frac{i}{\hbar} \Big( \int \chi(x) \dd{x} - C \Big) \bigg)
+\end{aligned}$$
+
+Where $C = \int \chi(x) \dd{x} |_{x = 0}$ is the initial point of the definite integral.
+For simplicity, we absorb the constant $C$ into $\psi(0)$.
+We can now clearly see that:
+
+$$\begin{aligned}
+ \psi'(x) = \pm \frac{i}{\hbar} \chi(x) \psi(x)
+ \quad \implies \quad
+ \chi(x) = \pm \frac{\hbar}{i} \frac{\psi'(x)}{\psi(x)}
+\end{aligned}$$
+
+Next, we insert this ansatz for $\psi(x)$ into the Schrödinger equation
+to get:
+
+$$\begin{aligned}
+ 0
+ &= \pm \frac{i}{\hbar} \dv{(\chi \psi)}{x} + \frac{p^2}{\hbar^2} \psi
+ = \pm \frac{i}{\hbar} \chi' \psi \pm \frac{i}{\hbar} \chi \psi' + \frac{p^2}{\hbar^2} \psi
+ = \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi
+\end{aligned}$$
+
+Dividing out $\psi$ and rearranging gives us the following, which is
+still exact:
+
+$$\begin{aligned}
+ \pm \frac{\hbar}{i} \chi'
+ = p^2 - \chi^2
+\end{aligned}$$
+
+Next, we expand this as a power series of $\hbar$. This is why it is
+called *semiclassical*: so far we have been using full quantum mechanics,
+but now we are treating $\hbar$ as a parameter which controls the
+strength of quantum effects:
+
+$$\begin{aligned}
+ \chi(x) = \chi_0(x) + \frac{\hbar}{i} \chi_1(x) + \frac{\hbar^2}{i^2} \chi_2(x) + ...
+\end{aligned}$$
+
+The heart of the WKB approximation is its assumption that quantum effects are
+sufficiently weak (i.e. $\hbar$ is small enough) that we only need to
+consider the first two terms, or, more specifically, that we only go up to
+$\hbar$, not $\hbar^2$ or higher. Inserting the first two terms of this
+expansion into the equation:
+
+$$\begin{aligned}
+ \pm \frac{\hbar}{i} \chi_0'
+ &= p^2 - \chi_0^2 - 2 \frac{\hbar}{i} \chi_0 \chi_1
+\end{aligned}$$
+
+Where we have discarded all terms containing $\hbar^2$. At order
+$\hbar^0$, we then get the expected classical result for $\chi_0(x)$:
+
+$$\begin{aligned}
+ 0 = p^2 - \chi_0^2
+ \quad \implies \quad
+ \chi_0(x) = p(x)
+\end{aligned}$$
+
+While at order $\hbar$, we get the following quantum-mechanical
+correction:
+
+$$\begin{aligned}
+ \pm \frac{\hbar}{i} \chi_0'
+ = - 2 \frac{\hbar}{i} \chi_0 \chi_1
+ \quad \implies \quad
+ \chi_1(x) = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)}
+\end{aligned}$$
+
+Therefore, our approximated wave function $\psi(x)$ currently looks like
+this:
+
+$$\begin{aligned}
+ \psi(x)
+ &\approx \psi(0) \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int \chi_0(x) \dd{x} \Big) \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big)
+\end{aligned}$$
+
+We can reduce the latter exponential using integration by substitution:
+
+$$\begin{aligned}
+ \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big)
+ &= \exp\!\Big( \!-\! \frac{1}{2} \int \frac{\chi_0'(x)}{\chi_0(x)} \dd{x} \Big)
+ = \exp\!\Big( \!-\! \frac{1}{2} \int \frac{1}{\chi_0}\:d\chi_0 \Big)
+ \\
+ &= \exp\!\Big( \!-\! \frac{1}{2} \ln\!\big(\chi_0(x)\big) \Big)
+ = \frac{1}{\sqrt{\chi_0(x)}}
+ = \frac{1}{\sqrt{p(x)}}
+\end{aligned}$$
+
+In the WKB approximation for $E > V$, the solution $\psi(x)$ is thus
+given by:
+
+$$\begin{aligned}
+ \boxed{
+ \psi(x) \approx \frac{A}{\sqrt{p(x)}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big)
+ }
+\end{aligned}$$
+
+What if $E < V$? In classical mechanics, this is not allowed; a ball
+cannot simply go through a potential bump without the necessary energy.
+However, in quantum mechanics, particles can **tunnel** through barriers.
+
+Conveniently, all we need to change for the WKB approximation is to let
+the momentum take imaginary values:
+
+$$\begin{aligned}
+ p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)}
+\end{aligned}$$
+
+And then take the absolute value in the appropriate place in front of
+$\psi(x)$:
+
+$$\begin{aligned}
+ \boxed{
+ \psi(x) \approx \frac{A}{\sqrt{|p(x)|}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big)
+ }
+\end{aligned}$$
+
+In the classical region ($E > V$), the wave function oscillates, and
+in the quantum-mechanical region ($E < V$) it is exponential. Note that for
+$E \approx V$ the approximation breaks down, due to the appearance of
+$p(x)$ in the denominator.
