Thanks for the help, everyone. I think the definition that was given by Halls and mathwonk was probably what matt meant to write. Thanks for clearing that up. :smile:
Right. I guess what I had in mind was a linear map parametrized by |h|, if that makes any sense at all. But that wouldn't be a linear map since it operates differently on different choices of h. Thanks.
Ah, okay. Can you explain?
edit: I guess it's because, if it were linear, you could always add more vectors to the sum so that
o(|h|)(Sum of vectors) > |h| as h->0?
You're right, I was assuming that. I know it doesn't really matter what the order is, but it might be more suggestive to write it o(|h|)h or something like that, to make it more clear that o(|h|) is a linear map. Thanks for the help :smile:
I could live with R(h) as the remainder if it is a vector in R^m. But it seems that h o(|h|) is a vector in R^n, and you can't add that to two vectors in R^m, can you?
Yes, that makes sense. What I'm getting at is the term proportional to h at the end. If h is a vector in R^n and f(h) is a vector in R^m, how do you make that particular definition work?
& int ; [ sub ] a [ /sub ] [ sup ] b [ /sup ]
without the spaces gives ∫ab.
I haven't figured out how to make the superscript go directly over the subscript. You can also make greek letters:
& alpha ;
& beta ;
& gamma ;
gives
α, β, γ, etc.
Yeah, as others pointed out, the empty set can't be a vector space because it has no zero vector. However, the empty set does span the vector space consisting of the zero vector, according to the definition of span: The span of a set of vectors is the smallest subspace containing those vectors.
I've seen it used in the context of integration by parts:
∫ab u dv = uv|ab - ∫ab v du
in which case the first term on the right is called the "surface term".
If you have a collection of more than two objects, with a correspondence between each pair, then that correspondence is not one to one. I still don't get what you mean. It can't be that every creation operator is isomorphic to every other creation operator. If you define a relation
a ~ b...
So the answer is yes? In your post, what is U? Are you assuming (or is it somehow obviously true) that there is a transformation b = UaU*, and then from there showing that U must be unitary? Also, how do you build an irreducible representation for a and a*? The most natural representation I...
You guessed it. The particular "a" that I'm talking about is the annihilation operator from quantum mechanics. So we are dealing with a Hilbert space, which comes equipped with an inner product. And in order for the relation [a, a*] = 1 to be possible, it has to be an infinite dimensional...