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Diffstat (limited to 'content')
| -rw-r--r-- | content/know/concept/binomial-distribution/index.pdc | 31 | ||||
| -rw-r--r-- | content/know/concept/boltzmann-equation/index.pdc | 363 | ||||
| -rw-r--r-- | content/know/concept/central-limit-theorem/index.pdc | 14 | ||||
| -rw-r--r-- | content/know/concept/ritz-method/index.pdc | 11 |
4 files changed, 397 insertions, 22 deletions
diff --git a/content/know/concept/binomial-distribution/index.pdc b/content/know/concept/binomial-distribution/index.pdc index e644164..183e7e9 100644 --- a/content/know/concept/binomial-distribution/index.pdc +++ b/content/know/concept/binomial-distribution/index.pdc @@ -115,7 +115,8 @@ By inserting $q = 1 - p$, we arrive at the desired expression. </div> As $N \to \infty$, the binomial distribution -turns into the continuous normal distribution: +turns into the continuous normal distribution, +a fact that is sometimes called the **de Moivre-Laplace theorem**: $$\begin{aligned} \boxed{ @@ -142,9 +143,9 @@ We use Stirling's approximation to calculate the factorials in $D_m$: $$\begin{aligned} \ln\!\big(P_N(n)\big) - &= \ln(N!) - \ln(n!) - \ln\!\big((N - n)!\big) + n \ln(p) + (N - n) \ln(q) + &= \ln\!(N!) - \ln\!(n!) - \ln\!\big((N - n)!\big) + n \ln\!(p) + (N - n) \ln\!(q) \\ - &\approx \ln(N!) - n \big( \ln(n)\!-\!\ln(p)\!-\!1 \big) - (N\!-\!n) \big( \ln(N\!-\!n)\!-\!\ln(q)\!-\!1 \big) + &\approx \ln\!(N!) - n \big( \ln\!(n)\!-\!\ln\!(p)\!-\!1 \big) - (N\!-\!n) \big( \ln\!(N\!-\!n)\!-\!\ln\!(q)\!-\!1 \big) \end{aligned}$$ For $D_0(\mu)$, we need to use a stronger version of Stirling's approximation @@ -152,16 +153,16 @@ to get a non-zero result. We take advantage of $N - N p = N q$: $$\begin{aligned} D_0(\mu) - &= \ln(N!) - \ln\!\big((N p)!\big) - \ln\!\big((N q)!\big) + N p \ln(p) + N q \ln(q) + &= \ln\!(N!) - \ln\!\big((N p)!\big) - \ln\!\big((N q)!\big) + N p \ln\!(p) + N q \ln\!(q) \\ - &= \Big( N \ln(N) - N + \frac{1}{2} \ln(2\pi N) \Big) - - \Big( N p \ln(N p) - N p + \frac{1}{2} \ln(2\pi N p) \Big) \\ - &\qquad - \Big( N q \ln(N q) - N q + \frac{1}{2} \ln(2\pi N q) \Big) - + N p \ln(p) + N q \ln(q) + &= \Big( N \ln\!(N) - N + \frac{1}{2} \ln\!(2\pi N) \Big) + - \Big( N p \ln\!(N p) - N p + \frac{1}{2} \ln\!(2\pi N p) \Big) \\ + &\qquad - \Big( N q \ln\!(N q) - N q + \frac{1}{2} \ln\!(2\pi N q) \Big) + + N p \ln\!(p) + N q \ln\!(q) \\ - &= N \ln(N) - N (p + q) \ln(N) + N (p + q) - N - \frac{1}{2} \ln(2\pi N p q) + &= N \ln\!(N) - N (p + q) \ln\!(N) + N (p + q) - N - \frac{1}{2} \ln\!(2\pi N p q) \\ - &= - \frac{1}{2} \ln(2\pi N p q) + &= - \frac{1}{2} \ln\!(2\pi N p q) = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big) \end{aligned}$$ @@ -170,17 +171,17 @@ This is indeed the case: $$\begin{aligned} D_1(n) - &= - \big( \ln(n)\!-\!\ln(p)\!-\!1 \big) + \big( \ln(N\!