From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Thu, 20 Oct 2022 18:25:31 +0200 Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex' --- source/know/concept/alfven-waves/index.md | 70 +++---- source/know/concept/archimedes-principle/index.md | 26 +-- source/know/concept/bb84-protocol/index.md | 50 ++--- source/know/concept/bell-state/index.md | 24 +-- source/know/concept/bells-theorem/index.md | 84 ++++----- source/know/concept/beltrami-identity/index.md | 48 ++--- source/know/concept/bernoullis-theorem/index.md | 26 +-- .../concept/bernstein-vazirani-algorithm/index.md | 34 ++-- source/know/concept/berry-phase/index.md | 88 ++++----- source/know/concept/binomial-distribution/index.md | 50 ++--- .../know/concept/blasius-boundary-layer/index.md | 34 ++-- source/know/concept/bloch-sphere/index.md | 30 +-- source/know/concept/blochs-theorem/index.md | 42 ++--- source/know/concept/boltzmann-equation/index.md | 85 +++++---- 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+-- source/know/concept/self-steepening/index.md | 40 ++-- source/know/concept/shors-algorithm/index.md | 196 +++++++++---------- source/know/concept/sigma-algebra/index.md | 44 ++--- source/know/concept/simons-algorithm/index.md | 96 +++++----- source/know/concept/slater-determinant/index.md | 14 +- .../concept/sokhotski-plemelj-theorem/index.md | 32 ++-- source/know/concept/spherical-coordinates/index.md | 34 ++-- source/know/concept/spitzer-resistivity/index.md | 52 +++--- source/know/concept/step-index-fiber/index.md | 208 ++++++++++----------- source/know/concept/stochastic-process/index.md | 48 ++--- source/know/concept/stokes-law/index.md | 109 +++++------ .../know/concept/sturm-liouville-theory/index.md | 118 ++++++------ source/know/concept/superdense-coding/index.md | 94 ++++++---- .../know/concept/thermodynamic-potential/index.md | 76 ++++---- .../time-dependent-perturbation-theory/index.md | 64 +++---- .../time-independent-perturbation-theory/index.md | 144 +++++++------- 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layout: "concept" --- In the [magnetohydrodynamic](/know/concept/magnetohydrodynamics/) description of a plasma, -we split the velocity $\vb{u}$, electric current $\vb{J}$, -[magnetic field](/know/concept/magnetic-field/) $\vb{B}$ -and [electric field](/know/concept/electric-field/) $\vb{E}$ like so, -into a constant uniform equilibrium (subscript $0$) -and a small unknown perturbation (subscript $1$): +we split the velocity $$\vb{u}$$, electric current $$\vb{J}$$, +[magnetic field](/know/concept/magnetic-field/) $$\vb{B}$$ +and [electric field](/know/concept/electric-field/) $$\vb{E}$$ like so, +into a constant uniform equilibrium (subscript $$0$$) +and a small unknown perturbation (subscript $$1$$): $$\begin{aligned} \vb{u} @@ -41,7 +41,7 @@ $$\begin{aligned} \end{aligned}$$ We do this for the momentum equation too, -assuming that $\vb{J}_0 \!=\! 0$ (to be justified later). +assuming that $$\vb{J}_0 \!=\! 0$$ (to be justified later). Note that the temperature is set to zero, such that the pressure vanishes: $$\begin{aligned} @@ -49,8 +49,8 @@ $$\begin{aligned} = \vb{J}_1 \cross \vb{B}_0 \end{aligned}$$ -Where $\rho$ is the uniform equilibrium density. -We would like an equation for $\vb{J}_1$, +Where $$\rho$$ is the uniform equilibrium density. +We would like an equation for $$\vb{J}_1$$, which is provided by the magnetohydrodynamic form of Ampère's law: $$\begin{aligned} @@ -62,14 +62,14 @@ $$\begin{aligned} \end{aligned}$$ Substituting this into the momentum equation, -and differentiating with respect to $t$: +and differentiating with respect to $$t$$: $$\begin{aligned} \rho \pdvn{2}{\vb{u}_1}{t} = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \pdv{}{\vb{B}1}{t} \Big) \cross \vb{B}_0 \bigg) \end{aligned}$$ -For which we can use Faraday's law to rewrite $\ipdv{\vb{B}_1}{t}$, +For which we can use Faraday's law to rewrite $$\ipdv{\vb{B}_1}{t}$$, incorporating Ohm's law too: $$\begin{aligned} @@ -78,7 +78,7 @@ $$\begin{aligned} = \nabla \cross (\vb{u}_1 \cross \vb{B}_0) \end{aligned}$$ -Inserting this into the momentum equation for $\vb{u}_1$ +Inserting this into the momentum equation for $$\vb{u}_1$$ thus yields its final form: $$\begin{aligned} @@ -86,9 +86,9 @@ $$\begin{aligned} = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vb{B}_0) \big) \Big) \cross \vb{B}_0 \bigg) \end{aligned}$$ -Suppose the magnetic field is pointing in $z$-direction, -i.e. $\vb{B}_0 = B_0 \vu{e}_z$. -Then Faraday's law justifies our earlier assumption that $\vb{J}_0 = 0$, +Suppose the magnetic field is pointing in $$z$$-direction, +i.e. $$\vb{B}_0 = B_0 \vu{e}_z$$. +Then Faraday's law justifies our earlier assumption that $$\vb{J}_0 = 0$$, and the equation can be written as: $$\begin{aligned} @@ -96,7 +96,7 @@ $$\begin{aligned} = v_A^2 \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg) \end{aligned}$$ -Where we have defined the so-called **Alfvén velocity** $v_A$ to be given by: +Where we have defined the so-called **Alfvén velocity** $$v_A$$ to be given by: $$\begin{aligned} \boxed{ @@ -105,24 +105,24 @@ $$\begin{aligned} } \end{aligned}$$ -Now, consider the following plane-wave ansatz for $\vb{u}_1$, -with wavevector $\vb{k}$ and frequency $\omega$: +Now, consider the following plane-wave ansatz for $$\vb{u}_1$$, +with wavevector $$\vb{k}$$ and frequency $$\omega$$: $$\begin{aligned} \vb{u}_1(\vb{r}, t) &= \vb{u}_1 \exp(i \vb{k} \cdot \vb{r} - i \omega t) \end{aligned}$$ -Inserting this into the above differential equation for $\vb{u}_1$ leads to: +Inserting this into the above differential equation for $$\vb{u}_1$$ leads to: $$\begin{aligned} \omega^2 \vb{u}_1 = v_A^2 \bigg( \Big( \vb{k} \cross \big( \vb{k} \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg) \end{aligned}$$ -To evaluate this, we rotate our coordinate system around the $z$-axis -such that $\vb{k} = (0, k_\perp, k_\parallel)$, -i.e. the wavevector's $x$-component is zero. +To evaluate this, we rotate our coordinate system around the $$z$$-axis +such that $$\vb{k} = (0, k_\perp, k_\parallel)$$, +i.e. the wavevector's $$x$$-component is zero. Calculating the cross products: $$\begin{aligned} @@ -149,7 +149,7 @@ $$\begin{aligned} \end{aligned}$$ We rewrite this equation in matrix form, -using that $k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$: +using that $$k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$$: $$\begin{aligned} \begin{bmatrix} @@ -161,9 +161,9 @@ $$\begin{aligned} = 0 \end{aligned}$$ -This has the form of an eigenvalue problem for $\omega^2$, +This has the form of an eigenvalue problem for $$\omega^2$$, meaning we must find non-trivial solutions, -where we cannot simply choose the components of $\vb{u}_1$ to satisfy the equation. +where we cannot simply choose the components of $$\vb{u}_1$$ to satisfy the equation. To achieve this, we demand that the matrix' determinant is zero: $$\begin{aligned} @@ -171,12 +171,12 @@ $$\begin{aligned} = 0 \end{aligned}$$ -This equation has three solutions for $\omega^2$, +This equation has three solutions for $$\omega^2$$, one for each of its three factors being zero. -The simplest case $\omega^2 = 0$ is of no interest to us, +The simplest case $$\omega^2 = 0$$ is of no interest to us, because we are looking for waves. -The first interesting case is $\omega^2 = v_A^2 k_\parallel^2$, +The first interesting case is $$\omega^2 = v_A^2 k_\parallel^2$$, yielding the following dispersion relation: $$\begin{aligned} @@ -188,10 +188,10 @@ $$\begin{aligned} The resulting waves are called **shear Alfvén waves**. From the eigenvalue problem, we see that in this case -$\vb{u}_1 = (u_{1x}, 0, 0)$, meaning $\vb{u}_1 \cdot \vb{k} = 0$: +$$\vb{u}_1 = (u_{1x}, 0, 0)$$, meaning $$\vb{u}_1 \cdot \vb{k} = 0$$: these waves are **transverse**. -The phase velocity $v_p$ and group velocity $v_g$ are as follows, -where $\theta$ is the angle between $\vb{k}$ and $\vb{B}_0$: +The phase velocity $$v_p$$ and group velocity $$v_g$$ are as follows, +where $$\theta$$ is the angle between $$\vb{k}$$ and $$\vb{B}_0$$: $$\begin{aligned} v_p @@ -204,7 +204,7 @@ $$\begin{aligned} = v_A \end{aligned}$$ -The other interesting case is $\omega^2 = v_A^2 k^2$, +The other interesting case is $$\omega^2 = v_A^2 k^2$$, which leads to so-called **compressional Alfvén waves**, with the simple dispersion relation: @@ -215,10 +215,10 @@ $$\begin{aligned} } \end{aligned}$$ -Looking at the eigenvalue problem reveals that $\vb{u}_1 = (0, u_{1y}, 0)$, -meaning $\vb{u}_1 \cdot \vb{k} = u_{1y} k_\perp$, -so these waves are not necessarily transverse, nor longitudinal (since $k_\parallel$ is free). -The phase velocity $v_p$ and group velocity $v_g$ are given by: +Looking at the eigenvalue problem reveals that $$\vb{u}_1 = (0, u_{1y}, 0)$$, +meaning $$\vb{u}_1 \cdot \vb{k} = u_{1y} k_\perp$$, +so these waves are not necessarily transverse, nor longitudinal (since $$k_\parallel$$ is free). +The phase velocity $$v_p$$ and group velocity $$v_g$$ are given by: $$\begin{aligned} v_p diff --git a/source/know/concept/archimedes-principle/index.md b/source/know/concept/archimedes-principle/index.md index dc02d12..364461f 100644 --- a/source/know/concept/archimedes-principle/index.md +++ b/source/know/concept/archimedes-principle/index.md @@ -23,7 +23,7 @@ which has a pressure and thus affects it. The right thing to do is treat the entire body as being submerged in a fluid with varying properties. -Let us consider a volume $V$ completely submerged in such a fluid. +Let us consider a volume $$V$$ completely submerged in such a fluid. This volume will experience a downward force due to gravity, given by: $$\begin{aligned} @@ -31,10 +31,10 @@ $$\begin{aligned} = \int_V \va{g} \rho_\mathrm{b} \dd{V} \end{aligned}$$ -Where $\va{g}$ is the gravitational field, -and $\rho_\mathrm{b}$ is the density of the body. -Meanwhile, the pressure $p$ of the surrounding fluid exerts a force -on the entire surface $S$ of $V$: +Where $$\va{g}$$ is the gravitational field, +and $$\rho_\mathrm{b}$$ is the density of the body. +Meanwhile, the pressure $$p$$ of the surrounding fluid exerts a force +on the entire surface $$S$$ of $$V$$: $$\begin{aligned} \va{F}_p @@ -44,7 +44,7 @@ $$\begin{aligned} Where we have used the divergence theorem. Assuming [hydrostatic equilibrium](/know/concept/hydrostatic-pressure/), -we replace $\nabla p$, +we replace $$\nabla p$$, leading to the definition of the **buoyant force**: $$\begin{aligned} @@ -54,7 +54,7 @@ $$\begin{aligned} } \end{aligned}$$ -For the body to be at rest, we require $\va{F}_g + \va{F}_p = 0$. +For the body to be at rest, we require $$\va{F}_g + \va{F}_p = 0$$. Concretely, the equilibrium condition is: $$\begin{aligned} @@ -64,8 +64,8 @@ $$\begin{aligned} } \end{aligned}$$ -It is commonly assumed that $\va{g}$ is constant everywhere, with magnitude $\mathrm{g}$. -If we also assume that $\rho_\mathrm{f}$ is constant on the "submerged" side, +It is commonly assumed that $$\va{g}$$ is constant everywhere, with magnitude $$\mathrm{g}$$. +If we also assume that $$\rho_\mathrm{f}$$ is constant on the "submerged" side, and zero on the "non-submerged" side, we find: $$\begin{aligned} @@ -73,12 +73,12 @@ $$\begin{aligned} = \mathrm{g} (m_\mathrm{b} - m_\mathrm{f}) \end{aligned}$$ -In other words, the mass $m_\mathrm{b}$ of the entire body -is equal to the mass $m_\mathrm{f}$ of the fluid it displaces. +In other words, the mass $$m_\mathrm{b}$$ of the entire body +is equal to the mass $$m_\mathrm{f}$$ of the fluid it displaces. This is the best-known version of Archimedes' principle. -Note that if $\rho_\mathrm{b} > \rho_\mathrm{f}$, -then the displaced mass $m_\mathrm{f} < m_\mathrm{b}$ +Note that if $$\rho_\mathrm{b} > \rho_\mathrm{f}$$, +then the displaced mass $$m_\mathrm{f} < m_\mathrm{b}$$ even if the entire body is submerged, and the object will therefore continue to sink. diff --git a/source/know/concept/bb84-protocol/index.md b/source/know/concept/bb84-protocol/index.md index 1091773..0f75930 100644 --- a/source/know/concept/bb84-protocol/index.md +++ b/source/know/concept/bb84-protocol/index.md @@ -26,8 +26,8 @@ because the later stages of the protocol involve revealing parts of the data over the (insecure) classical channel. For each bit, Alice randomly chooses a qubit basis, -either $\{ \Ket{0}, \Ket{1} \}$ (eigenstates of the $z$-spin $\hat{\sigma}_z$) -or $\{ \Ket{-}, \Ket{+} \}$ (eigenstates of the $x$-spin $\hat{\sigma}_x$). +either $$\{ \Ket{0}, \Ket{1} \}$$ (eigenstates of the $$z$$-spin $$\hat{\sigma}_z$$) +or $$\{ \Ket{-}, \Ket{+} \}$$ (eigenstates of the $$x$$-spin $$\hat{\sigma}_x$$). Using the basis she chose, she then transmits the bits to Bob over the quantum channel, encoding them as follows: @@ -38,7 +38,7 @@ $$\begin{aligned} \end{aligned}$$ Crucially, Bob has no idea which basis Alice used for any of the bits. -For every bit, he chooses $\hat{\sigma}_z$ or $\hat{\sigma}_x$ at random, +For every bit, he chooses $$\hat{\sigma}_z$$ or $$\hat{\sigma}_x$$ at random, and makes a measurement of the qubit, yielding 0 or 1. If he guessed the basis correctly, he gets the bit value intended by Alice, but if he guessed incorrectly, he randomly gets 0 or 1 with a 50-50 probability: @@ -67,7 +67,7 @@ Suppose that Eve is performing an *intercept-resend attack* (not very effective, but simple), where she listens on the quantum channel. For each qubit received from Alice, Eve chooses -$\hat{\sigma}_z$ or $\hat{\sigma}_x$ at random and measures it. +$$\hat{\sigma}_z$$ or $$\hat{\sigma}_x$$ at random and measures it. She records her results and resends the qubits to Bob using the basis she chose, which may or may not be what Alice intended. @@ -105,17 +105,17 @@ In practice, even without Eve, quantum channels are imperfect, and will introduce some errors in the qubits received by Bob. Suppose that after basis reconciliation, Alice and Bob have the strings -$\{a_1, ..., a_N\}$ and $\{b_1, ..., b_N\}$, respectively. -We define $p$ as the probability that Alice and Bob agree on the $n$th bit, +$$\{a_1, ..., a_N\}$$ and $$\{b_1, ..., b_N\}$$, respectively. +We define $$p$$ as the probability that Alice and Bob agree on the $$n$$th bit, which we assume to be greater than 50%: $$\begin{aligned} p = P(a_n = b_n) > \frac{1}{2} \end{aligned}$$ -Ideally, $p = 1$. To improve $p$, the following simple scheme can be used: -starting at $n = 1$, Alice and Bob reveal $A$ and $B$ over the classical channel, -where $\oplus$ is an XOR: +Ideally, $$p = 1$$. To improve $$p$$, the following simple scheme can be used: +starting at $$n = 1$$, Alice and Bob reveal $$A$$ and $$B$$ over the classical channel, +where $$\oplus$$ is an XOR: $$\begin{aligned} A = a_n \oplus a_{n+1} @@ -123,12 +123,12 @@ $$\begin{aligned} B = b_n \oplus b_{n+1} \end{aligned}$$ -If $A = B$, then $a_{n+1}$ and $b_{n+1}$ are discarded to prevent +If $$A = B$$, then $$a_{n+1}$$ and $$b_{n+1}$$ are discarded to prevent a listener on the classical channel from learning anything about the string. -If $A \neq B$, all of $a_n$, $b_n$, $a_{n+1}$ and $b_{n+1}$ are discarded, -and then Alice and Bob move on to $n = 3$, etc. +If $$A \neq B$$, all of $$a_n$$, $$b_n$$, $$a_{n+1}$$ and $$b_{n+1}$$ are discarded, +and then Alice and Bob move on to $$n = 3$$, etc. -Given that $A = B$, the probability that $a_n = b_n$, +Given that $$A = B$$, the probability that $$a_n = b_n$$, which is what we want, is given by: $$\begin{aligned} @@ -140,7 +140,7 @@ $$\begin{aligned} &= \frac{P(a_{n} = b_{n}) \: P(a_{n+1} = b_{n+1})}{P(a_{n} = b_{n}) \: P(a_{n+1} = b_{n+1}) + P(a_{n} \neq b_{n}) \: P(a_{n+1} \neq b_{n+1})} \end{aligned}$$ -We use the definition of $p$ to get the following inequality, +We use the definition of $$p$$ to get the following inequality, which can be verified by plotting: $$\begin{aligned} @@ -150,9 +150,9 @@ $$\begin{aligned} \end{aligned}$$ Alice and Bob can repeat this error correction scheme multiple times, -until their estimate of $p$ is satisfactory. +until their estimate of $$p$$ is satisfactory. This involves discarding many bits, -so the length $N_\mathrm{new}$ of the string they end up with +so the length $$N_\mathrm{new}$$ of the string they end up with after one iteration is given by: $$\begin{aligned} @@ -166,11 +166,11 @@ More efficient schemes exist, which do not consume so many bits. ## Privacy amplification -Suppose that after the error correction step, $p = 1$, +Suppose that after the error correction step, $$p = 1$$, so Alice and Bob fully agree on the random string. However, in the meantime, Eve has been listening, and has been doing a good job -building up her own string $\{e_1, ..., e_N\}$, +building up her own string $$\{e_1, ..., e_N\}$$, such that she knows more that 50% of the bits: $$\begin{aligned} @@ -178,9 +178,9 @@ $$\begin{aligned} \end{aligned}$$ **Privacy amplification** is an optional final step of the BB84 protocol -which aims to reduce Eve's $q$. +which aims to reduce Eve's $$q$$. Alice and Bob use their existing strings to generate a new one -$\{a_1', ..., a_M'\}$: +$$\{a_1', ..., a_M'\}$$: $$\begin{aligned} a_1' @@ -197,8 +197,8 @@ more efficient schemes exist, which consume less. To see why this improves Alice and Bob's privacy, suppose that Eve is following along, -and creates a new string $\{e_1', ..., e_M'\}$ -where $e_m' = e_{2m - 1} \oplus e_{2m}$. +and creates a new string $$\{e_1', ..., e_M'\}$$ +where $$e_m' = e_{2m - 1} \oplus e_{2m}$$. The probability that Eve's result agrees with Alice and Bob's string is given by: @@ -209,7 +209,7 @@ $$\begin{aligned} &= P(e_1 = a_1) \: P(e_2 = a_2) + P(e_1 \neq a_1) \: P(e_2 \neq a_2) \end{aligned}$$ -Recognizing $q$ 's definition, +Recognizing $$q$$ 's definition, we find the following inequality, which can be verified by plotting: @@ -219,8 +219,8 @@ $$\begin{aligned} < q \end{aligned}$$ -After repeating this step several times, $q$ will be close to 1/2, -which is the ideal value: for $q =$ 0.5, +After repeating this step several times, $$q$$ will be close to 1/2, +which is the ideal value: for $$q =$$ 0.5, Eve would only know 50% of the bits, which is equivalent to her guessing at random. diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md index 5f333a2..f454264 100644 --- a/source/know/concept/bell-state/index.md +++ b/source/know/concept/bell-state/index.md @@ -24,14 +24,14 @@ $$\begin{aligned} } \end{aligned}$$ -Where e.g. $\Ket{0}_A \Ket{1}_B = \Ket{0}_A \otimes \Ket{1}_B$ -is the tensor product of qubit $A$ in state $\Ket{0}$ and $B$ in $\Ket{1}$. +Where e.g. $$\Ket{0}_A \Ket{1}_B = \Ket{0}_A \otimes \Ket{1}_B$$ +is the tensor product of qubit $$A$$ in state $$\Ket{0}$$ and $$B$$ in $$\Ket{1}$$. These