Schrodinger's equation via Spectral Theorem [closed]

Question

How do you prove basic facts on the Schrodinger equation using the spectral theorem? More precisely, here is what I have in mind.

The version of the Spectral Theorem I am familiar with is the following.

Spectral Theorem for normal operators: Let $H$ be a Hilbert space and $T\in \mathcal{B}(H)$ be a normal operator. Then there is a unique resolution $P$ of the identity of $H$ over $\sigma(T)$ (the spectrum of $T$) such that $$ T=\int_{\sigma(T)} \lambda\, dP. $$

I would like a theorem like the following.

Theorem: Let $\psi$ satisfy the 1-D Schrodinger equation $$\left(-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} +u(x)\right)\psi=i\hbar \frac{\partial}{\partial t}\psi$$ where $u$ is some "nicely behaved" function, and let $f=\psi(x,0)$. Then $$(1)\qquad f=\sum_{\lambda} c_{\lambda}\phi_{\lambda}+\int_{\lambda} c_{\lambda}\phi_{\lambda}$$ where the first sum is over the discrete spectrum, the integral is over the continuous spectrum, $\phi_{\lambda}$ is the eigenfunction for $\lambda$, and $c_{\lambda}=\int f\overline{\phi_{\lambda}}\,dx$. Therefore, $$\psi=\sum_{\lambda} c_{\lambda}\phi_{\lambda}e^{-\frac{i\lambda}{\hbar}t}+\int_{\lambda} c_{\lambda}\phi_{\lambda}e^{-\frac{i\lambda}{\hbar}t}.$$

What is the mathematically correct way to state the above, and how can it be proved?

For instance, if $u_{\infty}=\lim_{x\to \pm \infty} u(x)$, for generic $u$, the discrete spectrum would consist of some finite set in $(\min u, u_{\infty})$ and the continuous spectrum would be $(u_{\infty},\infty)$.

Two issues are the following.

1. The Spectral Theorem as stated above only works for bounded operators, and the operator is not bounded. (We probably also need growth conditions on $f$.) I know there is a Spectral Theorem for unbounded operators but it seems to require more technicalities; if that is the way to go, how are those technicalities satisfied in this case?
2. The theorem doesn't describe what the measure $dP$ is explicitly, so how do I know that the resolution of the identity gives (1)?

Answer 1

There is a spectral theorem for unbounded operators. It is not simple. But if you have an unbounded operator, and want to use the spectral theorem, you have to use it:-) The difficult thing is to actually define the operator (using the boundary conditions) and prove that it is self-adjoint.

Verification of all facts you mentioned is not simple, but there are many good books on this, some of them are:

MR1186643 Berezin, F. A.; Shubin, M. A. The Schrödinger equation. Kluwer , Dordrecht, 1991. xviii+555.

Reed and Simon, Methods of modern mathematical physics.

Answer 2

At first you need to understand the spectral theorem for self-adjoint unbounded operators. There are many good textbooks: Yosida, Kato, Reed & Simon, Lax. Think of (1) as solution by separation of variables. But, there is no easy way to make it rigorous!