-\!n)\!-\!\ln(q)\!-\!1 \big) - 1 + 1 + &= - \big( \ln\!(n)\!-\!\ln\!(p)\!-\!1 \big) + \big( \ln\!(N\!-\!n)\!-\!\ln\!(q)\!-\!1 \big) - 1 + 1 \\ - &= - \ln(n) + \ln(N - n) + \ln(p) - \ln(q) + &= - \ln\!(n) + \ln\!(N - n) + \ln\!(p) - \ln\!(q) \\ D_1(\mu) - &= \ln(N q) - \ln(N p) + \ln(p) - \ln(q) - = \ln(N p q) - \ln(N p q) + &= \ln\!(N q) - \ln\!(N p) + \ln\!(p) - \ln\!(q) + = \ln\!(N p q) - \ln\!(N p q) = 0 \end{aligned}$$ -For the same reason, we expect that $D_2(\mu)$ is negative +For the same reason, we expect that $D_2(\mu)$ is negative. We find the following expression: $$\begin{aligned} diff --git a/content/know/concept/boltzmann-equation/index.pdc b/content/know/concept/boltzmann-equation/index.pdc new file mode 100644 index 0000000..84e75cd --- /dev/null +++ b/content/know/concept/boltzmann-equation/index.pdc @@ -0,0 +1,363 @@ +--- +title: "Boltzmann equation" +firstLetter: "B" +publishDate: 2022-10-02 +categories: +- Physics +- Thermodynamics +- Fluid mechanics + +date: 2022-09-25T17:32:30+02:00 +draft: false +markup: pandoc +--- + +# Boltzmann equation + +Consider a collection of particles, +each with its own position $\vb{r}$ and velocity $\vb{v}$. +We can thus define a probability density function $f(\vb{r}, \vb{v}, t)$ +describing the expected number of particles at $(\vb{r}, \vb{v})$ at time $t$. +Let the total number of particles $N$ be conserved, then clearly: + +$$\begin{aligned} + N = \iint_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{r}} \dd{\vb{v}} +\end{aligned}$$ + +At equilibrium, all processes affecting the particles +no longer have a net effect, so $f$ is fixed: + +$$\begin{aligned} + \dv{f}{t} + = 0 +\end{aligned}$$ + +If each particle's momentum only changes due to collisions, +then a non-equilibrium state can be described as follows, very generally: + +$$\begin{aligned} + \dv{f}{t} + = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col} +\end{aligned}$$ + +Where the right-hand side simply means "all changes in $f$ due to collisions". +Applying the chain rule to the left-hand side then yields: + +$$\begin{aligned} + \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col} + &= \pdv{f}{t} + \bigg( \pdv{f}{x} \dv{x}{t} \!+\! \pdv{f}{y} \dv{y}{t} \!+\! \pdv{f}{z} \dv{z}{t} \bigg) + + \bigg( \pdv{f}{v_x} \dv{v_x}{t} \!+\! \pdv{f}{v_y} \dv{v_y}{t} \!+\! \pdv{f}{v_z} \dv{v_z}{t} \bigg) + \\ + &= \pdv{f}{t} + \bigg( v_x \pdv{f}{x} \!+\! v_y \pdv{f}{y} \!+\! v_z \pdv{f}{z} \bigg) + + \bigg( a_x \pdv{f}{v_x} \!+\! a_y \pdv{f}{v_y} \!