diff options
Diffstat (limited to 'content')
| -rw-r--r-- | content/know/concept/capillary-action/index.pdc | 129 | ||||
| -rw-r--r-- | content/know/concept/ghz-paradox/index.pdc | 121 | ||||
| -rw-r--r-- | content/know/concept/quantum-gate/index.pdc | 289 | ||||
| -rw-r--r-- | content/know/concept/wetting/index.pdc | 124 | ||||
| -rw-r--r-- | content/know/concept/wkb-approximation/index.pdc | 27 |
5 files changed, 676 insertions, 14 deletions
diff --git a/content/know/concept/capillary-action/index.pdc b/content/know/concept/capillary-action/index.pdc new file mode 100644 index 0000000..e76b88a --- /dev/null +++ b/content/know/concept/capillary-action/index.pdc @@ -0,0 +1,129 @@ +--- +title: "Capillary action" +firstLetter: "C" +publishDate: 2021-03-29 +categories: +- Physics +- Fluid mechanics + +date: 2021-03-07T20:42:28+01:00 +draft: false +markup: pandoc +--- + +# Capillary action + +**Capillary action** refers to the movement of liquid +through narrow spaces due to surface tension, often against gravity. +It occurs when the [Laplace pressure](/know/concept/young-laplace-law/) +from surface tension is much larger in magnitude than the +[hydrostatic pressure](/know/concept/hydrostatic-pressure/) from gravity. + +Consider a spherical droplet of liquid with radius $R$. +The hydrostatic pressure difference +between the top and bottom of the drop +is much smaller than the Laplace pressure: + +$$\begin{aligned} + 2 R \rho g \ll 2 \frac{\alpha}{R} +\end{aligned}$$ + +Where $\rho$ is the density of the liquid, +$g$ is the acceleration due to gravity, +and $\alpha$ is the energy cost per unit surface area. +Rearranging the inequality yields: + +$$\begin{aligned} + R^2 \ll \frac{\alpha}{\rho g} +\end{aligned}$$ + +From the right-hand side we define the **capillary length** $L_c$, +so gravity is negligible if $R \ll L_c$: + +$$\begin{aligned} + \boxed{ + L_c + = \sqrt{\frac{\alpha}{\rho g}} + } +\end{aligned}$$ + +In general, for a system with characteristic length $L$, +the relative strength of gravity compared to surface tension +is described by the **Bond number** $\mathrm{Bo}$ +or **Eötvös number** $\mathrm{Eo}$: + +$$\begin{aligned} + \boxed{ + \mathrm{Bo} + = \mathrm{Eo} + = \frac{L^2}{L_c^2} + = \frac{m g}{\alpha L} + } +\end{aligned}$$ + +The right-most side gives an alternative way of understanding $\mathrm{Bo}$: +$m$ is the mass of a cube with side $L$, such that the numerator is the weight force, +and the denominator is the tension force of the surface. +In any case, capillary action can be observed when $\mathrm{Bo \ll 1}$. + +The most famous example of capillary action is **capillary rise**, +where a liquid "climbs" upwards in a narrow vertical tube with radius $R$, +apparently defying gravity. +Assuming the liquid-air interface is a spherical cap +with constant [curvature](/know/concept/curvature/) radius $R_c$, +then we know that the liquid is at rest +when the hydrostatic pressure equals the Laplace pressure: + +$$\begin{aligned} + \rho g h + \approx \alpha \frac{2}{R_c} + = 2 \alpha \frac{\cos\theta}{R} +\end{aligned}$$ + +Where $\theta$ is the liquid-tube contact angle, +and we are neglecting variations of the height $h$ due to the curvature +(i.e. the [meniscus](/know/concept/meniscus/)). +By isolating the above equation for $h$, +we arrive at **Jurin's law**, +which predicts the height climbed by a liquid in a tube with radius $R$: + +$$\begin{aligned} + \boxed{ + h + = 2 \frac{L_c^2}{R} \cos\theta + } +\end{aligned}$$ + +Depending on $\theta$, $h$ can be negative, +i.e. the liquid might descend below the ambient level. + + +An alternative derivation of Jurin's law balances the forces instead of the pressures. +On the right, we have the gravitational force +(i.e. the