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diff --git a/content/know/concept/selection-rules/index.pdc b/content/know/concept/selection-rules/index.pdc new file mode 100644 index 0000000..5da97e7 --- /dev/null +++ b/content/know/concept/selection-rules/index.pdc @@ -0,0 +1,546 @@ +--- +title: "Selection rules" +firstLetter: "S" +publishDate: 2021-06-02 +categories: +- Physics +- Quantum mechanics + +date: 2021-05-29T14:42:08+02:00 +draft: false +markup: pandoc +--- + +# Selection rules + +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: + +$$\begin{aligned} + \matrixel{f}{\hat{O}}{i} + = \matrixel{n_f \ell_f m_f}{\hat{O}}{n_i \ell_i m_i} +\end{aligned}$$ + +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. +Note that $n_f$ and $n_i$ typically do not matter in this context, +so they will be omitted from now on. + + +## 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: + +$$\begin{aligned} + \boxed{ + \Delta m = 0 \:\:\mathrm{or}\: \pm 1 + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-dipole-m"/> +<label for="proof-dipole-m">Proof</label> +<div class="hidden"> +<label for="proof-dipole-m">Proof.</label> +We know that the angular momentum $z$-component operator $\hat{L}_z$ satisfies: + +$$\begin{aligned} + \comm*{\hat{L}_z}{\hat{x}} = i \hbar \hat{y} + \qquad + \comm*{\hat{L}_z}{\hat{y}} = - i \hbar \hat{x} + \qquad + \comm*{\hat{L}_z}{\hat{z}} = 0 +\end{aligned}$$ + +We take the first relation, +and wrap it in $\bra{\ell_f m_f}$ and $\ket{\ell_i m_i}$, giving: + +$$\begin{aligned} + i \hbar \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} + &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{x}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{x} \hat{L}_z}{\ell_i m_i} + \\ + &= \hbar m_f \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} - \hbar m_i \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} + \\ + &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} +\end{aligned}$$ + +Next, we do the same thing with the second relation, for $[\hat{L}_z, \hat{y}]$, giving: + +$$\begin{aligned} + - i \hbar \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} + &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{y}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{y} \hat{L}_z}{\ell_i m_i} + \\ + &= \hbar m_f \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} - \hbar m_i \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} + \\ + &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} +\end{aligned}$$ + +Respectively isolating the two above results for $\hat{x}$ and $\hat{y}$, +we arrive at these equations: + +$$\begin{aligned} + \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} + &= i (m_f - m_i) \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} + \\ + \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} + &= - i (m_f - m_i) \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} +\end{aligned}$$ + +By inserting the first into the second, +we find (part of) the selection rule: + +$$\begin{aligned} + \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} + &= (m_f - m_i)^2 \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} +\end{aligned}$$ + +This can only be true if $\Delta m = \pm 1$, +unless the inner products of $\hat{x}$ and $\hat{y}$ are zero, +in which case we cannot say anything about $\Delta m$ yet. +Assuming the latter, we take the inner product of +the commutator $\comm*{\hat{L}_z}{\hat{z}} = 0$, and find: + +$$\begin{aligned} + 0 + &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{z}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{z} \hat{L}_z}{\ell_i m_i} + \\ + &= \hbar m_f \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i} - \hbar m_i \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i} + \\ + &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i} +\end{aligned}$$ + +If $\matrixel{f}{\hat{z}}{i} \neq 0$, we require $\Delta m = 0$. +The previous requirement was $\Delta m = \pm 1$, +implying that $\matrixel{f}{\hat{x}}{i} = \matrixel{f}{\hat{y}}{i} = 0$ +whenever $\matrixel{f}{\hat{z}}{i} \neq 0$. +Only if $\matrixel{f}{\hat{z}}{i} = 0$ +does the previous rule $\Delta m = \pm 1$ hold, +in which case the inner products of $\hat{x}$ and $\hat{y}$ are nonzero. +</div> +</div> + +Meanwhile, for the total angular momentum $\ell$ we have the following: + +$$\begin{aligned} + \boxed{ + \Delta \ell = \pm 1 + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-dipole-l"/> +<label for="proof-dipole-l">Proof</label> +<div class="hidden"> +<label for="proof-dipole-l">Proof.</label> +We start from the following relation +(which is already quite a chore to prove): + +$$\begin{aligned} + \comm{\hat{L}^2}{\comm*{\hat{L}^2}{\vu{r}}} + = 2 \hbar^2 (\vu{r} \hat{L}^2 + \hat{L}^2 \vu{r}) +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-dipole-l-comm"/> +<label for="proof-dipole-l-comm">Proof</label> +<div class="hidden"> +<label for="proof-dipole-l-comm">Proof.</label> +To begin with, we want to find the commutator of $\hat{L}^2$ and $\hat{x}$: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\hat{x}} + &= \comm*{\hat{L}_x^2}{\hat{x}} + \comm*{\hat{L}_y^2}{\hat{x}} + \comm*{\hat{L}_z^2}{\hat{x}} + = \comm*{\hat{L}_y^2}{\hat{x}} + \comm*{\hat{L}_z^2}{\hat{x}} + \\ + &= \hat{L}_y \comm*{\hat{L}_y}{\hat{x}} + \comm*{\hat{L}_y}{\hat{x}} \hat{L}_y + + \hat{L}_z \comm*{\hat{L}_z}{\hat{x}} + \comm*{\hat{L}_z}{\hat{x}} \hat{L}_z +\end{aligned}$$ + +Evaluating these commutators gives us: + +$$\begin{aligned} + \comm*{\hat{L}_y}{\hat{x}} + &= \comm*{\hat{z} \hat{p}_x}{\hat{x}} - \comm*{\hat{x} \hat{p}_z}{\hat{x}} + = \hat{z} \comm*{\hat{p}_x}{\hat{x}} + \comm*{\hat{z}}{\hat{x}} \hat{p}_x + - \hat{x} \comm*{\hat{p}_z}{\hat{x}} - \comm*{\hat{x}}{\hat{x}} \hat{p}_z + = - i \hbar \hat{z} + \\ + \comm*{\hat{L}_z}{\hat{x}} + &= \comm*{\hat{x} \hat{p}_y}{\hat{x}} - \comm*{\hat{y} \hat{p}_x}{\hat{x}} + = \hat{x} \comm*{\hat{p}_y}{\hat{x}} + \comm*{\hat{x}}{\hat{x}} \hat{p}_y + - \hat{y} \comm*{\hat{p}_x}{\hat{x}} - \comm*{\hat{y}}{\hat{x}} \hat{p}_x + = i \hbar \hat{y} +\end{aligned}$$ + +Which we then insert back into the original equation, yielding: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\hat{x}} + &= i \hbar (- \hat{L}_y \hat{z} - \hat{z} \hat{L}_y + \hat{L}_z \hat{y} + \hat{y} \hat{L}_z) +\end{aligned}$$ + +This can be simplified by introducing some more commutators: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\hat{x}} + &= i \hbar \big( \!