diff options
Diffstat (limited to 'source/know/concept')
| -rw-r--r-- | source/know/concept/convolution-theorem/index.md | 18 | ||||
| -rw-r--r-- | source/know/concept/debye-length/index.md | 2 | ||||
| -rw-r--r-- | source/know/concept/electric-dipole-approximation/index.md | 4 | ||||
| -rw-r--r-- | source/know/concept/gronwall-bellman-inequality/index.md | 17 | ||||
| -rw-r--r-- | source/know/concept/lagrange-multiplier/index.md | 187 | ||||
| -rw-r--r-- | source/know/concept/material-derivative/index.md | 4 | ||||
| -rw-r--r-- | source/know/concept/parsevals-theorem/index.md | 20 | ||||
| -rw-r--r-- | source/know/concept/rabi-oscillation/index.md | 41 | ||||
| -rw-r--r-- | source/know/concept/self-steepening/index.md | 15 | ||||
| -rw-r--r-- | source/know/concept/shors-algorithm/index.md | 9 | ||||
| -rw-r--r-- | source/know/concept/thermodynamic-potential/index.md | 10 | ||||
| -rw-r--r-- | source/know/concept/toffoli-gate/index.md | 4 |
12 files changed, 193 insertions, 138 deletions
diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md index 510417a..d10d85d 100644 --- a/source/know/concept/convolution-theorem/index.md +++ b/source/know/concept/convolution-theorem/index.md @@ -36,9 +36,9 @@ rearrange the integrals: $$\begin{aligned} \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} - &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp(i s k x') \dd{x'} \Big) \exp(-i s k x) \dd{k} + &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \: e^{i s k x'} \dd{x'} \Big) e^{-i s k x} \dd{k} \\ - &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k (x - x')) \dd{k} \Big) \dd{x'} + &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \: e^{-i s k (x - x')} \dd{k} \Big) \dd{x'} \\ &= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'} = A \cdot (f * g)(x) @@ -49,9 +49,9 @@ this time starting from a product in the $$x$$-domain: $$\begin{aligned} \hat{\mathcal{F}}\{f(x) \: g(x)\} - &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x} + &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \: e^{-i s x k'} \dd{k'} \Big) e^{i s k x} \dd{x} \\ - &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'} + &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \: e^{i s x (k - k')} \dd{x} \Big) \dd{k'} \\ &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'} = B \cdot (\tilde{f} * \tilde{g})(k) @@ -86,20 +86,20 @@ because we set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$: $$\begin{aligned} \hat{\mathcal{L}}\{(f * g)(t)\} - &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp(- s t) \dd{t} + &= \int_0^\infty \Big( \int_0^\infty g(t') \: f(t - t') \dd{t'} \Big) e^{-s t} \dd{t} \\ - &= \int_0^\infty \Big( \int_0^\infty f(t - t') \exp(- s t) \dd{t} \Big) g(t') \dd{t'} + &= \int_0^\infty \Big( \int_0^\infty f(t - t') \: e^{-s t} \dd{t} \Big) g(t') \dd{t'} \end{aligned}$$ Then we define a new integration variable $$\tau = t - t'$$, yielding: $$\begin{aligned} \hat{\mathcal{L}}\{(f * g)(t)\} - &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'} + &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s (\tau + t')} \dd{\tau} \Big) g(t') \dd{t'} \\ - &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'} + &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s \tau} \dd{\tau} \Big) g(t') \: e^{-s t'} \dd{t'} \\ - &= \int_0^\infty \tilde{f}(s) \: g(t') \exp(- s t') \dd{t'} + &= \int_0^\infty \tilde{f}(s) \: g(t') \: e^{-s t'} \dd{t'} = \tilde{f}(s) \: \tilde{g}(s) \end{aligned}$$ {% include proof/end.html id="proof-laplace" %} diff --git a/source/know/concept/debye-length/index.md b/source/know/concept/debye-length/index.md index e226ad9..5961c4f 100644 --- a/source/know/concept/debye-length/index.md +++ b/source/know/concept/debye-length/index.md @@ -123,7 +123,7 @@ This treatment only makes sense if the plasma is sufficiently dense, such that there is a large number of particles in a sphere with radius $$\lambda_D$$. -This corresponds to a large [Coulomb logarithm](/know/concept/coulomb-logarithm/) $$\ln\!