Showing $K_\pm$ are raising/lowering operators 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.
 A: 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 $(\star)$.
A: Are $p$ and $q$ the standard momentum and position operators in $L^2(\mathbb R)$? If the answer is positive, then:
$$K_\pm := K_1 \pm iK_2 = \frac{1}{2}\left(\frac{1}{\sqrt{2}}(p\pm iq) \right)^2\:.$$
In other words, introducing the standard operators $a = \frac{1}{\sqrt{2}}(p- iq) $ and $a^\dagger = \frac{1}{\sqrt{2}}(p+ iq) $ for the harmonic oscillator:
$$K_+ = \frac{1}{2}a^\dagger a^\dagger \:,\quad K_- =  \frac{1}{2}aa\:.$$
Similarly:
$$J_3  =  \frac 14 (p^2+q^2) = \frac{1}{2} \left(a^\dagger a + a a^\dagger\right) = \frac{1}{2} (a^\dagger a + \frac{1}{2}I)\:.$$
This last identity implies that, in the eigenvalues equation $$J_3\psi_m = m\psi_m\:, $$ it must be $m= \frac{1}{2}(n+1/2)$ for $n=0,1,2\ldots$ and $$\psi_m  = |(4m-1)/2\rangle\:,$$ where $|n\rangle$ is the standard basis of the harmonic oscillator with $n=0,1,2,\ldots$.
Let us come to the action of $K_\pm$ on the vectors $\psi_m$.
$$K_+ \psi_m = \frac{1}{2}a^\dagger a^\dagger |(4m-1)/2\rangle = \frac{1}{2} a^\dagger \sqrt{(4m-1)/2+1}|(4m-1)/2+1\rangle $$
$$K_+ \psi_m = \frac{1}{2} \sqrt{\frac{4m+3}{2}}\sqrt{\frac{4m+1}{2}}|\frac{4m+3}{2}\rangle = \frac{1}{4}\sqrt{(4m+3)(4m+1)}\psi_{m+1}\:.$$
We have found that:
$$K_+ \psi_m =  \frac{1}{2}\sqrt{\left(m+\frac{3}{4}\right)\left(m+\frac{1}{4}\right)}\psi_{m+1}\:.$$
Similarly
$$K_- \psi_m = \frac{1}{2}a a |(4m-1)/2\rangle = \frac{1}{2} a \sqrt{(4m-1)/2}|(4m-1)/2-1\rangle $$
$$K_- \psi_m = \frac{1}{2} \sqrt{\frac{4m-1}{2}}\sqrt{\frac{4m-3}{2}}|\frac{4m-5}{2}\rangle = \frac{1}{4}\sqrt{(4m-1)(4m-3)}\psi_{m-1}\:,$$
so that:
$$K_- \psi_m = \frac{1}{2}\sqrt{\left(m-\frac{1}{4}\right)\left(m-\frac{3}{4}\right)}\psi_{m-1}\:,$$
where $K_- \psi_m=0$ if $m=1/4,3/4$.
