strangerep said:
It depends on the details of the operator. This opens a large can of worms.
Yes, convergence issues are critical -- one must specify the topology in which one is working. But I won't give you a course in General Topology here.
For compact s.a. operators on (inf-dim) Hilbert space (in which one usually assumes the standard norm topology when discussing convergence issues), one finds a set of eigenvectors that constitute an orthormal basis. Further, the sequence of eigenvalues tends to 0.
For noncompact bounded s.a. operators on (inf-dim) Hilbert space, things are trickier. Basically, one constructs an integral measure over the space of eigenvalues, which generalizes the formulas you wrote.
For unbounded operators, things are trickier still, since one must specify domains carefully. (Such operators are not well-defined everywhere on the Hilbert space.) That's why some authors generalize to use a so-called "Rigged Hilbert Space" (kinda like a generalization of the Schwarz theory of distributions). Within that framework, there's another theorem, the so-called "nuclear spectral theorem" that establishes the existence of an integral measure that allows one to do QM.
Ballentine gives a brief, but useful, introduction to the notions (and motivation for) Rigged Hilbert Space. (Every serious student of QM should have a copy of Ballentine...)
I was thinking about this problem on my daily commute, and here's my analysis that I came up with, tell me if I'm going in the wrong direction:
So, we have two different (but hopefully equivalent) ways of expressing L^2, either through its complete and orthonormal set of eigenfunctions (spherical harmonics) or through the spatial derivatives (using spherical coordinates):
$$L^2=\sum_{l=0}^\infty\sum_{m=-l}^l \hbar^2l(l+1)\left|l,m\right>\left<l,m\right|$$
Or:
$$L^2=-\hbar^2\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta} \frac{\partial^2}{\partial\phi^2}\right)$$
Let us use the first representation, and look at the expectation value of this operator given some valid wavefunction:
$$\left<L^2\right>=\left<\Psi\right|L^2\left|\Psi\right>=\sum_{l=0}^∞ \sum_{m=-l}^l \hbar^2l(l+1)\left<\Psi|l,m\right>\left<l,m|\Psi\right>$$
Define the (complex) numbers:
$$a_{l,m}\equiv\left<l,m|\Psi\right>$$
And we have then:
$$\left<L^2\right>=\sum_{l=0}^\infty\sum_{m=-l}^l \hbar^2l(l+1)|a_{l,m}|^2$$
In order for this sum to converge, we must require that ##|a_{l,m}|^2\rightarrow 0## as ##l\rightarrow\infty##. Moreover, we must require that ##|a_{l,m}|^2## goes to zero faster than ##l^2## diverges by at least more than 1 power of l (e.g. it goes to zero as a power of 1/l^4, 1/l^3 wouldn't work I think since the harmonic series does not converge).
However, our usual definition of a Hilbert space, the space from which ##\left|\Psi\right>## can be picked from only requires that ##\left|\Psi\right>## is normalizable. In other words, our definition of a Hilbert space only requires:
$$\left<\Psi|\Psi\right><\infty$$
This constraint is equivalent to only the constraint:
$$\sum_{l=0}^\infty\sum_{m=-l}^l |a_{l,m}|^2<\infty$$
This is not a sufficient constraint to guarantee that our angular momentum operator has a well defined expectation value over all of the Hilbert space, for consider if:
$$|a_{l,m}|^2=\frac{1}{1+l^2},\quad m=0$$
$$\qquad else\quad a_{l,m}=0$$
This series converges, meaning our wave function would be normalizable, but the series for the expectation value of L^2 does not converge:
$$\sum_{l=0}^\infty\sum_{m=-l}^l \hbar^2l(l+1)|a_{l,m}|^2=\infty$$
This finally leads me to the conclusion that the Hilbert space is "too big" of a space for our definition of L^2 (big might not be the right word to use here, as L^2 might be defined for particular functions not in our Hilbert space, I have to think about this scenario with more detailed analysis), and looking back at the derivative definition of L^2, this is obvious. L^2 will not work for non twice-differentiable functions, but there are certainly non twice-differentiable functions which are normalizable (e.g. a square wave). This leads me to conclude, therefore, that physical states should be at least twice-differentiable (else, they would have infinite L^2). In fact, if we require L^p to exist for all p, we must require physical states to be infinitely differentiable. Therefore, can we say "physical wave functions must be normalizable, as well as smooth"? What about adding the requirement that they be analytic?
This leads me to the curious side-conclusion that the more times our wave function is differentiable, the faster ##|a_{l,m}|## will approach 0. If we require our wave function to be smooth, then ##|a_{l,m}|## must approach 0 faster than any polynomial function diverges. Is this result correct? It appears to be non-trivial (or maybe it is, but I'm just not smart enough).