(\Lambda)$$: +This corresponds to a large [Coulomb logarithm](/know/concept/coulomb-logarithm/) $$\ln(\Lambda)$$: $$\begin{aligned} 1 \ll \frac{4 \pi}{3} n_0 \lambda_D^3 = \frac{2}{9} \Lambda diff --git a/source/know/concept/electric-dipole-approximation/index.md b/source/know/concept/electric-dipole-approximation/index.md index 7c710ec..35cf00c 100644 --- a/source/know/concept/electric-dipole-approximation/index.md +++ b/source/know/concept/electric-dipole-approximation/index.md @@ -30,6 +30,8 @@ so that $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$: $$\begin{aligned} \comm{\vb{A}}{\vu{P}} \psi + &= (\vb{A} \cdot \vu{P} - \vu{P} \cdot \vb{A}) \psi + \\ &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi) \\ &= i \hbar (\nabla \cdot \vb{A}) \psi @@ -75,7 +77,7 @@ $$\begin{aligned} Where $$\vb{E}_0 = \omega \vb{A}_0$$. Let us restrict ourselves to visible light, whose wavelength $$2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$$. -Meanwhile, an atomic orbital is several Bohr $$\sim 10^{-10} \:\mathrm{m}$$, +Meanwhile, an atomic orbital is several Bohr radii $$\sim 10^{-10} \:\mathrm{m}$$, so $$\vb{k} \cdot \vb{x}$$ is negligible: $$\begin{aligned} diff --git a/source/know/concept/gronwall-bellman-inequality/index.md b/source/know/concept/gronwall-bellman-inequality/index.md index da1bcad..0d6db71 100644 --- a/source/know/concept/gronwall-bellman-inequality/index.md +++ b/source/know/concept/gronwall-bellman-inequality/index.md @@ -7,8 +7,8 @@ categories: layout: "concept" --- -Suppose we have a first-order ordinary differential equation -for some function $$u(t)$$, and that it can be shown from this equation +Suppose we have a first-order ordinary differential equation for some function $$u(t)$$, +and assume that we can prove from this equation that the derivative $$u'(t)$$ is bounded as follows: $$\begin{aligned} @@ -28,7 +28,7 @@ $$\begin{aligned} {% include proof/start.html id="proof-original" -%} -We define $$w(t)$$ to equal the upper bounds above +We define $$w(t)$$ as equal to the upper bounds above on both $$w'(t)$$ and $$w(t)$$ itself: $$\begin{aligned} @@ -40,7 +40,7 @@ $$\begin{aligned} \end{aligned}$$ Where $$w(0) = u(0)$$. -The goal is to show the following for all $$t$$: +Then the goal is to show the following for all $$t$$: $$\begin{aligned} \frac{u(t)}{w(t)} \le 1 @@ -102,7 +102,7 @@ $$\begin{aligned} \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg) \end{aligned}$$ -The parenthesized expression it bounded from above by $$\alpha(t)$$, +The parenthesized expression is bounded from above by $$\alpha(t)$$, thanks to the condition that $$u(t)$$ is assumed to satisfy, for the Grönwall-Bellman inequality to be true: @@ -131,7 +131,8 @@ $$\begin{aligned} &\le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \end{aligned}$$ -Insert this into the condition under which the Grönwall-Bellman inequality holds. +This yields the desired result after inserting it +into the condition under which the Grönwall-Bellman inequality holds. {% include proof/end.html id="proof-integral" %} @@ -158,14 +159,14 @@ $$\begin{aligned} &\le \alpha(t) + \alpha(t) \int_0^t \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s} \end{aligned}$$ -Now, consider the following straightfoward identity, involving the exponential: +Now, consider the following straightforward identity, involving the exponential: $$\begin{aligned} \dv{}{s}\exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) &= - \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \end{aligned}$$ -By inserting this into Grönwall-Bellman inequality, we arrive at: +By inserting this into normal Grönwall-Bellman inequality, we arrive at: $$\begin{aligned} u(t) diff --git a/source/know/concept/lagrange-multiplier/index.md b/source/know/concept/lagrange-multiplier/index.md index 8ee1054..a0b22aa 100644 --- a/source/know/concept/lagrange-multiplier/index.md +++ b/source/know/concept/lagrange-multiplier/index.md @@ -1,7 +1,7 @@ --- title: "Lagrange multiplier" sort_title: "Lagrange multiplier" -date: 2021-03-02 +date: 2022-12-17 # Originally 2021-03-02, major rewrite categories: - Mathematics - Physics @@ -9,108 +9,145 @@ layout: "concept" --- The method of **Lagrange multipliers** or **undetermined multipliers** -is a technique for optimizing (i.e. finding the extrema of) -a function $$f(x, y, z)$$, -subject to a given constraint $$\phi(x, y, z) = C$$, -where $$C$$ is a constant. - -If we ignore the constraint $$\phi$$, -optimizing $$f$$ simply comes down to finding stationary points: +is a technique for optimizing (i.e. finding extrema of) +a function $$f$$ subject to **equality constraints**. +For example, in 2D, the goal is to maximize/minimize $$f(x, y)$$ +while satisfying $$g(x, y) = 0$$. +We assume that $$f$$ and $$g$$ are both continuous +and have continuous first derivatives, +and that their domain is all of $$\mathbb{R}$$. + +Side note: many authors write that Lagrange multipliers +can be used for constraints of the form $$g(x, y) = c$$ for a constant $$c$$. +However, this method technically requires $$c = 0$$. +This issue is easy to solve: given $$g = c$$, +simply define $$\tilde{g} \equiv g - c = 0$$ +and use that as constraint instead. + +Before introducing $$g$$, +optimizing $$f$$ comes down to finding its stationary points: $$\begin{aligned} - 0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z} + 0 + &= \nabla f + = \bigg( \pdv{f}{x}, \pdv{f}{y} \bigg) \end{aligned}$$ -This problem is easy: -$$\dd{x}$$, $$\dd{y}$$, and $$\dd{z}$$ are independent and arbitrary, -so all we need to do is find the roots of -the partial derivatives $$f_x$$, $$f_y$$ and $$f_z$$, -which we respectively call $$x_0$$, $$y_0$$ and $$z_0$$, -and then the extremum is simply $$(x_0, y_0, z_0)$$. - -But the constraint $$\phi$$, over which we have no control, -adds a relation between $$\dd{x}$$, $$\dd{y}$$, and $$\dd{z}$$, -so if two are known, the third is given by $$\phi = C$$. -The problem is then a system of equations: +This problem is easy: the two dimensions can be handled independently, +so all we need to do is find the roots of the partial derivatives. + +However, adding $$g$$ makes the problem much more complicated: +points with $$\nabla f = 0$$ might not satisfy $$g = 0$$, +and points where $$g = 0$$ might not have $$\nabla f = 0$$. +The dimensions also cannot be handled independently anymore, +since they are implicitly related by $$g$$. + +Imagine a contour plot of $$g(x, y)$$. +The trick is this: if we follow a contour of $$g = 0$$, +the highest and lowest values of $$f$$ along the way +are the desired local extrema. +Recall our assumption that $$\nabla f$$ is continuous: +hence *along our contour* $$f$$ is slowly-varying +in the close vicinity of each such point, +and stationary at the point itself. +We thus have two categories of extrema: + +1. $$\nabla f = 0$$ there, + i.e. $$f$$ is slowly-varying along *all* directions around the point. + In other words, a stationary point of $$f$$ + coincidentally lies on a contour of $$g = 0$$. + +2. The contours of $$f$$ and $$g$$ are parallel around the point. + By definition, $$f$$ is stationary along each of its contours, + so when we find that $$f$$ is stationary at a point on our $$g = 0$$ path, + it means we touched a contour of $$f$$. + Obviously, each point of $$f$$ lies on *some* contour, + but if they are not parallel, + then $$f$$ is increasing or decreasing along our path, + so this is not an extremum and we must continue our search. + +What about the edge case that $$g = 0$$ and $$\nabla g = 0$$ in the same point, +i.e. we locally have no contour to follow? +Do we just take whatever value $$f$$ has there? +No, by convention, we do not, +because this does not really count as *optimizing* $$f$$. + +Now, in the 2nd category, parallel contours imply parallel gradients, +i.e. $$\nabla f$$ and $$\nabla g$$ differ only in magnitude, not direction. +Formally: $$\begin{aligned} - 0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z} - \\ - 0 &= \dd{\phi} = \phi_x \dd{x} + \phi_y \dd{y} + \phi_z \dd{z} + \nabla f = -\lambda \nabla g \end{aligned}$$ -Solving this directly would be a delicate balancing act -of all the partial