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| -rw-r--r-- | content/know/concept/bells-theorem/index.pdc | 378 | ||||
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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/wkb-approximation/index.pdc index cf44fc8..985bcec 100644 --- a/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc +++ b/content/know/concept/wkb-approximation/index.pdc @@ -1,5 +1,5 @@ --- -title: "Wentzel-Kramers-Brillouin approximation" +title: "WKB approximation" firstLetter: "W" publishDate: 2021-02-22 categories: @@ -11,7 +11,7 @@ draft: false markup: pandoc --- -# Wentzel-Kramers-Brillouin approximation +# 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 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. |