How Does Orthogonal Projection Work in L^2 Spaces with a Self-Adjoint Operator?

[AxiomOfChoice]

Suppose you have a self-adjoint operator $A$ on $L^2(\mathbb R^n)$ that has exactly one discrete, non-degenerate eigenvalue $a$ with associated normalized eigenfunction $f_a$. What does the projection onto the associated eigenspace look like? My guess would be:

$$P(f) = (f,f_a)f_a = f_a(x) \int_{\mathbb R^n} f(x) \cdot \overline{f_a(x)}\ dx.$$

This certainly satisfies $P^2(f) = P(f)$ and $P(f_a) = f_a$...

 

[micromass]

Suppose you have a self-adjoint operator $A$ on $L^2(\mathbb R^n)$ that has exactly one discrete, non-degenerate eigenvalue $a$ with associated normalized eigenfunction $f_a$. What does the projection onto the associated eigenspace look like? My guess would be:

$$P(f) = (f,f_a)f_a = f_a(x) \int_{\mathbb R^n} f(x) \cdot \overline{f_a(x)}\ dx.$$

This certainly satisfies $P^2(f) = P(f)$ and $P(f_a) = f_a$...

You still need to check that $P$ is self-adjoint. But this is true.

So indeed, the projection operator has the form you indicate.

In physics texts, they denote this operator often by $\left|f_a\right\rangle \left\langle f_a\right|$. In math texts, it is denoted by $f_a\otimes f_a$.

 

[AxiomOfChoice]

You still need to check that $P$ is self-adjoint.

I'd only need to do this if I were interested in showing $P$ was an orthogonal projection, right? Is that somehow crucial here? If so, I guess I don't see why.

 

[micromass]

I'd only need to do this if I were interested in showing $P$ was an orthogonal projection, right? Is that somehow crucial here? If so, I guess I don't see why.

Two remarks on this:

1) If you are working in a Hilbert space and if you say the word projection, then people will automatically assume orthogonal projections. The word "projection operator" in Hilbert space implies for the most people that it is orthogonal. So although you are right, you should specify when you are working with other kind of projections.

2) You said in your post

What does the projection onto ...

The word "the" indicates that the projection is unique. But if you allow non-orthogonal projections, then this is not the case. There are many projections that do the job.

You are correct though, and I'm just nitpicking. But I feel that this was important enough to mention anway.

 

[AxiomOfChoice]

The word "the" indicates that the projection is unique. But if you allow non-orthogonal projections, then this is not the case. There are many projections that do the job.

This is interesting. My intuition tells me that whenever you're considering any kind of eigenvalue problem for a (bounded, unbounded, self-adjoint, non-self-adjoint, etc.) linear operator in a Hilbert space, then for any eigenvalue there is an associated eigenspace and (at least) one projection onto that space. But is it *always* true that the projection has to be an orthogonal projection?

I guess this could be a very deep question with a textbook-chapter-length answer, or not very deep at all...

 

[micromass]

Yes, on any closed subspace of the Hilbert space, there is an orthogonal projection.

So any eigenspace of a bounded operator admits an orthogonal projection.