+\! a_z \pdv{f}{v_z} \bigg) + \\ + &= \pdv{f}{t} + \vb{v} \cdot \nabla f + \vb{a} \cdot \pdv{f}{\vb{v}} +\end{aligned}$$ + +Where we have introduced the shorthand $\pdv*{f}{\vb{v}}$. +Inserting Newton's second law $\vb{F} = m \vb{a}$ +leads us to the **Boltzmann equation** or +**Boltzmann transport equation** (BTE): + +$$\begin{aligned} + \boxed{ + \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}} + = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col} + } +\end{aligned}$$ + +But what about the collision term? +Expressions for it exist, which are almost exact in many cases, +but unfortunately also quite difficult to work with. +In addition, $f$ is a 7-dimensional function, +so the BTE is already hard to solve without collisions. +We only present the simplest case, +known as the **Bhatnagar-Gross-Krook approximation**: +if the equilibrium state $f_0(\vb{r}, \vb{v})$ is known, +then each collision brings the system closer to $f_0$: + +$$\begin{aligned} + \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}} + = \frac{f_0 - f}{\tau} +\end{aligned}$$ + +Where $\tau$ is the average collision period. +The right-hand side is called the **Krook term**. + + + +## Moment equations + +From the definition of $f$, +we see that integrating over all $\vb{v}$ yields the particle density $n$: + +$$\begin{aligned} + n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}} +\end{aligned}$$ + +Consequently, a purely velocity-dependent quantity $Q(\vb{v})$ can be averaged like so: + +$$\begin{aligned} + \expval{Q} + = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}} +\end{aligned}$$ + +With that in mind, we multiply the collisionless BTE equation by $Q(\vb{v})$ and integrate, +assuming that $\vb{F}$ does not depend on $\vb{v}$: + +$$\begin{aligned} + 0 + &= \int_{-\infty}^\infty Q \bigg( \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}} \bigg) \dd{\vb{v}} + \\ + &= \int Q \pdv{f}{t} \dd{\vb{v}} + \int (\vb{v} \cdot \nabla f) \: Q \dd{\vb{v}} + \frac{\vb{F}}{m} \cdot \int Q \pdv{f}{\vb{v}} \dd{\vb{v}} + \\ + &= \pdv{t} \int Q f \dd{\vb{v}} + \int \Big( \nabla \cdot (\vb{v} f) - f (\nabla \cdot \vb{v}) \Big) Q \dd{\vb{v}} + + \frac{\vb{F}}{m} \cdot \int \bigg( \pdv{\vb{v}} (Q f) - f \pdv{Q}{\vb{v}} \bigg) \dd{\vb{v}} +\end{aligned}$$ + +The first integral is simply $n \expval{Q}$. +In the second integral, note that $\vb{v}$ is a coordinate +and hence not dependent on $\vb{r}$, so $\nabla \cdot \vb{v} = 0$. +Since $f$ is a probability density, $f \to 0$ for $\vb{v} \to \pm\infty$, +so the first term in the third integral vanishes after it is integrated: + +$$\begin{aligned} + 0 + &= \pdv{t} \big(n \expval{Q}\big) + \int \nabla \cdot (\vb{v} f) \: Q \dd{\vb{v}} + + \frac{\vb{F}}{m} \cdot \bigg( \Big[ Q f \Big]_{-\infty}^\infty - \int f \pdv{Q}{\vb{v}} \dd{\vb{v}} \bigg) + \\ + &= \pdv{t} \big(n \expval{Q}\big) + \nabla \cdot \int Q \vb{v} f \dd{\vb{v}} + - \frac{\vb{F}}{m} \cdot \int f \pdv{Q}{\vb{v}} \dd{\vb{v}} +\end{aligned}$$ + +We thus arrive at the prototype of the BTE's so-called **moment equations**: + +$$\begin{aligned} + \boxed{ + 0 + = \pdv{t} \big(n \expval{Q}\big) + \nabla \cdot \big(n \expval{Q \vb{v}}\big) - \frac{\vb{F}}{m} \cdot \bigg( n \expval{\pdv{Q}{\vb{v}}} \bigg) + } +\end{aligned}$$ + +If we set $Q = m$, then the mass density $\rho = n \expval{Q}$, +and we find that the **zeroth moment** of the BTE describes conservation of mass, +where $\vb{V} \equiv \expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$ is the fluid velocity: + +$$\begin{aligned} + \boxed{ + 0 + = \pdv{\rho}{t} + \nabla \cdot \big(\rho \vb{V}\big) + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-moment0"/> +<label for="proof-moment0">Proof</label> +<div class="hidden"> +<label for="proof-moment0">Proof.</label> +We insert $Q = m$ into our prototype, +and since $m$ is constant, the rest is trivial: + +$$\begin{aligned} + 0 + &= \pdv{t} \big(n \expval{m}\big) + \nabla \cdot \big(n \expval{m \vb{v}}\big) - \frac{\vb{F}}{m} \cdot \bigg( n \expval{\pdv{m}{\vb{v}}} \bigg) + \\ + &= \pdv{\rho}{t} + \nabla \cdot \big(\rho \expval{\vb{v}}\big) - 0 +\end{aligned}$$ +</div> +</div> + +If we instead choose the momentum $Q = m \vb{v}$, +we find that the **first moment** of the BTE describes conservation of momentum, +where $\hat{P}$ is the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/): + +$$\begin{aligned} + \boxed{ + 0 + = \pdv{t} \big(\rho \vb{V}\big) + \rho \vb{V} (\nabla \cdot \vb{V}) + \nabla \cdot \hat{P} - n \vb{F} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-moment1"/> +<label for="proof-moment1">Proof</label> +<div class="hidden"> +<label for="proof-moment1">Proof.</label> +We insert $Q = m \vb{v}$ into our prototype and recognize $\rho$ wherever possible: + +$$\begin{aligned} + 0 + &= \pdv{t} \big(n \expval{m \vb{v}}\big) + \nabla \cdot \big(n \expval{m \vb{v} \vb{v}}\big) + - \frac{\vb{F}}{m} \cdot \bigg( n \expval{\pdv{(m \vb{v})}{\vb{v}}} \bigg) + \\ + &= \pdv{t} \big(\rho \expval{\vb{v}}\big) + \nabla \cdot \big(\rho \expval{\vb{v} \vb{v}}\big) + - \vb{F} \cdot \bigg( n \expval{\pdv{\vb{v}}{\vb{v}}} \bigg) +\end{aligned}$$ + +With $\vb{v} \vb{v}$ being a dyadic product. +To give it a physical interpretation, +we split $\vb{v} = \vb{V} \!+\! \vb{w}$, +where $\vb{V}$ is the average velocity vector, +and $\vb{w}$ is the local deviation from $\vb{V}$: + +$$\begin{aligned} + \expval{\vb{v} \vb{v}} + &= \expval{(\vb{V} \!+\! \vb{w}) (\vb{V} \!+\! \vb{w})} + = \expval{\vb{V} \vb{V} + 2 \vb{V} \vb{w} + \vb{w} \vb{w}} + = \vb{V} \vb{V} + 2 \vb{V} \expval{\vb{w}} + \expval{\vb{w} \vb{w}} +\end{aligned}$$ + +Since $\vb{w}$ represents a deviation from the mean, $\expval{\vb{w}} = 0$. +We define the pressure tensor: + +$$\begin{aligned} + \hat{P} + \equiv \rho \expval{\vb{w} \vb{w}} + = \rho \expval{(\vb{v} \!-\! \vb{V}) (\vb{v} \!