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is exp (-kx) an eigenfunction?
That's like asking "Is 23 an answer?" - it might be, but we have to know what the question is first.is exp (-kx) an eigenfunction?
In quantum mechanics, the set of possible wave functions depends on the boundary conditions. With the usual one-dimensional case in which space is the real numbers [itex]-\infty < x < +\infty[/itex], the boundary condition is that [itex]\int_{-\infty}^{+\infty} |\psi(x)|^2 dx < \infty[/itex]. In the case of [itex]e^{-kx}[/itex], this integral is not defined (because that function goes to infinity as [itex]x \rightarrow -\infty[/itex].is exp (-kx) an eigenfunction?
if the operator O acts on the function F, then the function will be mapped to another function which is the same function f times an eigenvalue E.OK, so what is the general relationship that defines the eigenfunctions and eigenvalues of an operator? If function f is an eigenfunction of operator O, with eigenvalue E, what relationship has to be true?
Correct!so if d/dx exp(-kx) = -k.exp(-kx) the eigenvalue E is -k and the function f remains exp(-kx)