states form an orthonormal basis for the two-qubit [Hilbert space](/know/concept/hilbert-space/). More importantly, however, is that the Bell states are maximally entangled, -which we prove here for $\ket{\Phi^{+}}$. +which we prove here for $$\ket{\Phi^{+}}$$. Consider the following pure [density operator](/know/concept/density-operator/): $$\begin{aligned} @@ -40,7 +40,7 @@ $$\begin{aligned} &= \frac{1}{2} \Big( \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big) \Big( \Bra{0}_A \Bra{0}_B + \Bra{1}_A \Bra{1}_B \Big) \end{aligned}$$ -The reduced density operator $\hat{\rho}_A$ of qubit $A$ is then calculated as follows: +The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated as follows: $$\begin{aligned} \hat{\rho}_A @@ -54,14 +54,14 @@ $$\begin{aligned} = \frac{1}{2} \hat{I} \end{aligned}$$ -This result is maximally mixed, therefore $\ket{\Phi^{+}}$ is maximally entangled. +This result is maximally mixed, therefore $$\ket{\Phi^{+}}$$ is maximally entangled. The same holds for the other three Bell states, -and is equally true for qubit $B$. +and is equally true for qubit $$B$$. -This means that a measurement of qubit $A$ -has a 50-50 chance to yield $\Ket{0}$ or $\Ket{1}$. +This means that a measurement of qubit $$A$$ +has a 50-50 chance to yield $$\Ket{0}$$ or $$\Ket{1}$$. However, due to the entanglement, -measuring $A$ also has consequences for qubit $B$: +measuring $$A$$ also has consequences for qubit $$B$$: $$\begin{aligned} \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 @@ -81,10 +81,10 @@ $$\begin{aligned} = \frac{1}{2} \end{aligned}$$ -As an example, if $A$ collapses into $\Ket{0}$ due to a measurement, -then $B$ instantly also collapses into $\Ket{0}$, never $\Ket{1}$, +As an example, if $$A$$ collapses into $$\Ket{0}$$ due to a measurement, +then $$B$$ instantly also collapses into $$\Ket{0}$$, never $$\Ket{1}$$, even if it was not measured. -This was a specific example for $\ket{\Phi^{+}}$, +This was a specific example for $$\ket{\Phi^{+}}$$, but analogous results can be found for the other Bell states. diff --git a/source/know/concept/bells-theorem/index.md b/source/know/concept/bells-theorem/index.md index 3b71dbf..a01bf9e 100644 --- a/source/know/concept/bells-theorem/index.md +++ b/source/know/concept/bells-theorem/index.md @@ -13,7 +13,7 @@ layout: "concept" cannot be explained by theories built on so-called **local hidden variables** (LHVs). -Suppose that we have two spin-1/2 particles, called $A$ and $B$, +Suppose that we have two spin-1/2 particles, called $$A$$ and $$B$$, in an entangled [Bell state](/know/concept/bell-state/): $$\begin{aligned} @@ -22,23 +22,23 @@ $$\begin{aligned} \end{aligned}$$ 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 $\Ket{\downarrow}$. +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 $B$. +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 $B$ +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 $\lambda$, +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: +along arbitrary directions $$\vec{a}$$ and $$\vec{b}$$, +so the outcomes for particles $$A$$ and $$B$$ are: $$\begin{aligned} A(\vec{a}, \lambda) = \pm 1 @@ -46,8 +46,8 @@ $$\begin{aligned} B(\vec{b}, \lambda) = \pm 1 \end{aligned}$$ -Where $\pm 1$ are the eigenvalues of the Pauli matrices -in the chosen directions $\vec{a}$ and $\vec{b}$: +Where $$\pm 1$$ are the eigenvalues of the Pauli matrices +in the chosen directions $$\vec{a}$$ and $$\vec{b}$$: $$\begin{aligned} \hat{\sigma}_a @@ -59,8 +59,8 @@ $$\begin{aligned} = b_x \hat{\sigma}_x + b_y \hat{\sigma}_y + b_z \hat{\sigma}_z \end{aligned}$$ -Whether $\lambda$ is a scalar or a vector does not matter; -we simply demand that it follows an unknown probability distribution $\rho(\lambda)$: +Whether $$\lambda$$ is a scalar or a vector does not matter; +we simply demand that it follows an unknown probability distribution $$\rho(\lambda)$$: $$\begin{aligned} \int \rho(\lambda) \dd{\lambda} = 1 @@ -68,8 +68,8 @@ $$\begin{aligned} \rho(\lambda) \ge 0 \end{aligned}$$ -The product of the outcomes of $A$ and $B$ then has the following expectation value. -Note that we only multiply $A$ and $B$ for shared $\lambda$-values: +The product of the outcomes of $$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} @@ -83,7 +83,7 @@ which both prove Bell's theorem. ## Bell inequality -If $\vec{a} = \vec{b}$, then we know that $A$ and $B$ always have opposite spins: +If $$\vec{a} = \vec{b}$$, then we know that $$A$$ and $$B$$ always have opposite spins: $$\begin{aligned} A(\vec{a}, \lambda) @@ -98,8 +98,8 @@ $$\begin{aligned} = - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda} \end{aligned}$$ -Next, we introduce an arbitrary third direction $\vec{c}$, -and use the fact that $( A(\vec{b}, \lambda) )^2 = 1$: +Next, we introduce an arbitrary third direction $$\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} @@ -109,7 +109,7 @@ $$\begin{aligned} \end{aligned}$$ Inside the integral, the only factors that can be negative -are the last two, and their product is $\pm 1$. +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: @@ -121,7 +121,7 @@ $$\begin{aligned} &\le \int \rho(\lambda) \dd{\lambda} - \int \rho(\lambda) A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \dd{\lambda} \end{aligned}$$ -Since $\rho(\lambda)$ is a normalized probability density function, +Since $$\rho(\lambda)$$ is a normalized probability density function, we arrive at the **Bell inequality**: $$\begin{aligned} @@ -131,18 +131,18 @@ $$\begin{aligned} } \end{aligned}$$ -Any theory involving an LHV $\lambda$ must obey this inequality. +Any theory involving an LHV $$\lambda$$ must obey this inequality. The problem, however, is that quantum mechanics dictates the expectation values -for the state $\Ket{\Psi^{-}}$: +for the state $$\Ket{\Psi^{-}}$$: $$\begin{aligned} \Expval{A_a B_b} = - \vec{a} \cdot \vec{b} \end{aligned}$$ Finding directions which violate the Bell inequality is easy: -for example, if $\vec{a}$ and $\vec{b}$ are orthogonal, -and $\vec{c}$ is at a $\pi/4$ angle to both of them, -then the left becomes $0.707$ and the right $0.293$, +for example, if $$\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$$, which clearly disagrees