derivatives. - -To help us solve this, we introduce a "dummy" parameter $$\lambda$$, -the so-called **Lagrange multiplier**, -and contruct a new function $$L$$ given by: - -$$\begin{aligned} - L(x, y, z) = f(x, y, z) + \lambda \phi(x, y, z) -\end{aligned}$$ +Where $$\lambda$$ is the **Lagrange multiplier** +that quantifies the difference in magnitude between the gradients. +By setting $$\lambda = 0$$, this equation also handles the 1st category $$\nabla f = 0$$. +Some authors define $$\lambda$$ with the opposite sign. -At the extremum, $$\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$$, -so now the problem is a "single" equation again: +The method of Lagrange multipliers uses these facts +to rewrite a constrained $$N$$-dimensional optimization problem +as an unconstrained $$(N\!+\!1)$$-dimensional optimization problem +by defining the **Lagrangian function** $$\mathcal{L}$$ as follows: $$\begin{aligned} - 0 = \dd{L} - = (f_x + \lambda \phi_x) \dd{x} + (f_y + \lambda \phi_y) \dd{y} + (f_z + \lambda \phi_z) \dd{z} + \boxed{ + \mathcal{L}(x, y, \lambda) + \equiv f(x, y) + \lambda g(x, y) + } \end{aligned}$$ -Assuming $$\phi_z \neq 0$$, we now choose $$\lambda$$ such that $$f_z + \lambda \phi_z = 0$$. -This choice represents satisfying the constraint, -so now the remaining $$\dd{x}$$ and $$\dd{y}$$ are independent again, -and we simply have to find the roots of $$f_x + \lambda \phi_x$$ and $$f_y + \lambda \phi_y$$. - -In effect, after introducing $$\lambda$$, -we have four unknowns $$(x, y, z, \lambda)$$, -but also four equations: +Let us do an unconstrained optimization of $$\mathcal{L}$$ as usual, +by demanding it is stationary: $$\begin{aligned} - L_x = L_y = L_z = 0 - \qquad \quad - \phi = C + 0 + = \nabla \mathcal{L} + &= \bigg( \pdv{\mathcal{L}}{x}, \pdv{\mathcal{L}}{y}, \pdv{\mathcal{L}}{\lambda} \bigg) + \\ + &= \bigg( \pdv{f}{x} + \lambda \pdv{g}{x}, \:\:\: \pdv{f}{y} + \lambda \pdv{g}{y}, \:\:\: g \bigg) \end{aligned}$$ -We are only really interested in the first three unknowns $$(x, y, z)$$, -so $$\lambda$$ is sometimes called the **undetermined multiplier**, -since it is just an algebraic helper whose value is irrelevant. - -This method generalizes nicely to multiple constraints or more variables: -suppose that we want to find the extrema of $$f(x_1, ..., x_N)$$ +The last item in this vector represents $$g = 0$$, +and the others $$\nabla f = -\lambda \nabla g$$ as discussed earlier. +To solve this equation, +we assign $$\lambda$$ a value that agrees with it +(such a value exists for each local extremum +according to our above discussion of the two categories), +and then find the locations $$(x, y)$$ that satisfy it. +However, as usual for optimization problems, +this method only finds *local* extrema *and* saddle points; +it is a necessary condition for optimality, but not sufficient. + +We often assign $$\lambda$$ an algebraic expression rather than a value, +usually without even bothering to calculate its final actual value. +In fact, in some cases, $$\lambda$$'s only function is to help us reason +about the interdependence of a system of equations +(see [example 3](https://en.wikipedia.org/wiki/Lagrange_multiplier#Example_3:_Entropy) on Wikipedia); +then $$\lambda$$ is not even given an expression! +Hence it is sometimes also called an *undetermined multiplier*. + +This method generalizes nicely to multiple constraints or more variables. +Suppose that we want to find the extrema of $$f(x_1, ..., x_N)$$ subject to $$M < N$$ conditions: $$\begin{aligned} - \phi_1(x_1, ..., x_N) = C_1 \qquad \cdots \qquad \phi_M(x_1, ..., x_N) = C_M -\end{aligned}$$ - -This once again turns into a delicate system of $$M+1$$ equations to solve: - -$$\begin{aligned} - 0 &= \dd{f} = f_{x_1} \dd{x_1} + ... + f_{x_N} \dd{x_N} - \\ - 0 &= \dd{\phi_1} = \phi_{1, x_1} \dd{x_1} + ... + \phi_{1, x_N} \dd{x_N} - \\ - &\vdots - \\ - 0 &= \dd{\phi_M} = \phi_{M, x_1} \dd{x_1} + ... + \phi_{M, x_N} \dd{x_N} + g_1(x_1, ..., x_N) = c_1 + \qquad \cdots \qquad + g_M(x_1, ..., x_N) = c_M \end{aligned}$$ Then we introduce $$M$$ Lagrange multipliers $$\lambda_1, ..., \lambda_M$$ -and define $$L(x_1, ..., x_N)$$: +and define $$\mathcal{L}(x_1, ..., x_N)$$: $$\begin{aligned} - L = f + \sum_{m = 1}^M \lambda_m \phi_m + \mathcal{L} \equiv f + \sum_{m = 1}^M \lambda_m g_m \end{aligned}$$ -As before, we set $$\dd{L} = 0$$ and choose the multipliers $$\lambda_1, ..., \lambda_M$$ -to eliminate $$M$$ of its $$N$$ terms: +As before, we set $$\nabla \mathcal{L} = 0$$ and choose the multipliers +$$\lambda_1, ..., \lambda_M$$ to satisfy the resulting system of $$(N\!