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/selection-rules/index.md | 155 ++++++++++++++------------- 1 file changed, 78 insertions(+), 77 deletions(-) (limited to 'source/know/concept/selection-rules') diff --git a/source/know/concept/selection-rules/index.md b/source/know/concept/selection-rules/index.md index 84227ba..373486e 100644 --- a/source/know/concept/selection-rules/index.md +++ b/source/know/concept/selection-rules/index.md @@ -10,33 +10,33 @@ layout: "concept" In quantum mechanics, it is often necessary to evaluate matrix elements of the following form, -where $\ell$ and $m$ respectively represent -the total angular momentum and its $z$-component: +where $$\ell$$ and $$m$$ respectively represent +the total angular momentum and its $$z$$-component: $$\begin{aligned} \matrixel{f}{\hat{O}}{i} = \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} \end{aligned}$$ -Where $\hat{O}$ is an operator, $\Ket{i}$ is an initial state, and -$\Ket{f}$ is a final state (usually at least; $\Ket{i}$ and $\Ket{f}$ +Where $$\hat{O}$$ is an operator, $$\Ket{i}$$ is an initial state, and +$$\Ket{f}$$ is a final state (usually at least; $$\Ket{i}$$ and $$\Ket{f}$$ can be any states). **Selection rules** are requirements on the relations -between $\ell_i$, $\ell_f$, $m_i$ and $m_f$, which, if not met, +between $$\ell_i$$, $$\ell_f$$, $$m_i$$ and $$m_f$$, which, if not met, guarantee that the above matrix element is zero. ## Parity rules -Let $\hat{O}$ denote any operator which is odd under spatial inversion +Let $$\hat{O}$$ denote any operator which is odd under spatial inversion (parity): $$\begin{aligned} \hat{\Pi}^\dagger \hat{O} \hat{\Pi} = - \hat{O} \end{aligned}$$ -Where $\hat{\Pi}$ is the parity operator. -We wrap this property of $\hat{O}$ -in the states $\Ket{\ell_f m_f}$ and $\Ket{\ell_i m_i}$: +Where $$\hat{\Pi}$$ is the parity operator. +We wrap this property of $$\hat{O}$$ +in the states $$\Ket{\ell_f m_f}$$ and $$\Ket{\ell_i m_i}$$: $$\begin{aligned} \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} @@ -48,7 +48,7 @@ $$\begin{aligned} \end{aligned}$$ Which clearly can only be true if the exponent is even, -so $\Delta \ell \equiv \ell_f - \ell_i$ must be odd. +so $$\Delta \ell \equiv \ell_f - \ell_i$$ must be odd. This leads to the following selection rule, often referred to as **Laporte's rule**: @@ -60,9 +60,9 @@ $$\begin{aligned} If this is not the case, then the only possible way that the above equation can be satisfied -is if the matrix element vanishes $\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} = 0$. +is if the matrix element vanishes $$\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} = 0$$. We can derive an analogous rule for -any operator $\hat{E}$ which is even under parity: +any operator $$\hat{E}$$ which is even under parity: $$\begin{aligned} \hat{\Pi}^\dagger \hat{E} \hat{\Pi} = \hat{E} @@ -76,10 +76,10 @@ $$\begin{aligned} ## Dipole rules Arguably the most common operator found in such matrix elements -is a position vector operator, like $\vu{r}$ or $\hat{x}$, +is a position vector operator, like $$\vu{r}$$ or $$\hat{x}$$, and the associated selection rules are known as **dipole rules**. -For the $z$-component of angular momentum $m$ we have the following: +For the $$z$$-component of angular momentum $$m$$ we have the following: $$\begin{aligned} \boxed{ @@ -92,7 +92,7 @@ $$\begin{aligned} -Meanwhile, for the total angular momentum $\ell$ we have the following: +Meanwhile, for the total angular momentum $$\ell$$ we have the following: $$\begin{aligned} \boxed{ @@ -195,7 +195,7 @@ $$\begin{aligned} -We then multiply this relation by $\Bra{f} = \Bra{\ell_f m_f}$ on the left -and $\Ket{i} = \Ket{\ell_i m_i}$ on the right, +We then multiply this relation by $$\Bra{f} = \Bra{\ell_f m_f}$$ on the left +and $$\Ket{i} = \Ket{\ell_i m_i}$$ on the right, so the right-hand side becomes: $$\begin{aligned} @@ -450,21 +450,22 @@ $$\begin{aligned} &= \big( (\ell_f + \ell_i + 1)^2 - 1 \big) \big( (\ell_f - \ell_i)^2 - 1 \big) \end{aligned}$$ -The first factor is zero if $\ell_f = \ell_i = 0$, -in which case the matrix element $\matrixel{f}{\vu{r}}{i} = 