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?

  • $\begingroup$ $X$ 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. $\endgroup$ – Will Russell Sep 24 '18 at 1:30
  • $\begingroup$ 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. $\endgroup$ – Will Russell Sep 24 '18 at 12:16
  • $\begingroup$ 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 $x\psi' + 2\psi = 0$. $\endgroup$ – Winther Sep 24 '18 at 13:57
  • $\begingroup$ 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. $\endgroup$ – Will Russell Sep 24 '18 at 23:05
  • $\begingroup$ If it's not in the book take a look here: en.wikipedia.org/wiki/Position_operator $\endgroup$ – Winther Sep 24 '18 at 23:08

Your operator X may be thought of as a constant, since it does nothing on "numbers", which include the parameter x, so it commutes with everything just like a constant.

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.


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