From f9cce7d563d0ea2ac591c31ff7d248ad3d02d1ac Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 3 Jun 2021 19:30:38 +0200
Subject: Expand knowledge base

---
 content/know/concept/selection-rules/index.pdc | 177 ++++++++++++++++++++++++-
 1 file changed, 172 insertions(+), 5 deletions(-)

(limited to 'content/know/concept/selection-rules')

diff --git a/content/know/concept/selection-rules/index.pdc b/content/know/concept/selection-rules/index.pdc
index 5da97e7..22dfd64 100644
--- a/content/know/concept/selection-rules/index.pdc
+++ b/content/know/concept/selection-rules/index.pdc
@@ -20,7 +20,7 @@ the total angular momentum and its $z$-component:
 
 $$\begin{aligned}
     \matrixel{f}{\hat{O}}{i}
-    = \matrixel{n_f \ell_f m_f}{\hat{O}}{n_i \ell_i m_i}
+    = \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i}
 \end{aligned}$$
 
 Where $\hat{O}$ is an operator, $\ket{i}$ is an initial state, and
@@ -28,8 +28,6 @@ $\ket{f}$ 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,
 guarantee that the above matrix element is zero.
-Note that $n_f$ and $n_i$ typically do not matter in this context,
-so they will be omitted from now on.
 
