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Diffstat (limited to 'content/know/concept/selection-rules/index.pdc')
| -rw-r--r-- | content/know/concept/selection-rules/index.pdc | 177 |
1 files changed, 172 insertions, 5 deletions
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: |