with the inequality, meaning that LHVs are impossible. @@ -152,8 +152,8 @@ meaning that LHVs are impossible. The **Clauser-Horne-Shimony-Holt** or simply **CHSH inequality** takes a slightly different approach, and is more useful in practice. -Consider four spin directions, two for $A$ called $\vec{a}_1$ and $\vec{a}_2$, -and two for $B$ called $\vec{b}_1$ and $\vec{b}_2$. +Consider four spin directions, two for $$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} @@ -196,7 +196,7 @@ $$\begin{aligned} + \bigg|\! \int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg| \end{aligned}$$ -Using the fact that the product of $A$ and $B$ is always either $-1$ or $+1$, +Using the fact that the product of $$A$$ and $$B$$ is always either $$-1$$ or $$+1$$, we can reduce this to: $$\begin{aligned} @@ -209,7 +209,7 @@ $$\begin{aligned} \end{aligned}$$ Evaluating these integrals gives us the following inequality, -which holds for both choices of $\pm$: +which holds for both choices of $$\pm$$: $$\begin{aligned} \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big| @@ -235,8 +235,8 @@ $$\begin{aligned} &\ge \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} + \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big| \end{aligned}$$ -The quantity on the right-hand side is sometimes called the **CHSH quantity** $S$, -and measures the correlation between the spins of $A$ and $B$: +The quantity on the right-hand side is sometimes called the **CHSH quantity** $$S$$, +and measures the correlation between the spins of $$A$$ and $$B$$: $$\begin{aligned} \boxed{ @@ -244,7 +244,7 @@ $$\begin{aligned} } \end{aligned}$$ -The CHSH inequality places an upper bound on the magnitude of $S$ +The CHSH inequality places an upper bound on the magnitude of $$S$$ for LHV-based theories: $$\begin{aligned} @@ -258,15 +258,15 @@ $$\begin{aligned} Quantum physics can violate the CHSH inequality, but by how much? Consider the following two-particle operator, -whose expectation value is the CHSH quantity, i.e. $S = \expval{\hat{S}}$: +whose expectation value is the CHSH quantity, i.e. $$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}$$ -Where $\otimes$ is the tensor product, -and e.g. $\hat{A}_1$ is the Pauli matrix for the $\vec{a}_1$-direction. +Where $$\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} @@ -292,15 +292,15 @@ $$\begin{aligned} \end{aligned}$$ Spin operators are unitary, so their square is the identity, -e.g. $\hat{A}_1^2 = \hat{I}$. Therefore $\hat{S}^2$ reduces to: +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 $\expval{\hat{S}^2}$, +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: @@ -321,7 +321,7 @@ $$\begin{aligned} \le 2 \end{aligned}$$ -And $\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$ for the same reason. +And $$\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$$ for the same reason. The norm is the largest eigenvalue, therefore: $$\begin{aligned} @@ -336,7 +336,7 @@ $$\begin{aligned} We thus arrive at **Tsirelson's bound**, which states that quantum mechanics can violate -the CHSH inequality by a factor of $\sqrt{2}$: +the CHSH inequality by a factor of $$\sqrt{2}$$: $$\begin{aligned} \boxed{ @@ -359,8 +359,8 @@ $$\begin{aligned} \hat{B}_2 = \frac{\hat{\sigma}_z - \hat{\sigma}_x}{\sqrt{2}} \end{aligned}$$ -Using the fact that $\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}$, -it can then be shown that $S = 2 \sqrt{2}$ in this case. +Using the fact that $$\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}$$, +it can then be shown that $$S = 2 \sqrt{2}$$ in this case. diff --git a/source/know/concept/beltrami-identity/index.md b/source/know/concept/beltrami-identity/index.md index 3fa566c..be9a344 100644 --- a/source/know/concept/beltrami-identity/index.md +++ b/source/know/concept/beltrami-identity/index.md @@ -8,26 +8,26 @@ categories: layout: "concept" --- -Consider a general functional $J[f]$ of the following form, -with $f(x)$ an unknown function: +Consider a general functional $$J[f]$$ of the following form, +with $$f(x)$$ an unknown function: $$\begin{aligned} J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x} \end{aligned}$$ -Where $L$ is the Lagrangian. -To find the $f$ that maximizes or minimizes $J[f]$, +Where $$L$$ is the Lagrangian. +To find the $$f$$ that maximizes or minimizes $$J[f]$$, the [calculus of variations](/know/concept/calculus-of-variations/) -states that the Euler-Lagrange equation must be solved for $f$: +states that the Euler-Lagrange equation must be solved for $$f$$: $$\begin{aligned} 0 = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f'} \Big) \end{aligned}$$ -We now want to know exactly how $L$ depends on the free variable $x$, -since it is a function of $x$, $f(x)$ and $f'(x)$. +We now want to know exactly how $$L$$ depends on the free variable $$x$$, +since it is a function of $$x$$, $$f(x)$$ and $$f'(x)$$. Using the chain rule: $$\begin{aligned} @@ -44,16 +44,16 @@ $$\begin{aligned} &= \dv{}{x} \bigg( f' \pdv{L}{f'} \bigg) + \pdv{L}{x} \end{aligned}$$ -Although we started from the "hard" derivative $\idv{L}{x}$, -we arrive at an expression for the "soft" derivative $\ipdv{L}{x}$, -describing the *explicit* dependence of $L$ on $x$: +Although we started from the "hard" derivative $$\idv{L}{x}$$, +we arrive at an expression for the "soft" derivative $$\ipdv{L}{x}$$, +describing the *explicit* dependence of $$L$$ on $$x$$: $$\begin{aligned} - \pdv{L}{x} = \dv{}{x} \bigg( f' \pdv{L}{f'} - L \bigg) \end{aligned}$$ -What if $L$ does not explicitly depend on $x$, i.e. $\ipdv{L}{x} = 0$? +What if $$L$$ does not explicitly depend on $$x$$, i.e. $$\ipdv{L}{x} = 0$$? In that case, the equation can be integrated to give the **Beltrami identity**: $$\begin{aligned} @@ -63,19 +63,19 @@ $$\begin{aligned} } \end{aligned}$$ -Where $C$ is a constant. -This says that the left-hand side is a conserved quantity in $x$, +Where $$C$$ is a constant. +This says that the left-hand side is a conserved quantity in $$x$$, which could be useful to know. -If we insert a concrete expression for $L$, -the Beltrami identity might be easier to solve for $f$ than the full Euler-Lagrange equation. -The assumption $\ipdv{L}{x} = 0$ is justified; -for