-\! ( \comm*{\hat{L}_y}{\hat{z}} + \hat{z} \hat{L}_y ) - \hat{z} \hat{L}_y + + ( \comm*{\hat{L}_z}{\hat{y}} + \hat{y} \hat{L}_z ) + \hat{y} \hat{L}_z \big) +\end{aligned}$$ + +Evaluating these commutators gives us: + +$$\begin{aligned} + \comm*{\hat{L}_y}{\hat{z}} + &= \comm*{\hat{z} \hat{p}_x}{\hat{z}} - \comm*{\hat{x} \hat{p}_z}{\hat{z}} + = \hat{z} \comm*{\hat{p}_x}{\hat{z}} + \comm*{\hat{z}}{\hat{z}} \hat{p}_x + - \hat{x} \comm*{\hat{p}_z}{\hat{z}} - \comm*{\hat{x}}{\hat{z}} \hat{p}_z + = i \hbar \hat{x} + \\ + \comm*{\hat{L}_z}{\hat{y}} + &= \comm*{\hat{x} \hat{p}_y}{\hat{y}} - \comm*{\hat{y} \hat{p}_x}{\hat{y}} + = \hat{x} \comm*{\hat{p}_y}{\hat{y}} + \comm*{\hat{x}}{\hat{y}} \hat{p}_y + - \hat{y} \comm*{\hat{p}_x}{\hat{y}} - \comm*{\hat{y}}{\hat{y}} \hat{p}_x + = - i \hbar \hat{x} +\end{aligned}$$ + +Substituting these then leads us to the first milestone of this proof: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\hat{x}} + &= i \hbar \big( \!-\! i \hbar \hat{x} - \hat{z} \hat{L}_y - \hat{z} \hat{L}_y + - i \hbar \hat{x} + \hat{y} \hat{L}_z + \hat{y} \hat{L}_z \big) + \\ + &= 2 i \hbar (\hat{y} \hat{L}_z - \hat{z} \hat{L}_y - i \hbar \hat{x}) +\end{aligned}$$ + +Repeating this process for $\comm*{\hat{L}^2}{\hat{y}}$ and $\comm*{\hat{L}^2}{\hat{z}}$, +we find analogous expressions: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\hat{y}} + &= 2 i \hbar (\hat{z} \hat{L}_x - \hat{x} \hat{L}_z - i \hbar \hat{y}) + \\ + \comm*{\hat{L}^2}{\hat{z}} + &= 2 i \hbar (\hat{x} \hat{L}_y - \hat{y} \hat{L}_x - i \hbar \hat{z}) +\end{aligned}$$ + +Next, we take the commutator with $\hat{L}^2$ of the commutator we just found: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\hat{x}}} + &= 2 i \hbar \big(\comm*{\hat{L}^2}{\hat{y} \hat{L}_z} - \comm*{\hat{L}^2}{\hat{z} \hat{L}_y} - i \hbar \comm*{\hat{L}^2}{\hat{x}}\big) + \\ + &= 2 i \hbar \big( \hat{y} \comm*{\hat{L}^2}{\hat{L}_z} + \comm*{\hat{L}^2}{\hat{y}} \hat{L}_z + - \hat{z} \comm*{\hat{L}^2}{\hat{L}_y} - \comm*{\hat{L}^2}{\hat{z}} \hat{L}_y + - i \hbar \comm*{\hat{L}^2}{\hat{x}} \big) +\end{aligned}$$ + +Where we used that $\comm*{\hat{L}^2}{\hat{L}_y} = \comm*{\hat{L}^2}{\hat{L}_z} = 0$. +The other commutators look familiar: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\hat{x}}} + &= 2 i \hbar \big( \comm*{\hat{L}^2}{\hat{y}} \hat{L}_z + - \comm*{\hat{L}^2}{\hat{z}} \hat{L}_y + - i \hbar \comm*{\hat{L}^2}{\hat{x}} \big) +\end{aligned}$$ + +By inserting the expressions we found earlier for these commutators, we get: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\hat{x}}} + %&= - 2 \hbar^2 \big( 2 (\hat{z} \hat{L}_x - \hat{x} \hat{L}_z - i \hbar \hat{y}) \hat{L}_z + %- 2 (\hat{x} \hat{L}_y - \hat{y} \hat{L}_x - i \hbar \hat{z}) \hat{L}_y + %- (\hat{L}^2 \hat{x} - \hat{x} \hat{L}^2) \big) + %\\ + &= - 4 \hbar^2 \big( \hat{z} \hat{L}_x \hat{L}_z - \hat{x} \hat{L}_z^2 - i \hbar \hat{y} \hat{L}_z + + \hat{y} \hat{L}_x \hat{L}_y - \hat{x} \hat{L}_y^2 + i \hbar \hat{z} \hat{L}_y \big) \\ + &\qquad\qquad + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big) +\end{aligned}$$ + +Substituting the well-known commutators +$i \hbar \hat{L}_y = \comm*{\hat{L}_z}{\hat{L}_x}$ and +$i \hbar \hat{L}_z = \comm*{\hat{L}_x}{\hat{L}_y}$: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\hat{x}}} + &= - 4 \hbar^2 \big( \hat{z} \hat{L}_x \hat{L}_z + \hat{y} \hat{L}_x \hat{L}_y + - \hat{x} \hat{L}_y^2 - \hat{x} \hat{L}_z^2 + + \hat{z} \comm*{\hat{L}_z}{\hat{L}_x} - \hat{y} \comm*{\hat{L}_x}{\hat{L}_y} \big) \\ + &\qquad\qquad + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big) + \\ + &= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + - \hat{x} \hat{L}_y^2 - \hat{x} \hat{L}_z^2 \big) + + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big) +\end{aligned}$$ + +By definition, $\hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 = \hat{L}^2$, +which we use to arrive at: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\hat{x}}} + %&= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 + %- \hat{x} \hat{L}_x^2 - \hat{x} \hat{L}_y^2 - \hat{x} \hat{L}_z^2 \big) + %+ 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big) + %\\ + &= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 - \hat{x} \hat{L}^2 \big) + + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big) + \\ + &= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 \big) + + 2 \hbar^2 \big( \hat{L}^2 \hat{x} + \hat{x} \hat{L}^2 \big) +\end{aligned}$$ + +The second term is what we want to prove, +so the first term must vanish: + +$$\begin{aligned} + \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 + = (\vu{r} \cdot \vu{L}) \hat{L}_x + = (\vu{r} \cdot (\vu{r} \cross \vu{p})) \hat{L}_x + = (\vu{p} \cdot (\vu{r} \cross \vu{r})) \hat{L}_x + = 0 +\end{aligned}$$ + +Where $\vu{L} = \vu{r} \cross \vu{p}$ by definition, +and the cross product of a vector with itself is zero. + +This process can be repeated for +$\comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\hat{y}}}$ and +$\comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\hat{z}}}$, +leading us to: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\hat{x}}} + &= 2 \hbar^2 (\hat{x} \hat{L}^2 + \hat{L}^2 \hat{x}) + \\ + \comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\hat{y}}} + &= 2 \hbar^2 (\hat{y} \hat{L}^2 + \hat{L}^2 \hat{y}) + \\ + \comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\hat{z}}} + &= 2 \hbar^2 (\hat{z} \hat{L}^2 + \hat{L}^2 \hat{z}) +\end{aligned}$$ + +At last, this brings us to the desired equation for $\comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\vu{r}}}$, +with $\vu{r} = (\hat{x}, \hat{y}, \hat{z})$. +</div> +</div> + +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: + +$$\begin{aligned} + 2 \hbar^2 \matrixel{f}{\vu{r} \hat{L}^2 \!\!+\! \hat{L}^2 \!\vu{r}}{i} + %\matrixel{\ell_f m_f}{\comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\vu{r}}}}{\ell_i m_i} + &= 2 \hbar^2 \big( \matrixel{f}{\vu{r} \hat{L}^2}{i} + \matrixel{f}{\hat{L}^2 \vu{r}}{i} \big) + \\ + &= 2 \hbar^2 \big( \hbar^2 \ell_i (\ell_i \!