0$ anyway. +The first factor is zero if $$\ell_f = \ell_i = 0$$, +in which case the matrix element $$\matrixel{f}{\vu{r}}{i} = 0$$ anyway. The other, non-trivial option is therefore: $$\begin{aligned} (\ell_f - \ell_i)^2 = 1 \end{aligned}$$ + ## Rotational rules -Given a general (pseudo)scalar operator $\hat{s}$, +Given a general (pseudo)scalar operator $$\hat{s}$$, which, by nature, must satisfy the following relations with the angular momentum operators: @@ -476,8 +477,8 @@ $$\begin{aligned} \comm{\hat{L}_{\pm}}{\hat{s}} = 0 \end{aligned}$$ -Where $\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y$. -The inner product of any such $\hat{s}$ must obey these selection rules: +Where $$\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y$$. +The inner product of any such $$\hat{s}$$ must obey these selection rules: $$\begin{aligned} \boxed{ @@ -490,9 +491,9 @@ $$\begin{aligned} \end{aligned}$$ It is common to write this in the following more complete way, where -$\matrixel{\ell_f}{|\hat{s}|}{\ell_i}$ is the **reduced matrix element**, -which is identical to $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}$, but -with a different notation to say that it does not depend on $m_f$ or $m_i$: +$$\matrixel{\ell_f}{|\hat{s}|}{\ell_i}$$ is the **reduced matrix element**, +which is identical to $$\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}$$, but +with a different notation to say that it does not depend on $$m_f$$ or $$m_i$$: $$\begin{aligned} \boxed{ @@ -506,7 +507,7 @@ $$\begin{aligned} -Similarly, given a general (pseudo)vector operator $\vu{V}$, +Similarly, given a general (pseudo)vector operator $$\vu{V}$$, which, by nature, must satisfy the following commutation relations, -where $\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$: +where $$\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$$: $$\begin{gathered} \comm{\hat{L}_z}{\hat{V}_z} = 0 @@ -596,7 +597,7 @@ $$\begin{gathered} \comm{\hat{L}_{\pm}}{\hat{V}_{\mp}} = \pm 2 \hbar \hat{V}_z \end{gathered}$$ -The inner product of any such $\vu{V}$ must obey the following selection rules: +The inner product of any such $$\vu{V}$$ must obey the following selection rules: $$\begin{aligned} \boxed{ @@ -637,17 +638,17 @@ Selection rules are not always about atomic electron transitions, or angular mom According to the **principle of indistinguishability**, permuting identical particles never leads to an observable difference. In other words, the particles are fundamentally indistinguishable, -so for any observable $\hat{O}$ and multi-particle state $\Ket{\Psi}$, we can say: +so for any observable $$\hat{O}$$ and multi-particle state $$\Ket{\Psi}$$, we can say: $$\begin{aligned} \matrixel{\Psi}{\hat{O}}{\Psi} = \matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi} \end{aligned}$$ -Where $\hat{P}$ is an arbitrary permutation operator. -Indistinguishability implies that $\comm{\hat{P}}{\hat{O}} = 0$ -for all $\hat{O}$ and $\hat{P}$, -which lets us prove the above equation, using that $\hat{P}$ is unitary: +Where $$\hat{P}$$ is an arbitrary permutation operator. +Indistinguishability implies that $$\comm{\hat{P}}{\hat{O}} = 0$$ +for all $$\hat{O}$$ and $$\hat{P}$$, +which lets us prove the above equation, using that $$\hat{P}$$ is unitary: $$\begin{aligned} \matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi} @@ -656,9 +657,9 @@ $$\begin{aligned} = \matrixel{\Psi}{\hat{O}}{\Psi} \end{aligned}$$ -Consider a symmetric state $\Ket{s}$ and an antisymmetric state $\Ket{a}$ +Consider a symmetric state $$\Ket{s}$$ and an antisymmetric state $$\Ket{a}$$ (see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)), -which obey the following for a permutation $\hat{P}$: +which obey the following for a permutation $$\hat{P}$$: $$\begin{aligned} \hat{P} \Ket{s} @@ -668,8 +669,8 @@ $$\begin{aligned} = - \Ket{a} \end{aligned}$$ -Any obervable $\hat{O}$ then satisfies the equation below, -again thanks to the fact that $\hat{P} = \hat{P}^{-1}$: +Any obervable $$\hat{O}$$ then satisfies the equation below, +again thanks to the fact that $$\hat{P} = \hat{P}^{-1}$$: $$\begin{aligned} \matrixel{s}{\hat{O}}{a} -- cgit v1.2.3