---
title: "Selection rules"
sort_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:

$$\begin{aligned}
    \boxed{
        \Delta m = 0 \:\:\mathrm{or}\: \pm 1
    }
\end{aligned}$$


{% include proof/start.html id="proof-dipole-m" -%}
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.
{% include proof/end.html id="proof-dipole-m" %}


Meanwhile, for the total angular momentum $$\ell$$ we have the following:

$$\begin{aligned}
    \boxed{
        \Delta \ell = \pm 1
    }
\end{aligned}$$


{% include proof/start.html id="proof-dipole-l" -%}
We start from the following relation
(which is already quite a chore to prove):

$$\begin{aligned}
    \Comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}
    = 2 \hbar^2 (\vu{r} \hat{L}^2 + \hat{L}^2 \vu{r})
\end{aligned}$$


{% include proof/start.html id="proof-dipole-l-commutator" -%}
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 $$\vu{r} = (\hat{x}, \hat{y}, \hat{z})$$.
{% include proof/end.html id="proof-dipole-l-commutator" %}


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:

$$\begin{aligned}
    (\ell_f - \ell_i)^2
    = 1
\end{aligned}$$
{% include proof/end.html id="proof-dipole-l" %}



## 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 $$m_i$$:

$$\begin{aligned}
    \boxed{
        \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
        = \delta_{\ell_f \ell_i} \delta_{m_f m_i} \matrixel{\ell_f}{|\hat{s}|}{\ell_i}
    }
\end{aligned}$$


{% include proof/start.html id="proof-rotation-scalar" -%}
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.
{% include proof/end.html id="proof-rotation-scalar" %}


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.