Relationship between simultaneous eigenfunctions of angular momentum and ladder operators
Summary
Let’s denote the angular momentum operators $L^{2}$ and $L_{z}$, and their eigenvalues as $\ell(\ell+1)\hbar^{2}$ and $m\hbar$. The normalized simultaneous eigenfunctions corresponding to each eigenvalue are referred to as $\ket{\ell, m}$.
$$ \begin{align*} L^{2} \ket{\ell, m} &= \ell(\ell+1)\hbar^{2}\ket{\ell, m} \\ L_{z}\ket{\ell, m} &= m\hbar\ket{\ell, m} \end{align*} $$
For the ladder operators for angular momentum $L_{\pm}$ and the eigenfunctions $\ket{\ell, m}$, the following relational expression holds.
$$ \begin{align*} L_{+}\ket{\ell, m} &= \sqrt{(l-m)(l+m+1)}\hbar\ket{\ell, m+1} \\ L_{-}\ket{\ell, m} &= \sqrt{(l+m)(l-m+1)}\hbar\ket{\ell, m-1} \end{align*} $$
Explanation
Applying the ladder operators $L_{\pm}$ to $\ket{\ell, m}$ results in the eigenvalue equation for $$ \begin{align*} L_{+}\ket{\ell, m} &= \sqrt{(l-m)(l+m+1)}\hbar\ket{\ell, m+1} \\ L_{-}\ket{\ell, m} &= \sqrt{(l+m)(l-m+1)}\hbar\ket{\ell, m-1} \end{align*} $$, increasing (or decreasing) the eigenvalue by $\hbar$.
$$ L_{z} \ket{\ell, m} = m\hbar \ell\ket{\ell, m} \quad \implies L_{z}L_{+}\ket{\ell, m} = (m+1) \hbar \ket{\ell, m+1} $$
Therefore, $L_{+}\ket{\ell, m}$ is also an eigenfunction corresponding to the eigenvalue $(m+1)\hbar$. This theorem tells us about the relational expression between $L_{+}\ket{\ell, m}$ and $\ket{\ell, m+1}$, where the normalized state among the eigenfunctions corresponding to $(m++1)$ is referred to as $\ket{\ell, m+1}$.
Proof
First, let’s start from the eigenvalue equation of $L_{z}$. The possible values of $m$ differ by $1$, thus the following holds.
$$ \begin{align*} L_{z}\ket{\ell, m} &= m\hbar \ket{\ell, m} \\ L_{z}\ket{\ell, m+1} &= (m+1) \hbar \ket{\ell, m+1} \end{align*} $$
Additionally, since $L_{+}$ increases the eigenvalue of $L_{z}$ by $\hbar$,
$$ L_{z} L_{+} \ket{\ell, m} = (m+1) \hbar L_{+} \ket{\ell, m} $$
Thus, $\ket{\ell, m+1}$ is also an eigenfunction corresponding to the eigenvalue $(m + 1)\hbar$, and $L_{+} \ket{\ell, m}$ is also an eigenfunction corresponding to the eigenvalue $(m + 1)\hbar$. Therefore, for some constant $C_{+}$, the following holds.
$$ L_{+}\ket{\ell, m}=C_{+}\ket{\ell, m+1} \tag{1} $$
Similarly, we obtain the following equation for $L_{-}$.
$$ L_{-}\ket{\ell, m} = C_{-}\ket{\ell, m-1} \tag{2} $$
Note that $(1)$ and $(2)$ are not eigenvalue equations. To determine the value of $C_{+}$, we compute the inner product of $L_{+}\ket{\ell, m}$ with itself.
$$ \begin{align*} \bra{\psi}L_{+}^{\ast}L_{+}\ket{\psi} &= \bra{\ell, m+1}C_{+}^{\ast}C_{+}\ket{\ell, m+1} \\ &= \left| C_{+} \right|^{2} \braket{\ell, m+1 | \ell, m+1} \\ &= \left| C_{+} \right|^{2} \end{align*} $$
$$ L_{-}L_{+} = L^{2} - L_{z}^{2} - \hbar L_{z} $$
The left-hand side of the above equation yields the following upon direct calculation.
$$ \begin{align*} \bra{\psi}L_{+}^{\ast}L_{+}\ket{\psi} &= \bra{\ell, m} (L_{+})^{\ast}L_{+} \ket{\ell, m} \\ &= \bra{\ell, m} L_{-}L_{+} \ket{\ell, m} \\ &= \bra{\ell, m} L^{2} -{L_{z}}^{2} - \hbar L_{z} \ket{\ell, m}\\ &= \left[ l(l+1)\hbar ^{2} -m^{2}\hbar^{2} -m\hbar^{2} \right]\braket{\ell, m | \ell, m} \\ &= \hbar^{2} (\ell^{2}+\ell-m^{2}-m) \\ &= \hbar^{2} [(\ell^{2}-m^{2})+(\ell-m)] \\ &= \hbar^{2} (\ell-m)(\ell+m+1) \end{align*} $$
Therefore, we obtain the following.
$$ C_{+} = \hbar\sqrt{(\ell-m)(\ell+m+1)} $$
Similarly, calculating $C_{-}$ yields the following.
$$ C_{-} = \hbar\sqrt{(\ell+m)(\ell-m+1)} $$