From 6ce0bb9a8f9fd7d169cbb414a9537d68c5290aae Mon Sep 17 00:00:00 2001 From: Prefetch Date: Fri, 14 Oct 2022 23:25:28 +0200 Subject: Initial commit after migration from Hugo --- source/know/concept/selection-rules/index.md | 698 +++++++++++++++++++++++++++ 1 file changed, 698 insertions(+) create mode 100644 source/know/concept/selection-rules/index.md (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 new file mode 100644 index 0000000..2ce5748 --- /dev/null +++ b/source/know/concept/selection-rules/index.md @@ -0,0 +1,698 @@ +--- +title: "Selection rules" +date: 2021-06-02 +categories: +- Physics +- Quantum mechanics +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: + +$$\begin{aligned} + \matrixel{f}{\hat{O}}{i} + = \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} +\end{aligned}$$ + +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 $\Ket{f}$ +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. + + +## Parity rules + +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 $\Ket{\ell_i m_i}$: + +$$\begin{aligned} + \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} + &= - \matrixel{\ell_f m_f}{\hat{\Pi}^\dagger \hat{O} \hat{\Pi}}{\ell_i m_i} + \\ + &= - \matrixel{\ell_f m_f}{(-1)^{\ell_f} \hat{O} (-1)^{\ell_i}}{\ell_i m_i} + \\ + &= (-1)^{\ell_f + \ell_i + 1} \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} +\end{aligned}$$ + +Which clearly can only be true if the exponent is even, +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**: + +$$\begin{aligned} + \boxed{ + \Delta \ell \:\:\text{is odd} + } +\end{aligned}$$ + +If this is not the case, +then the only possible way that the above equation can be satisfied +is if the matrix element vanishes $\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} = 0$. +We can derive an analogous rule for +any operator $\hat{E}$ which is even under parity: + +$$\begin{aligned} + \hat{\Pi}^\dagger \hat{E} \hat{\Pi} = \hat{E} + \quad \implies \quad + \boxed{ + \Delta \ell \:\:\text{is even} + } +\end{aligned}$$ + + +## Dipole rules + +Arguably the most common operator found in such matrix elements +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: + +$$\begin{aligned} + \boxed{ + \Delta m = 0 \:\:\mathrm{or}\: \pm 1 + } +\end{aligned}$$ + +
+ + + +
+ +Meanwhile, for the total angular momentum $\ell$ we have the following: + +$$\begin{aligned} + \boxed{ + \Delta \ell = \pm 1 + } +\end{aligned}$$ + +
+ + + +
+ + +## 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 $m_i$: + +$$\begin{aligned} + \boxed{ + \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} + = \delta_{\ell_f \ell_i} \delta_{m_f m_i} \matrixel{\ell_f}{|\hat{s}|}{\ell_i} + } +\end{aligned}$$ + +
+ + + +
+ +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 are not always about atomic electron transitions, or angular momenta even. + +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: + +$$\begin{aligned} + \matrixel{\Psi}{\hat{O}}{\Psi} + = \matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi} +\end{aligned}$$ + +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: + +$$\begin{aligned} + \matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi} + = \matrixel{\Psi}{\hat{P}^{-1} \hat{O} \hat{P}}{\Psi} + = \matrixel{\Psi}{\hat{P}^{-1} \hat{P} \hat{O}}{\Psi} + = \matrixel{\Psi}{\hat{O}}{\Psi} +\end{aligned}$$ + +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 $\hat{P}$: + +$$\begin{aligned} + \hat{P} \Ket{s} + = \Ket{s} + \qquad + \hat{P} \Ket{a} + = - \Ket{a} +\end{aligned}$$ + +Any obervable $\hat{O}$ then satisfies the equation below, +again thanks to the fact that $\hat{P} = \hat{P}^{-1}$: + +$$\begin{aligned} + \matrixel{s}{\hat{O}}{a} + = \matrixel{\hat{P} s}{\hat{O}}{a} + = \matrixel{s}{\hat{P}^{-1} \hat{O}}{a} + = \matrixel{s}{\hat{O} \hat{P}}{a} + = \matrixel{s}{\hat{O}}{\hat{P} a} + = - \matrixel{s}{\hat{O}}{a} +\end{aligned}$$ + +This leads us to the **superselection rule**, +which states that there can never be any interference +between states of different permutation symmetry: + +$$\begin{aligned} + \boxed{ + \matrixel{s}{\hat{O}}{a} + = 0 + } +\end{aligned}$$ + + + +## References +1. D.J. Griffiths, D.F. Schroeter, + *Introduction to quantum mechanics*, 3rd edition, + Cambridge. -- cgit v1.2.3