example, if $x$ is time, it means that the potential is time-independent. +If we insert a concrete expression for $$L$$, +the Beltrami identity might be easier to solve for $$f$$ than the full Euler-Lagrange equation. +The assumption $$\ipdv{L}{x} = 0$$ is justified; +for example, if $$x$$ is time, it means that the potential is time-independent. ## Higher dimensions -Above, a 1D problem was considered, i.e. $f$ depended only on a single variable $x$. -Consider now a 2D problem, such that $J[f]$ is given by: +Above, a 1D problem was considered, i.e. $$f$$ depended only on a single variable $$x$$. +Consider now a 2D problem, such that $$J[f]$$ is given by: $$\begin{aligned} J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y} @@ -87,7 +87,7 @@ $$\begin{aligned} 0 = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f_x} \Big) - \dv{}{y} \Big( \pdv{L}{f_y} \Big) \end{aligned}$$ -Once again, we calculate the hard $x$-derivative of $L$ (the $y$-derivative is analogous): +Once again, we calculate the hard $$x$$-derivative of $$L$$ (the $$y$$-derivative is analogous): $$\begin{aligned} \dv{L}{x} @@ -99,7 +99,7 @@ $$\begin{aligned} &= \dv{}{x} \Big( f_x \pdv{L}{f_x} \Big) + \dv{}{y} \Big( f_x \pdv{L}{f_y} \Big) + \pdv{L}{x} \end{aligned}$$ -This time, we arrive at the following expression for the soft derivative $\ipdv{L}{x}$: +This time, we arrive at the following expression for the soft derivative $$\ipdv{L}{x}$$: $$\begin{aligned} - \pdv{L}{x} @@ -109,9 +109,9 @@ $$\begin{aligned} Due to the derivatives, this cannot be cleanly turned into an analogue of the 1D Beltrami identity, and therefore we use that name only in the 1D case. -However, if $\ipdv{L}{x} = 0$, this equation is still useful. +However, if $$\ipdv{L}{x} = 0$$, this equation is still useful. For an off-topic demonstration of this fact, -let us choose $x$ as the transverse coordinate, and integrate over it to get: +let us choose $$x$$ as the transverse coordinate, and integrate over it to get: $$\begin{aligned} 0 @@ -123,7 +123,7 @@ $$\begin{aligned} \end{aligned}$$ If our boundary conditions cause the boundary term to vanish (as is often the case), -then the integral on the right is a conserved quantity with respect to $y$. +then the integral on the right is a conserved quantity with respect to $$y$$. While not as elegant as the 1D Beltrami identity, the above 2D counterpart still fulfills the same role. diff --git a/source/know/concept/bernoullis-theorem/index.md b/source/know/concept/bernoullis-theorem/index.md index 12bd0ca..6b933d2 100644 --- a/source/know/concept/bernoullis-theorem/index.md +++ b/source/know/concept/bernoullis-theorem/index.md @@ -10,8 +10,8 @@ layout: "concept" --- For inviscid fluids, **Bernuilli's theorem** states -that an increase in flow velocity $\va{v}$ is paired -with a decrease in pressure $p$ and/or potential energy. +that an increase in flow velocity $$\va{v}$$ is paired +with a decrease in pressure $$p$$ and/or potential energy. For a qualitative argument, look no further than one of the [Euler equations](/know/concept/euler-equations/), with a [material derivative](/know/concept/material-derivative/): @@ -22,16 +22,16 @@ $$\begin{aligned} = \va{g} - \frac{\nabla p}{\rho} \end{aligned}$$ -Assuming that $\va{v}$ is constant in $t$, -it becomes clear that a higher $\va{v}$ requires a lower $p$. +Assuming that $$\va{v}$$ is constant in $$t$$, +it becomes clear that a higher $$\va{v}$$ requires a lower $$p$$. ## Simple form For an incompressible fluid -with a time-independent velocity field $\va{v}$ (i.e. **steady flow**), +with a time-independent velocity field $$\va{v}$$ (i.e. **steady flow**), Bernoulli's theorem formally states that the -**Bernoulli head** $H$ is constant along a streamline: +**Bernoulli head** $$H$$ is constant along a streamline: $$\begin{aligned} \boxed{ @@ -40,8 +40,8 @@ $$\begin{aligned} } \end{aligned}$$ -Where $\Phi$ is the gravitational potential, such that $\va{g} = - \nabla \Phi$. -To prove this theorem, we take the material derivative of $H$: +Where $$\Phi$$ is the gravitational potential, such that $$\va{g} = - \nabla \Phi$$. +To prove this theorem, we take the material derivative of $$H$$: $$\begin{aligned} \frac{\mathrm{D} H}{\mathrm{D} t} @@ -63,7 +63,7 @@ $$\begin{aligned} + \va{v} \cdot \big( \va{g} + \nabla \Phi \big) + \va{v} \cdot \Big( \frac{\nabla p}{\rho} - \frac{\nabla p}{\rho} \Big) \end{aligned}$$ -Using the fact that $\va{g} = - \nabla \Phi$, +Using the fact that $$\va{g} = - \nabla \Phi$$, we are left with the following equation: $$\begin{aligned} @@ -72,12 +72,12 @@ $$\begin{aligned} \end{aligned}$$ Assuming that the flow is steady, both derivatives vanish, -leading us to the conclusion that $H$ is conserved along the streamline. +leading us to the conclusion that $$H$$ is conserved along the streamline. In fact, there exists **Bernoulli's stronger theorem**, -which states that $H$ is constant *everywhere* in regions with -zero [vorticity](/know/concept/vorticity/) $\va{\omega} = 0$. -For a proof, see the derivation of $\va{\omega}$'s equation of motion. +which states that $$H$$ is constant *everywhere* in regions with +zero [vorticity](/know/concept/vorticity/) $$\va{\omega} = 0$$. +For a proof, see the derivation of $$\va{\omega}$$'s equation of motion. ## References diff --git a/source/know/concept/bernstein-vazirani-algorithm/index.md b/source/know/concept/bernstein-vazirani-algorithm/index.md index af49841..f91c0ba 100644 --- a/source/know/concept/bernstein-vazirani-algorithm/index.md +++ b/source/know/concept/bernstein-vazirani-algorithm/index.md @@ -17,10 +17,10 @@ It is extremely similar to the and even uses the same circuit. It solves a very artificial problem: -we are given a "black box" function $f(x)$ -that takes an $N$-bit $x$ and returns a single bit, +we are given a "black box" function $$f(x)$$ +that takes an $$N$$-bit $$x$$ and returns a single bit, which we are promised is the lowest bit of the bitwise dot product -of $x$ with an unknown $N$-bit string $s$: +of $$x$$ with an unknown $$N$$-bit string $$s$$: $$\begin{aligned} f(x) @@ -28,10 +28,10 @@ $$\begin{aligned} = (s_1 x_1 + s_2 x_2 + \:...