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# is (x*d/dx)^2 an eigenfunction of exp(-ikx)/x ?

according to wolframalpha it should be an eigenfunction, however when I actually work it out i end up with

((2*k^2*x^2 + 1)*exp(-ikx))/2x^2

by the way, k is just a constant

### 2 Answers

- Elizabeth MLv 78 years agoFavorite Answer
According to Wikipedia f(x) is an eigenfunction of the linear operator A

if Af=δf where δ is a constant (an eigenvalue). (x*d/dx)^2 seems to be the operator

and exp(-ikx)/x the eigenfunction. I will use D for d/dx.

D(exp(-ikx)/x)=(-1/x^2)exp(-ikx) -(ik/x)exp(-ikx),

so xD(exp(-ikx)/x)= -(1/x)exp(-ikx) -ikexp(-ikx)=( - 1/x -ik)exp(-ikx)

(xD)^2(exp(-ikx)/x)= xD[(-1/x-ik)exp(-ikx)]

=x[(1/x^2)exp(-ikx)-ik(-1/x-ik)exp(-ikx)]

=exp(-ikx)/x+(ik-k^2)exp(-ikx)