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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
- 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.
so xD(exp(-ikx)/x)= -(1/x)exp(-ikx) -ikexp(-ikx)=( - 1/x -ik)exp(-ikx)
- Michael joshuaLv 68 years ago
my heads just exploded