\: + s_N x_N) \:\:(\bmod \: 2) \end{aligned}$$ -The goal is to find $s$. +The goal is to find $$s$$. To solve this problem, -a classical computer would need to call $f(x)$ exactly $N$ times -with $x = 2^n$ for $n \in \{ 0, ..., N \!-\! 1\}$. +a classical computer would need to call $$f(x)$$ exactly $$N$$ times +with $$x = 2^n$$ for $$n \in \{ 0, ..., N \!-\! 1\}$$. However, the Bernstein-Vazirani algorithm allows a quantum computer to do it with only a single query. It uses the following circuit: @@ -40,9 +40,9 @@ It uses the following circuit: -Where $U_f$ is a phase oracle, +Where $$U_f$$ is a phase oracle, whose action is defined as follows, -where $\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$: +where $$\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$$: $$\begin{aligned} \Ket{x} @@ -51,14 +51,14 @@ $$\begin{aligned} = (-1)^{s \cdot x} \Ket{x} \end{aligned}$$ -That is, it introduces a phase flip based on the value of $f(x)$. +That is, it introduces a phase flip based on the value of $$f(x)$$. For an example implementation of such an oracle, see the Deutsch-Jozsa algorithm: its circuit is identical to this one, -but describes $U_f$ in a different (but equivalent) way. +but describes $$U_f$$ in a different (but equivalent) way. -Starting from the state $\Ket{0}^{\otimes N}$, -applying the [Hadamard gate](/know/concept/quantum-gate/) $H$ +Starting from the state $$\Ket{0}^{\otimes N}$$, +applying the [Hadamard gate](/know/concept/quantum-gate/) $$H$$ to all qubits yields: $$\begin{aligned} @@ -68,7 +68,7 @@ $$\begin{aligned} = \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \Ket{x} \end{aligned}$$ -This is an equal superposition of all candidates $\Ket{x}$, +This is an equal superposition of all candidates $$\Ket{x}$$, which we feed to the oracle: $$\begin{aligned} @@ -78,7 +78,7 @@ $$\begin{aligned} \end{aligned}$$ Then, thanks to the definition of the Hadamard transform, -a final set of $H$-gates leads us to: +a final set of $$H$$-gates leads us to: $$\begin{aligned} \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{s \cdot x} \Ket{x} @@ -87,9 +87,9 @@ $$\begin{aligned} = \Ket{s_1} \cdots \Ket{s_N} \end{aligned}$$ -Which, upon measurement, gives us the desired binary representation of $s$. -For comparison, the Deutsch-Jozsa algorithm only cares whether $s = 0$ or $s \neq 0$, -whereas this algorithm is interested in the exact value of $s$. +Which, upon measurement, gives us the desired binary representation of $$s$$. +For comparison, the Deutsch-Jozsa algorithm only cares whether $$s = 0$$ or $$s \neq 0$$, +whereas this algorithm is interested in the exact value of $$s$$. diff --git a/source/know/concept/berry-phase/index.md b/source/know/concept/berry-phase/index.md index eedc548..d237ea5 100644 --- a/source/know/concept/berry-phase/index.md +++ b/source/know/concept/berry-phase/index.md @@ -8,8 +8,8 @@ categories: layout: "concept" --- -Consider a Hamiltonian $\hat{H}$ that does not explicitly depend on time, -but does depend on a given parameter $\vb{R}$. +Consider a Hamiltonian $$\hat{H}$$ that does not explicitly depend on time, +but does depend on a given parameter $$\vb{R}$$. The Schrödinger equations then read: $$\begin{aligned} @@ -20,9 +20,9 @@ $$\begin{aligned} &= E_n(\vb{R}) \Ket{\psi_n(\vb{R})} \end{aligned}$$ -The general full solution $\Ket{\Psi_n}$ has the following form, -where we allow $\vb{R}$ to evolve in time, -and we have abbreviated the traditional phase of the "wiggle factor" as $L_n$: +The general full solution $$\Ket{\Psi_n}$$ has the following form, +where we allow $$\vb{R}$$ to evolve in time, +and we have abbreviated the traditional phase of the "wiggle factor" as $$L_n$$: $$\begin{aligned} \Ket{\Psi_n(t)} @@ -31,11 +31,11 @@ $$\begin{aligned} L_n(t) \equiv \int_0^t E_n(\vb{R}(t')) \dd{t'} \end{aligned}$$ -The **geometric phase** $\gamma_n(t)$ is more interesting. -It is not included in $\Ket{\psi_n}$, -because it depends on the path $\vb{R}(t)$ -rather than only the present $\vb{R}$ and $t$. -Its dynamics can be found by inserting the above $\Ket{\Psi_n}$ +The **geometric phase** $$\gamma_n(t)$$ is more interesting. +It is not included in $$\Ket{\psi_n}$$, +because it depends on the path $$\vb{R}(t)$$ +rather than only the present $$\vb{R}$$ and $$t$$. +Its dynamics can be found by inserting the above $$\Ket{\Psi_n}$$ into the time-dependent Schrödinger equation: $$\begin{aligned} @@ -58,8 +58,8 @@ $$\begin{aligned} &= \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t} \end{aligned}$$ -Front-multiplying by $i \Bra{\Psi_n}$ gives us -the equation of motion of the geometric phase $\gamma_n$: +Front-multiplying by $$i \Bra{\Psi_n}$$ gives us +the equation of motion of the geometric phase $$\gamma_n$$: $$\begin{aligned} \boxed{ @@ -68,7 +68,7 @@ $$\begin{aligned} } \end{aligned}$$ -Where we have defined the so-called **Berry connection** $\vb{A}_n$ as follows: +Where we have defined the so-called **Berry connection** $$\vb{A}_n$$ as follows: $$\begin{aligned} \boxed{ @@ -77,9 +77,9 @@ $$\begin{aligned} } \end{aligned}$$ -Importantly, note that $\vb{A}_n$ is real, -provided that $\Ket{\psi_n}$ is always normalized for all $\vb{R}$. -To prove this, we start from the fact that $\nabla_\vb{R} 1 = 0$: +Importantly, note that $$\vb{A}_n$$ is real, +provided that $$\Ket{\psi_n}$$ is always normalized for all $$\vb{R}$$. +To prove this, we start from the fact that $$\nabla_\vb{R} 1 = 0$$: $$\begin{aligned} 0 @@ -91,14 +91,14 @@ $$\begin{aligned} = 2 \Imag\{ \vb{A}_n \} \end{aligned}$$ -Consequently, $\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is always real, -because $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary. +Consequently, $$\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is always real, +because $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary. -Suppose now that the parameter $\vb{R}(t)$ is changed adiabatically +Suppose now that the parameter $$\vb{R}(t)$$ is changed adiabatically (i.e. so slow that the system stays in the same eigenstate) -for $t \in [0, T]$, along a circuit $C$ with $\vb{R}(0) \!=\! \vb{R}(T)$. -Integrating the phase $\gamma_n(t)$ over this contour $C$ then yields -the **Berry phase** $\gamma_n(C)$: +for $$t \in [0, T]$$, along a circuit $$C$$ with $$\vb{R}(0) \!