Ladder Operators of Lie Algebras
Definition
Let a Lie algebra $\mathfrak{g}$ be given. For $H\in \mathfrak{g}$, an element $E_{\alpha} \in \mathfrak{g}\setminus\left\{ 0 \right\}$ that satisfies the following relation for some nonzero scalar $\alpha$ is called a ladder operator of $H$.
$$ [H, E_{\alpha}] = \alpha E_{\alpha}, \quad (\alpha \ne 0,\ E_{\alpha} \ne 0) \tag{1} $$
Explanation
It is called an operator because the place where it actually acts meaningfully is a representation space of $\mathfrak{g}$. It is called a ladder operator because, given a representation $\pi$, applying $\pi(E_{\alpha})$ to an eigenvector of $\pi(H)$ moves it over to another eigenvector. So, strictly speaking, it is $\pi(E_{\alpha})$ that is the ladder operator. The term ladder operator is especially widely used in quantum mechanics.
From here on, when dealing with the action on the representation space, we omit $\pi$ for convenience as follows.
$$ H = \pi(H), \quad E_{\alpha} = \pi(E_{\alpha}) $$
Let $\lambda$ be an eigenvalue of $H$, and let $v_{\lambda}$ be a corresponding eigenvector.
$$ Hv_{\lambda} = \lambda v_{\lambda} $$
Then, when $E_{\alpha}v_{\lambda}$ is nonzero, it can be shown as follows that it is an eigenvector corresponding to the eigenvalue $\lambda+\alpha$.
$$ \begin{align*} H(E_{\alpha}v_{\lambda}) &= (HE_{\alpha})v_{\lambda} \\ &= (E_{\alpha}H + \alpha E_{\alpha})v_{\lambda} \\ &= (\lambda E_{\alpha} + \alpha E_{\alpha})v_{\lambda} \\ &= (\lambda + \alpha)(E_{\alpha}v_{\lambda}) \end{align*} $$
In particular, when $\alpha$ is real, if $\alpha > 0$ the eigenvalue keeps going up, so $E_{\alpha}$ is called a raising operator. Conversely, if $\alpha \lt 0$, it is called a lowering operator.
Eigenvectors of $\ad_{H}$
Meanwhile, expressing $(1)$ in terms of the adjoint map of the Lie algebra gives the following.
$$ [H, E_{\alpha}] = \ad_{H} (E_{\alpha}) = \alpha E_{\alpha} $$
That is, $E_{\alpha}$ satisfying $(1)$ means that $E_{\alpha}$ is an eigenvector of the linear transformation $\ad_{H}$ corresponding to the eigenvalue $\alpha$. One must be careful not to confuse the two: $E_{\alpha}$ is an eigenvector belonging to the vector space $\mathfrak{g}$, whereas $v_{\lambda}$, which appeared in the explanation above, belongs to the vector space $V$.
Ladder Operators in the Opposite Direction
Let $\mathfrak{g}$ be a finite-dimensional semisimple Lie algebra. Let $E_{\alpha} \in \mathfrak{g}$ be an eigenvector of $\ad_{H}$ corresponding to the eigenvalue $\alpha$. Since $\mathfrak{g}$ is 🔒(26/10/04)semisimple, the Killing form $B$ is nondegenerate. Let us express the property of the Killing form $B(\ad_{H}X, Y) = - B(X, \ad_{H}Y)$ in terms of matrices. If $G$ is the matrix corresponding to $B$ and $A$ is the matrix corresponding to $\ad_{H}$, this property is expressed as follows.
$$ A^{\mathsf{T}} G = - GA $$
Since $B$ is nondegenerate, $G$ is invertible. Therefore, as below, $A^{\mathsf{T}}$ and $-A$ are similar.
$$ G^{-1}A^{\mathsf{T}} G = -A $$
Denote the set of eigenvalues of $A$ by $\sigma (A)$. Transposition does not change eigenvalues. And since similar matrices have the same eigenvalues, the following holds.
$$ \sigma(A) = \sigma(A^{\mathsf{T}}) = \sigma(-A) = -\sigma(A) $$
Therefore, if $\alpha$ is an eigenvalue of $\ad_{H}$, then $-\alpha$ is also an eigenvalue.
$$ \alpha \in \sigma(\ad_{H}) \implies -\alpha \in \sigma(\ad_{H}) $$
Hence there exists an eigenvector $E_{-\alpha} \in \mathfrak{g}$ of $\ad_{H}$ corresponding to the eigenvalue $-\alpha$ such that the following holds.