energy-per-distance to lift the liquid), +and on the left, the surface tension force +(i.e. the energy-per-distance of the liquid-tube interface): + +$$\begin{aligned} + \pi R^2 \rho g h + \approx 2 \pi R (\alpha_{sg} - \alpha_{sl}) +\end{aligned}$$ + +Where $\alpha_{sg}$ and $\alpha_{sl}$ are the energy costs +of the solid-gas and solid-liquid interfaces. +Thanks to the [Young-Dupré relation](/know/concept/young-dupre-relation/), +we can rewrite this as follows: + +$$\begin{aligned} + R \rho g h + = 2 \alpha \cos\theta +\end{aligned}$$ + +Isolating this for $h$ simply yields Jurin's law again, as expected. + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/ghz-paradox/index.pdc b/content/know/concept/ghz-paradox/index.pdc new file mode 100644 index 0000000..4b872d8 --- /dev/null +++ b/content/know/concept/ghz-paradox/index.pdc @@ -0,0 +1,121 @@ +--- +title: "GHZ paradox" +firstLetter: "G" +publishDate: 2021-03-29 +categories: +- Physics +- Quantum mechanics +- Quantum information + +date: 2021-03-29T15:15:41+02:00 +draft: false +markup: pandoc +--- + +# GHZ Paradox + +The **Greenberger-Horne-Zeilinger** or **GHZ paradox** +is an alternative proof of [Bell's theorem](/know/concept/bells-theorem/) +that does not use inequalities, +but the three-particle entangled **GHZ state** $\ket{\mathrm{GHZ}}$ instead, + +$$\begin{aligned} + \boxed{ + \ket{\mathrm{GHZ}} + = \frac{1}{\sqrt{2}} \Big( \ket{000} + \ket{111} \Big) + } +\end{aligned}$$ + +Where $\ket{0}$ and $\ket{1}$ are qubit states, +for example, the eigenvalues of the Pauli matrix $\hat{\sigma}_z$. + +If we now apply certain products of the Pauli matrices $\hat{\sigma}_x$ and $\hat{\sigma}_y$ +to the three particles, we find: + + +$$\begin{aligned} + \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \ket{\mathrm{GHZ}} + &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0} + + \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \Big) + \\ + &= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes \ket{1} \otimes \ket{1} + \ket{0} \otimes \ket{0} \otimes \ket{0} \Big) + = \ket{\mathrm{GHZ}} + \\ + \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \ket{\mathrm{GHZ}} + &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_y \ket{0} \otimes \hat{\sigma}_y \ket{0} + + \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_y \ket{1} \otimes \hat{\sigma}_y \ket{1} \Big) + \\ + &= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes i \ket{1} \otimes i \ket{1} + \ket{0} \otimes i \ket{0} \otimes i \ket{0} \Big) + = - \ket{\mathrm{GHZ}} +\end{aligned}$$ + +In other words, the GHZ state is a simultaneous eigenstate of these composite operators, +with eigenvalues $+1$ and $-1$, respectively. +Let us introduce two other product operators, +such that we have a set of four observables, +for which $\ket{\mathrm{GHZ}}$ gives these eigenvalues: + +$$\begin{aligned} + \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x + \quad &\implies \quad +1 + \\ + \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y + \quad &\implies \quad -1 + \\ + \hat{\sigma}_y \otimes \hat{\sigma}_x \otimes \hat{\sigma}_y + \quad &\implies \quad -1 + \\ + \hat{\sigma}_y \otimes \hat{\sigma}_y \otimes \hat{\sigma}_x + \quad &\implies \quad -1 +\end{aligned}$$ + +According to any local hidden variable (LHV) theory, +the measurement outcomes of the operators are predetermined, +and the three particles $A$, $B$ and $C$ can be measured separately, +or in other words, the eigenvalues can be factorized: + +$$\begin{aligned} + \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x + \quad &\implies \quad +1 = m_x^A m_x^B m_x^C + \\ + \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y + \quad &\implies \quad -1 = m_x^A m_y^B m_y^C + \\ + \hat{\sigma}_y \otimes \hat{\sigma}_x \otimes \hat{\sigma}_y + \quad &\implies \quad -1 = m_y^A m_x^B m_y^C + \\ + \hat{\sigma}_y \otimes \hat{\sigma}_y \otimes \hat{\sigma}_x + \quad &\implies \quad -1 = m_y^A m_y^B m_x^C +\end{aligned}$$ + +Where $m_x^A = \pm 1$ etc. +Let us now multiply both sides of these four equations together: + +$$\begin{aligned} + (+1) (-1) (-1) (-1) + &= (m_x^A m_x^B m_x^C) (m_x^A m_y^B m_y^C) (m_y^A m_x^B m_y^C) (m_y^A m_y^B m_x^C) + \\ + -1 + &= (m_x^A)^2 (m_x^B)^2 (m_x^C)^2 (m_y^A)^2 (m_y^B)^2 (m_y^C)^2 +\end{aligned}$$ + +This is a contradiction: the left-hand side is $-1$, +but all six factors on the right are $+1$. +This means that we must have made an incorrect assumption along the way. + +Our only assumption was that we could factorize the eigenvalues, +so that e.g. particle $A$ could be measured on its own +without an "action-at-a-distance" effect on $B$ or $C$. +However, because that leads us to a contradiction, +we must conclude that action-at-a-distance exists, +and that therefore all LHV-based theories are invalid. + + + +## References +1. N. Brunner, + *Quantum information theory: lecture notes*, + 2019, unpublished. +2. J.B. Brask, + *Quantum information: lecture notes*, + 2021, unpublished. diff --git a/content/know/concept/quantum-gate/index.pdc b/content/know/concept/quantum-gate/index.pdc new file mode 100644 index 0000000..cd09094 --- /dev/null +++ b/content/know/concept/quantum-gate/index.pdc @@ -0,0 +1,289 @@ +--- +title: "Quantum gate" +firstLetter: "Q" +publishDate: 2021-03-29 +categories: +- Quantum information + +date: 2021-03-29T21:37:57+02:00 +draft: false +markup: pandoc +--- + + +# Quantum gate + +In quantum computing, **quantum gates** are the equivalent +of classical binary logic gates such as $\mathrm{NOT}$, $\mathrm{AND}$, etc. +Because of the continuous nature of qubits, +the number of possible quantum gates is uncountably infinite, +so we only consider the most important examples here. + + +## One-qubit gates + +As an example, consider the following must general single-qubit state $\ket{\psi}$: + +$$\begin{aligned} + \ket{\psi} + = \alpha \ket{0} + \beta \ket{1} + = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} +\end{aligned}$$ + +Arguably the most famous and/or most fundamental quantum gates are the **Pauli matrices**: + +$$\begin{aligned} + \boxed{ + X = + \begin{bmatrix} + 0 & 1 \\ + 1 & 0 + \end{bmatrix} + } + \qquad + \boxed{ + Y = + \begin{bmatrix} + 0 & -i \\ + i & 0 + \end{bmatrix} + } + \qquad + \boxed{ + Z = + \begin{bmatrix} + 1 & 0 \\ + 0 & -1 + \end{bmatrix} + } +\end{aligned}$$ + +They have the following effect on $\ket{\psi}$. +Note that $X$ is equivalent to the classical $\mathrm{NOT}$ gate +(and is often given that name), +and $Z$ is sometimes called the **phase-flip gate**: + +$$\begin{aligned} + X \ket{\psi} + = \begin{bmatrix} \beta \\ \alpha \end{bmatrix} + \qquad + Y \ket{\psi} + = \begin{bmatrix} -i \beta \\ i \alpha \end{bmatrix} + \qquad + Z \ket{\psi} + = \begin{bmatrix} \alpha \\ -\beta \end{bmatrix} +\end{aligned}$$ + +In fact, $Z$ is a specific case of the **phase shift gate** $R_\phi$, +which modifies the qubit's phase without changing its amplitudes. +For an angle $\phi$, it is given by: + +$$\begin{aligned} + \boxed{ + R_\phi = + \begin{bmatrix} + 1 & 0 \\ + 0 & e^{i \phi} + \end{bmatrix} + } +\end{aligned}$$ + +For $\phi = \pi$, we recover the Pauli-$Z$ gate. +In general, the action of $R_\phi$ is as follows: + +$$\begin{aligned} + R_\phi \ket{\psi} + = \begin{bmatrix} \alpha \\ e^{i \phi} \beta \end{bmatrix} +\end{aligned}$$ + +Two common special cases of $R_\phi$ +are $\phi = \pi/2$ and $\phi = \pi/4$, +respectively called $S$ and $T$: + +$$\begin{aligned} + \boxed{ + S = R_{\pi/2} = + \begin{bmatrix} + 1 & 0 \\ + 0 & i + \end{bmatrix} + } + \qquad \quad + \boxed{ + T = R_{\pi/4} = + \frac{1}{\sqrt{2}} + \begin{bmatrix} + \sqrt{2} & 0 \\ + 0 & 1 + i + \end{bmatrix} + } +\end{aligned}$$ + +Finally, we have the **Hadamard gate** $H$, +which is defined as follows: + +$$\begin{aligned} + \boxed{ + H = \frac{1}{\sqrt{2}} + \begin{bmatrix} + 1 & 1 \\ + 1 & -1 + \end{bmatrix} + } +\end{aligned}$$ + +Its action consists of rotating the qubit +by $\pi$ around the axis $(X + Z) / \sqrt{2}$ of the Bloch sphere: + +$$\begin{aligned} + H \ket{\psi} + = \frac{1}{\sqrt{2}} \begin{bmatrix} \alpha + \beta \\ \alpha - \beta \end{bmatrix} +\end{aligned}$$ + +Notably, it maps the eigenstates of $X$ and $Z$ to each other, +and is its own inverse (i.e. unitary): + +$$\begin{aligned} + H \ket{0} = \ket{+} + \qquad + H \ket{1} = \ket{-} + \qquad + H \ket{+} = \ket{0} + \qquad + H \ket{-} = \ket{1} +\end{aligned}$$ + +The **Clifford gates** are a set including $X$, $Y$, $Z$, $H$ and $S$, +or more generally any gates that rotate +by multiples of $\pi/2$ around the Bloch sphere. +This set is **not universal**, meaning that if we start from $\ket{0}$, +we can only reach $\ket{0}$, $\ket{1}$, $\ket{+}$, $\ket{-}$, $\ket{+i}$ $\ket{-i}$ using these gates. + +If we add *any* non-Clifford gate, for example $T$, +then we can reach any point on the Bloch sphere, +which means that the set is **universal**. + +However, there is a problem: a qubit has an uncountable infinity of states, +but a quantum circuit consists of a countably infinite sequence of gates, at most. +Therefore, technically, we can never reach the whole Bloch sphere, +but we *can* come up with circuits that approximate a target state to some degree $\varepsilon$. +This is the definition of universality: +any state can be approximated. + + +## Two-qubit gates + +As an example, let us consider +the following two pure one-qubit states $\ket{\psi_1}$ and $\ket{\psi_2}$: + +$$\begin{aligned} + \ket{\psi_1} + = \alpha_1 \ket{0} + \beta_1 \ket{1} + = \begin{bmatrix} \alpha_1 \\ \beta_1 \end{bmatrix} + \qquad \quad + \ket{\psi_2} + = \alpha_2 \ket{0} + \beta_2 \ket{1} + = \begin{bmatrix} \alpha_2 \\ \beta_2 \end{bmatrix} +\end{aligned}$$ + +The composite state of both qubits, assuming they are pure, +is then their tensor product $\otimes$: + +$$\begin{aligned} + \ket{\psi_1 \psi_2} + = \ket{\psi_1} \otimes \ket{\psi_2} + &= \alpha_1 \alpha_2 \ket{00} + \alpha_1 \beta_2 \ket{01} + \beta_1 \alpha_2 \ket{10} + \beta_1 \beta_2 \ket{11} + \\ + &= c_{00} \ket{00} + c_{01} \ket{01} + c_{10} \ket{10} + c_{11} \ket{11} +\end{aligned}$$ + +Note that a two-qubit system may be [entangled](/know/concept/quantum-entanglement/), +in which case the coefficients $c_{00}$ etc. cannot be written as products, +i.e. $\ket{\psi_2}$ cannot be expressed separately from $\ket{\psi_1}$, and vice versa. + +In other words, the general action of a two-qubit quantum gate +can be expressed in the basis of $\ket{00}$, $\ket{01}$, $\ket{10}$ and $\ket{11}$, +but not always in the basis of $\ket{0}_1$, $\ket{1}_1$, $\ket{0}_2$ and $\ket{1}_2$. + +With that said, the first two-qubit gate is $\mathrm{SWAP}$, +which simply swaps $\ket{\psi_1}$ and $\ket{\psi_2}$: + +$$\begin{aligned} + \boxed{ + \mathrm{SWAP} = + \begin{bmatrix} + 1 & 0 & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 0 & 1 + \end{bmatrix} + } +\end{aligned}$$ + +This matrix is given in the basis of $\ket{00}$, $\ket{01}$, $\ket{10}$ and $\ket{11}$. +Note that $\mathrm{SWAP}$ cannot generate entanglement, +so if its input is separable, its output is too. +In any case, its effect is clear: + +$$\begin{aligned} + \mathrm{SWAP} \ket{\psi_1 \psi_2} + &= c_{00} \ket{00} + c_{10} \ket{01} + c_{01} \ket{10} + c_{11} \ket{11} +\end{aligned}$$ + +Next, there is the **controlled NOT gate** $\mathrm{CNOT}$, +which "flips" (applies $X$ to) $\ket{\psi_2}$ if $\ket{\psi_1}$ is true: + +$$\begin{aligned} + \boxed{ + \mathrm{CNOT} = + \begin{bmatrix} + 1 & 0 & 0 & 0 \\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 0 & 1 \\ + 0 & 0 & 1 & 0 + \end{bmatrix} + } +\end{aligned}$$ + +That is, it swaps the last two coefficients $c_{10}$ and $c_{11}$ in the composite state vector: + +$$\begin{aligned} + \mathrm{CNOT} \ket{\psi_1 \psi_2} + &= c_{00} \ket{00} + c_{01} \ket{01} + c_{11} \ket{10} + c_{10} \ket{11} +\end{aligned}$$ + +More generally, each one-qubit gate $U$ can be turned into a **controlled** $U$ **gate**: + +$$\begin{aligned} + \boxed{ + \mathrm{CU} = + \begin{bmatrix} + 1 & 0 & 0 & 0 \\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & u_{00} & u_{01} \\ + 0 & 0 & u_{10} & u_{11} + \end{bmatrix} + } +\end{aligned}$$ + +Where the lower-right 2x2 block is simply $U$. +The general action of this gate is given by: + +$$\begin{aligned} + \mathrm{CU} \ket{\psi_1 \psi_2} + &= c_{00} \ket{00} + c_{01} \ket{01} + (c_{10} u_{00} + c_{11} u_{01}) \ket{10} + (c_{10} u_{10} + c_{11} u_{11}) \ket{11} +\end{aligned}$$ + +A set of gates is **universal** if all possible mappings +from $n$ to $n$ qubits can be approximated using only these gates. +A minimal universal set is $\{\mathrm{CNOT}, T, S\}$, +and there exist many others. + + +## References +1. J.S. Neergaard-Nielsen, + *Quantum information: lectures notes*, + 2021, unpublished. +2. S. Aaronson, + *Introduction to quantum information science: lecture notes*, + 2018, unpublished. diff --git a/content/know/concept/wetting/index.pdc b/content/know/concept/wetting/index.pdc new file mode 100644 index 0000000..e5bbadf --- /dev/null +++ b/content/know/concept/wetting/index.pdc @@ -0,0 +1,124 @@ +--- +title: "Wetting" +firstLetter: "W" +publishDate: 2021-03-29 +categories: +- Physics +- Fluid mechanics + +date: 2021-03-29T16:20:44+02:00 +draft: false +markup: pandoc +--- + +# Wetting + +In fluid statics, **wetting** is the ability +of a given liquid to touch a given surface. +When a droplet of the liquid is placed on the surface, +the **wettability** determines the contact angle $\theta$. + +If $\theta = 0$, we have **perfect** or **complete wetting**: +the droplet spreads out over the entire surface. +The other extreme is **dewetting** or **non-wetting**, +where $\theta = \pi$, such that the droplet "floats" on the surface, +which in the specific case of water is called **hydrophobia**. +Furthermore, $\theta < \pi/2$ is **high wettability**, +and $\pi/2 < \theta < \pi$ is **low wettability**. + +For a perfectly smooth homogeneous surface, +$\theta$ is determined by +the [Young-Dupré relation](/know/concept/young-dupre-relation/): + +$$\begin{aligned} + \alpha_{sg} - \alpha_{sl} + = \alpha_{gl} \cos\theta +\end{aligned}$$ + +In practice, however, surfaces can be rough and/or inhomogeneous. +We start with the former. + +A rough surface has some structure, which may contain "gaps". +There are two options: +either the droplet fills those gaps (a **Wenzel state**), +or it floats over them (a **Cassie-Baxter state**). + +For a Wenzel state, we define the **roughness ratio** $r$ +as the relative increase of the surface's area due to its rough structure, +where $A_{real}$ and $A_{app}$ are the real and apparent areas: + +$$\begin{aligned} + r = \frac{A_{real}}{A_{app}} +\end{aligned}$$ + +The net energy cost $E$ of spreading the droplet over the surface is then given by: + +$$\begin{aligned} + E_{sl} + &= (\alpha_{sg} - \alpha_{sl}) A_{real} + = \alpha_{gl} A_{real} \cos\theta + \\ + &= \alpha_{gl} A_{app} r \cos\theta + = \alpha_{gl} A_{app} \cos\theta^* +\end{aligned}$$ + +Where we have defined the **apparent contact angle** $\theta^*$ +as the correction to $\theta$ to account for the roughness. +It is expressed as follows: + +$$\begin{aligned} + \boxed{ + \cos\theta^* + = r \cos\theta + } +\end{aligned}$$ + +For Cassie-Baxter states, where the gaps remain air-filled, +we define $f$ as the "non-gap" fraction of the apparent surface, such that: + +$$\begin{aligned} + E + &= A_{app} \big( f (\alpha_{sg} - \alpha_{sl}) - (1 - f) \alpha_{gl} \big) + \\ + &= A_{app} \alpha_{gl} \big( f \cos\theta + f - 1 \big) +\end{aligned}$$ + +Note the signs: for the solid-liquid interface, +we "spend" $\alpha_{sg}$ and "get back" $\alpha_{sl}$, +while for the gas-liquid interface, we spend nothing, +but get $\alpha_{gl}$. +The apparent angle $\theta^*$ is therefore: + +$$\begin{aligned} + \boxed{ + \cos\theta^* + = f (\cos\theta + 1) - 1 + } +\end{aligned}$$ + +We generalize this equation to inhomogeneous surfaces +consisting of two materials with contact angles $\theta_1$ and $\theta_2$. +The energy cost of the interface is then given by: + +$$\begin{aligned} + E + &= A \big( f_1 (\alpha_{s1g} - \alpha_{s1l}) + (1 - f_1) (\alpha_{s2g} - \alpha_{s2l}) \big) + \\ + &= A \alpha_{gl} \big( f_1 \cos\theta_1 + (1 - f_1) \cos\theta_2 \big) +\end{aligned}$$ + +Such that $\theta^*$ for an inhomogeneous surface is given by this equation, +called **Cassie's law**: + +$$\begin{aligned} + \boxed{ + \cos\theta^* + = f_1 \cos\theta_1 + (1 - f_1) \cos\theta_2 + } +\end{aligned}$$ + +Note that the materials need not be solids, +for example, if one is air, we recover the previous case for rough surfaces. +Cassie's law can also easily be generalized to three or more materials, +and to include Wenzel-style roughness ratios $r_1$, $r_2$, etc. + diff --git a/content/know/concept/wkb-approximation/index.pdc b/content/know/concept/wkb-approximation/index.pdc index 985bcec..b8ace29 100644 --- a/content/know/concept/wkb-approximation/index.pdc +++ b/content/know/concept/wkb-approximation/index.pdc @@ -29,17 +29,17 @@ $$\begin{aligned} m^2 (x')^2 = 2 m (E - V(x)) \end{aligned}$$ -The left-hand side of the rearrangement is simply the momentum squared, +The left-hand side of the rearranged version 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: +Note that this is under the assumption that $E > V$, +which is always true in classical mechanics, +but not necessarily in quantum mechanics. +We rewrite the Schrödinger equation: $$\begin{aligned} 0 @@ -172,19 +172,18 @@ $$\begin{aligned} } \end{aligned}$$ -What if $E < V$? In classical mechanics, this is not allowed; a ball +What if $E < V$? In classical mechanics, this is just not allowed; a ball cannot simply go through a potential bump without the necessary energy. -However, in quantum mechanics, particles can **tunnel** through barriers. +On the other hand, in quantum physics, particles can **tunnel** through barriers. -Conveniently, all we need to change for the WKB approximation is to let -the momentum take imaginary values: +Luckily, the only thing we need to change for the WKB approximation +is to let the momentum take imaginary values: $$\begin{aligned} p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)} \end{aligned}$$ -And then take the absolute value in the appropriate place in front of -$\psi(x)$: +And then take the absolute value in the appropriate place in front of $\psi(x)$: $$\begin{aligned} \boxed{ @@ -193,9 +192,9 @@ $$\begin{aligned} \end{aligned}$$ 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. +in the quantum-physical region ($E < V$) it is exponential. +Note that for $E \approx V$ the approximation breaks down, +because of the appearance of $p(x)$ in the denominator. ## References |