+\!M)$$ 1D equations, +and then find the coordinates of the extrema. -$$\begin{aligned} - 0 = \dd{L} - = \sum_{n = 1}^N \Big( f_{x_n} + \sum_{m = 1}^M \lambda_m \phi_{x_n} \Big) \dd{x_n} -\end{aligned}$$ ## References diff --git a/source/know/concept/material-derivative/index.md b/source/know/concept/material-derivative/index.md index 93e8ad0..7225053 100644 --- a/source/know/concept/material-derivative/index.md +++ b/source/know/concept/material-derivative/index.md @@ -16,9 +16,9 @@ e.g. the temperature or pressure, represented by a scalar field $$f(\va{r}, t)$$. If the fluid is static, the evolution of $$f$$ is simply $$\ipdv{f}{t}$$, -since each point of the fluid is motionless. +since each point is motionless. However, if the fluid is moving, we have a problem: -the fluid molecules at position $$\va{r} = \va{r}_0$$ are not necessarily +the fluid molecules at position $$\va{r} = \va{r}_0$$ are not the same ones at time $$t = t_0$$ and $$t = t_1$$. Those molecules take $$f$$ with them as they move, so we need to account for this transport somehow. diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md index 377f3a1..41e8fed 100644 --- a/source/know/concept/parsevals-theorem/index.md +++ b/source/know/concept/parsevals-theorem/index.md @@ -26,20 +26,21 @@ $$\begin{aligned} {% include proof/start.html id="proof-fourier" -%} -We insert the inverse FT into the defintion of the inner product: +We insert the inverse FT into the definition of the inner product: $$\begin{aligned} \Inprod{f}{g} &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}\big)^* \: \hat{\mathcal{F}}^{-1}\{\tilde{g}(k)\} \dd{x} \\ &= B^2 \int - \Big( \int \tilde{f}^*(k_1) \exp(i s k_1 x) \dd{k_1} \Big) - \Big( \int \tilde{g}(k) \exp(- i s k x) \dd{k} \Big) + \Big( \int \tilde{f}^*(k') \: e^{i s k' x} \dd{k'} \Big) + \Big( \int \tilde{g}(k) \: e^{- i s k x} \dd{k} \Big) \dd{x} \\ - &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \tilde{g}(k) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s x (k_1 - k)) \dd{x} \Big) \dd{k_1} \dd{k} + &= 2 \pi B^2 \iint \tilde{f}^*(k') \: \tilde{g}(k) \Big( \frac{1}{2 \pi} + \int_{-\infty}^\infty e^{i s x (k' - k)} \dd{x} \Big) \dd{k'} \dd{k} \\ - &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k} + &= 2 \pi B^2 \iint \tilde{f}^*(k') \: \tilde{g}(k) \: \delta\big(s (k' \!-\! k)\big) \dd{k'} \dd{k} \\ &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k} = \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}}{\tilde{g}} @@ -54,13 +55,14 @@ $$\begin{aligned} &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k} \\ &= A^2 \int - \Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big) - \Big( \int g(x) \exp(i s k x) \dd{x} \Big) + \Big( \int f^*(x') \: e^{- i s k x'} \dd{x'} \Big) + \Big( \int g(x) \: e^{i s k x} \dd{x} \Big) \dd{k} \\ - &= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x} + &= 2 \pi A^2 \iint f^*(x') \: g(x) \Big( \frac{1}{2 \pi} + \int_{-\infty}^\infty e^{i s k (x - x')} \dd{k} \Big) \dd{x'} \dd{x} \\ - &= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x} + &= 2 \pi A^2 \iint f^*(x') \: g(x) \: \delta\big(s (x \!