=\! \vb{R}(T)$$. +Integrating the phase $$\gamma_n(t)$$ over this contour $$C$$ then yields +the **Berry phase** $$\gamma_n(C)$$: $$\begin{aligned} \boxed{ @@ -107,9 +107,9 @@ $$\begin{aligned} } \end{aligned}$$ -But we have a problem: $\vb{A}_n$ is not unique! +But we have a problem: $$\vb{A}_n$$ is not unique! Due to the Schrödinger equation's gauge invariance, -any function $f(\vb{R}(t))$ can be added to $\gamma_n(t)$ +any function $$f(\vb{R}(t))$$ can be added to $$\gamma_n(t)$$ without making an immediate physical difference to the state. Consider the following general gauge transformation: @@ -118,9 +118,9 @@ $$\begin{aligned} \equiv \exp(i f(\vb{R})) \: \Ket{\psi_n(\vb{R})} \end{aligned}$$ -To find $\vb{A}_n$ for a particular choice of $f$, +To find $$\vb{A}_n$$ for a particular choice of $$f$$, we need to evaluate the inner product -$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$: +$$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$$: $$\begin{aligned} \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n} @@ -131,16 +131,16 @@ $$\begin{aligned} &= i \nabla_\vb{R} f + \inprod{\psi_n}{\nabla_\vb{R} \psi_n} \end{aligned}$$ -Unfortunately, $f$ does not vanish as we would have liked, -so $\vb{A}_n$ depends on our choice of $f$. +Unfortunately, $$f$$ does not vanish as we would have liked, +so $$\vb{A}_n$$ depends on our choice of $$f$$. However, the curl of a gradient is always zero, -so although $\vb{A}_n$ is not unique, -its curl $\nabla_\vb{R} \cross \vb{A}_n$ is guaranteed to be. -Conveniently, we can introduce a curl in the definition of $\gamma_n(C)$ +so although $$\vb{A}_n$$ is not unique, +its curl $$\nabla_\vb{R} \cross \vb{A}_n$$ is guaranteed to be. +Conveniently, we can introduce a curl in the definition of $$\gamma_n(C)$$ by applying Stokes' theorem, under the assumption -that $\vb{A}_n$ has no singularities in the area enclosed by $C$ -(fortunately, $\vb{A}_n$ can always be chosen to satisfy this): +that $$\vb{A}_n$$ has no singularities in the area enclosed by $$C$$ +(fortunately, $$\vb{A}_n$$ can always be chosen to satisfy this): $$\begin{aligned} \boxed{ @@ -149,10 +149,10 @@ $$\begin{aligned} } \end{aligned}$$ -Where we defined $\vb{B}_n$ as the curl of $\vb{A}_n$. -Now $\gamma_n(C)$ is guaranteed to be unique. -Note that $\vb{B}_n$ is analogous to a magnetic field, -and $\vb{A}_n$ to a magnetic vector potential: +Where we defined $$\vb{B}_n$$ as the curl of $$\vb{A}_n$$. +Now $$\gamma_n(C)$$ is guaranteed to be unique. +Note that $$\vb{B}_n$$ is analogous to a magnetic field, +and $$\vb{A}_n$$ to a magnetic vector potential: $$\begin{aligned} \vb{B}_n(\vb{R}) @@ -160,9 +160,9 @@ $$\begin{aligned} = \Imag\!\Big\{ \nabla_\vb{R} \cross \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} \Big\} \end{aligned}$$ -Unfortunately, $\nabla_\vb{R} \psi_n$ is difficult to evaluate explicitly, -so we would like to rewrite $\vb{B}_n$ such that it does not enter. -We do this as follows, inserting $1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m}$ along the way: +Unfortunately, $$\nabla_\vb{R} \psi_n$$ is difficult to evaluate explicitly, +so we would like to rewrite $$\vb{B}_n$$ such that it does not enter. +We do this as follows, inserting $$1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m}$$ along the way: $$\begin{aligned} i \vb{B}_n @@ -172,9 +172,9 @@ $$\begin{aligned} &= \sum_{m} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n} \end{aligned}$$ -The fact that $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary +The fact that $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary means it is parallel to its complex conjugate, -and thus the cross product vanishes, so we exclude $n$ from the sum: +and thus the cross product vanishes, so we exclude $$n$$ from the sum: $$\begin{aligned} \vb{B}_n @@ -200,8 +200,8 @@ $$\begin{aligned} } \end{aligned}$$ -Which only involves $\nabla_\vb{R} \hat{H}$, -and is therefore easier to evaluate than any $\Ket{\nabla_\vb{R} \psi_n}$. +Which only involves $$\nabla_\vb{R} \hat{H}$$, +and is therefore easier to evaluate than any $$\Ket{\nabla_\vb{R} \psi_n}$$. diff --git a/source/know/concept/binomial-distribution/index.md b/source/know/concept/binomial-distribution/index.md index 14ba4cb..1193a93 100644 --- a/source/know/concept/binomial-distribution/index.md +++ b/source/know/concept/binomial-distribution/index.md @@ -9,11 +9,11 @@ layout: "concept" --- The **binomial distribution** is a discrete probability distribution -describing a **Bernoulli process**: a set of independent $N$ trials where +describing a **Bernoulli process**: a set of independent $$N$$ trials where each has only two possible outcomes, "success" and "failure", -the former with probability $p$ and the latter with $q = 1 - p$. +the former with probability $$p$$ and the latter with $$q = 1 - p$$. The binomial distribution then gives the probability -that $n$ out of the $N$ trials succeed: +that $$n$$ out of the $$N$$ trials succeed: $$\begin{aligned} \boxed{ @@ -22,8 +22,8 @@ $$\begin{aligned} \end{aligned}$$ The first factor is known as the **binomial coefficient**, which describes the -number of microstates (i.e. permutations) that have $n$ successes out of $N$ trials. -These happen to be the coefficients in the polynomial $(a + b)^N$, +number of microstates (i.e. permutations) that have $$n$$ successes out of $$N$$ trials. +These happen to be the coefficients in the polynomial $$(a + b)^N$$, and can be read off of Pascal's triangle. It is defined as follows: @@ -33,10 +33,10 @@ $$\begin{aligned} } \end{aligned}$$ -The remaining factor $p^n (1 - p)^{N - n}$ is then just the +The remaining factor $$p^n (1 - p)^{N - n}$$ is then just the probability of attaining each microstate. -The expected or mean number of successes $\mu$ after $N$ trials is as follows: +The expected or mean number of successes $$\mu$$ after $$N$$ trials is as follows: $$\begin{aligned} \boxed{ @@ -49,7 +49,7 @@ $$\begin{aligned} -Meanwhile, we find the following variance $\sigma^2$, -with $\sigma$ being the standard deviation: +Meanwhile, we find the following variance $$\sigma^2$$, +with $$\sigma$$ being the standard deviation: $$\begin{aligned} \boxed{ @@ -79,7 +79,7 @@ $$\begin{aligned}