 
 ## Parity rules
@@ -478,11 +476,180 @@ $$\begin{aligned}
 </div>
 
 
+## Rotational rules
+
+Given a general (pseudo)scalar operator $\hat{s}$,
+which, by nature, must satisfy the
+following relations with the angular momentum operators:
+
+$$\begin{aligned}
+    \comm*{\hat{L}^2}{\hat{s}} = 0
+    \qquad
+    \comm*{\hat{L}_z}{\hat{s}} = 0
+    \qquad
+    \comm*{\hat{L}_{\pm}}{\hat{s}} = 0
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta \ell = 0
+    }
+    \qquad \quad
+    \boxed{
+        \Delta m = 0
+    }
+\end{aligned}$$
+
+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 $m_i$:
+
+$$\begin{aligned}
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
+        = \delta_{\ell_f \ell_i} \delta_{m_f m_i} \matrixel{\ell_f}{|\hat{s}|}{\ell_i}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-rot-scalar"/>
+<label for="proof-rot-scalar">Proof</label>
+<div class="hidden">
+<label for="proof-rot-scalar">Proof.</label>
+Firstly, we look at the commutator of $\hat{s}$ with the $z$-component $\hat{L}_z$:
+$$\begin{aligned}
+    0
+    = \matrixel{\ell_f m_f}{\comm*{\hat{L}_z}{\hat{s}}}{\ell_i m_i}
+    &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}_z}{\ell_i m_i}
+    \\
+    &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
+\end{aligned}$$
+
+Which can only be true if $m_f \!-\! m_i = 0$, unless,
+of course, $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} = 0$ by itself.
+
+Secondly, we look at the commutator of $\hat{s}$ with the total angular momentum $\hat{L}^2$:
+
+$$\begin{aligned}
+    0
+    = \matrixel{\ell_f m_f}{\comm*{\hat{L}^2}{\hat{s}}}{\ell_i m_i}
+    &= \matrixel{\ell_f m_f}{\hat{L}^2 \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}^2}{\ell_i m_i}
+    \\
+    &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
+\end{aligned}$$
+
+Assuming $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \neq 0$,
+this can only be satisfied if the following holds:
+
+$$\begin{aligned}
+    0
+    = \ell_f^2 + \ell_f - \ell_i^2 - \ell_i
+    = (\ell_f + \ell_i) (\ell_f - \ell_i) + (\ell_f - \ell_i)
+\end{aligned}$$
+
+If $\ell_f = \ell_i = 0$ this equation is trivially satisfied.
+Otherwise, the only option is $\ell_f \!-\! \ell_i = 0$,
+which is another part of the selection rule.
+
+Thirdly, we look at the commutator of $\hat{s}$ with the ladder operators $\hat{L}_\pm$:
+
+$$\begin{aligned}
+    0
+    = \matrixel{\ell_f m_f}{\comm*{\hat{L}_\pm}{\hat{s}}}{\ell_i m_i}
+    &= \matrixel{\ell_f m_f}{\hat{L}_\pm \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}_\pm}{\ell_i m_i}
+    \\
+    &= C_f \matrixel{\ell_f (m_f\!\mp\!1)}{\hat{s}}{\ell_i m_i} - C_i \matrixel{\ell_f m_f}{\hat{s}}{\ell_i (m_i\!\pm\!1)}
+\end{aligned}$$
+
+Where $C_f$ and $C_i$ are constants given below.
+We already know that $\Delta \ell = 0$ and $\Delta m = 0$,
+so the above matrix elements are only nonzero if $m_f = m_i \pm 1$.
+Therefore:
+
+$$\begin{aligned}
+    C_i
+    &= \hbar \sqrt{\ell_i (\ell_i + 1) - m_i (m_i \pm 1)}
+    \\
+    C_f
+    &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_f (m_f \!\mp\! 1)}
+    \\
+    &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - (m_i \!\pm\! 1) (m_i \!\pm\! 1 \!\mp\! 1)}
+    \\
+    &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_i (m_i \!\pm\! 1)}
+\end{aligned}$$
+
+In other words, $C_f = C_i$. The above equation therefore reduces to:
+
+$$\begin{aligned}
+    \matrixel{\ell_f m_i}{\hat{s}}{\ell_i m_i}
+    &= \matrixel{\ell_f (m_i \!\pm\! 1)}{\hat{s}}{\ell_i (m_i\!\pm\!1)}
+\end{aligned}$$
+
+Which means that the value of the matrix element
+does not depend on $m_i$ (or $m_f$) at all.
+</div>
+</div>
+
+Similarly, given a general (pseudo)vector operator $\vu{V}$,
+which, by nature, must satisfy the following commutation relations,
+where $\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$:
+
+$$\begin{gathered}
+    \comm*{\hat{L}_z}{\hat{V}_z} = 0
+    \qquad
+    \comm*{\hat{L}_z}{\hat{V}_{\pm}} = \pm \hbar \hat{V}_{\pm}
+    \qquad
+    \comm*{\hat{L}_{\pm}}{\hat{V}_z} = \mp \hbar \hat{V}_{\pm}
+    \\
+    \comm*{\hat{L}_{\pm}}{\hat{V}_{\pm}} = 0
+    \qquad
+    \comm*{\hat{L}_{\pm}}{\hat{V}_{\mp}} = \pm 2 \hbar \hat{V}_z
+\end{gathered}$$
+
+The inner product of any such $\vu{V}$ must obey the following selection rules:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta \ell
+        = 0 \:\:\mathrm{or}\: \pm 1
+    }
+    \qquad
+    \boxed{
+        \Delta m
+        = 0 \:\:\mathrm{or}\: \pm 1
+    }
+\end{aligned}$$
+
+In fact, the complete result involves the Clebsch-Gordan coefficients (from spin addition):
+
+$$\begin{gathered}
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{V}_{z}}{\ell_i m_i}
+        = C^{\ell_i \: 1 \: \ell_f}_{m_i \: 0 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i}
+    }
+    \\
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{V}_{+}}{\ell_i m_i}
+        = - \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: 1 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i}
+    }
+    \\
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{V}_{-}}{\ell_i m_i}
+        = \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: -1 \:m_f} \matrixel{\ell_f}{|\hat{V}}{|\ell_i}
+    }
+\end{gathered}$$
+
+
 ## Superselection rule
 
-Selection rules need not always be about atomic electron transitions.
+Selection rules are not always about atomic electron transitions, or angular momenta even.
+
 According to the **principle of indistinguishability**,
-permutating identical particles never leads to an observable difference.
+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:
 
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