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author | Prefetch | 2022-10-20 18:25:31 +0200 |
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committer | Prefetch | 2022-10-20 18:25:31 +0200 |
commit | 16555851b6514a736c5c9d8e73de7da7fc9b6288 (patch) | |
tree | 76b8bfd30f8941d0d85365990bcdbc5d0643cabc /source/know/concept/dyson-equation | |
parent | e5b9bce79b68a68ddd2e51daa16d2fea73b84fdb (diff) |
Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'
Diffstat (limited to 'source/know/concept/dyson-equation')
-rw-r--r-- | source/know/concept/dyson-equation/index.md | 62 |
1 files changed, 31 insertions, 31 deletions
diff --git a/source/know/concept/dyson-equation/index.md b/source/know/concept/dyson-equation/index.md index 962b64a..ae9eb35 100644 --- a/source/know/concept/dyson-equation/index.md +++ b/source/know/concept/dyson-equation/index.md @@ -9,7 +9,7 @@ layout: "concept" --- Consider the time-dependent Schrödinger equation, -describing a wavefunction $\Psi_0(\vb{r}, t)$: +describing a wavefunction $$\Psi_0(\vb{r}, t)$$: $$\begin{aligned} i \hbar \pdv{}{t}\Psi_0(\vb{r}, t) @@ -18,50 +18,50 @@ $$\begin{aligned} By definition, this equation's [fundamental solution](/know/concept/fundamental-solution/) -$G_0(\vb{r}, t; \vb{r}', t')$ satisfies the following: +$$G_0(\vb{r}, t; \vb{r}', t')$$ satisfies the following: $$\begin{aligned} \Big( i \hbar \pdv{}{t}- \hat{H}_0(\vb{r}) \Big) G_0(\vb{r}, t; \vb{r}', t') = \delta(\vb{r} - \vb{r}') \: \delta(t - t') \end{aligned}$$ -From this, we define the inverse $\hat{G}{}_0^{-1}(\vb{r}, t)$ -as follows, so that $\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$: +From this, we define the inverse $$\hat{G}{}_0^{-1}(\vb{r}, t)$$ +as follows, so that $$\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$$: $$\begin{aligned} \hat{G}{}_0^{-1}(\vb{r}, t) &\equiv i \hbar \pdv{}{t}- \hat{H}_0(\vb{r}) \end{aligned}$$ -Note that $\hat{G}{}_0^{-1}$ is an operator, while $G_0$ is a function. +Note that $$\hat{G}{}_0^{-1}$$ is an operator, while $$G_0$$ is a function. For the sake of consistency, we thus define -the operator $\hat{G}_0(\vb{r}, t)$ -as a multiplication by $G_0$ -and integration over $\vb{r}'$ and $t'$: +the operator $$\hat{G}_0(\vb{r}, t)$$ +as a multiplication by $$G_0$$ +and integration over $$\vb{r}'$$ and $$t'$$: $$\begin{aligned} \hat{G}_0(\vb{r}, t) \: f \equiv \iint_{-\infty}^\infty G_0(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \: \dd{\vb{r}}' \dd{t'} \end{aligned}$$ -For an arbitrary function $f(\vb{r}, t)$, -so that $\hat{G}{}_0^{-1} \hat{G}_0 = \hat{G}_0 \hat{G}{}_0^{-1} = 1$. +For an arbitrary function $$f(\vb{r}, t)$$, +so that $$\hat{G}{}_0^{-1} \hat{G}_0 = \hat{G}_0 \hat{G}{}_0^{-1} = 1$$. Moving on, the Schrödinger equation can be rewritten like so, -using $\hat{G}{}_0^{-1}$: +using $$\hat{G}{}_0^{-1}$$: $$\begin{aligned} \hat{G}{}_0^{-1}(\vb{r}, t) \: \Psi_0(\vb{r}, t) = 0 \end{aligned}$$ -Let us assume that $\hat{H}_0$ is simple, -such that $G_0$ and $\hat{G}{}_0^{-1}$ can be found without issues +Let us assume that $$\hat{H}_0$$ is simple, +such that $$G_0$$ and $$\hat{G}{}_0^{-1}$$ can be found without issues by solving the defining equation above. Suppose we now add a more complicated and -possibly time-dependent term $\hat{H}_1(\vb{r}, t)$, +possibly time-dependent term $$\hat{H}_1(\vb{r}, t)$$, in which case the corresponding fundamental solution -$G(\vb{r}, \vb{r}', t, t')$ satisfies: +$$G(\vb{r}, \vb{r}', t, t')$$ satisfies: $$\begin{aligned} \delta(\vb{r} - \vb{r}') \: \delta(t - t') @@ -71,8 +71,8 @@ $$\begin{aligned} \end{aligned}$$ This equation is typically too complicated to solve, -so we would like an easier way to calculate this new $G$. -The perturbed wavefunction $\Psi(\vb{r}, t)$ +so we would like an easier way to calculate this new $$G$$. +The perturbed wavefunction $$\Psi(\vb{r}, t)$$ satisfies the Schrödinger equation: $$\begin{aligned} @@ -80,9 +80,9 @@ $$\begin{aligned} = 0 \end{aligned}$$ -We know that $\hat{G}{}_0^{-1} \Psi_0 = 0$, +We know that $$\hat{G}{}_0^{-1} \Psi_0 = 0$$, which we put on the right, -and then we apply $\hat{G}_0$ in front: +and then we apply $$\hat{G}_0$$ in front: $$\begin{aligned} \hat{G}_0^{-1} \Psi - \hat{H}_1 \Psi @@ -110,7 +110,7 @@ $$\begin{aligned} \end{aligned}$$ The parenthesized expression clearly has the same recursive pattern, -so we denote it by $\hat{G}$ and write the so-called **Dyson equation**: +so we denote it by $$\hat{G}$$ and write the so-called **Dyson equation**: $$\begin{aligned} \boxed{ @@ -119,9 +119,9 @@ $$\begin{aligned} } \end{aligned}$$ -Such an iterative scheme is excellent for approximating $\hat{G}(\vb{r}, t)$. +Such an iterative scheme is excellent for approximating $$\hat{G}(\vb{r}, t)$$. Once a satisfactory accuracy is obtained, -the perturbed wavefunction $\Psi$ can be calculated from: +the perturbed wavefunction $$\Psi$$ can be calculated from: $$\begin{aligned} \boxed{ @@ -131,10 +131,10 @@ $$\begin{aligned} \end{aligned}$$ This relation is equivalent to the Schrödinger equation. -So now we have the operator $\hat{G}(\vb{r}, t)$, -but what about the fundamental solution function $G(\vb{r}, t; \vb{r}', t')$? -Let us take its definition, multiply it by an arbitrary $f(\vb{r}, t)$, -and integrate over $G$'s second argument pair: +So now we have the operator $$\hat{G}(\vb{r}, t)$$, +but what about the fundamental solution function $$G(\vb{r}, t; \vb{r}', t')$$? +Let us take its definition, multiply it by an arbitrary $$f(\vb{r}, t)$$, +and integrate over $$G$$'s second argument pair: $$\begin{aligned} \iint \big( \hat{G}{}_0^{-1} \!-\! \hat{H}_1 \big) G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'} @@ -142,9 +142,9 @@ $$\begin{aligned} = f \end{aligned}$$ -Where we have hidden the arguments $(\vb{r}, t)$ for brevity. -We now apply $\hat{G}_0(\vb{r}, t)$ to this equation -(which contains an integral over $t''$ independent of $t'$): +Where we have hidden the arguments $$(\vb{r}, t)$$ for brevity. +We now apply $$\hat{G}_0(\vb{r}, t)$$ to this equation +(which contains an integral over $$t''$$ independent of $$t'$$): $$\begin{aligned} \hat{G}_0 f @@ -154,8 +154,8 @@ $$\begin{aligned} \end{aligned}$$ Here, the shape of Dyson's equation is clearly recognizable, -so we conclude that, as expected, the operator $\hat{G}$ -is defined as multiplication by the function $G$ followed by integration: +so we conclude that, as expected, the operator $$\hat{G}$$ +is defined as multiplication by the function $$G$$ followed by integration: $$\begin{aligned} \hat{G}(\vb{r}, t) \: f(\vb{r}, t) |