-\! x')\big) \dd{x'} \dd{x} \\ &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x} = \frac{2 \pi A^2}{|s|} \Inprod{f}{g} diff --git a/source/know/concept/rabi-oscillation/index.md b/source/know/concept/rabi-oscillation/index.md index 07f8b25..2fcdea8 100644 --- a/source/know/concept/rabi-oscillation/index.md +++ b/source/know/concept/rabi-oscillation/index.md @@ -15,11 +15,11 @@ In quantum mechanics, from the derivation of we know that a time-dependent term $$\hat{H}_1$$ in the Hamiltonian affects the state as follows, where $$c_n(t)$$ are the coefficients of the linear combination -of basis states $$\Ket{n} \exp(-i E_n t / \hbar)$$: +of basis states $$\Ket{n} e^{-i E_n t / \hbar}$$: $$\begin{aligned} i \hbar \dv{c_m}{t} - = \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1}{n} \exp(i \omega_{mn} t) + = \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1}{n} e^{i \omega_{mn} t} \end{aligned}$$ Where $$\omega_{mn} \equiv (E_m \!-\! E_n) / \hbar$$ @@ -31,10 +31,10 @@ in which case the above equation can be expanded to the following: $$\begin{aligned} \dv{c_a}{t} - &= - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{b} \exp(- i \omega_0 t) \: c_b - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{a} \: c_a + &= - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{b} e^{-i \omega_0 t} \: c_b - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{a} c_a \\ \dv{c_b}{t} - &= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp(i \omega_0 t) \: c_a - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{b} \: c_b + &= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} e^{i \omega_0 t} \: c_a - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{b} c_b \end{aligned}$$ Where $$\omega_0 \equiv \omega_{ba}$$ is positive. @@ -44,10 +44,10 @@ states that the diagonal matrix elements vanish, leaving: $$\begin{aligned} \dv{c_a}{t} - &= - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{b} \exp(- i \omega_0 t) \: c_b + &= - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{b} e^{-i \omega_0 t} \: c_b \\ \dv{c_b}{t} - &= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp(i \omega_0 t) \: c_a + &= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} e^{i \omega_0 t} \: c_a \end{aligned}$$ We now choose $$\hat{H}_1$$ to be as follows, @@ -56,7 +56,7 @@ sinusoidally oscillating with a spatially odd $$V(\vec{r})$$: $$\begin{aligned} \hat{H}_1(t) = V \cos(\omega t) - = \frac{V}{2} \Big( \exp(i \omega t) + \exp(-i \omega t) \Big) + = \frac{V}{2} \Big( e^{i \omega t} + e^{-i \omega t} \Big) \end{aligned}$$ We insert this into the equations for $$c_a$$ and $$c_b$$, @@ -64,16 +64,16 @@ and define $$V_{ab} \equiv \matrixel{a}{V}{b}$$, leading us to: $$\begin{aligned} \dv{c_a}{t} - &= - i \frac{V_{ab}}{2 \hbar} \Big( \exp\!\big(i (\omega \!-\! \omega_0) t\big) + \exp\!\big(\!-\! i (\omega \!+\! \omega_0) t\big) \Big) \: c_b + &= - i \frac{V_{ab}}{2 \hbar} \Big( e^{i (\omega - \omega_0) t} + e^{-i (\omega + \omega_0) t} \Big) \: c_b \\ \dv{c_b}{t} - &= - i \frac{V_{ab}}{2 \hbar} \Big( \exp\!\big(i (\omega \!+\! \omega_0) t\big) + \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t\big) \Big) \: c_a + &= - i \frac{V_{ab}}{2 \hbar} \Big( e^{i (\omega + \omega_0) t} + e^{-i (\omega - \omega_0) t} \Big) \: c_a \end{aligned}$$ Here, we make the [rotating wave approximation](/know/concept/rotating-wave-approximation/): assuming we are close to resonance $$\omega \approx \omega_0$$, -we argue that $$\exp(i (\omega \!+\! \omega_0) t)$$ +we argue that $$e^{i (\omega + \omega_0) t}$$ oscillates so fast that its effect is negligible when the system is observed over a reasonable time interval. Dropping those terms leaves us with: @@ -82,10 +82,10 @@ $$\begin{aligned} \boxed{ \begin{aligned} \dv{c_a}{t} - &= - i \frac{V_{ab}}{2 \hbar} \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b + &= - i \frac{V_{ab}}{2 \hbar} \: e^{i (\omega - \omega_0) t} \: c_b \\ \dv{c_b}{t} - &= - i \frac{V_{ba}}{2 \hbar} \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t \big) \: c_a + &= - i \frac{V_{ba}}{2 \hbar} \: e^{-i (\omega - \omega_0) t} \: c_a \end{aligned} } \end{aligned}$$ @@ -96,13 +96,12 @@ and then substitute $$\idv{c_b}{t}$$ for the second equation: $$\begin{aligned} \dvn{2}{c_a}{t} - &= - i \frac{V_{ab}}{2 \hbar} \bigg( i (\omega - \omega_0) \: c_b + \dv{c_b}{t} \bigg) \exp\!\big(i (\omega \!