-\! \vb{V})} +\end{aligned}$$ + +This leads to the expected result, +where $\nabla \cdot (\rho \vb{V}\vb{V})$ represents the fluid momentum, +and $\nabla \cdot \hat{P}$ the viscous/pressure momentum: + +$$\begin{aligned} + 0 + &= \pdv{t} \big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F} +\end{aligned}$$ +</div> +</div> + +Finally, if we choose the kinetic energy $Q = m |\vb{v}|^2 / 2$, +we find that the **second moment** gives conservation of energy, +where $U$ is the thermal energy density and $\vb{J}$ is the heat flux: + +$$\begin{aligned} + \boxed{ + 0 + = \pdv{t} \bigg(\frac{\rho}{2} |\vb{V}|^2 + U \bigg) + + \nabla \cdot \bigg(\frac{\rho}{2} |\vb{V}|^2 \vb{V} + \vb{V} \cdot \hat{P} + U \vb{V} + \vb{J} \bigg) + - \vb{F} \cdot \big( n \vb{V} \big) + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-moment2"/> +<label for="proof-moment2">Proof</label> +<div class="hidden"> +<label for="proof-moment2">Proof.</label> +We insert $Q = m |\vb{v}|^2 / 2$ into our prototype and recognize $\rho$ wherever possible: + +$$\begin{aligned} + 0 + &= \pdv{t} \bigg(n \expval{\frac{m |\vb{v}|^2}{2}}\bigg) + + \nabla \cdot \bigg(n \expval{\frac{m |\vb{v}|^2}{2} \vb{v}}\bigg) + - \frac{\vb{F}}{m} \cdot \bigg( n \expval{\pdv{\vb{v}} \frac{m |\vb{v}|^2}{2}} \bigg) + \\ + &= \pdv{t} \bigg(\frac{\rho}{2} \expval{|\vb{v}|^2}\bigg) + + \nabla \cdot \bigg(\frac{\rho}{2} \expval{|\vb{v}|^2 \vb{v}}\bigg) + - \frac{\vb{F}}{2} \cdot \bigg( n \expval{\pdv{|\vb{v}|^2}{\vb{v}}} \bigg) +\end{aligned}$$ + +We handle these terms one by one. Substituting $\vb{v} = \vb{V} + \vb{w}$ in the first gives: + +$$\begin{aligned} + \expval{|\vb{v}|^2} + &= \expval{(\vb{V} \!+\! \vb{w}) \cdot (\vb{V} \!+\! \vb{w})} + = \expval{|\vb{V}|^2 + 2 \vb{V} \cdot \vb{w} + |\vb{w}|^2} + \\ + &= |\vb{V}|^2 + 2 \vb{V} \cdot \expval{\vb{w}} + \expval{|\vb{w}|^2} + = |\vb{V}|^2 + \expval{|\vb{w}|^2} +\end{aligned}$$ + +And likewise for the second term, +where we recognize the stress tensor $\expval{\vb{w} \vb{w}}$: + +$$\begin{aligned} + \expval{|\vb{v}|^2 \vb{v}} + &= \expval{(\vb{V} \!+\! \vb{w}) \cdot (\vb{V} \!+\! \vb{w}) (\vb{V} \!+\! \vb{w})} + = \expval{(|\vb{V}|^2 + 2 \vb{V} \cdot \vb{w} + |\vb{w}|^2) (\vb{V} \!+\! \vb{w})} + \\ + &= \expval{|\vb{V}|^2 \vb{V} + |\vb{V}|^2 \vb{w} + + 2 (\vb{V} \cdot \vb{w}) \vb{V} + 2 (\vb{V} \cdot \vb{w}) \vb{w} + + |\vb{w}|^2 \vb{V} + |\vb{w}|^2 \vb{w}} + \\ + &= |\vb{V}|^2 \vb{V} + |\vb{V}|^2 \expval{\vb{w}} + + 2 (\vb{V} \cdot \expval{\vb{w}}) \vb{V} + 2 \expval{(\vb{V} \cdot \vb{w}) \vb{w}} + + \expval{|\vb{w}|^2} \vb{V} + \expval{|\vb{w}|^2 \vb{w}} + \\ + &= |\vb{V}|^2 \vb{V} + 0 + 0 + 2 \vb{V} \cdot \expval{\vb{w} \vb{w}} + + \expval{|\vb{w}|^2} \vb{V} + \expval{|\vb{w}|^2 \vb{w}} +\end{aligned}$$ + +The third term is fairly obvious, but we calculate it rigorously just to be safe: + +$$\begin{aligned} + \pdv{|\vb{v}|^2}{\vb{v}} + &= \pdv{\vb{v}} \big( v_x^2 + v_y^2 + v_z^2 \big) + = \vu{e}_x \pdv{v_x^2}{v_x} + \vu{e}_y \pdv{v_y^2}{v_y} + \vu{e}_z \pdv{v_z^2}{v_z} + = 2 \vb{v} +\end{aligned}$$ + +To clarify the physical interpretation, +we define $U$, $\vb{J}$ and $\hat{P}$ as follows: + +$$\begin{aligned} + U + &\equiv \frac{\rho}{2} \expval{|\vb{w}|^2} + = \frac{\rho}{2} \expval{(\vb{v} \!