Angular momentum Operators and Commutation

strangerep
Given that there is an operator L^2, with eigenvalue l(l+1), can you define an operator L with eigenvalue square root of l(l+1) when applied to an eigenfunction, thus telling you the total angular momentum of the system? [...]

[...] this is frustrating me that I don't have a yes or no answer.
Matterwave already gave you an answer in post #18. That's why I didn't elaborate.

If you still don't understand why post #18 answers your question, then feel free to ask further questions, quoting the context of that post.

Btw, in case you think I'm being too tough on you, let me say that I actually understand your situation, having been through it myself. In one u'grad course, they presumed knowledge of group theory even though the course sequence I'd been allowed to choose did not include group theory up to that point. I was close to a nervous breakdown by the end, but got through eventually.

Take a few deep breaths, and then continue...

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strangerep
[...]

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).

Oh, I'm sure you're smart enough -- since you've essentially arrived at the standard motivation for using so-called "Rigged Hilbert Spaces" instead of ordinary Hilbert spaces. The basic idea is that we need a "small" space ##\Omega## on which all the observables are well-defined everywhere. This means that we must be able to operate with any observable arbitrarily many times on any element of ##\Omega##, and still get a result that's also in ##\Omega##. One may complete ##\Omega## in norm topology to get a Hilbert space ##H##. One can also construct the dual space of ##\Omega##, denoted ##\Omega'##. This triple of spaces, i.e., ##\Omega \subset H \subset \Omega'## is known as a "Gel'fand triple" or "Rigged Hilbert Space".

I think you'd really enjoy ch1 of Ballentine, especially section 1.4. Can you access a copy?

Have you studied distribution theory yet? The Schwarz theory of distributions is essentially a prototypical case of Rigged Hilbert Spaces, applicable to the usual position/momentum observables.

Matterwave
Gold Member
Sounds interesting. I have not really studied distribution theory (I know roughly what they are since I've encountered them several times, but not much more than that). Unfortunately I only have finite time in which to study, and right now I'm busy reviewing other material. Hopefully when I'm not so busy I'll get back to look into this some more. In the mean time, I'll see if I can locate a copy of Ballentine, thanks for the recommendation. :)

dextercioby
Homework Helper
@Post #24 above: One trivially gets that the L^2 angular momentum operator in L^2(R^3) is unbounded, hence cannot be defined everywhere in the Hilbert space. See sections 10.2 and 11.5 of Blank et al's book "Hilbert Space Operators in Quantum Physics".

Matterwave
Gold Member
@Post #24 above: One trivially gets that the L^2 angular momentum operator in L^2(R^3) is unbounded, hence cannot be defined everywhere in the Hilbert space. See sections 10.2 and 11.5 of Blank et al's book "Hilbert Space Operators in Quantum Physics".

Ok, thanks for calling my argument trivial, I put a lot of thought and effort into it! It wasn't trivial to me anyways lol.

I'll eventually get around to reading some of this stuff...