-\! \omega_0) t \big) + &= - i \frac{V_{ab}}{2 \hbar} \bigg( i (\omega - \omega_0) \: c_b + \dv{c_b}{t} \bigg) e^{i (\omega - \omega_0) t} \\ - &= - i \frac{V_{ab}}{2 \hbar} \bigg( i (\omega - \omega_0) \: c_b - - i \frac{V_{ba}}{2 \hbar} \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t \big) \: c_a \bigg) - \exp\!\big(i (\omega \!-\! \omega_0) t \big) + &= - i \frac{V_{ab}}{2 \hbar} \bigg( i (\omega - \omega_0) \: c_b + - i \frac{V_{ba}}{2 \hbar} \: e^{-i (\omega - \omega_0) t} \: c_a \bigg) e^{i (\omega - \omega_0) t} \\ - &= \frac{V_{ab}}{2 \hbar} (\omega - \omega_0) \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b - \frac{|V_{ab}|^2}{(2 \hbar)^2} c_a + &= \frac{V_{ab}}{2 \hbar} (\omega - \omega_0) \: e^{i (\omega - \omega_0) t} \: c_b - \frac{|V_{ab}|^2}{(2 \hbar)^2} \: c_a \end{aligned}$$ In the first term, we recognize $$\idv{c_a}{t}$$, @@ -113,7 +112,7 @@ $$\begin{aligned} = \dvn{2}{c_a}{t} - i (\omega - \omega_0) \dv{c_a}{t} + \frac{|V_{ab}|^2}{(2 \hbar)^2} \: c_a \end{aligned}$$ -To solve this, we make the ansatz $$c_a(t) = \exp(\lambda t)$$, +To solve this, we make the ansatz $$c_a(t) = e^{\lambda t}$$, which, upon insertion, gives us: $$\begin{aligned} @@ -148,7 +147,7 @@ to be determined from initial conditions (and normalization): $$\begin{aligned} \boxed{ c_a(t) - = \Big( A \sin(\tilde{\Omega} t / 2) + B \cos(\tilde{\Omega} t / 2) \Big) \exp\!\big(i (\omega \!-\! \omega_0) t / 2 \big) + = \Big( A \sin(\tilde{\Omega} t / 2) + B \cos(\tilde{\Omega} t / 2) \Big) e^{i (\omega - \omega_0) t / 2} } \end{aligned}$$ @@ -173,7 +172,7 @@ Note that the period was halved by squaring. This periodic "flopping" of the particle between $$\Ket{a}$$ and $$\Ket{b}$$ is known as **Rabi oscillation**, **Rabi flopping** or the **Rabi cycle**. This is a more accurate treatment -of the flopping found from first-order perturbation theory. +of the flopping found from first-order perturbation theory in textbooks. The name **generalized Rabi frequency** suggests that there is a non-general version. @@ -185,6 +184,8 @@ $$\begin{aligned} \equiv \frac{V_{ba}}{\hbar} \end{aligned}$$ +Some authors use $$|V_{ba}|$$ instead, +but not doing that lets us use $$\Omega$$ as a nice abbreviation. As an example, Rabi oscillation arises in the [electric dipole approximation](/know/concept/electric-dipole-approximation/), where $$\hat{H}_1$$ is: diff --git a/source/know/concept/self-steepening/index.md b/source/know/concept/self-steepening/index.md index 9666167..e06b0b5 100644 --- a/source/know/concept/self-steepening/index.md +++ b/source/know/concept/self-steepening/index.md @@ -48,15 +48,16 @@ $$\begin{aligned} \end{aligned}$$ The phase $$\phi$$ is not so interesting, so we focus on the latter equation for $$P$$. -As it turns out, it has a general solution of the form below, which shows that -more intense parts of the pulse will tend to lag behind compared to the rest: +As it turns out, it has a general solution of the form below (you can verify this yourself), +which shows that more intense parts of the pulse +will lag behind compared to the rest: $$\begin{aligned} P(z,t) = f(t - 3 \varepsilon z P) \end{aligned}$$ Where $$f$$ is the initial power profile: $$f(t) = P(0,t)$$. -The derivatives $$P_t$$ and $$P_z$$ are then given by: +The derivatives $$P_t$$ and $$P_z$$ are given by: $$\begin{aligned} P_t @@ -76,12 +77,15 @@ These derivatives both go to infinity when their denominator is zero, which, since $$\varepsilon$$ is positive, will happen earliest where $$f'$$ has its most negative value, called $$f_\mathrm{min}'$$, which is located on the trailing edge of the pulse. -At the propagation distance where this occurs, $$L_\mathrm{shock}$$, +At the propagation distance $$z$$ where this occurs, $$L_\mathrm{shock}$$, the pulse will "tip over", creating a discontinuous shock: $$\begin{aligned} + 0 + = 1 + 3 \varepsilon z f_\mathrm{min}' + \qquad \implies \qquad \boxed{ - L_\mathrm{shock} = -\frac{1}{3 \varepsilon f_\mathrm{min}'} + L_\mathrm{shock} \equiv -\frac{1}{3 \varepsilon f_\mathrm{min}'} } \end{aligned}$$ @@ -135,5 +139,6 @@ $$\begin{aligned} \end{aligned}$$ + ## References 