Categories: Physics, Quantum mechanics.

# 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{\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):

$\begin{gathered} \boxed{ \matrixel{\ell_f m_f}{\hat{V}_{z}}{\ell_i m_i} = C^{\ell_i \: 1 \: \ell_f}_{m_i \: 0 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i} } \\ \boxed{ \matrixel{\ell_f m_f}{\hat{V}_{+}}{\ell_i m_i} = - \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: 1 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i} } \\ \boxed{ \matrixel{\ell_f m_f}{\hat{V}_{-}}{\ell_i m_i} = \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: -1 \:m_f} \matrixel{\ell_f}{|\hat{V}}{|\ell_i} } \end{gathered}$