$$ [H, E_{-\alpha}] = \ad_{H} E_{-\alpha} = -\alpha E_{-\alpha} $$
Therefore, applying $E_{-\alpha}E_{\alpha}$ or $E_{\alpha}E_{-\alpha}$ to an eigenvector $v_{\lambda} \in V_{\lambda}$ brings it back to the original eigenspace $V_{\lambda}$. Note here that it returns to the original eigenspace; this does not mean the eigenvector itself is preserved. If $\dim V_{\lambda} \ge 2$, it may possibly turn into a different eigenvector.
$$ \begin{align*} E_{-\alpha}E_{\alpha} v_{\lambda} &= v_{\lambda} &\mathsf{(X)} \\ E_{-\alpha}E_{\alpha} v_{\lambda} &\in V_{\lambda} &\mathsf{(O)} \end{align*} $$
Stated rigorously using the representation, it is as follows.
$$ \pi(E_{\alpha}) : V_{\lambda} \to V_{\lambda+\alpha}, \quad \pi(E_{-\alpha}) : V_{\lambda+\alpha} \to V_{\lambda} $$
$$ \pi(E_{-\alpha})\pi(E_{\alpha})(V_{\lambda}) \subset V_{\lambda} $$
In Finite Dimensions
Suppose $V$ is finite-dimensional. Applying $E_{\alpha}$ to an eigenvector $v_{\lambda}$ of $H$, it seems we could keep obtaining eigenvectors corresponding to $\lambda + \alpha$, $\lambda + 2\alpha$, $\cdots$. This is because repeating the computation above shows that the following holds for each $k = 0, 1, 2, \cdots$.
$$ H \left( (E_{\alpha})^{k} v_{\lambda} \right) = (\lambda + k\alpha) \left( (E_{\alpha})^{k} v_{\lambda} \right) $$
Since $\alpha \ne 0$, the values $\lambda$, $\lambda + \alpha$, $\lambda + 2\alpha$, $\cdots$ are all distinct. However, eigenvectors corresponding to distinct eigenvalues are linearly independent, and since $V$ is finite-dimensional, we cannot obtain such linearly independent eigenvectors indefinitely. In other words, if $(E_{\alpha})^{k}v_{\lambda} \ne 0$ for all $k$, there would be infinitely many linearly independent vectors in $V$, which is a contradiction. Therefore, there exists a positive integer $m$ such that the following holds.
$$ (E_{\alpha})^{m} v_{\lambda} = 0 $$
Choose the smallest such $m$. Then, by the minimality of $m$, $w^{+} = (E_{\alpha})^{m-1}v_{\lambda}$ is nonzero and satisfies the following.
$$ E_{\alpha}w^{+} = (E_{\alpha})^{m} v_{\lambda} = 0, \qquad Hw^{+} = \left( \lambda + (m-1)\alpha \right) w^{+} $$
That is, as we climb the ladder, we must eventually reach a last eigenvector from which we can no longer move to a new eigenvector and which is sent to the zero vector. As seen in the previous section, if $\mathfrak{g}$ is semisimple then $-\alpha$ is also an eigenvalue of $\ad_{H}$, so the same argument applies verbatim to the opposite-direction operator $E_{-\alpha}$. Likewise, choose the smallest positive integer $\ell$ such that $(E_{-\alpha})^{\ell} v_{\lambda} = 0$, and let $w^{-} = (E_{-\alpha})^{\ell - 1} v_{\lambda}$. Then it satisfies the following.
$$ E_{-\alpha}w^{-} = (E_{-\alpha})^{\ell} v_{\lambda} = 0, \qquad Hw^{-} = \left( \lambda - (\ell-1)\alpha \right) w^{-} $$
In the end, the ladder of eigenvectors obtained by starting from $v_{\lambda}$ and repeatedly applying $E_{\alpha}$ or $E_{-\alpha}$, respectively, terminates at both ends, and the vectors at the two ends, $w^{+}$ and $w^{-}$, are sent to $0$ by $E_{\alpha}$ and $E_{-\alpha}$, respectively.
$$ \begin{align*} &v_{\lambda} \xrightarrow{\ E_{\alpha}\ } E_{\alpha}v_{\lambda} \xrightarrow{\ E_{\alpha}\ } \cdots \xrightarrow{\ E_{\alpha}\ } w^{+} \xrightarrow{\ E_{\alpha}\ } 0 \\ 0 \xleftarrow{\ E_{-\alpha}\ } w^{-} \xleftarrow{\ E_{-\alpha}\ } \cdots \xleftarrow{\ E_{-\alpha}\ } E_{-\alpha}v_{\lambda} \xleftarrow{\ E_{-\alpha}\ }\enspace &v_{\lambda} \end{align*} $$