+\! 1) \matrixel{f}{\vu{r}}{i} + + \hbar^2 \ell_f (\ell_f \!+\! 1) \matrixel{f}{\vu{r}}{i} \big) + \\ + &= 2 \hbar^4 \big(\ell_f (\ell_f \!+\! 1) + \ell_i (\ell_i \!+\! 1)\big) \matrixel{f}{\vu{r}}{i} +\end{aligned}$$ + +And, likewise, the left-hand side becomes: + +$$\begin{aligned} + \matrixel{f}{\comm*{\hat{L}^2}{\comm*{\hat{L}^2}{\vu{r}}}}{i} + &= \matrixel{f}{\hat{L}^2 \comm*{\hat{L}^2}{\vu{r}}}{i} + - \matrixel{f}{\comm*{\hat{L}^2}{\vu{r}} \hat{L}^2}{i} + \\ + &= \hbar^2 \ell_f (\ell_f \!+\! 1) \matrixel{f}{\comm*{\hat{L}^2}{\vu{r}}}{i} + - \hbar^2 \ell_i (\ell_i \!+\! 1) \matrixel{f}{\comm*{\hat{L}^2}{\vu{r}}}{i} + \\ + &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \matrixel{f}{\comm*{\hat{L}^2}{\vu{r}}}{i} + \\ + &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) + \big( \matrixel{f}{\hat{L}^2 \vu{r}}{i} - \matrixel{f}{\vu{r} \hat{L}^2}{i} \big) + \\ + &= \hbar^4 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big)^2 \matrixel{f}{\vu{r}}{i} +\end{aligned}$$ + +Obviously, both sides are equal to each other, +leading to the following equation: + +$$\begin{aligned} + 2 \ell_f (\ell_f \!+\! 1) + 2 \ell_i (\ell_i \!+\! 1) + &= \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big)^2 +\end{aligned}$$ + +To proceed, we rewrite the right-hand side like so: + +$$\begin{aligned} + \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big)^2 + &= \big( \ell_f^2 - \ell_i^2 + \ell_f - \ell_i \big)^2 + \\ + &= \big( (\ell_f + \ell_i) (\ell_f - \ell_i) + (\ell_f - \ell_i) \big)^2 + \\ + &= (\ell_f + \ell_i)^2 (\ell_f - \ell_i)^2 + 2 (\ell_f + \ell_i) (\ell_f - \ell_i)^2 + (\ell_f - \ell_i)^2 + \\ + &= \big( (\ell_f + \ell_i)^2 + 2 (\ell_f + \ell_i) + 1 \big) (\ell_f - \ell_i)^2 + \\ + &= (\ell_f + \ell_i + 1)^2 (\ell_f - \ell_i)^2 +\end{aligned}$$ + +And then we do the same to the left-hand side, yielding: + +$$\begin{aligned} + 2 (\ell_f^2 + \ell_i^2 + \ell_f + \ell_i) + &= 2 \ell_f^2 + 2 \ell_i^2 + 2 \ell_f \ell_i - 2 \ell_f \ell_i + 2 \ell_f + 2 \ell_i + 1 - 1 + \\ + &= (\ell_f + \ell_i + 1)^2 + \ell_f^2 + \ell_i^2 - 2 \ell_f \ell_i - 1 + \\ + &= (\ell_f + \ell_i + 1)^2 + (\ell_f - \ell_i)^2 - 1 +\end{aligned}$$ + +The equation above has thus been simplified to the following form: + +$$\begin{aligned} + (\ell_f + \ell_i + 1)^2 + (\ell_f - \ell_i)^2 - 1 + &= (\ell_f + \ell_i + 1)^2 (\ell_f - \ell_i)^2 +\end{aligned}$$ + +Rearranging yields a product equal to zero, +so one or both of the factors must vanish: + +$$\begin{aligned} + 0 + &= (\ell_f + \ell_i + 1)^2 (\ell_f - \ell_i)^2 - (\ell_f + \ell_i + 1)^2 - (\ell_f - \ell_i)^2 + 1 + \\ + &= \big( (\ell_f + \ell_i + 1)^2 - 1 \big) \big( (\ell_f - \ell_i)^2 - 1 \big) +\end{aligned}$$ + +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 other, non-trivial option is therefore: + +$$\begin{aligned} + (\ell_f - \ell_i)^2 + = 1 +\end{aligned}$$ +</div> +</div> + + +## Superselection rule + +Selection rules need not always be about atomic electron transitions. +According to the **principle of indistinguishability**, +permutating 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. |