In this post, I have the following operators defined: $$K_1=\frac 14(p^2-q^2)$$ $$K_2=\frac 14 (pq+qp)$$ $$J_3 = \frac 14 (p^2+q^2)$$ I am given $ J_3|m\rangle = m|m\rangle$ and asked to show that $K_\pm \equiv K_1 \pm i K_2$ are ladder operators.
My approach (raising operator): $$K_+|m\rangle=K_1|m\rangle+iK_2|m\rangle$$ $$=K_1|m\rangle+[J_3,K1]|m\rangle$$ $$=K_1|m\rangle+ (J_3K_1-K_1J_3)|m\rangle$$ $$=K_1|m\rangle+J_3K_1|m\rangle-K_1m|m\rangle$$
First off, I'm unsure if this is the correct approach, and then I'm also lost on what to do next.
Answer
Here's the basic idea behind ladder operators in a bit of generality.
Let's say that I have a self-adjoint operator $J$ on the Hilbert space $\mathcal H$ of a given system, and suppose that $\{|m\rangle\}$ were an orthonormal basis for $\mathcal H$ consisting of eigenvectors of $J$, namely \begin{align} J|m\rangle = m|m\rangle. \end{align} Now, suppose you were to also find an operator $O_+$ that has the following commutation relation with $J$: \begin{align} [J,O_+] = cO_+ \tag{$\star$} \end{align} for some number $c$, then notice that an interesting thing happens when we apply $O_+$ to the states $|m\rangle$; \begin{align} J(O_+|m\rangle) &= (O_+ J +[J,O_+])|m\rangle\\ &= (O_+J+cO_+)|m\rangle \\ &= (m+c)(O_+|m\rangle). \end{align} In other words, $O_+|m\rangle$ is an eigenvector of $J$ with eigenvalue $m+c$, so $O_+$ raises the eigenvalues of a given state by $c$.
In your case, $J_3$ is analogous to $J$, and you simply need to show that $K_\pm$ have commutation relations with $J_3$ that are analogous to $(\star)$.
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