Hermiticity Question

• November 3rd 2008, 08:49 AM
joker_900
Hermiticity Question

Consider the set of functions {f(x)} of the real variable x defined on the interval −inf < x < inf
that go to zero faster than 1/x for x --> ±inf, i.e.
lim(x-->±inf) xf(x) = 0
For unit weight function, determine which of the following linear operators is Hermitian when acting upton {f(x)}:
(a) d/dx + x
(b) −i d/dx + x^2
(c) ix d/dx
(d) i d3/dx3

I tried using definition of a hermitian operator as <u, Hv> = (<v, Hu>)* but I don't get what u and v are, unless they are just general members of the set {f(x)} in which case I'm still confused!

Thanks!
• November 3rd 2008, 10:35 AM
Opalg
Quote:

Originally Posted by joker_900

Consider the set of functions {f(x)} of the real variable x defined on the interval −inf < x < inf
that go to zero faster than 1/x for x --> ±inf, i.e.
lim(x-->±inf) xf(x) = 0
For unit weight function, determine which of the following linear operators is Hermitian when acting upon {f(x)}:
(a) d/dx + x
(b) −i d/dx + x^2
(c) ix d/dx
(d) i d3/dx3

I tried using definition of a hermitian operator as <u, Hv> = (<v, Hu>)* but I don't get what u and v are, unless they are just general members of the set {f(x)} in which case I'm still confused!

In these example, the space V on which H acts is the set of functions f(x) on the real line for which $\textstyle\lim_{|x|\to\infty} xf(x) = 0$. (They had better be differentiable functions for the question to make sense, though this is not specified.) The definition of H being hermitian is: $\langle Hf,g\rangle = \langle f,Hg\rangle$, for all f, g in V. The inner product $\langle f,g\rangle$ means the integral (with respect to the unit weight function) of the product of f(x) and the complex conjugate of g(x).

In part (a), for example, the operator H is differentiation plus multiplication by x, so that $Hf(x) = f'(x) + xf(x)$. You have to check whether the integrals of $\bigl(f'(x) + xf(x)\bigr)g(x)^*$ and $f(x)\bigl(g'(x) + xg(x)\bigr)^*$ (where the asterisk denotes the complex conjugate) are equal. It looks as though integration by parts will figure somewhere in the calculation.