Recent content by HeisenbergsDog

If ##U,V## are finite dimensional inner product spaces and ##T:U\to V## a linear map, then by definition, ##T^\dagger: V \to U##. Thus if ##T=T^\dagger## then we must have ##U=V##. Suppose ##U## is a subspace of ##V##, then we may have <v,Tu> = <Tv,u> for all ##u,v \in U## so that ##T^\dagger...

Ok, so thank you Svein for making me understand this a little better. I'm not quite there yet, though. In linear algebra (with finite dimensional vector spaces) we define an endomorphism to be self-adjoint if <u,Lv> = <Lu,v>. It doesn't make sense for an operator ##T: U\to V## to be...

I see. But ##V \subset L_2[a,b]## only if ##a,b## are finite, right? That is, we get into trouble if the interval is all of ##\mathbb{R}##.

Oh ok. So ##\hat{L}## is in general not an endomorphism. I thought the term "linear operator" on ##V## was the same as endomorphism on ##V##. That's the definition I've seen on wikipedia, and in Axler's linear algebra book, among others. Is the same true for "operators" in quantum mechanics...

This I can buy. Perfectly good vector space. Let's call it ##V##. Here I get into trouble. Let ##u \in V##. Then ##u## is twice differentiable and ##u(a)=u(b)=0##. We have no other restrictions on ##u##. Now let ##\hat{L}## be defined as above. If ##\hat{L}## were to be an operator, then we...

Consider a general Sturm-Liouiville problem ##[p(x) y'(x) ]' + [q(x) + \lambda \omega (x) ] y(x) = 0 ## ##\quad \Leftrightarrow \quad \hat{L} y(x) = \lambda \omega(x) y(x) \quad \text{with} \quad \hat{L} y(x) = -[p(x) y'(x)]' - q(x) y(x) ## where ##p(x), p'(x), q(x), \omega(x)## are...

Thank you!

Hello fellow physicists/physics students! Im in my 3rd year of university studying physics. Finally took the time to register here.