-\! \vb{V}) \cdot (\vb{v} \!-\! \vb{V})} + \\ + \vb{J} + &\equiv \frac{\rho}{2} \expval{|\vb{w}|^2 \vb{w}} + = \frac{\rho}{2} \expval{(\vb{v} \!-\! \vb{V}) \cdot (\vb{v} \!-\! \vb{V})(\vb{v} \!-\! \vb{V})} + \\ + \hat{P} + &\equiv \rho \expval{\vb{w} \vb{w}} + = \rho \expval{(\vb{v} \!-\! \vb{V}) (\vb{v} \!-\! \vb{V})} +\end{aligned}$$ + +Putting it all together, we arrive at the expected result, namely: + +$$\begin{aligned} + 0 + &= \pdv{t} \bigg(\frac{\rho}{2} |\vb{V}|^2 + U \bigg) + + \nabla \cdot \bigg(\frac{\rho}{2} |\vb{V}|^2 \vb{V} + \vb{V} \cdot \hat{P} + U \vb{V} + \vb{J} \bigg) + - \vb{F} \cdot \big( n \vb{V} \big) +\end{aligned}$$ + +For the sake of clarity, we write out the pressure term, including the outer divergence: + +$$\begin{aligned} + \nabla \cdot (\vb{V} \cdot \hat{P}) + &= (\nabla \cdot \hat{P}{}^{\mathrm{T}}) \cdot \vb{V} + = \nabla \cdot + \begin{bmatrix} + P_{xx} & P_{xy} & P_{xz} \\ + P_{yx} & P_{yy} & P_{yz} \\ + P_{zx} & P_{zy} & P_{zz} + \end{bmatrix} + \cdot + \begin{bmatrix} + V_x \\ V_y \\ V_z + \end{bmatrix} + \\ + &= + \begin{bmatrix} + \displaystyle \pdv{P_{xx}}{x} + \pdv{P_{xy}}{y} + \pdv{P_{xz}}{z} \\ + \displaystyle \pdv{P_{yx}}{x} + \pdv{P_{yy}}{y} + \pdv{P_{yz}}{z} \\ + \displaystyle \pdv{P_{zx}}{x} + \pdv{P_{zy}}{y} + \pdv{P_{zz}}{z} + \end{bmatrix}^{\mathrm{T}} + \cdot + \begin{bmatrix} + V_x \\ V_y \\ V_z + \end{bmatrix} + = \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i +\end{aligned}$$ +</div> +</div> + + + +## References +1. M. Salewski, A.H. Nielsen, + *Plasma physics: lecture notes*, + 2021, unpublished. diff --git a/content/know/concept/central-limit-theorem/index.pdc b/content/know/concept/central-limit-theorem/index.pdc index a957736..518c224 100644 --- a/content/know/concept/central-limit-theorem/index.pdc +++ b/content/know/concept/central-limit-theorem/index.pdc @@ -53,7 +53,7 @@ $$\begin{aligned} \phi(k) = \int_{-\infty}^\infty p(x) \exp(i k x) \dd{x} \end{aligned}$$ -Note that $\phi(k)$ can be interpreted as the average of $\exp(i k x)$. +Note that $\phi(k)$ can be interpreted as the average of $\exp\!(i k x)$. We take its Taylor expansion in two separate ways, where an overline denotes the mean: @@ -62,7 +62,7 @@ $$\begin{aligned} = \sum_{n = 0}^\infty \frac{k^n}{n!} \: \phi^{(n)}(0) \qquad \phi(k) - = \overline{\exp(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n} + = \overline{\exp\!(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n} \end{aligned}$$ By comparing the coefficients of these two power series, @@ -88,12 +88,12 @@ using our earlier relation: $$\begin{aligned} C^{(1)} &= - i \dv{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0} - = - i \frac{\phi'(0)}{\exp(0)} + = - i \frac{\phi'(0)}{\exp\!