+
+
+## References
+1. D.J. Griffiths, D.F. Schroeter,
+ *Introduction to quantum mechanics*, 3rd edition,
+ Cambridge.
+2. R. Shankar,
+ *Principles of quantum mechanics*, 2nd edition,
+ Springer.
diff --git a/content/know/concept/young-dupre-relation/index.pdc b/content/know/concept/young-dupre-relation/index.pdc
new file mode 100644
index 0000000..6b6d89a
--- /dev/null
+++ b/content/know/concept/young-dupre-relation/index.pdc
@@ -0,0 +1,102 @@
+---
+title: "Young-Dupré relation"
+firstLetter: "Y"
+publishDate: 2021-03-07
+categories:
+- Physics
+- Fluid mechanics
+
+date: 2021-03-07T15:05:50+01:00
+draft: false
+markup: pandoc
+---
+
+# Young-Dupré relation
+
+In fluid mechanics, the **Young-Dupré relation** relates the contact
+angle of a droplet at rest on a surface to the surface tensions of the interfaces.
+Let $\alpha_{gl}$, $\alpha_{sl}$ and $\alpha_{sg}$ respectively be
+the energy costs of the liquid-gas, solid-liquid and solid-gas interfaces:
+
+$$\begin{aligned}
+ \boxed{
+ \alpha_{sg} - \alpha_{sl}
+ = \alpha_{gl} \cos\theta
+ }
+\end{aligned}$$
+
+The derivation is simple:
+this is the only expression that maintains the droplet's boundaries
+when you account for the surface tension force pulling along each interface.
+
+A more general derivation is possible by using the
+[calculus of variations](/know/concept/calculus-of-variations/).
+In 2D, the upper surface of the droplet is denoted by $y(x)$.
+Consider the following Lagrangian $\mathcal{L}$,
+with the two first terms respectively being the energy costs
+of the top and bottom surfaces:
+
+$$\begin{aligned}
+ \mathcal{L}
+ = \alpha_{gl} \sqrt{1 + (y')^2} + (\alpha_{sl} - \alpha_{sg}) + \lambda y
+\end{aligned}$$
+
+And the last term comes from the constraint
+that the volume $V$ of the droplet must be constant:
+
+$$\begin{aligned}
+ V = \int_0^L y \dd{x}
+\end{aligned}$$
+
+The total energy to be minimized is thus given by the following functional,
+where the endpoints of the droplet are $x = 0$ and $x = L$:
+
+$$\begin{aligned}
+ E[y(x)]
+ = \int_0^L \Big( \alpha_{gl} \sqrt{1 + (y')^2} + (\alpha_{sl} - \alpha_{sg}) + \lambda y \Big) \dd{x}
+\end{aligned}$$
+
+In this optimization problem, the endpoint $L$ is a free parameter,
+i.e. the $L$-value of the optimum is unknown and must be found.
+In such cases, the optimum $y(x)$ needs to satisfy the so-called *transversality condition*
+at the variable endpoint, in this case $x = L$:
+
+$$\begin{aligned}
+ 0
+ &= \Big( \mathcal{L} - y' \pdv{\mathcal{L}}{y'} \Big)_{x = L}
+ \\
+ &= \bigg( \alpha_{gl} \sqrt{1 + (y')^2} + (\alpha_{sl} - \alpha_{sg}) + \lambda y - \frac{(y')^2}{\sqrt{1 + (y')^2}} \bigg)_{x = L}
+ \\
+ &= \bigg( \alpha_{gl} \frac{1}{\sqrt{1 + (y')^2}} + (\alpha_{sl} - \alpha_{sg}) + \lambda y \bigg)_{x = L}
+\end{aligned}$$
+
+Due to the droplet's shape, we have the boundary condition $y(L) = 0$,
+so the last term vanishes.
+We are thus left with the following equation:
+
+$$\begin{aligned}
+ \alpha_{gl} \frac{1}{\sqrt{1 + (y'(L))^2}}
+ = \alpha_{sg} - \alpha_{sl}
+\end{aligned}$$
+
+At the edge of the droplet, imagine a small rectangular triangle
+with one side $\dd{x}$ on the $x$-axis,
+the hypotenuse on $y(x)$ having length $\dd{x} \sqrt{1 + (y')^2}$,
+and the corner between them being the contact point with angle $\theta$.
+Then, from the definition of the cosine:
+
+$$\begin{aligned}
+ \cos\theta
+ = \frac{\dd{x}}{\dd{x} \sqrt{1 + (y'(L))^2}}
+ = \frac{1}{\sqrt{1 + (y'(L))^2}}
+\end{aligned}$$
+
+When inserted into the above transversality condition,
+this yields the Young-Dupré relation.
+
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
--
cgit v1.2.3