- #1
AxiomOfChoice
- 533
- 1
Suppose you have a self-adjoint operator [itex]A[/itex] on [itex]L^2(\mathbb R^n)[/itex] that has exactly one discrete, non-degenerate eigenvalue [itex]a[/itex] with associated normalized eigenfunction [itex]f_a[/itex]. What does the projection onto the associated eigenspace look like? My guess would be:
[tex]
P(f) = (f,f_a)f_a = f_a(x) \int_{\mathbb R^n} f(x) \cdot \overline{f_a(x)}\ dx.
[/tex]
This certainly satisfies [itex]P^2(f) = P(f)[/itex] and [itex]P(f_a) = f_a[/itex]...
[tex]
P(f) = (f,f_a)f_a = f_a(x) \int_{\mathbb R^n} f(x) \cdot \overline{f_a(x)}\ dx.
[/tex]
This certainly satisfies [itex]P^2(f) = P(f)[/itex] and [itex]P(f_a) = f_a[/itex]...