(0)} = \overline{x} \\ C^{(2)} &= - \dv[2]{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0} - = \frac{\big(\phi'(0)\big)^2}{\exp(0)^2} - \frac{\phi''(0)}{\exp(0)} + = \frac{\big(\phi'(0)\big)^2}{\exp\!(0)^2} - \frac{\phi''(0)}{\exp\!(0)} = - \overline{x}^2 + \overline{x^2} = \sigma^2 \end{aligned}$$ @@ -153,7 +153,7 @@ with $\sqrt{N}$ appearing in the arguments of $\phi_n$: $$\begin{aligned} \phi_z(k) &= \idotsint - \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp(i k z) + \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp\!(i k z) \dd{x_1} \cdots \dd{x_N} \\ &= \idotsint @@ -184,7 +184,7 @@ $$\begin{aligned} = i k \overline{z} - \frac{k^2}{2} \sigma_z^2 \\ \phi_z(k) - &\approx \exp(i k \overline{z}) \exp\!(- k^2 \sigma_z^2 / 2) + &\approx \exp\!(i k \overline{z}) \exp\!(- k^2 \sigma_z^2 / 2) \end{aligned}$$ We take its inverse Fourier transform to get the density $p(z)$, @@ -194,7 +194,7 @@ which is even already normalized: $$\begin{aligned} p(z) = \hat{\mathcal{F}}^{-1} \{\phi_z(k)\} - &= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \overline{z})\big) \exp(- k^2 \sigma_z^2 / 2) \dd{k} + &= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \overline{z})\big) \exp\!(- k^2 \sigma_z^2 / 2) \dd{k} \\ &= \frac{1}{\sqrt{2 \pi \sigma_z^2}} \exp\!\Big(\!-\! \frac{(z - \overline{z})^2}{2 \sigma_z^2} \Big) \end{aligned}$$ diff --git a/content/know/concept/ritz-method/index.pdc b/content/know/concept/ritz-method/index.pdc index 320771a..1fe0d31 100644 --- a/content/know/concept/ritz-method/index.pdc +++ b/content/know/concept/ritz-method/index.pdc @@ -229,6 +229,10 @@ $$\begin{aligned} } \end{aligned}$$ +In the context of quantum mechanics, this is not surprising, +since any superposition of multiple states +is guaranteed to have a higher energy than the ground state. + Note that the convergence to $\lambda_0$ goes as $|c_n|^2$, while $u$ converges to $u_0$ as $|c_n|$ by definition, so even a fairly bad guess $u$ will give a decent estimate for $\lambda_0$. @@ -348,10 +352,17 @@ in a limited basis would yield a matrix $\overline{H}$ giving rough eigenvalues. The point of this discussion is to rigorously show the validity of this approach. +If we only care about the ground state, +then we already know $\lambda$ from $R[u]$, +so all we need to do is solve the above matrix equation for $a_n$. +Keep in mind that $\overline{M}$ is singular, +and $a_n$ are only defined up to a constant factor. + Nowadays, there exist many other methods to calculate eigenvalues of complicated operators $\hat{H}$, but an attractive feature of the Ritz method is that it is single-step, whereas its competitors tend to be iterative. +That said, the Ritz method cannot recover from a poorly chosen basis. |