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