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.

  • $\begingroup$ I assume your $K_\pm$ are raising/lowering operators of angular momentum? In that case, given some state $|\psi\rangle$ try to find out the angular momentum of $K_\pm|\psi\rangle$ $\endgroup$
    – Danu
    Feb 13, 2014 at 17:44
  • $\begingroup$ This question is not entirely well-posed unless you provide the commutation relations you are assuming between the $K$s or the $p$s and $q$s. $\endgroup$ Feb 14, 2014 at 16:38

2 Answers 2


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 $(\star)$.

  • $\begingroup$ Thanks! The previous part of the problem was to figure out the commutator relations. I never even though of acting $J_3$ on the operator. $\endgroup$ Feb 13, 2014 at 19:48
  • $\begingroup$ @yankeefan11 Sure thing. Yeah it's quite a clever little construction. $\endgroup$ Feb 13, 2014 at 19:49
  • $\begingroup$ So can we actually operate on a state with $K_\pm$? The next part is to show there is a lowest eigenvalue such that $K_-|m\rangle=0$ $\endgroup$ Feb 13, 2014 at 19:55
  • $\begingroup$ @yankeefan11 Yes you can certainly operate on a given state with $K_\pm$. $\endgroup$ Feb 13, 2014 at 19:58
  • 1
    $\begingroup$ @yankeefan11 That sounds like a good separate question ;) But you should really try to think about it for a while. $\endgroup$ Feb 13, 2014 at 20:04

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:

$$K_+ \psi_m = \frac{1}{2}\sqrt{\left(m+\frac{3}{4}\right)\left(m+\frac{1}{4}\right)}\psi_{m+1}\:.$$


$$K_- \psi_m = \frac{1}{2}a a |(4m-1)/2\rangle = \frac{1}{2} a \sqrt{(4m-1)/2}|(4m-1)/2-1\rangle $$

$$K_- \psi_m = \frac{1}{2} \sqrt{\frac{4m-1}{2}}\sqrt{\frac{4m-3}{2}}|\frac{4m-5}{2}\rangle = \frac{1}{4}\sqrt{(4m-1)(4m-3)}\psi_{m-1}\:,$$

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 $m=1/4,3/4$.

  • $\begingroup$ Yes, the OP must have meant the definition with $K_{\pm}= K_1 \pm i K_2$ (and one line later in the question it is used like that). I correct this in the question now. $\endgroup$ Feb 13, 2014 at 21:43
  • $\begingroup$ In accordance with our homework policy, I'm temporarily deleting this. $\endgroup$
    – Qmechanic
    Feb 13, 2014 at 21:47

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