---
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:
$$\begin{aligned}
\boxed{
\Delta m = 0 \:\:\mathrm{or}\: \pm 1
}
\end{aligned}$$
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.
Meanwhile, for the total angular momentum $\ell$ we have the following:
$$\begin{aligned}
\boxed{
\Delta \ell = \pm 1
}
\end{aligned}$$
We start from the following relation
(which is already quite a chore to prove):
$$\begin{aligned}
\Comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}
= 2 \hbar^2 (\vu{r} \hat{L}^2 + \hat{L}^2 \vu{r})
\end{aligned}$$
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:
$$\begin{aligned}
\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}}
&= - 4 \hbar^2 \big( \hat{z} \hat{L}_x \hat{L}_z - \hat{x} \hat{L}_z^2 - i \hbar \hat{y} \hat{L}_z
+ \hat{y} \hat{L}_x \hat{L}_y - \hat{x} \hat{L}_y^2 + i \hbar \hat{z} \hat{L}_y \big) \\
&\qquad\qquad + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big)
\end{aligned}$$
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:
$$\begin{aligned}
\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}}
&= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 - \hat{x} \hat{L}^2 \big)
+ 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big)
\\
&= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 \big)
+ 2 \hbar^2 \big( \hat{L}^2 \hat{x} + \hat{x} \hat{L}^2 \big)
\end{aligned}$$
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 $\vu{r} = (\hat{x}, \hat{y}, \hat{z})$.
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:
$$\begin{aligned}
2 \hbar^2 \matrixel{f}{\vu{r} \hat{L}^2 \!\!+\! \hat{L}^2 \!\vu{r}}{i}
&= 2 \hbar^2 \big( \matrixel{f}{\vu{r} \hat{L}^2}{i} + \matrixel{f}{\hat{L}^2 \vu{r}}{i} \big)
\\
&= 2 \hbar^2 \big( \hbar^2 \ell_i (\ell_i \!+\! 1) \matrixel{f}{\vu{r}}{i}
+ \hbar^2 \ell_f (\ell_f \!+\! 1) \matrixel{f}{\vu{r}}{i} \big)
\\
&= 2 \hbar^4 \big(\ell_f (\ell_f \!+\! 1) + \ell_i (\ell_i \!+\! 1)\big) \matrixel{f}{\vu{r}}{i}
\end{aligned}$$
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:
$$\begin{aligned}
(\ell_f - \ell_i)^2
= 1
\end{aligned}$$
## 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 $m_i$:
$$\begin{aligned}
\boxed{
\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
= \delta_{\ell_f \ell_i} \delta_{m_f m_i} \matrixel{\ell_f}{|\hat{s}|}{\ell_i}
}
\end{aligned}$$
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.
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.