# How do I compute the eigenfunctions of an operator that contains another operator?

Given the operator $$A = (X\frac{d}{dx}+2)$$, where $$X$$ is a linear operator, how can I find the eigenfunction of $$A$$ corresponding to a zero eigenvalue?

In general, this is just a matter of solving the differential equation for $$AF(x) = 0$$, however, in this case, that leaves me with the differential equation $$X \frac{d}{dx} F(x) + 2 F(x) = 0$$, and I'm just not really sure what to do with that $$X$$ operator.

Anyone out there that can get me past this step?

• $X$ isn't specified in the problem, though we could probably assume that it is the position operator $i\hbar\frac{d}{dp_x}$. Obviously a general solution without this assumption is preferred if it is possible. – Will Russell Sep 24 '18 at 1:30
• Yep, that's all the details. This is an exercise from chapter 2 of Quantum Mechanics Concepts and Applications by Nouredine Zettili. As I look at it more, I'm wondering if I can just treat the X operator as a constant for the purpose of solving the differential. – Will Russell Sep 24 '18 at 12:16
• I found the problem in that textbook. $X$ is the position operator! If you included the rest it would be clear as you are asked to compute commutators like $[A,X]$ and $[A,P]$ in the next subproblem. When acting on a real space wavefunction $X\psi = x\psi$ (you are quoting the momentum space representation of it above) so you are left with the ODE $x\psi' + 2\psi = 0$. – Winther Sep 24 '18 at 13:57
• I apologize, I thought I had provided enough context for the problem without complicating it with unnecessary minutia, and clearly didn't provide enough. This is actually much simpler than I was trying to make it, thank you for that. If you don't mind a followup "why?" type question, can you explain why you are allowed to do this when $X$ is acting on the derivative of $\psi$ rather than on $\psi$ itself? I'm assuming that there is an operator rule that I somehow missed. – Will Russell Sep 24 '18 at 23:05
• If it's not in the book take a look here: en.wikipedia.org/wiki/Position_operator – Winther Sep 24 '18 at 23:08

It is then evident that $$X\frac{d}{dx} F(x,X)= -2 F(x,X)$$ is solved by $$F(x,X)=c e^{{-2x}X^{-1}} ,$$ where $$X^{-1}$$ is the inverse operator to X.