1. B.R. Suydam, [Self-steepening of optical pulses](https://doi.org/10.1007/0-387-25097-2_6), 2006, Springer. diff --git a/source/know/concept/shors-algorithm/index.md b/source/know/concept/shors-algorithm/index.md index a47151a..5ae5077 100644 --- a/source/know/concept/shors-algorithm/index.md +++ b/source/know/concept/shors-algorithm/index.md @@ -33,9 +33,10 @@ Shor's algorithm can solve practically every such problem. ## Integer factorization -Originally, Shor's algorithm was designed to factorize an integer $$N$$, -in which case the goal is to find the period $$s$$ of -the modular exponentiation function $$f$$ (for reasons explained later): +Originally, Shor's algorithm was designed to factorize an integer $$N$$. +For reasons explained later, +this means our goal is to find the period $$s$$ of +the modular exponentiation function $$f$$: $$\begin{aligned} f(x) @@ -79,7 +80,7 @@ $$\begin{aligned} \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{f(x)} \end{aligned}$$ -Then we measure $$f(x)$$, causing it collapse as follows, +Then we measure $$f(x)$$, causing it collapse as follows for an unknown arbitrary value of $$x_0$$: $$\begin{aligned} diff --git a/source/know/concept/thermodynamic-potential/index.md b/source/know/concept/thermodynamic-potential/index.md index ece1551..b3bedda 100644 --- a/source/know/concept/thermodynamic-potential/index.md +++ b/source/know/concept/thermodynamic-potential/index.md @@ -15,8 +15,8 @@ Such functions are either energies (hence *potential*) or entropies. Which potential (of many) decides the equilibrium states for a given system? That depends which variables are assumed to already be in automatic equilibrium. Such variables are known as the **natural variables** of that potential. -For example, if a system can freely exchange heat with its surroundings, -and is consequently assumed to be at the same temperature $$T = T_{\mathrm{sur}}$$, +For example, if a system can freely exchange heat with its environment, +and is consequently assumed to be at the same temperature $$T = T_{\mathrm{env}}$$, then $$T$$ must be a natural variable. The link from natural variables to potentials @@ -32,6 +32,7 @@ Mathematically, the potentials are related to each other by [Legendre transformation](/know/concept/legendre-transform/). + ## Internal energy The **internal energy** $$U$$ represents @@ -76,6 +77,7 @@ to help keep track of which function depends on which variables. They are meaningless; these are normal partial derivatives. + ## Enthalpy The **enthalpy** $$H$$ of a system, in units of energy, @@ -115,6 +117,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Helmholtz free energy The **Helmholtz free energy** $$F$$ represents @@ -154,6 +157,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Gibbs free energy The **Gibbs free energy** $$G$$ represents @@ -192,6 +196,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Landau potential The **Landau potential** or **grand potential** $$\Omega$$, in units of energy, @@ -230,6 +235,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Entropy The **entropy** $$S$$, in units of energy over temperature, diff --git a/source/know/concept/toffoli-gate/index.md b/source/know/concept/toffoli-gate/index.md index b9d9528..9a99e69 100644 --- a/source/know/concept/toffoli-gate/index.md +++ b/source/know/concept/toffoli-gate/index.md @@ -66,10 +66,10 @@ it swaps the last two coefficients: $$\begin{aligned} \mathrm{CCNOT} \Ket{\psi} - &= \mathrm{CCNOT} \big( c_{000} \Ket{000} + c_{001} \Ket{001} + c_{010} \Ket{010} + c_{011} \Ket{011} \\ + &= \mathrm{CCNOT} \big( c_{000} \Ket{000} + c_{001} \Ket{001} + c_{010} \Ket{010} + c_{011} \Ket{011} + \\ &\qquad\qquad\quad\:\; c_{100} \Ket{100} + c_{101} \Ket{101} + c_{110} \Ket{110} + c_{111} \Ket{111} \big) \\ - &= c_{000} \Ket{000} + c_{001} \Ket{001} + c_{010} \Ket{010} + c_{011} \Ket{011} \\ + &= c_{000} \Ket{000} + c_{001} \Ket{001} + c_{010} \Ket{010} + c_{011} \Ket{011} + \\ &\quad\,\, c_{100} \Ket{100} + c_{101} \Ket{101} + c_{111} \Ket{110} + c_{110} \Ket{111} \end{aligned}$$ |